Experiment-Based Structural Mechanics [1st ed.] 9789811583100, 9789811583117

Table of contents :
Front Matter ....Pages i-xxv
Basic Experiment (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 1-18
Tension Member (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 19-32
Flexural Member (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 33-69
Compressive Member (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 71-97
Truss (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 99-117
Other Tests (Kyung-Jae Shin, Swoo-Heon Lee)....Pages 119-140
Back Matter ....Pages 141-143

Citation preview

Kyung-Jae Shin Swoo-Heon Lee

Experiment-Based Structural Mechanics

Experiment-Based Structural Mechanics

Kyung-Jae Shin Swoo-Heon Lee •

Experiment-Based Structural Mechanics

123

Kyung-Jae Shin Steel Structure and Structural Dynamics Department of Architectural Engineering Kyungpook National University (KNU) Daegu, Korea (Republic of)

Swoo-Heon Lee Steel Structural Engineering School of Convergence and Fusion System Engineering Kyungpook National University (KNU) Sangju-si, Korea (Republic of)

ISBN 978-981-15-8310-0 ISBN 978-981-15-8311-7 https://doi.org/10.1007/978-981-15-8311-7

(eBook)

© Springer Nature Singapore Pte Ltd. 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

Structural mechanics is the foundation of architectural or civil engineering and is taught during freshman or sophomore in college. The theoretical principle challenges the students as a formidable subject, and there are often many hard-tounderstand cases of how actual structures are applied. It is because the textbooks simply represent members such as beam, frame, truss as a line and support condition as a symbol, making it difficult to picture how actual structures are used. Moreover, although students may pass engineering tests such as moment distribution, characteristics of member section, analysis of a truss, analysis of a statically indeterminate structure, and principle of bending resistance of concrete section, they often have trouble in grasping the actual principle behind the tests. This textbook was conceived with the intention to help the students comprehend the resistance principle of a structure through simple model tests and understand how dynamic computational equation is used to represent the principles mathematically. It will be used for understanding of dynamics, reinforced concrete structure, steel frame structure, etc., during one semester course under the title of ‘structural tests’ or ‘structural and material tests’ through performing various tests described in the book. The tests generally require a laboratory equipped with various test settings. However, the contents of this textbook are such that the tests can be carried out at a lecture room; it does not require complex test equipment but can be performed with simple test devices. The authors acknowledge and appreciate the contributions of Dr. Hee-Du Lee, graduate students, Jun-Seop Lee, Da-Som Choo, Hye-Min Shin, Kang-Baek Park, Han-Min Park, So-Yeong Kim, Yu-Hyeon Lee, Seong-Geun Park, and undergraduates working and studying at our research center. Daegu, Korea (Republic of) Sangju-si, Korea (Republic of) July 2019

Kyung-Jae Shin Professor at Kyungpook National University Swoo-Heon Lee Assistant Professor at Kyungpook National University

v

Experimental Equipment

Structural tests involve measuring load and deformation (or displacement) under such loading condition that the strength and behavior of the structures can be found out. The displacement or load-measuring devices are selected according to measurement range (maximum displacement or maximum load) and measurement precision (0.01 mm or 0.01 N). The measurement range and precision are in trade-off relationship; the higher the measurement range, the lower measurement precision, and the lower measurement range, the higher measurement precision. Accordingly, the following displacement or load measurement devices with adequate precision are used for the structural tests described in this textbook.

1. Displacement-Measuring Devices (1) Ruler ① Measurement range: 300 mm, 600 mm, 1000 mm ② Precision: 0.5 mm * 1 mm ③ Ruler is used to measure the size of a section or a long member or a displacement of over 10 mm. (2) Vernier calipers ① Measurement range: 200 mm, 300 mm ② Precision: 0.01 mm * 0.05 mm ③ It is useful for measuring more precise measurement of a section size or short member, compared to a ruler.

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Experimental Equipment

▲ Vernier calipers

▲ Digital vernier calipers

(3) Dial indicator or Dial gauge ① Measurement range: 10 mm * 50 mm ② Precision: 0.001 mm * 0.01 mm ③ It amplifies the displacement using the principle of mechanic gear and was used a great deal in the past. However, the following linear variable differential transformer (LVDT) has been used more frequently. ▲ Dial gauge

(4) Linear variable differential transformer (LVDT) ① Measurement range: 10 mm * 200 mm ② Precision: 0.001 mm * 0.01 mm

Experimental Equipment

ix

③ Cantilever is embedded inside LVDT, and the strain of cantilever is measured by four strain gauges in full bridge circuit. An appropriate coefficient is multiplied to convert it to displacement for output. ④ It is mainly used for precision test. A data logger is necessary to read the displacement value, and the measured data can be automatically stored by the data logger. ▲ Electronic LVDT

▲ Electronic LVDT of wire type

(5) Strain gauge ① Measurement range: 0.001% * 20% ② Precision: 1 lm/m or 110−6 m/m ③ A gauge measures strain (relative displacement, m/m). The strain gauge is used in architectural, civil, and mechanical engineering fields. ④ Its precision is very high, and it is a very stable sensor. It is used inside load cell and LVDT. ⑤ A section is reduced according to Poisson’s ratio as the length gets larger due to the strain on an electric resistance member caused by the external force P. The resistance of a uniform conductor with a length L and cross-sectional area A is changed by the changed length DL and cross-sectional area A'. The changed resistance is precisely measured to convert it to strain. It is fabricated in the form of a thin foil and is applied on the surface of a material to measure the strain of its surface. Three-axis strain gauge is used to measure the direction (or angle) of principle strain or shear strain.

x

Experimental Equipment

A'

A

P

ΔL/2

P

L

ΔL/2

▲ Strain

▲ Uniaxial strain gauge

▲ Three axis strain gauge

2. Load-Measuring Devices The test described in this textbook uses an electric scale with load cell to measure the load (or weight) and, actually, the load cell is mainly utilized for a precision test. Only small scales are used in this text contents. The load cells are used for larger structural experiment.

Experimental Equipment

(1) Scale ▲ Scale with low capacity

▲ Scale with high capacity

(2) Load cell ▲ Load cell for compression and tension

▲ Load cell for compression

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Experimental Equipment

▲ Load cell with high capacity for compression and tension

3. Other Devices The following measurement devices are not used in this text test. (1) Pressure cell

▲ Pressure cell

▲ Earth pressure cell

Experimental Equipment

(2) Accelerometer ▲ Three-axis accelerometer

▲ Earthquake accelerometer

(3) Clinometer and Goniometer ▲ Digital clinometer

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Experimental Equipment

▲ Digital goniometer

4. Loading Devices In most tests, the loading is applied by hand or simple weight such as papers or coins. (1) Weight ▲ Papers

▲ Weight

▲ Coins

Experimental Equipment

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▲ Loading by hand

College

Department

ID No.

Test Team

Name

Test Date

Test Time

Report Date

mm/dd/yy

Start ~ End

mm/dd/yy

Test topic 1. Purpose 2. Scope - Test material and technique 3. Test equipment and method - Equipment and description (use figure) - Procedure description - Standard test method - Test environmental condition (Date, Place, Weather, Temperature, Humidity) 4. Test results - Basic principle (Refer to textbooks on structural mechanics, RC, Steel structure design) - Description of applied equation, theoretical equation - Data collection and computation - Preparation of tables and figures - Comparison of theoretic and test result 5. Examination of test result - Description of test result - Suggest questions - Things to consider during additional future test 6. Conclusion and Review 7. Reference

Materials and Tools Used in Each Section

The structural tests are to be carried out by referring to the following table after finding out if it is a necessity to purchase materials or tools and whether they can be used according to the test topic. Section

Materials

Tools

1. Strength of flexural member made of paper

–A4 or letter size papers –Paper or plastic cups –Foam board (t = 5 mm) –Glue stick or paste –A4 or letter size papers –Foam board (t = 5 mm) –Glue stick or paste –Spaghetti noodle –Play dough (or clay)

–Coins

2. Strength of compressive member made of paper 3. Bending, compression, and tensile test of a spaghetti noodle

4. Centroid of section

–Styrofoam (t = 5 mm) –Packing tape

5. Test and design of tension member

–Styrofoam (t = 5 mm) –Packing tape –Large plastic bag –Paper (around 160 gf/m2) –Packing tape

6. Stress concentration test

7. Flexural strength test of beam

8. Bending test and design of flexural member

–Foam board (t = 5 mm) –Isopink (t = 30 mm) –Instant adhesive –Packing tape –Isopink (t = 30 mm) –Adhesive for Styrofoam

–Box cutter –Ruler –Scale (60 N) –Ruler –Permanent pen –Clips –Scale –Box cutter –Ruler –Push-pins –Box cutter –Ruler –Weight scale –Box cutter –Ruler –Weight scale –Box cutter –Ruler –Scale (30 N) –Box cutter –Ruler –Weight scale (continued)

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Materials and Tools Used in Each Section

(continued) Section

Materials

Tools

9. Deflection test of beam

–Paper or plastic cups –Packing tape –Coins –Foam board (t = 3 mm) –Isopink (t = 50 mm) –Glue

–Steel ruler –Scale (10 N) –Vernier calipers –Glue gun –Box cutter –Ruler –Scale (3 N) –Box cutter –Ruler –Weight –Box cutter –Ruler –Scale (60 N) –Box cutter –Ruler –Vernier calipers –Scale (3 N, 30 N) –Box cutter –Ruler –Scale (3 N, 30 N) –Box cutter –Ruler –A4 or letter papers –Scale –Box cutter –Ruler –Glue gun and glue –Scale and weight scale –Box cutter –Ruler –Awl –Box cutter –Ruler –Glue gun –Scale (60 N) –Glue gun –U-bolt –Loading board –Load device –Weight (or steel plate) –Glue gun –Box cutter –Ruler –Scale (1 N) (continued)

10. Lateral buckling test of beam

11. Shear center test (focused on channel section)

–Styrofoam (t = 5 mm) –Adhesive for Styrofoam

12. Principle of reinforced concrete beam

–Wooden stick –Foam board (t = 1 mm) –Instant adhesive –Spaghetti noodle

13. Buckling test of spaghetti noodle

14. Buckling test of Styrofoam column

–Styrofoam (t = 5 mm)

15. Buckling test of long column (depending on boundary condition)

–Styrofoam (t = 5 mm) –Foam board (t = 5 mm) –Adhesive for Styrofoam –Packing tape

16. Competition for buckling strength of column

–Foam board (t = 5 mm) –Styrofoam (t = 5 mm) –A4 or letter papers (for loading) –Instant adhesive

17. Determination of tensile and compressive members in truss

–Foam board (t = 2 or 3 mm) –Bolts and nuts (/2 mm)

18. Strength test of truss made of spaghetti noodle

–Spaghetti noodle –Foam board (t = 3 mm) –Glue

19. Strength competition for spaghetti truss

–Spaghetti –Glue

20. Competition for effective reinforcement of truss

–Foam board (t = 5 mm) –Glue

Materials and Tools Used in Each Section

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(continued) Section

Materials

Tools

21. Vibration test (dynamics)

–Rubber band –Plastic bag –Thumbtack –Styrofoam (t = 5 mm) –Glue –Packing tape

–Ruler –Coins –Scale (10 N) –Glue gun –Box cutter –Ruler –Clip –Permanent pen –Strain meter –Laptop –Universal Test Machine (UTM) –Lead wire

22. Arch and dome structures

23. Creating a load cell

–Cylinder steel with yield strength exceeding 400 MPa –Four strain gauges (R = 120 X or 350 X) –CN bond –M-coat A coating –Sand paper –Acetone and cotton

Contents

1 Basic Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Strength of Flexural Member Made of Paper . . . . . . . . 1.1.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 1.1.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.4 Test Results . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.5 Writing a Report . . . . . . . . . . . . . . . . . . . . . . . 1.1.6 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Strength of Compressive Member Made of Paper . . . . . 1.2.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 1.2.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.4 Writing a Report . . . . . . . . . . . . . . . . . . . . . . . 1.2.5 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Bending, Compression, and Tensile Test of a Spaghetti Noodle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 1.3.3 Summary of Test Method . . . . . . . . . . . . . . . . . 1.3.4 Detailed Test Method . . . . . . . . . . . . . . . . . . . . 1.3.5 Test Results and Analysis . . . . . . . . . . . . . . . . 1.3.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Centroid of Section . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 1.4.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Computational Method by Theory . . . . . . . . . . 1.4.5 Theory on Centroid . . . . . . . . . . . . . . . . . . . . . 1.4.6 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Flexural Member . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Flexural Strength Test of a Beam . . . . . . . . . . . 3.1.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.1.2 Materials and Tools . . . . . . . . . . . . . . . . 3.1.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.1.4 Reviews . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Theory on Flexural Member . . . . . . . . . . 3.2 Bending Test and Design of a Flexural Member . 3.2.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.2.2 Materials and Tools . . . . . . . . . . . . . . . . 3.2.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.2.4 Test Results and Applied Computation . . 3.2.5 Applied Test of Section Design . . . . . . . 3.2.6 Example of Design . . . . . . . . . . . . . . . . 3.2.7 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Deflection Test of a Beam . . . . . . . . . . . . . . . . 3.3.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.3.2 Materials and Tools . . . . . . . . . . . . . . . . 3.3.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.3.4 Test Results and Deflection Computation 3.3.5 Case Study . . . . . . . . . . . . . . . . . . . . . . 3.3.6 Theory on Beam Deflection . . . . . . . . . . 3.3.7 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Lateral Buckling Test of a Beam . . . . . . . . . . . . 3.4.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.4.2 Materials and Tools . . . . . . . . . . . . . . . . 3.4.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.4.4 Test Results . . . . . . . . . . . . . . . . . . . . . 3.4.5 Theory on Lateral Buckling of a Beam . .

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2 Tension Member . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Test and Design of a Tension Member . . . . . 2.1.1 Purpose of Test . . . . . . . . . . . . . . . . . 2.1.2 Materials and Tools . . . . . . . . . . . . . . 2.1.3 Test Method . . . . . . . . . . . . . . . . . . . 2.1.4 Test Results and Applied Computation 2.1.5 Case Study . . . . . . . . . . . . . . . . . . . . 2.1.6 Theory on Tensile Stress . . . . . . . . . . 2.1.7 Task . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Stress Concentration Test . . . . . . . . . . . . . . . 2.2.1 Purpose of Test . . . . . . . . . . . . . . . . . 2.2.2 Materials and Tools . . . . . . . . . . . . . . 2.2.3 Test Method . . . . . . . . . . . . . . . . . . . 2.2.4 Theory on Stress Concentration . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

3.5 Shear Center Test (Focused on Channel Section) 3.5.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.5.2 Materials and Tools . . . . . . . . . . . . . . . . 3.5.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.5.4 Theory on Shear Center . . . . . . . . . . . . . 3.5.5 Examples . . . . . . . . . . . . . . . . . . . . . . . 3.6 Principle of a Reinforced Concrete Beam . . . . . . 3.6.1 Purpose of Test . . . . . . . . . . . . . . . . . . . 3.6.2 Materials and Tools . . . . . . . . . . . . . . . . 3.6.3 Test Method . . . . . . . . . . . . . . . . . . . . . 3.6.4 Test Results and Applied Computation . . 3.6.5 Test Case . . . . . . . . . . . . . . . . . . . . . . . 3.6.6 Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.7 Principle of an RC Beam . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Compressive Member . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Buckling Test of a Spaghetti Noodle . . . . . . . . . . . . . . . . . 4.1.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Test Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.5 How to Prepare a Report . . . . . . . . . . . . . . . . . . . . 4.1.6 Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.7 Buckling Theory on a Column . . . . . . . . . . . . . . . . 4.1.8 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Buckling Test of a Styrofoam Column . . . . . . . . . . . . . . . . 4.2.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Test Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 How to Prepare a Report . . . . . . . . . . . . . . . . . . . . 4.2.6 Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Buckling Test of a Long Column (Depending on Boundary Condition) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.4 Test Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.5 Applied Column Design . . . . . . . . . . . . . . . . . . . . . 4.3.6 Buckling Characteristics of a Long Column . . . . . . 4.4 Competition for Buckling Strength of a Column . . . . . . . . . 4.4.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . . . .

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xxiv

Contents

4.4.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Consideration of Key Point . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Determination of Tensile and Compressive Members in a Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Materials and Tools . . . . . . . . . . . . . . . . . . . 5.1.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Strength Test of a Truss Made of Spaghetti Noodles 5.2.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Materials and Tools . . . . . . . . . . . . . . . . . . . 5.2.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . 5.2.4 How to Prepare a Report . . . . . . . . . . . . . . . 5.2.5 Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.6 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Strength Competition of a Spaghetti Truss . . . . . . . . 5.3.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Materials and Tools . . . . . . . . . . . . . . . . . . . 5.3.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . 5.3.4 How to Prepare a Report . . . . . . . . . . . . . . . 5.4 Competition for Effective Reinforcement of a Truss . 5.4.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Materials and Tools . . . . . . . . . . . . . . . . . . . 5.4.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 Test Results . . . . . . . . . . . . . . . . . . . . . . . . 5.4.5 Efficiency of Material in a Truss . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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94 96 97 99

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99 99 100 100 103 103 104 104 107 107 110 110 110 110 111 113 113 113 113 113 116 117 117

6 Other Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Vibration Test (Dynamics) . . . . . . . . . . . . . . . . . . . . . 6.1.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 6.1.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.4 Comparison of Test Frequency and Theoretical Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.5 Applied Vibration Test . . . . . . . . . . . . . . . . . . . 6.1.6 Vibration Theory . . . . . . . . . . . . . . . . . . . . . . . 6.1.7 How to Prepare a Report . . . . . . . . . . . . . . . . . 6.2 Arch and Dome Structures . . . . . . . . . . . . . . . . . . . . . 6.2.1 Purpose of Catenary Test . . . . . . . . . . . . . . . . . 6.2.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . 6.2.3 Test Method . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 Catenary Theory . . . . . . . . . . . . . . . . . . . . . . .

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119 119 119 119 119

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121 122 122 126 127 127 127 127 130

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Contents

6.3 Creating a Load Cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Purpose of Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Materials and Tools . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 How to Make a Load Cell . . . . . . . . . . . . . . . . . . . . . 6.3.4 Calculating Potential Difference of Wheatstone Bridge (Example) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.5 Creating a Load Cell (Example) . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xxv

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132 132 132 132

. . . 136 . . . 138 . . . 140

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Chapter 1

Basic Experiment

1.1 Strength of Flexural Member Made of Paper 1.1.1 Purpose of Test • Presenting a flexural member (i.e., beam) section which can resist the ultimate flexural strength by using one piece of paper – A4 (or letter) size paper is used as a horizontal flexural member to suggest a way to withstand the maximum flexural resistance. – It is to affirm that different shapes of the cross sections can increase the flexural capacityeven if the quantity of material is the same. – It is verified that an efficient construction is possible depending on the shape of the member section. .

1.1.2 Materials and Tools (1) (2) (3) (4) (5)

Ten sheets of A4 (or letter) size paper (Fig. 1.1) Three paper (or plastic) cups (Fig. 1.1) A sheet of foam board with size of about 50 mm × 50 mm × 5 mm (Fig. 1.1) Glue stick or paste Coins such as quarter (USA), ten pence (UK), e0.2–e1 (Euro), and 100 (South Korea)

© Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_1

1

2

1 Basic Experiment

Fig. 1.1 Materials

1.1.3 Test Method (1) The paper is freely folded in the lengthwise direction so that the span length of this folded paper beam is to be the long length of paper [1]. (2) Two cups are placed in the interval of 300 mm between the centers, and both ends of the folded paper beam [3] are laid over the center of two cups as shown in Fig. 1.2a (3) A foam board with size of 50 mm × 50 mm × 5 mm is placed over the center of the paper beam to disperse a load and balance a cup as shown in Fig. 1.2b, and a cup is laid over the foam board. (4) Load is applied by putting coins into the cup one by one as shown in Fig. 1.2c. (5) Coins are added until the paper beam collapses, and then the number of coins thrown in the cup is counted. (6) Glue can be once applied lengthwise to make the paper beam a closed section member.

(a) Paper beam is setup

(b) Cup for loading is placed

Fig. 1.2 Strength test for folded paper beam

(c) Coins are added into cup

1.1 Strength of Flexural Member Made of Paper

3

Table 1.1 Summary of test results Shape of section

Image of section

Number of coins

Section #1 Section #2 Section #3 Section #4 Section #5 Section #6

1.1.4 Test Results (1) A minimum of three section shapes are planned for the test. (2) The number of added coins before the collapse of paper beam is recorded by the shape of the sections, and the shape of the section model with the most coins is analyzed. (3) The strong section model and the weak section model [2] are compared for finding out what makes the difference. (4) After analyzing several models, a better (improved) section model is proposed by repeating the process of Sect. 1.1.3 above.

1.1.5 Writing a Report (1) The report should be prepared by including test overview, test purpose and method, test result, analysis, conclusion, and review referring to ‘Test Report (Sample)’ written in page XV of this book. (2) The test results should include the image of the paper beam model and the number of coins, as shown in the example of Table 1.1. (3) The review and conclusion section should include your opinion of what should be done to make an efficient section shape for a beam.

1.1.6 Task • Prove the ultimate load of a horizontal member (i.e., beam) section through a test and present the result.

4

1 Basic Experiment

1.2 Strength of Compressive Member Made of Paper 1.2.1 Purpose of Test • Presenting a compressive member (i.e., column) section which can resist the ultimate load by using one sheet of paper – A4 (or letter) size paper is used as a vertical member (column) subjected to compressive force, and the maximum strength is compared through variable tests. – Although the test specimens would have the same materials and height, the type of the test specimen section is to be varied. – A4 (or letter) papers are loaded onto the prepared vertical test specimen one by one to determine which type of vertical member is the most advantageous.

1.2.2 Materials and Tools (1) (2) (3) (4)

A4 (or letter) size papers (12 sheets for compressive specimen and weight) A sheet of foam board of 450 mm × 600 mm × 5 mm size Glue stick or paste, box cutter, ruler Electronic scale of about 60 N capacity.

1.2.3 Test Method (1) Specimen preparation ➀ Prepare two compressive members (i.e., columns) made of paper like Fig. 1.4, according to drawings of Fig. 1.3. ➁ Cut the foam board for loading plate in A4 size of 210 mm × 297 mm or letter size of 216 mm × 279 mm.

1.2 Strength of Compressive Member Made of Paper

5

: Fold outward : Fold inward : Paste

(a) Circular shape

(b) Triangular shape

(c) Rectangular shape

(d) Hexagonal shape

(e) Octangular shape

(f) Polygonal shape

Fig. 1.3 Section shape of compressive members

Fig. 1.4 Folded paper columns

6

1 Basic Experiment

(2) Test method ➀ Two columns are placed in the distance of 150 mm between the centers as shown in Fig. 1.5, and the foam board is laid over two columns as shown in Fig. 1.6. ➁ Load is applied with A4 (or letter) papers on the foam board (Fig. 1.7). Fig. 1.5 Setup of two column

Fig. 1.6 Setup of loading plate

Fig. 1.7 Stacking papers

1.2 Strength of Compressive Member Made of Paper

7

Fig. 1.8 Measurement of weight

➂ Record the number of papers and the weight when the paper columns collapse (Fig. 1.8).

1.2.4 Writing a Report (1) The report should be prepared by including test overview, test purpose and method, test result, analysis, conclusion, and review referring to ‘Test Report (Sample)’ written in page XV of this book. (2) The test result should be summarized as shown in the example of Table 1.2, Table 1.2 Summary of test results

Shape of section

Load Number of papers (sheets)

Circular section Triangular section Rectangular section Hexagonal section Octangular section Polygonal section

Weight (N)

8

1 Basic Experiment

and the section of the compressive member which withstood the largest load is discussed [1–3]. (3) Notice which section was the most efficient by comparing the measured weight.

1.2.5 Task • Propose another section to enhance the load-carrying capacity and describe the test results.

1.3 Bending, Compression, and Tensile Test of a Spaghetti Noodle 1.3.1 Purpose of Test • Experimentally understand the resistance through bending, compression, and tensile tests using spaghetti noodle as structural members – The spaghetti noodle is used as the bending material, and the weight is measured at the bending failure by applying the play dough. – The spaghetti noodle is used as the compression material, and the weight is measured at the compressive failure by applying the play dough. – The spaghetti noodle is used as the tensile material, and the weight is measured at the tensile failure by applying the play dough.

1.3.2 Materials and Tools (1) 1 bag of spaghetti (φ 2 mm and 250 mm length), play dough (or clay) (2) Ruler, permanent pens, clips, scale (3 N).

1.3.3 Summary of Test Method (1) Specimen preparation ➀ Spaghetti noodle is marked at 100 and 150 mm in length with a permanent pen as shown in Fig. 1.9. ➁ The load is applied by attaching the paly dough to one end of the spaghetti noodle.

1.3 Bending, Compression, and Tensile Test of a Spaghetti Noodle

9

Fig. 1.9 Spaghetti specimen preparation

Fig. 1.10 Test methods

W L=100 or 150 mm

L=100 or 150 mm

W

L=100 or 150 mm

(a) Bending test

W

(b) Compression test

(c) Tensile test

(2) Summary of test method ➀ The test sequence is the bending test, compression test, and tensile test sequence as shown in Fig. 1.10.

10

1 Basic Experiment

➁ In the bending test as shown in Fig. 1.10a, the clay as a concentrated load is fixed to the free end of the spaghetti noodle. ➂ In the compression test, the clay as a compression load is fastened to the top of the spaghetti noodle as shown in Fig. 1.10b. ➃ The tensile test is performed by attaching the clay to be a tensile load to the bottom of the spaghetti noodle as shown in Fig. 1.10c. ➄ Review how the spaghetti noodle resists most effectively through three tests: bending, compression, and tension.

1.3.4 Detailed Test Method (1) Bending test ➀ Fasten the clay to the free end of each spaghetti noodle of 100 and 150 mm lengths as shown in Fig. 1.11. ➁ Hold the opposite fixed end by hand so that the spaghetti noodle is kept in cantilever shape. ➂ Increase the amount of clay gradually until the spaghetti noodle breaks. Keep the time interval of at least 10 s after clay is added. ➃ After the spaghetti noodle is destroyed, weigh the clay load on the scale. (2) Compressive test ➀ Fasten the clay to the top of each spaghetti noodle of lengths 100 and 150 mm as shown in Fig. 1.12. ➁ Hold the opposite end by hand so that the spaghetti noodle is formed vertically like a cantilever column. ➂ Increase the amount of clay gradually until the spaghetti noodle breaks. ➃ After the spaghetti noodle is destroyed, measure the weight of the clay on the scale. ➄ Analyze the difference in failure shape of the spaghetti noodle of 100 and 150 mm in length.

(a) L = 100 mm Fig. 1.11 Bending test

(b) L = 150 mm

1.3 Bending, Compression, and Tensile Test of a Spaghetti Noodle

11

Fig. 1.12 Compression test

(a) L = 100 mm

(b) L = 150 mm

(3) Tensile test ➀ Fasten the clay to the bottom of each spaghetti noodle of 100 and 150 mm lengths as shown in Fig. 1.13. ➁ Clip the bottom side of the noodle to prevent the clay from flowing down as shown in Fig. 1.13a. ➂ Increase the amount of clay gradually until the spaghetti noodle breaks. ➃ After the spaghetti noodle is destroyed, weigh the paper clay on the scale. ➄ It can be found that the tensile test specimens are not broken well even under a large load as compared with the other two tests. At this time, the appropriate load is assumed to be the failure load even if no failure occurs (e.g., 150 gf).

(a) Clipping to bottom Fig. 1.13 Tensile test

(b) L = 100 mm

(c) L = 150 mm

12

1 Basic Experiment

1.3.5 Test Results and Analysis (1) It can be confirmed that the resistance of spaghetti noodle is strong in the order of bending < compression < tension [1, 2]. (2) The test results are as shown in Table 1.3. (3) The average value of each test is plotted in Fig. 1.14. (4) In the case of the bending test, it is observed that the shorter the spaghetti noodle is, the more the resistance is improved. (5) The shorter the length of the spaghetti noodles, the better the resistance, in the compression test. (6) The tensile test shows very high resistance when compared to other tests. Table 1.3 Summary of test data L = 100 mm

Test

Bending

L = 150 mm

Measured load (N)

Average (N)

Measured load (N)

Average (N)

0.208

0.209

0.131

0.132

Bending failure

0.242

Compressive failure

0.211

0.133

0.209 Compression

0.133

0.455

0.455

0.242

0.457 Tension

Failure mode

0.244

0.452

0.241

1.5 or above

1.5 or above

No failure

Weight of clay (N)

1.5 1.2

Length of 150 mm Length of 100 mm

0.9 0.6 0.3 0

Bending

Compression

Type of test Fig. 1.14 Comparison by three test types

Tensile

1.3 Bending, Compression, and Tensile Test of a Spaghetti Noodle

13

1.3.6 Conclusions (1) As a result of the bending test for the spaghetti noodle, flexural failure occurred at about 0.209 N (20.95 gf) and 0.132 N (13.23 gf) when the specimen lengths were 100 mm and 150 mm, respectively. The shorter the length of the spaghetti noodle, the better the ability to resist bending. (2) Compression test of spaghetti noodle resulted in compressive failure at the load of about 0.455 N (45.46 gf) for 100 mm length and about 0.242 N (24.23 gf) for 150 mm length. The shorter the length of the spaghetti noodle, the better the ability to resist compression. (3) As a result of the tensile test for the spaghetti noodle, no failure was observed because it was able to withstand more than 1.5 N (150 gf) load of clay in both 100 and 150 mm lengths. It can be seen that the resistance of the spaghetti noodle to the tensile force is very large. (4) As a result of the three tests, the resistance of the spaghetti noodle was found to be strong in the order of bending < compression < tension. (5) It can be said that it is most effective when members are used as tensile material owing to its high resistance to tension [3, 4].

1.4 Centroid of Section 1.4.1 Purpose of Test • Comprehending the centroid of a section and finding it out experimentally and theoretically – When a moment is applied to a flexural member, the flexural stress takes place inside the member, and the same tensile stress and compressive stress as couple of forces act on the section. – Flexural stress increases outward from the neutral axis on cross section. – The neutral axis of the flexural member is regarded as the centroid of a section, and it is used for the computation of geometric properties such as moment of inertia and section modulus for a given cross section. The centroid of a section can be regarded as the same concept as the center of gravity. – Given a section of Styrofoam is prepared, the centroid of a section is verified with a theoretical approach after it is empirically tested.

14

1 Basic Experiment

1.4.2 Materials and Tools (1) A sheet of Styrofoam of 600 mm × 900 mm × 5 mm size (2) Box cutter, ruler, push-pins, packing tape.

1.4.3 Test Method (1) Sections of Fig. 1.15 are cut from a given Styrofoam. (2) Test method. ➀ A push-pin is pushed through the spot presumed to be the centroid of a section, and both sides of the push-pin are placed lightly on both hands to be rotated. ➁ If the stop angle due to a heavier side is consistent, the position of the push-pin is changed to be closer to the center of gravity. ➂ If the center of gravity is not in the section but outside, packing tape is attached to both sides of the section as to continue the experiment with push-pin, as shown in Fig. 1.16.

C

C

Fig. 1.15 Sample section

C

1.4 Centroid of Section

15

Fig. 1.16 Preparation and test

1.4.4 Computational Method by Theory (1) Compute the centroid of a section for the given sections of Fig. 1.17. In the following examples, x¯ is the distance from the left end of the section, and y¯ is the distance from the bottom of the section. ➀ The T-shape section of Fig. 1.17(a) is symmetrical to y-axis, and accordingly, x¯ does not need to be computed. The y¯ is obtained as follows: y¯ =

120 × 40 × 180 + 40 × 160 × 80 A1 y1 + A2 y2 = A1 + A2 120 × 40 + 40 × 160 2

40

A1

1

40

A2

200

A1

160

1

200

150

C C

C

2

A1

130 150

40

A2 50

50 40

(a) T-shape Fig. 1.17 Variable sections

100

(b) L-shape #1

20

100 120

(c) L-shape #2

20

50

A2

16

1 Basic Experiment

=

1,376,000 ≈ 122.86 11,200

➁ With respect to section of Fig. 1.17(b), x¯ =

A1 x 1 + A2 x 2 437,500 50 × 200 × 25 + 50 × 50 × 25 = = 35 = A1 + A2 150 × 200 + 50 × 50 12,500

y¯ =

A1 y1 + A2 y2 50 × 200 × 100 + 50 × 50 × 25 1,062,500 = = = 85 A1 + A2 50 × 200 + 50 × 50 12,500

(3) ➂ and with respect to section of Fig. 1.17(c), x¯ =

y¯ =

A1 x 1 + A2 x 2 20 × 150 × 10 + 100 × 20 × 70 172,000 = = = 34.4 A1 + A2 20 × 150 + 100 × 20 5000

245,000 A1 y1 + A2 y2 20 × 150 × 75 + 100 × 20 × 10 = = 49 = A1 + A2 20 × 150 + 100 × 20 5000

1.4.5 Theory on Centroid (1) The moment of inertia can be computed only when the location of centroid is certain. Then, the flexural deformation can also be analyzed. If the shape of a section is symmetric, the location of the centroid can be easily obtained because it is the center of the section. If not, the centroid of a section should be approached with an understanding of the concept of geometric moment of area [1–3]. (2) The geometric moment of area represents the sums of the products of the differential area and its coordinate. The geometric moments of area S x and S y with respect to the x- and y-axes can be computed by the following Eqs. (1.1) and (1.2), respectively.  Sx =

yd A = y¯ A

(1.1)

xd A = x¯ A

(1.2)

A

 Sy = A

(3) Examining Eqs. (1.3)–(1.4) and Fig. 1.18, the geometric moment of area for each axis always becomes ‘0’. In other words, the coordinates of the centroid of a section can be computed by the following equations.

1.4 Centroid of Section

17

Fig. 1.18 Centroid of an arbitrary section

C

dA

 xd A Sy = A x¯ = A A dA  yd A Sx y¯ = = A A A dA

(1.3)

(1.4)

1.4.6 Task

30

130 160

183 200

17

• Obtain experimentally the centroids of the following two sections, and verify them through theoretical approach (unit: mm)

30

70 100

144.5

11 300

144.5

18

1 Basic Experiment

Centroid of simple sections

▲ Rectangular section

▲ Circular section

x¯ = b2 , y¯ =

x¯ = y¯ =

D 2

1

1

2

2

h 2

▲ Triangular section y¯1 =

h 3,

y¯2 =

2h 3

▲ Trapezoidal section y¯1 =

h 2a+b 3 a+b ,

y¯2 =

h a+2b 3 a+b

References 1. Gere, J.M.: Mechanics of Materials, 5th edn. Nelson Thornes Ltd (2001) 2. Hibbeler, R.C.: Structural Analysis, 10th edn. Pearson (2018) 3. Salmon, C.G., Johnson, J.E., Malhas, F.A.: Steel Structures—Design and Behavior (Emphasizing Load and Resistance Factor Design), 5th edn. Pearson Education, Inc (2009) 4. Salvadori, M., Heller, R.: Structure in Architecture, 2nd edn (1962)

Chapter 2

Tension Member

2.1 Test and Design of a Tension Member 2.1.1 Purpose of Test • Understanding the design concept of a tension member – Fracture stress is comprehended through a tensile test using Styrofoam. – Then, the section design is performed in consideration of a safety factor with respect to the required load. – Safety of the proposed section is verified through a tensile test.

2.1.2 Materials and Tools (1) A sheet of Styrofoam of 600 mm × 900 mm × 5 mm size. (2) Box cutter, ruler, packing tape, packing string, large plastic bag, electronic weight scale.

2.1.3 Test Method (1) Tensile test specimen [1] ➀ A Styrofoam is cut in width B of 10–20 mm by each team to prepare tensile test specimens as shown in Fig. 2.1. ➁ Pulling rings are made of packing tape and attached to both ends of the specimen so that both ends of the specimen can be pulled from both directions, as shown in Fig. 2.2.

© Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_2

19

2 Tension Member

50

5

B=10~20

20

R20 80 450

Fig. 2.1 Tensile test specimen (unit: mm)

Fig. 2.2 Pulling rings at both ends

(2) Test method and procedure ➀ As shown in Fig. 2.3, one person (➊) stands on a weight scale to record his/her weight as the initial weight W 0 , and then the other person (➋) slowly pulls the specimen vertically downward from outside the scale until the specimen breaks. ➁ Another person (➌) records a video of the display on the weight scale with a mobile phone until the specimen breaks. The failure load of weight scale value W 1 is reviewed through video replay and recorded in the report. ➂ The failure load Pfr is the difference between the weight scale value W 1 at the time of fracture and the initial weight W 0 (i.e., weight of ➊ person). ➃ The tensile fracture stress Pfr is computed by dividing the fracture load Pfr by the sectional area A according to Eq. (2.1). In other words, the stress is the load applied to a unit area. σfr =

Pf r A

(2.1)

where Pfr is the fracture load (N). σ fr is the fracture stress (MPa or N/mm2 ). A is the cross-sectional area (mm2 ). ➄ The test results of five specimens are recorded in Table 2.1. ➅ Caution when the specimen is pulled; make the best effort to pull in the lengthwise direction of the specimen to avoid eccentric load.

2.1 Test and Design of a Tension Member

21

Fig. 2.3 Loading method

Table 2.1 Summary of tensile test No.

Section size t × B (mm × mm)

Area A (mm2 )

Initial weight W 0 (N)

Fracture weight W 1 (N)

Fracture load Pfr = W 1 –W 0 (N)

Fracture stress σ fr = Pfr /A (MPa)

1 2 3 4 5 Mean

2.1.4 Test Results and Applied Computation (1) The largest and the smallest values among the five tests are excluded, and the middle three data are averaged to compute the fracture load Pfr . (2) Fracture stress σ fr is computed by using Eq. (2.1).

22

2 Tension Member

(3) The σ fr is divided by safety factor (S = 1.5) to compute allowable stress σ allow , as expressed in Eq. (2.2). σallow =

σfr σfr = S 1.5

(2.2)

where σ allow is the allowable stress (MPa or N/mm2 ). is the fracture stress (MPa or N/mm2 ). σ fr S is the safety factor. (4) In actual structural design, the safety factor S of greater than 1.5 is used, and it varies with materials (steel: 3.0, concrete: 4.0) or type of load (S for long-term load > S for short-term load, S for dynamic load > S for static load).

2.1.5 Case Study (1) Based on tensile test results of Table 2.2, the design process of the tension member is as follows: ➀ The average of fracture stress is computed from three test data excluding the largest and the smallest values: σmean (= σ f r ) =

0.91 + 0.93 + 0.93 = 0.92 MPa 3

➁ The allowable stress is given as σallow =

σfr 0.92 = = 0.61 MPa S 1.5

(2) Design of tensile member Table 2.2 Summary of tensile test (sample) No.

Section size t × B (mm × mm)

Area A (mm2 )

Initial weight W 0 (N)

Fracture weight W 1 (N)

Fracture load Fracture Pfr = W 1 –W 0 stress σ fr = Pfr /A (MPa) (N)

1

5 × 17

85

710

787

77

2

5 × 20

100

710

803

93

0.93

3

5 × 17.5

87.5

710

792

82

0.93

4

5 × 17

85

710

785

75

0.85

5

5 × 20

100

710

805

95

0.95

Mean

–

–

–

–

–

0.92

0.91

2.1 Test and Design of a Tension Member

23

Fig. 2.4 Loading method

➀ Weight of 10 books is P = 5.3 kgf = 53 N ➁ Design required sectional area is given by Areq. =

P σallow

=

53 N = 87 mm2 0.61 MPa

➂ The required width of section is expressed as Breq. =

Areq. 87 mm2 = = 17.4 mm t 5 mm

➃ The test specimen is prepared with the width of B = 17.4 mm as shown in Fig. 2.1. ➄ After making a test specimen, the large plastic bag with the weight of 10 books P is hung down the specimen to verify its safety. Note that the top of the specimen is stabilized by holding the pulling ring so that it would not be subjected to eccentric force. The bottom of the specimen connected to the plastic bag is fastened at both sides with packing tape to be linked to the plastic bag with a packing string (Fig. 2.4). ➅ After the safety of the specimen against P is verified through the first test, the same book is added one by one to obtain the load leading to its fracture.

24

2 Tension Member

2.1.6 Theory on Tensile Stress (1) The internal resisting force taking place inside an object against external force is called stress. Unit stress refers to the stress acting on a unit area due to external force. The unit stress perpendicular to the length of a section in axis direction is called normal stress, which is composed of tensile stress σ t and compressive stress σ c [2, 4]. σt =

Pt Pc , σc = A A

(2.3)

where Pt and Pc are the tension and compression force (N), respectively. σ t and σ c are the tensile and compressive stress (MPa), respectively. A is the cross-sectional area. (2) Tensile stress and compressive stress can be understood as the case of distance between molecules of an object becomes further and closer, respectively (Fig. 2.5). Force applied to a member is transmitted through node of member end. If the node becomes further apart or closer, the member is subjected to tensile stress or compressive stress, respectively. As tensile load increases gradually, tensile stress also increases in proportion. When the external force reaches the ultimate value which the member can resist, the member will fracture (break). When this happens, the stress value is called fracture stress σ fr .

Fig. 2.5 Normal stresses

2.1 Test and Design of a Tension Member

2.1.7 Task • Design and fabricate a section, which can lift up 10 text books to be verified through a test. Assume that a safety factor S is 1.5. Tensile test of metallic material

– Tensile test coupon of Fig. 2.6 is installed between the top and bottom grips of the testing machine as shown in Fig. 2.7. Generally, strain gauge and/or extensometer are installed within the gauge length to measure the strain of a coupon. Strain gauge can directly read the strain, while strain from extensometer can be obtained by dividing the measured displacement by the gauge length.

Fig. 2.6 Coupon of flat and round type

Fig. 2.7 Examples of tensile test

– When strain gauge is used, the measurement can be less than the actual strain for the case of specimen fracture not taking place at the installation location of the gauge. There are often cases of impossible measurement to the point of fracture strain due to break-off of the gauge prior to reaching actual fracture of the specimen even if the fracture occurs at the location of the gauge. – Using extensometer, the gauge length becomes wider than the case of using strain gauge, and actual fracture strain can be measured, as shown in Fig. 2.8,

25

26

2 Tension Member

in many cases because it is not subjected to break-off like the case of a strain gauge. This curve shows the mechanical properties of steel material. 900 800

Specimen #1

Stress (MPa)

700 600 500

Specimen #2

400

Specimen #3

300 200 100 0

0

50,000

100,000 150,000 200,000 250,000 300,000 Strain (με)

Fig. 2.8 Stress–strain obtained through tensile test

– Yield point (yield strength): Yield point of a steel material refers to the point at which only strain increases with no increase of stress. The yield points of specimens #2 and #3 can be observed to be about 500 MPa and 535 MPa, respectively. Likewise, mild carbon steels clearly show their yield points. However, like the case of specimen #1, the yield point of high-strength steels is not clearly shown sometimes. Then, 0.2% offset method or 0.5% of stress of total strain is often defined as the yield strength. – Tensile strength: The maximum stress measured during a tensile test.

2.2 Stress Concentration Test 2.2.1 Purpose of Test • Understanding the strength reduction caused by stress concentration – When a member is subjected to a tensile force and there is a partial loss of area, the tensile stress distribution is not uniform and concentrated on the faulty spot or notch. – In case of faulty round hole, the stress can be concentrated three times more. Sharp angled fault of a tensile member can increase the stress concentration to further decrease its strength.

2.2 Stress Concentration Test

27

– The purpose of this test is to experimentally confirm the difference in tensile strength according to various cross-sectional defect types assuming paper as a tensile material.

2.2.2 Materials and Tools (1) Three pieces of thick paper with weight of about 1.6 N/m2 (= 160 gf/m2 ) and size of 250 mm × 420 mm. (2) Box cutter, ruler, packing tape, electronic weight scale.

2.2.3 Test Method (1) Specimen preparation ➀ Cut the papers in the same size of 50 mm × 420 mm, as shown in Fig. 2.9. ➁ Round and diamond shapes are cut from the center of the papers in size of about 24 mm (or 1 inch), as shown in Figs. 2.10 and 2.11. ➂ Pulling rings made of packing tape (Fig. 2.12) are fastened on both sides of the test specimen as shown in Fig. 2.13. (2) Test method ➀ One person (➊) stands on a weight scale and holds the top of the prepared specimen. ➁ The other person (➋) pulls the bottom of the specimen from outside the scale, and refer to Fig. 2.3 in Sect. 2.1. ➂ Another person (➌) records a video of the display on the weight scale with a mobile phone until the specimen breaks. The failure load is recorded through video replay. Fig. 2.9 Specimen w/o loss of area

28

2 Tension Member

Fig. 2.10 Specimen w/loss of area

420

13 12 12 13

50

P

13 24 13

P

P

50

P

420 Fig. 2.11 Detailed dimension of specimens

Fig. 2.12 Pulling ring

88

2.2 Stress Concentration Test

29

Fig. 2.13 Loading method

Table 2.3 Summary of test results

No.

Defect type

1

Circular

Load (N)

Diamond 2

Circular Diamond

3

Circular Diamond

4

Circular Diamond

➃ The loads are recorded according to faulty hole upon fracture of the specimen, and 4 sets of tensile test s are repeated in the same manner. The maximum applied loads until the fracture occurrence of the specimens are recorded and compared (Table 2.3).

2.2.4 Theory on Stress Concentration (1) As previously discussed in Sect. 2.1, the stress of the member subjected to axial force can be represented as σ = P/A, where P is the applied axial force and A denotes area of the member. This equation is valid under the assumption that the stress is uniformly distributed over the section of the member [8, 9]. However, if there is a change of the member section, the uniform distribution of stress also changes. Change of the member section means there is/are groove, hole (Figs. 2.14 and 2.15), crack, thread, etc., in the member section or the member, of which the section size changes (Figs. 2.16 and 2.17). This discontinuity of section results in high stress at the location of discontinuity, which is called stress concentration. Stress concentration is observed not only in a member

P

P d b

Fig. 2.14 Stress concentration at plate with circular hole

2 Tension Member

c/2

30

c/2

t

P max

Fig. 2.15 Stress concentration factor K for flat bars with circular holes

Fig. 2.16 Stress concentration factor K for flat bars with shoulder fillets

subjected to axial force but also in members subjected to torsion or bending moment [2, 4]. (2) Let us consider a case of a round hole forming at the center of a member with plate shape as shown in Fig. 2.14. The plate width b is relatively large in comparison with the thickness t, and the diameter of the hole is represented by the symbol of d. Maximum stress σ max , which amounts to several times greater than nominal stress σ nom = P/A, appears at the edge of the hole due to applied load [1, 3, 5–7]. As much as the distance of the width b from the edge, the stress

References

31

Fig. 2.17 Stress concentration factor K for round bars with shoulder fillets

Fig. 2.18 Saint-Venant’s principle

is uniformly distributed as nominal stress σ nom . If the size d of the hole becomes very small, maximum stress can be three times the nominal stress. Nonetheless, as the hole size gets larger, the stress concentration phenomenon is not clearly observed (Fig. 2.15). (3) Stress concentration depends also on how load is applied. It can occur when concentrated load is applied to a very small area as shown in Fig. 2.18. Stress amounting to several times nominal stress σ nom = P/A takes place at or around the point of concentrated load. The stress quickly decreases as the distance from the point of concentrated load increases. When the distance of the applied load is the width b of a plate member section, the stress is uniformly distributed and gets closer to nominal stress. This phenomenon is called Saint-Venant’s principle, which has an important significance in design and analysis of rod, beam, shaft, etc [10, 11].

32

2 Tension Member

References 1. AISC. Steel Construction Manual, 15th edn. American Institute of Steel Construction (AISC) (2017) 2. Gere, J.M.: Mechanics of Materials, 5th edn, Nelson Thornes Ltd (2001) 3. Hardash, S.G., Bjorhovde, R.: New design criteria for gusset plates in tension. Eng. J. AISC 22(2nd Quarter), 77–94 (1985) 4. Hibbeler, R.C.: Structural Analysis, 10th edn. Pearson (2018) 5. Kulak, G.L., Fisher, J.W., Struik, J.H.A.: Guide to Design Criteria for Bolted and Riveted Joints, 2nd edn, Wiley (1987) 6. Munse, W.H., Chesson, E.: Riveted and bolted joints: net section design. J. Struct. Div. ASCE 89(ST2), 107–126 (1963) 7. Ricles, J.M., Yurav, J.A.: Strength of double-row bolted-web connections. J. Struct. Eng. ASCE 109, 126–142 (1983) 8. Salmon, C.G., Johnson, J.E., Malhas, F.A.: Steel Structures—Design and Behavior (Emphasizing Load and Resistance Factor Design), 5th edn. Pearson Education, Inc (2009) 9. Salvadori, M, Heller, R.: Structure in Architecture, 2nd edn (1962) 10. Timoshenko, S.P., Goodier, J.N.: Theory of Elasticity, 3rd edn. McGraw-Hill, Inc (1970) 11. Ugural, A.C., Fenster, S.K.: Advanced Strength and Applied Elasticity, 4th edn. Pearson Education, Inc (2003)

Chapter 3

Flexural Member

3.1 Flexural Strength Test of a Beam 3.1.1 Purpose of Test • Understanding methods to increase the strength of a beam as a flexural member – The strength changes due to section enlargement from one layer to two layers by using foam boards. – A composite effect is created by using adhesives and packing tapes. – Present an effective build-up method for section design.

3.1.2 Materials and Tools (1) A sheet of foam board with 5 mm thickness (2) A sheet of extruded polystyrene (also known as Isopink) in 600 mm × 900 mm × 30 mm size (3) Instant adhesive, packing tape, box cutter, ruler (4) Electronic scale of about 30 N capacity.

3.1.3 Test Method (1) Eight pieces of 30 mm × 400 mm size are cut from a sheet of foam board, and five specimens are prepared in accordance with Table 3.1. (2) For the specimen of case ➂, packing tape is wrapped once referring to the drawing in order to convey horizontal shear force. The wrapped tape functions as partial shear connector between the top and bottom layers [8, 9]. © Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_3

33

34

3 Flexural Member

Table 3.1 Built-up method and support condition No.

Setup

Explanation

➀

Note

Simply supported beam using a Reference specimen piece of foam board

P

Support jig Scale

➁

P

Support jig Scale

➂

P

Support jig

Does the specimen exhibit twice the strength of case ➀?

Composite beam with once taping of two layers (※refer to the image for the taping location)

Does the strength increase more than the case ➁?

Simply supported beam with two layers bonded using instant adhesive

Does the strength increase more than the cases ➁ and ➂?

Both ends shall be inserted in groove of support jig.

Does the strength increase more than the case ➀?

30

Scale

Simply supported beam with freely placing two layers (i.e., no connection between the top and bottom layers)

50

20

400 t=5

➃

20

50

(unit : mm)

P

Support jig Scale

➄

P

Support jig Scale

(3) For the specimen of case ➃, the top and bottom layers are made to be one combined member by bonding them with instant adhesive (or glue) at the contact surface. (4) Fixing of two ends of specimen ➄ is achieved by inserting them into ready-made groove as shown in Fig. 3.1 so that the ends would not rotate during loading. (5) A support jig of Fig. 3.1 is made of Isopink, and it is placed on electronic scale. (6) Prepared specimens are laid on support jig, and a concentrated load with one hand is applied to the mid-span of specimens. A foam board with 30 mm × 50 mm is placed at loading point in order to prevent stress concentration and distribute the load. If the measured failure load is conspicuously different from the results of other teams, perform the failure test again. (7) Since failure of a foam board occurs rather suddenly, the loading should be applied slowly. The loading process should be video-recorded with a mobile phone to obtain accurate load value. (8) The measured failure loads as well as the failure shape should also be recorded in Table 3.2.

3.1 Flexural Strength Test of a Beam

35

5 30 30 30 30

Load distributor (30x50)

P

e Groov ort jig Supp

15 0

50 300 50 Fig. 3.1 Test setup and loading (unit: mm)

3.1.4 Reviews (1) The strength change of five tests in response to each test condition is discussed by using the data summarized in Table 3.2. (2) Case ➀ tests a simply supported beam under concentrated loading. This is a counterpart specimen and is compared with the strength of case ➁. Theoretically, specimen of case ➁ should exhibit twice the strength of specimen of case ➀, but it is to be empirically verified. (3) Specimen of case ➂ is made of two layers stacked together with a tape. It behaves like a partially composite beam due to its incomplete composite. Accordingly, it Table 3.2 Summary of test results No. ➀

Setup P

Failure shape

Load (N)

Comparison

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

300

➁

P 300

➂

P 300

➃

P 300

➄

P 300

36

3 Flexural Member Packing tape

Packing tape

Foam board (Top)

Foam board (Top)

Foam board (Bottom)

Foam board (Bottom)

(a) Before loading

(b) After loading it is skewed to the right

Fig. 3.2 Shear deformation

should exhibit greater strength than specimen of case ➁. Examining the deformation of tape before and after loading as shown in Fig. 3.2, sliding is observed due to shear of top and bottom layers. When flexural stress occurs at the section of the two stacked beams, maximum shear force takes place at the contact surface. Stud connectors are used between concrete slab and steel beam to increase the section performance of an actual structure by resisting this shear force [1]. (4) Specimen of case ➃ can theoretically manifest four times greater strength of specimen of case ➀. Its quantity is twice of the latter, but its strength can be theoretically four times greater. However, its manifested strength is seldom four times greater than the latter in actual tests. Its reason should be investigated. (5) Specimen of case ➄ is a beam fixed at both ends and should theoretically exhibit twice the strength of specimen of case ➀. Nonetheless, since the condition of fixing both ends is not perfect in a real test, there are many instances of not manifesting the strength twice that of specimen of case ➀.

3.1.5 Theory on Flexural Member (1) The bending of the member due to external force as shown in Fig. 3.3 is called flexural deformation, and the member resisting flexural deformation is referred as a flexural member. The moment occurring due to external force such as concentrated loading, uniformly distributed loading, and moment makes a member bend. When the member undergoes flexural deformation, it is classified into two parts of the convex part under tensile force and the concave part under compressive force. The center of a section exhibits a line called neutral axis (NA), which is not subjected to tensile force and compressive force. Generally,

Compression Tension

N.A.

(a) By positive (+) moment Fig. 3.3 Bending deformation

Tension Compression

N.A.

(b) By negative (–) moment

3.1 Flexural Strength Test of a Beam

37

σc c2

Fig. 3.4 Arbitrary section and stress distribution by positive moment

c1

N.A.

σ bending σt

(a) Section

(b) Stress distribution

the moment making the bottom convex as shown in Fig. 3.3a is called positive (+) moment, and the moment making the top concave as shown in Fig. 3.3b is called negative (–) moment [4, 5]. (2) Flexural member is simultaneously subjected to compressive stress σc and tensile stress σt centered on the neutral axis, and it fractures upon either of the stresses reaching fracture stress σfr . Flexural stress σbending is computed as follows (Fig. 3.4): M y I

(3.1)

M M c2 = − I Zc

(3.2)

M M c1 = I Zt

(3.3)

σbending = ∓ σc−max = − σt−max = where M I y c1 c2 Zt Zc

is the internal moment (N · mm); is the moment of inertia (mm4 ); is the distance from neutral axis (mm); is the distances between tensile extreme fiber and neutral axis (mm); is the distances between compressive extreme fiber and neutral axis (mm); is the section modulus for tension (mm3 ); and is the section modulus for compression (mm3 ).

(3) Flexural stress is linearly distributed from the neutral axis according to Eq. (3.1), and the largest stress occurs at the ends of a section in accordance with Eqs. (3.2) and (3.3). Fracture takes place at the location of the maximum flexural stress (σt-max or σc-max ), first reaching the fracture stress σfr [10, 11]

38

3 Flexural Member

Moment of inertia and sectional modulus of simple sections

h/3

h

h

r

h/2

D

r

2h/3

b

b

▲ Rectangular section I =

bh 3 12

Z=

bh 2 6

▲ Circular section I =

π D4 64

Z=

π D3 32

▲ Triangular section

=

πr 4 4

I =

=

πr 3 4

Z1 =

bh 3 36 bh 2 12 ,

Z2 =

bh 2 24

.

3.2 Bending Test and Design of a Flexural Member 3.2.1 Purpose of Test • Understanding the design concept and the design procedure of the flexural member – The failure load is measured through the bending test of a small member with Styrofoam. – The concept of flexural stress is to be understood, and the allowable flexural stress is computed in consideration of safety ratio after obtaining fracture stress from fracture load. – The design concept of a flexural member, which can resist larger load considering the allowable flexural stress of material, is to be understood, and a design method of a structure is comprehended through a basic bending test.

3.2.2 Materials and Tools (1) Two sheets of extruded polystyrene (also known as Isopink) with 600 mm × 900 mm × 30 mm (2) Box cutter, ruler, adhesives for Styrofoam, electronic weight scale.

3.2.3 Test Method (1) Cut the Styrofoam in different widths of 40, 50, and 60 mm but the same length of 400 mm as shown in Fig. 3.5. At this time of Styrofoam specimen preparation, the length should be in the lengthwise direction of rolling because the strength can be lowered if it is cut perpendicular to the grain (rolling direction).

3.2 Bending Test and Design of a Flexural Member

39

Fig. 3.5 Bending specimen for test (unit: mm) 0, 60

5 b= 40,

40 0

b

30

Fig. 3.6 Support jig (unit: mm)

15

50 0

300 50

Fig. 3.7 Loading and video-recording

(2) A support jig with effective span of 300 mm is fabricated with Styrofoam as shown in Fig. 3.6. The jig of Fig. 3.1 in Sect. 3.1 can be used again. (3) Put the support jig on the weight scale and the test specimen with 40–60 mm width as shown in Fig. 3.7. (4) One person (➊) slowly applies concentrated load on the mid-span with one hand. The other person (➋) video-records the process with a mobile phone focused on the load display of the scale. The video is replayed to read accurate failure load.

40

3 Flexural Member

3.2.4 Test Results and Applied Computation (1) Concentrated load is applied on the simple beams with span of 300 mm, and the flexural fracture load Pfr is measured for three flexural specimens (b = 40 mm, 50 mm, and 60 mm). (2) When concentrated load is applied on the mid-span of the simply supported beam as shown in Figs. 3.7 and 3.8, maximum moment M = PL/4 takes place at the mid-span [1, 4, 5]. Where P is the applied load, and L is the span length of the member. As the load P gradually increases, the moment increases along the dashed line of Fig. 3.8. When the external force exceeds the resisting capacity of material, the beam fails in flexure. (3) Computation of flexural fracture stress ➀ Flexural stress of flexural member is computed as follows: σb =

M Z

(3.4)

where M is the internal moment (N mm) σ b is the bending stress (MPa or N/mm2 ) and Z is the section modulus (mm3 ). ➁ The sectional modulus of rectangular section is expressed in the following: Z=

bh 2 6

Fig. 3.8 Moment increase with increased loading

(3.5)

P

P/2

P/2 L/2

L/2 L

Loading on simple beam

safe failure

limit of flexural resistance Bending moment diagram

3.2 Bending Test and Design of a Flexural Member

41

where Z is the section modulus (mm3 ); b is the width of section (mm); and h is the height of section (mm). ➂ For example, if the specimen of the specification b = 40 mm, h = 30 mm, and L = 300 mm is fractured at the concentrated center load of Pfr = 48 N, the fracture moment is given as Mfr =

Pf r L 48 N × 300 mm = = 3600 N · mm 4 4 Z=

σfr = σallow =

bh 2 40 × 302 = = 6000 mm3 6 6

Mfr 3600 N mm = = 0.6 N/mm2 Z 6000 mm3

σfr 0.6 N/mm2 = = 0.4 N/mm2 , S 1.5

S is a safety factor

➃ Carry out tests for three specimens, and fill in the computational results in Table 3.3. Note that, although this test uses safety ratio of S = 1.5 for the design of the member, greater safety ratio should be used for actual structures involving human lives. ➄ Record average of allowable flexural stress of above ➂. Use this average as a design data in the following Sect. 3.2.5.

Table 3.3 Summary of test results (sample) No.

Section size Sectional b × h (mm modulus Z × mm) (mm3 )

Fracture load Pfr (N)

Fracture moment M fr (N mm)

Fracture stress σ fr (N/mm2 )

Allowable stress σallow (N/mm2 )

1

40 × 30

6000

48

3600

0.60

0.40

2

50 × 30

7500

59

4425

0.59

0.39

3

60 × 30

9000

63

4725

0.52

0.34

Mean

–

–

–

–

0.38

42

3 Flexural Member

3.2.5 Applied Test of Section Design • Design and fabricate a beam with 800 mm span length (note that actual total length is 900 mm) by using the given Isopink material. The design condition is such that the specimen should be able to withstand the weight of one person (➊) standing on the mid-span of the beam for over 10 s, and then to have the specimen beam collapse upon standing of two persons (➋) simultaneously standing on the specimen.

▲ Safe for one person and collapse for two persons (unit: mm)

▲ Picture of actural bending test by a person standing

3.2.6 Example of Design This example shows a flexural member design method. Each team should suggest different type of beam section according to weight of each team member [1, 8]. (1) Assume that a section is designed as shown in Fig. 3.9 and the maximum moment M c at the mid-span of a simple beam is shown in Fig. 3.10. (2) Assume that the weight of a person is 800 N, and the maximum moment M c is

3.2 Bending Test and Design of a Flexural Member

43

160

30

100

30

Fig. 3.9 Assumed section (unit: mm)

30

90

30

150

Fig. 3.10 Moment distribution

P 800

Mc

Mc =

800 N × 800 mm PL = = 160 × 103 N · mm 4 4

(3) Performances of the proposed section are given as I =

bh 3 150 × 1603 90 × 1003 BH3 − = − = 43.7 × 106 mm4 12 12 12 12 Z= σb =

43.7 × 106 I = = 546 × 103 mm3 H/2 160/2 160 × 103 N · mm M = = 0.293 N/mm2 Z 546 × 103 mm3 σb 0.293 = 0.73 or 73% = σallow 0.4

(4) Since the specimen is designed with 73% of allowable stress σallow , adequate safety is secured. Nonetheless, if it is over safely designed, it is possible to reduce the section of a member [9–11).

44

3 Flexural Member

3.2.7 Task • Compute sectional modulus for neutral axis of the following sections as shown in the following figures (unit: mm).

6

30

20 6

30

30

25

20

10 10

30

20

20

30

3.3 Deflection Test of a Beam 3.3.1 Purpose of Test • Understanding deflection and its calculation method of a flexural member – The deflection of a simple beam and a cantilever made of steel ruler is measured when coins as uniformly distributed load are applied. – The measured deflection is compared with theoretical approach. – The deflection with respect to various spans and support condition s is discussed through simple deflection tests, and the deflection equation for a beam is to be understood.

3.3.2 Materials and Tools (1) Steel rulers with capacities of 300, 500, and 600 mm (2) Two paper (or plastic) cups, packing tape, vernier calipers (3) Thirty coins (e.g., quarter in USA, ten pence in UK, e0.2–1 in Europe, and 100 won in South Korea), electronic scale with about 10 N capacity.

3.3 Deflection Test of a Beam

45

3.3.3 Test Method (1) Weight one coin or ten coins to take the average weight, as shown Fig. 3.11 and measure the diameter of coin using vernier calipers. (2) Arrange several coins in line with a steel ruler as shown in Fig. 3.12, and apply the packing tape on front and back of the coins to make uniformly distributed load. Compute the uniformly distributed load of the coins arranged in a line. (3) Measure and record the width and thickness of the steel ruler with a vernier calipers in order to compute the moment of inertia I required for deflection computation. (4) Select a stainless steel ruler with measuring limit of 600 mm, and place its two ends on the two cups as shown in Fig. 3.13. Since the total length of ruler is longer than 600 mm, the effective span length between two cups should be 600 mm. The height at the mid-span of the specimen before placing the coins, which is applied by self-weight, is measured with a 300 mm steel ruler as shown in Fig. 3.14. This measured height is set to be the initial deflection δ 0 . Fig. 3.11 Measuring a coin weight

Fig. 3.12 Uniformly distributed load with taped coin arrangement

46

3 Flexural Member

0

600 Cup

Steel ruler Fig. 3.13 Setup of steel ruler and supports (unit: mm)

Fig. 3.14 Measuring an initial deflection

1

(5) After placing the coins made for uniformly distributed loading on the steel ruler as shown in Fig. 3.15, the deflection δ 1 is measured as shown in Fig. 3.16. (6) Deflection δ of the simple beam using steel ruler under uniformly distributed loading is computed by δ 0 –δ 1 . (7) Deflection of cantilever

Fig. 3.15 Deflection due to distributed coins

3.3 Deflection Test of a Beam

47

Fig. 3.16 Measuring the deflection by loading

Fig. 3.17 Cantilever beam

➀ Cantilever refers to a member, of which one end is fixed against moving and rotating, and the other end is freely moveable, as shown in Fig. 3.17. Since fixing against rotating is difficult to realize in an experiment, the cantilever test is carried out by using symmetrical condition of the specimen so that rotation of the fixed part (center of the specimen) would not take place. ➁ A cup is moved to the center of steel ruler to simulate fixing of one end of cantilever in this test as shown in Fig. 3.18, paying attention to balance both ends. Using this test setup, the initial deflection δ 0 of one free end is measured. ➂ Deflection δ 1 is induced by loading uniformly distributed coins as shown in Fig. 3.19. Care should be taken to balance both sides of cantilever during loading. ➃ Deflection δ of the cantilever subjected to uniformly distributed load is computed by δ 0 –δ 1 . (8) Applied test: Carry out aforementioned simple beam test and cantilever test with a ruler of 500 mm measuring capacity. Since deflection of a beam due to Fig. 3.18 Setup of steel ruler and support

48

3 Flexural Member

Fig. 3.19 Loading and deflection

uniformly distributed load is proportional in fourth power to span length, a span reduction results in less deflection than expected. On the contrary, because using a ruler of 700 mm measuring capacity increases deflection more than expected, this fact should be considered for structural design.

3.3.4 Test Results and Deflection Computation (1) Moment of inertia I is computed by using the given steel ruler section. I =

bt 3 12

(3.6)

where b is the width of section (mm); and t is the thickness of section (mm). (2) The load per unit length ω is computed by using the total number of coins placed on the ruler. ω (N/mm) =

total number of coins × weight of one coin × acceleration of gravity (3.7) span length

(3) The modulus of elasticity E of the stainless steel ruler can be found from internet search. (4) Deflection is computed by a theoretical equation, using span lengths and support conditions. δ=

5ωL 4 for simply supported beam 384E I

(3.8)

ωL 4 for cantilever beam 8E I

(3.9)

δ=

where assume that L = 500 mm and L = 600 mm for simple beam and L = 250 mm and L = 300 mm for cantilever beam, respectively. (5) Compare with theoretical deflections and test results for each case.

3.3 Deflection Test of a Beam

49

Table 3.4 Summary of test results Support condition

L (mm) b (mm) t (mm) I (mm4 ) δ theory (mm) δ 0 (mm) δ 1 (mm)

*δ

Simple beam

600

–

–

–

–

–

–

500

–

–

–

–

–

Cantilever 300 beam

–

–

–

–

Left Right

Right

250

–

–

–

Left

Left

Right

Right

test

(mm)

– Left

– –

Note that δ test = δ 0 –δ 1

3.3.5 Case Study (1) Weight of ten coins is 0.539 N → weight of one coin is p = 0.0539 N (2) If the diameter d of the coin is 24 mm, the uniformly distributed load due to coin is ω=

0.0539 N p = = 0.002246 N/mm d 24 mm

(3) Computation of moment of inertia I of a steel ruler and deflection (Table 3.4).

3.3.6 Theory on Beam Deflection (1) The concept of designing a structure so that the stress exerted to the member is to be less than the allowable stress of the material was examined through a flexural test of a beam in the previous chapter [1, 5, 10, 11]. However, there is still one more thing to consider besides stress design as far as actual structural design is concerned. That is deflection. Generally, it refers to deformation of a flexural member toward gravity direction due to gravity. (2) When you are lying on a bed, trying to go sleep, and find the ceiling drooping or trying to go across a rocking rope bridge, you will be psychologically scared. Therefore, an inspection to examine deflection is important to take away psychological presentiment and to prevent the installed equipment from being impeded [8, 9] (Fig. 3.20). (3) It will be ideal that a structure does not show a deflection, but there is no material with infinite stiffness. Thus, it is inevitable, and there will be deflection of a structure. Deflection increases in proportion to load (P, ω) and to length (L) of the member between cubic and fourth powered. It decreases in proportion to flexural stiffness (EI). (4) Methods to compute deflection of a member include elastic curve method, elastic load method, theorem of moment area, conjugated beam method, virtual work method, Castiliano’s theorem, etc [4, 5]. However, this textbook will not describe these methods but show deflection equation for a typical statically determinate structure according to the applied load.

50

3 Flexural Member

Fig. 3.20 Deflection

Deflection equations

δ=

P L3 3E I

δ=

ωL 4 8E I

δ=

δ=

P L3 48E I

δ=

5ωL 4 384E I

δx =

M L2 2E I

M 6E I



2L x − 3x 2 +

x3 L



3.3.7 Task • Find slope at both supports and maximum deflection at the mid-span when concentrated load or uniformly distributed load is applied to a simply supported beam (E= 205,000 MPa, I= 7 × 107 mm4 ).

3.4 Lateral Buckling Test of a Beam

51

3.4 Lateral Buckling Test of a Beam 3.4.1 Purpose of Test • Observing the strength change according to unbraced length of a beam. – Although an efficient section can be designed with thinner web and flange in order to enhance the flexural strength of a beam, given the same quantity of material, lowering of strength can take place due to local buckling in case of increased width to thickness ratio of the beam if the flange or web is too thin. – As the height of a beam section increases, the in-plane flexural capacity also increases. However, if the beam height reaches a certain level, the out-of -plane lateral buckling (or sideway movement) can occur. It results in rather decrease of the flexural capacity. – In this case, flexural capacity of the beam can be managed through a lateral support, which can prevent it from collapsing to a side. – Lateral support between two members is used to shorten the unbraced length. When the unbraced length is shortened to a certain length, the flexural capacity of the beam nears plastic moment M p . Flexural strength test is performed by varying the unbraced length using lateral support in order to investigate this phenomenon experimentally.

3.4.2 Materials and Tools (1) A sheet of foam board with 600 mm × 900 mm × 3 mm (2) ½ sheet of extruded polystyrene (also known as Isopink) with 600 mm × 900 mm × 50 mm 3) Glue gun and glue, ruler, box cutter, electronic scale.

3.4.3 Test Method (1) Specimen preparation ➀ Six H-shaped beams with section of H–66 × 20 × 3 × 3, and total length of 600 mm are prepared as shown in Fig. 3.21. Each team can vary 60 mm web height of to be 40 mm or 50 mm. ➁ Two beams are connected in a set; therefore, prepared H-shape beams make up three sets. ➂ Three sets are prepared in the following three cases. L b is the unbraced length.  a Case 1: Lateral support is located at A and A . L b is 500 mm.  b Case 2: Lateral support is located at A, A , and B. L b is 250 mm.

3 Flexural Member

66

60

52

3 20 Fig. 3.21 H-shape (unit: mm)  c Case 3: Lateral support is located at A, A , B, C, and C’. L b is 125 mm.

➃ Support jig is simply made of Isopink, as shown in Figs. 3.22 and 3.23.

C

A

P

Plate for loading

B

C'

X-brace

P

A'

50

Stiffener

50

125

50

Support jig

125

125

125

100

50

Plate for loading

Top view

Fig. 3.22 Test setup (unit: mm)

Fig. 3.23 Actual test setup

Side view

80 100

Front view

3.4 Lateral Buckling Test of a Beam

53

Fig. 3.24 Loading

(2) Test method ➀ Support jig is placed on a scale and the specimen is laid upon the support jig as shown in Fig. 3.23. ➁ Load is applied to the mid-span of the beam with Isopink of 50 mm × 50 mm × 450 mm as shown in Fig. 3.24. ➂ Load is kept applying even though the beam collapses laterally. When it collapses laterally, the loading is controlled by a fingertip so that only vertical load is applied. ➃ The strength at the fracture time is recorded by a mobile phone, taking a video of the load value. Compare the measured load values in the three cases with different unbraced length.

3.4.4 Test Results (1) Failure mode: Observation of failure mode as a test result is very important. Sudden fracture or joint failure results in inability to manifest maximum performance of used materials. The structure should be designed so that the strength does not drop sharply even if deformation continues after showing signs of collapse (Table 3.5).

54

3 Flexural Member

Table 3.5 Summary of test results (sample) Type of lateral support

Unbraced length, L b (mm)

Measured load, P (N)

Strength comparison

Failure mode

Case 1

500

21.12

–

–

Case 2

250

41.28

–

–

Case 3

125

46.28

–

–

3.4.5 Theory on Lateral Buckling of a Beam (1) Nominal flexural strength of beam [1, 3] ➀ The nominal strength M n of steel beam shows the relationship with unbraced length L b , as shown in Fig. 3.25. If the L b is short, the beam can show the maximum strength M p with plastic deformation. Plasticity zone is characterized by large deformation capacity and exhibits ductile failure behavior of not showing sudden failure. Therefore, it is possible to predict collapse of the beam, and the materials manifest strength as efficiently as possible. ➁ Actually, it is desirable to design the lateral support member to be within plasticity zone. If the L b is long, it belongs to inelasticity zone, and the complex design equation is required as shown in Fig. 3.25. As the L b increases more, it reaches elasticity zone, and the strength of the beam drops below its initial strength. The design equation becomes more complex, and sudden fracture occurs. Since economic value of the section decreases in Mn M n=M p M n = M p (M p 0.7F y S x)

Mp

Lb L p Lr L p

≤ Mp

M n with Cb>1 Mn = Mr

π2E Jc Lb 1+0.078 S xh0 rts Lb 2 rts

2

Sx ≤ M p

0.7F y S x M n with Cb=1

Lp Plastic moment

Lr Inelastic lateral-torsional buckling

Lb Elastic lateral-torsional buckling

Fig. 3.25 Nominal flexural strength as function of unbraced length and moment gradient

3.4 Lateral Buckling Test of a Beam

55

this case, lateral support should be used to design the beam at full plastic moment strength M p . ➂ Since most slabs of a building function as lateral support, design of full plasticity moment is possible. However, because lateral support is not possible until the slabs are cured, there should be a caution in this regard. (2) Lateral buckling failure of a beam and lateral support method [8, 9] ➀ Figure 3.26 shows failure type of a specimen above. Case 1 shows the specimen greatly bending laterally and then reaching failure. Case 2 exhibits the lateral deformation somewhat diminished. ➁ Figure 3.27 shows the behavior of purlins with haystack. Its structural type is such that the purlin with greater height than width is not laterally supported but merely spreads over the span of a girder. The deformation of lateral buckling, of which the section collapses laterally and the mid-span exhibits significant deformation, can be seen. ➂ Figure 3.28 shows an image of collapse accident during construction due to lack of lateral support. It is common to see collapse of a structure without lateral support during construction. The lateral support location of a girder should have the top and bottom flanges connected by a brace or small beam to prevent it from collapsing laterally. Fig. 3.26 Failure shape of tested specimen

Fig. 3.27 Lateral buckling of purlin

56

3 Flexural Member

Fig. 3.28 Collapse by lack of lateral support

➃ Figure 3.29 shows a case of lateral support with a small truss crossing a girder perpendicularly. ➄ Figure 3.30 shows building of a stable top plate with reduced span of the slab as well as lateral support by connecting a girder with a small beam.

Fig. 3.29 Honeycomb beam supported laterally by truss

Fig. 3.30 Girder supported laterally by several small beams

3.5 Shear Center Test (Focused on Channel Section)

57

3.5 Shear Center Test (Focused on Channel Section) 3.5.1 Purpose of Test • Comprehending the concept of shear center and verifying it experimentally after theoretical approach – General computation of stress at a beam is carried out under the condition of no torsion. If the section is symmetrical to the loading axis and load is applied to the section center, no torsion will occur. However, if it is not symmetrical, torsional stress takes place at the section. The location of the section with no torsional stress is referred to as shear center. – Shear center of a section is explored through a test, and its deviation from theoretical value is computed. A typical case of observing torsional stress even loading on center of the section is channel section. The task of this test is to prepare channel section and find the shear center through a test to compare it with theoretical value.

3.5.2 Materials and Tools (1) A sheet of Styrofoam with size of 600 mm × 900 mm × 5 mm (2) Box cutter, ruler, adhesive for Styrofoam, weight.

3.5.3 Test Method (1) The specimen specification with size of C–60 × 20 × 3 and 600 mm total length is shown in Fig. 3.31. Loading plate is attached at free end of A, and the top of the plate is overlaid over the center of the section (30 mm from the top). (2) The fixed end of B is placed on a desk and is fixed by holding it with a hand as shown in Fig. 3.32. (3) The loading by using a weight is applied to a random location to observe torsional deformation of the free end of A as shown in Fig. 3.33. The eccentricity location

30

60

20

Loading plate

A

Fig. 3.31 Specimen dimension (unit: mm)

B

600

58

3 Flexural Member

Fig. 3.32 Setup

e

e

P

Fig. 3.33 Loading with eccentricity

P

3.5 Shear Center Test (Focused on Channel Section)

59

e, where the vertical deflection occurs but no torsion occurs, when load P is applied to section, is recorded to compare it with theoretical computational value. (4) If a weight is not available, loading can be applied by the tip of a ball pen. Then, the location where no rotation of the loading plate occurs is investigated by varying the eccentric distance e. (5) After recording the eccentric distance, it is compared with the theoretical value as discussed in the following Sect. 5.4.

3.5.4 Theory on Shear Center (1) Since channel beam is not symmetrical to 2 axes but 1 axis, the load applied parallel to the non-symmetrical axis brings about bending and torsional deformation even if it passes through the section center. However, when the line of loading crosses a certain spot of the section, no torsional deformation takes place in the member. This spot is called shear center [1, 3–5]. (2) In case of channel as shown in Fig. 3.34, it is symmetrical to x-axis only, and the shear center coincides with the centroid on x-axis [8, 9]. However, shear center takes place at the location as much as x c apart from y-axis. The location of shear center at a section can be computed from equilibrium condition of internal moment due to shear stress and external moment owing to shear force. The moment equilibrium condition with respect to the centroid of section is shown in Eq. (3.10). Therefore, the distance from centroid to shear center can be computed from Eq. (3.11).  V xc − xc =

n(vt)ds = 0 1 V

(3.10)

 n(vt)ds

(3.11)

Fig. 3.34 Shear center of channel section

Shear Center

Centroid

V

60

3 Flexural Member

q2 V Sc

q1

C

Fig. 3.35 Shear flow at channel

where V is the shear force (N). (3) Equation (3.12) expresses shear stress v. The vt of Eq. (3.13) is shear stress multiplied by thickness of the section and represents shear force per unit length. It is called shear flow q [10, 11] (Fig. 3.35). v=

VS tI

q = vt =

(3.12)

VS I

(3.13)

where v is the shear stress (MPa or N/mm2 ); S and I are the geometric moment of area (mm3 ) and the moment of inertia (mm4 ); t is the web thickness (mm); and q is the shear flow (N/mm). (4) Fig. 3.36 shows the location of shear center depending on the shape of a section.

Sc Sc

Sc Sc

Fig. 3.36 Shear center in variable section

3.5 Shear Center Test (Focused on Channel Section)

61

3.5.5 Examples • Compute the shear center location for C–300 × 90 × 9 × 13 section (unit: mm). [Solution] (1) The section properties of C-300 × 90 × 9 × 13 are A = 4857 mm2 , C y = 22.3 mm,

Ix = 64.4 × 106 mm4

(2) The geometric moments of area for flange of ➀ and web plus flange of ➁, with respect to x-axis, are respectively given by S1 = 90 × 13 × (300/2 − 13/2) = 167.9 × 103 mm3 S2 = S1 + (300/2 − 13) × 9 × (300/2 − 13)/2 = 252.4 × 103 mm3 (3) Shear stress and shear flow owing to a shear force V are, respectively, given by Eqs. (3.12) and (3.13) v1 =

V S1 167.9 × 103 × V = = 0.29 × 10−3 V tw I x 9 × 64.4 × 106

q1 = tw v1 = 9 × 0.29 × 10−3 V = 2.61 × 10−3 V v2 =

V S2 252.4 × 103 × V = = 0.44 × 10−3 V tw I x 9 × 64.4 × 106

q2 = tw v2 = 9 × 0.44 × 10−3 V = 3.96 × 10−3 V (4) Shear flow is shown in Fig. 3.37, and the equilibrium condition of moment at web center of ➁ is given by 

 M = V ×m−

 1 × 2.61 × 10−3 V × 90 (300 − 13) = 0 2

m = 33.7 mm (5) Shear center is located 33.7 mm left from the center line of web, and it can be computed from the centroid as follows.  xc = m + C y − tw /2 = 33.7 + (22.3 − 9/2) = 51.5 mm

62

3 Flexural Member

13

q2

V

300

9

Sc

C

q1

90

Fig. 3.37 Shear flow at channel (unit: mm)

• Compute the shear center location for C–70 × 30 × 5 × 5 section (unit: mm). [Solution] (1) Neutral axis of C-70 × 30 × 5 × 5 section is Cy =

A1 x 1 + A2 x 2 (70 − 5 × 2) × 5 × 5/2 + (30 × 5 × 30/2) × 2 = = 8.75 mm A1 + A2 (70 − 5 × 2) × 5 + (30 × 5) × 2

(2) Moment of inertia and geometric moment of area for flange and flange plus web, with respect to x-axis are respectively calculated by Ix =

BH3 bh 3 30 × 703 (30 − 5) × (70 − 5 × 2)3 − = − = 407.5 × 103 mm4 12 12 12 12

S1 = 30 × 5 × (70/2 − 5/2) = 30 × 5 × 32.5 = 4875 mm3 S2 = S1 + (70/2 − 5) × 5 × (70/2 − 5)/2 = 4875 + 2250 = 7125 mm3 (3) Shear flow owing to a shear force V is computed from Eqs. (3.12) and (3.13). q1 = tw v1 =

V S1 4875V = = 11.96 × 10−3 V Ix 407.5 × 103

3.5 Shear Center Test (Focused on Channel Section)

63

Fig. 3.38 Shear flow at channel (unit: mm) 5

q2

5 70

V Sc

C

q1

30

q2 = tw v2 =

V S2 7125V = = 17.48 × 10−3 V Ix 407.5 × 103

(4) From shear flow as shown in Fig. 3.38, the equilibrium condition of moment at web center of ➁ is given by 

 M = V ×m−

 1 × 11.96 × 10−3 V × 30 (70 − 5) = 0 2

m = 11.7 mm (5) Shear center is located 11.7 mm left from the center line of web, and it can be computed from the centroid as follows.  xc = m + C y − tw /2 = 11.7 + (8.75 − 5/2) = 17.95 mm

3.6 Principle of a Reinforced Concrete Beam 3.6.1 Purpose of Test • Comprehending the principle of a reinforced concrete (RC) beam – RC beam is reinforced by inserting steel bar into tensile part of a concrete section so that the steel can resist pure tensile force, resulting in greater flexural strength of a beam – If a load perpendicular to the length of a beam is transferred to inside of the beam, the steel resists the external moment with horizontal force. – It is the most advantageous to increase the height of a beam to enhance the efficiency of a steel bar.

64

3 Flexural Member

– The test helps students to comprehend how to enhance the sectional efficiency with a steel bar and how the same sections manifest different moment resisting effect by height of a beam.

3.6.2 Materials and Tools (1) Wooden stick of three different sections (a) 30 mm in width × 30 mm in height × 600 mm in length (b) 30 mm in width × 45 mm in height × 600 mm in length (c) 30 mm in width × 60 mm in height × 600 mm in length (2) (3) (4) (5)

A sheet of foam board with size of 300 mm × 100 mm × 1 mm Support jig of Figs. 3.6 and 3.7 in Sect. 3.2. Box cutter, ruler, instant adhesive Electronic scale with 60 N capacity

3.6.3 Test Method (1) Specimen preparation ➀ Cut sixty foam boards to be used as steel bar in 5 mm width and 100 mm length. ➁ Place two wooden sticks of same section against each other, and affix four foam boards as steel bar with an instant adhesive. The foam boards are bonded so that the top and bottom are symmetrical (Figs. 3.39 and 3.40).

600 100

cut

45

300

300 30

Fig. 3.39 Wooden stick bonded with foam boards

3.6 Principle of a Reinforced Concrete Beam

65

30

Foam board (5 mm x 1 mm x 100 mm)

30

30

45

60

30

Fig. 3.40 Three different sections

Table 3.6 Tensile strength of foam board (Sample)

No. of coupon

T 1 (N)

1

12.3

2

10.2

3

14.8

4

15.3

5

11.2

Mean

12.8

(2) Test method ➀ Tensile tests similar to those of Sect. 2.1 are performed with foam boards of 1 mm thickness and 5 mm width by team, and the results are recorded in Table 3.6. Five tensile tests are carried out to measure tensile strength of a foam board. The average of three tensile test results excluding the maximum and minimum values is used to obtain tensile strength T 1 of one foam board with size of 1 mm × 5 mm. Packing tapes are applied to both sides of the foam board so that eccentric force would not take place during tensile strength test. ➁ Support jig for flexural loading is placed on a scale with 60 N capacity as shown in Fig. 3.41, and concentrated load at mid-span is applied to measure the experimental maximum load Ptest . The span length is set to be 300 mm. ➂ Experimental variables are the following. a. 30 mm in width × 30 mm in height × 600 mm in length b. 30 mm in width × 45 mm in height × 600 mm in length c. 30 mm in width × 60 mm in height × 600 mm in length d. Number of foam board bars in bottom (two foam boards in top): 2, 3, 4

66

3 Flexural Member

Fig. 3.41 Setup for loading

3.6.4 Test Results and Applied Computation (1) Moment resistance M n of a beam is computed by the following equation [2, 6, 7]. Mn = T1 × n × j × d

(3.14)

where T 1 is the tensile strength of foam board from tensile test (N), n is the number of foam board bars bonded on wooden stick beam, d is the distance from top of beam to center of steel bars in RC beam (mm), and j is the distance ratio between center of stress block and steel bars (about 0.9). (2) The concentrated load P applied at mid-span in simply supported beam can be calculated from equations [4, 5] M=

PL 4

(3.15)

P=

4M L

(3.16)

where M is the bending moment at mid-span in simply supported beam (N mm), P is the applied load at mid-span (N), and L is the span length (mm). (3) M n computed from Eq. (3.14) is substituted to Eq. (3.16) to obtain theoretical maximum load Ptheory .

3.6 Principle of a Reinforced Concrete Beam

67

3.6.5 Test Case (1) Strength of a foam board as replacement for steel bar obtained in Sect. 3.6.3 is recorded in Table 3.6. (2) The maximum load Ptest was found to be 12.0 N based on the bending test result of bonding two foam boards on 30 mm × 45 mm wooden beam. (3) Theoretical approach Mn = T1 × n × j × d = 12.8 × 2 × 0.9 × 46 = 1060 N · mm Ptheor y =

4 × 1060 4Mn = = 14.1 N L 300

Ptest 12.0 = 0.85 = Ptheor y 14.1 Therefore, there was an error of about 15% (Table 3.7).

Table 3.7 Test results Ptest Ptheor y

Size of beam

The number of foam boards

Ptest

Ptheory

b × h × L (mm)

Top (EA)

Bottom (EA)

(N)

(N)

30 × 30 × 600

2

2

–

–

–

2

3

–

–

–

30 × 45 × 600

30 × 60 × 600

2

4

–

–

2

2

12.0

14.1

2

3

–

–

– 0.85 –

2

4

–

–

–

2

2

–

–

–

2

3

–

–

–

2

4

–

–

–

68

3 Flexural Member

3.6.6 Tasks • Find values of Ptest and Ptheory for three sections of Table 3.7 and compare them. ➀ 30 mm × 30 mm section of Fig. 3.40a + 2, 3, or 4 tensile steel bars ➁ 30 mm × 45 mm section of Fig. 3.40b + 2, 3, or 4 tensile steel bars ➂ 30 mm × 60 mm section of Fig. 3.40c + 2, 3, or 4 tensile steel bars

3.6.7 Principle of an RC Beam (1) When an RC beam is reinforced with a rebar on tensile side only as shown in Fig. 3.42, it is called a beam with single reinforcement. When the RC section is subjected to positive (+) moment, the rebar under the section and the concrete over the neutral axis (NA) is subjected to tensile force and compressive force, respectively [6, 7]. (2) Nominal flexural strength M n refers to the flexural resistance, where the tensile rebar yields and the concrete strain at compressive extreme fiber of concrete reaches 0.003 as shown in Fig. 3.42b. The ultimate state, at which compressive force C of concrete and tensile force T of rebar is the same as shown in Fig. 3.42c, is considered in computation of nominal flexural strength M n . Here, C represents compressive force from equivalent rectangular stress block, and T denotes tensile force at the yield state of tensile rebar. Computation of nominal flexural strength is carried out in the following order.

0.85f c'

d

d

c

C

N.A.

d a/2

a

= 0.003

c

c

a/2

C=T

T b (a) Section

s

>

y

(b) Strain distribution

(c) Stress distribution

Fig. 3.42 Stress and strain distribution of beam with single reinforcement

3.6 Principle of a Reinforced Concrete Beam

69

0.85 f c ab = As f y As f y a , c=  0.85 f c b β1   a Mn = A s f y d − 2

a=

where C f’c a b c T As fy d

is the compressive force acting on concrete (N), is the compressive strength of concrete (MPa), is the height of equivalent rectangular stress block (mm), is the beam width of concrete (mm), is the depth of neutral axis from top (mm), is the tensile force acting on tension reinforcement (N), is the total area of tension reinforcements (mm2 ), is the yield strength of tension reinforcements (MPa), is the effective depth (mm), and

β1 = a/c

if f c ≤ 28 MPa = 0.85   = 0.85 − 0.007 f c − 28 ≥ 0.65 if f c ≥ 28 MPa

References 1. AISC. Steel Construction Manual, 15th edn. American Institute of Steel Construction (AISC) (2017) 2. American Concrete Institute Committee 318. Building Code Requirements for Structural Concrete (ACI 318-14) and Commentary on Building Code Requirements for Structural Concrete (ACI 318R-14). ACI, MI, USA 3. Chen, W.F., Lui, E.M.: Structural Stability—Theory and Implementation. Elsevier Science Publishing Co., Inc (1987) 4. Gere, J.M.: Mechanics of Materials, 5th edn. Nelson Thornes Ltd (2001) 5. Hibbeler, R.C.: Structural Analysis, 10th edn. Pearson (2018) 6. Macgregor, J.G., Wight, J.K.: Reinforced Concrete—Mechanics and Design, 4th edn. Pearson Education, Inc (2005) 7. Park, R., Paulay, T.: Reinforced Concrete Structures. Wiley (1975) 8. Salmon, C.G., Johnson, J.E., Malhas, F.A.: Steel Structures—Design and Behavior (Emphasizing Load and Resistance Factor Design), 5th edn. Pearson Education, Inc (2009) 9. Salvadori, M., Heller, R.: Structure in Architecture, 2nd edn (1962) 10. Timoshenko, S.P., Goodier, J.N.: Theory of Elasticity, 3rd edn. McGraw-Hill, Inc (1970) 11. Ugural, A.C., Fenster, S.K.: Advanced Strength and Applied Elasticity, 4th edn. Pearson Education, Inc (2003)

Chapter 4

Compressive Member

4.1 Buckling Test of a Spaghetti Noodle 4.1.1 Purpose of Test • Obtaining the modulus of elasticity of a spaghetti noodle through buckling test with assuming that the spaghetti noodle is a column – Buckling loads depending on the length of spaghetti noodles are obtained through simple buckling test of them. – The concept of a buckling equation is comprehended, and a method to obtain elasticity modulus of a slender member without tensile test is to be understood.

4.1.2 Materials and Tools (1) One bag of spaghetti noodles with around 2 mm diameter (2) Box cutter, Ruler, Vernier calipers, Electronic scales with 3 and 30 N capacities

4.1.3 Test Method (1) Spaghetti noodle specimens are prepared by the lengths of Table 4.1, as shown in Fig. 4.1. (2) Spaghetti specimens are vertically placed on a scale, as shown in Fig. 4.2. A load is vertically applied on the spaghetti section. Care should be taken so that the spaghetti sections will not tilt. (3) The failure load is video-recorded, and spaghetti noodles of each length are tested. © Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_4

71

72

4 Compressive Member

Table 4.1 Summary of test results (sample) L (mm)

E (MPa)

Pcr-theory (N)

255

Pcr-test (N) 0.44

3690.99

0.43

230

0.56

3821.68

0.53

200

0.75

3870.18

0.70

170

1.00

3728.28

0.97

140

1.34

3388.22

1.43

110

2.09

3262.44

2.32

86

3.70

3530.28

3.80

86

3.80

3625.69

3.80

86

3.30

3148.63

3.80

86

3.10

2957.80

3.80

80

6.00

4953.84

4.39

80

4.80

3963.07

4.39

75

4.10

2975.20

5.00

75

4.67

3388.83

5.74

70

8.40

5309.89

5.74

70

4.57

2888.83

5.74

70

4.32

2730.80

5.74

60

7.70

3576.05

7.81

60

10.60

4920.55

7.81

Fig. 4.1 Spaghetti noodles

4.1 Buckling Test of a Spaghetti Noodle

73

Fig. 4.2 Buckling test for spaghetti noodle

4.1.4 Test Results (1) The diameters of spaghetti sections are measured with vernier calipers to compute average diameter, and then moment of inertia is calculated. The moment of inertia for a cylindrical section is computed by I =

π d4 64

(4.1)

where I and d are the moment of inertia (mm4 ) and diameter (mm) of spaghetti section, respectively; (2) Modulus of elasticity E is computed by Euler’s buckling equation for a column. Since a short column is hard to apply the load, the accurate value of E is difficult to measure. Therefore, a minimum of three test data mainly composed of long members is used [3, 4]. Pcr =

π2E I (L e )2

(4.2)

where Pcr E I Le

is the Euler’s buckling load (N), is the modulus of elasticity (MPa), is the moment of inertia (mm4 ), and is the effective buckling length of column (mm)

(3) For example, if Pcr of a spaghetti noodle of 110 mm length is 2.09 N, then

74

4 Compressive Member

I = E=

π × 24 π d4 = = 0.786 mm4 64 64

Pcr (L e )2 2.09 × (1 × 110)2 = = 3260 N/mm2 or MPa 2 π I π 2 × 0.786

(4) Compute theoretical buckling load, and plot a graph of theoretical value along with test data.

4.1.5 How to Prepare a Report (1) Since short column is difficult to apply the load, its accurate test value is hard to measure. Therefore, elasticity modulus data are mainly obtained from long columns. (2) Test results are to be clearly presented with a table and a graph.

4.1.6 Samples (1) Test data: See to Table 4.1 (2) Elasticity modulus E was computed to be 3626.96 MPa through the average of those of five long columns as presented in Table 4.1. Theoretical buckling load along the column length was computed by using theoretical buckling equation π 2 EI/(L e )2 and average E of a spaghetti section, and the result is depicted in the graph of the following Fig. 4.3. (3) Calculation process ➀ Record test results in Table 4.2. 12

Experiment Euler's formula

10 8

Pcr (N)

Fig. 4.3 Comparison with test results and Euler’s formula

6 4 2 0

0

50

100

150

L (mm)

200

250

300

4.1 Buckling Test of a Spaghetti Noodle Table 4.2 Summary of test results

75

L (mm) 255

Pcr-test (N) 1 2 3

210

1 2 3

170

1 2 3

130

1 2 3

90

1 2 3

50

1 2 3

230

1 2 3

190

1 2 3

150

1 2 3

110

1 2 3

70

1 2 3

E (MPa)

76

4 Compressive Member

Fig. 4.4 Comparison of test results and Euler’s formula

12 10

Pcr (N)

8 6 4 2 0

0

50

100

150

200

250

300

L (mm)

➁ Compute the followings d=( I =

π( π d4 = 64 64

) mm )4

=(

) mm4

L e = 255, 230, 210, 190, 170, 150, 130, 110, 90, 70, 50 mm E=

( )×( Pcr-test (L e )2 = π2I π2 × (

)2 )

=(

) N/mm2 or MPa

➂ Average of the elasticity modulus of long columns with lengths of 255– 170 mm is E =(

) N/mm2 or MPa

➃ Plot the test data as shown in Fig. 4.4. ➄ Buckling load Pcr for each tested length L e of spaghetti section is computed by using the calculated average of elasticity modulus of ➂ and Pcr = π 2 EI/(L e )2 , and Pcr-test is marked on Fig. 4.4.

4.1.7 Buckling Theory on a Column (1) The member subjected to compressive force in axis direction is called a column (or compressive member). When the compressive force acts on a column having

4.1 Buckling Test of a Spaghetti Noodle

77

Fig. 4.5 Deformation of short column owing to compression

Fig. 4.6 Compressive test of concrete

relatively large cross section and a short length, the column is vertically shortened and laterally expanded, as shown in Fig. 4.5. As the compressive force increases gradually and reaches the ultimate strength of the material, compressive failure takes place as shown in Fig. 4.6. This column is called a short column [3, 4]. (2) On the contrary, when the compressive force is applied to a column with small cross section and relatively long length, it behaves like a short column under small compressive force. However, when the load increases beyond a certain critical load Pcr , the column loses its stability even the stress applied to the section does not reach ultimate strength, and it bends laterally. This column is called a long column, and the phenomenon of bending laterally is called buckling [2] (Fig. 4.7). (3) The compressive stress that takes place at short column similarly follows the theory of tensile member as discussed in Sect. 2.1. This section focuses on a buckling behavior of a long column. Buckling is related to bending moment due to eccentric force or slenderness ratio of compressive member. Buckling direction is to the strong axis, where the moment of inertia is the maximum as shown in Fig. 4.8. In other words, buckling takes place perpendicular to the

78

4 Compressive Member

Fig. 4.7 Deformation of long column owing to compression

Δ

Fig. 4.8 Buckling direction with respect to weak and strong axis

Strong axis Weak axis

weak axis, where the moment of inertia is the minimum [2–4]. Figures 4.9 and 4.10 show the buckling of structures. (4) When a long column is subjected to center compressive force, buckling load (or critical load) can be computed from buckling differential equation according to Euler’s formula. As shown in Fig. 4.11, the following equation can be expressed based on the equilibrium condition that states the external moment applied from outside (Eq. 4.3) and the internal moment reacting from inside (Eq. 4.4) of an arbitrary point A, where buckling takes place at the long column with both hinged ends, are the same. Equation (4.5) is the differential equation of the deflection curve [3, 4].

4.1 Buckling Test of a Spaghetti Noodle

79

Fig. 4.9 Buckling in structures

Fig. 4.10 Local buckling Fig. 4.11 Elastic buckling at a long column with hinges at both ends

P

EI

A

80

4 Compressive Member

M = Py

(4.3)

E I y  = −M

(4.4) Py =0 EI

(4.5)

P EI

(4.6)

y = c1 cos kx + c2 sin kx

(4.7)

−E I y  = P y,

y  +

y  + k 2 y = 0, k 2 =

where M P y EI c1 and c2

is the external moment (kN·m or N·mm), is the compressive force (kN or N), is the deflection at arbitrary point A (mm), is the flexural stiffness (N·mm2 ), and are integration constants (to be evaluated from the boundary condition)

(5) Considering the boundary condition [y(0) = 0 and y(L) = 0] at both ends of a simple support column from Eq. (4.7), c1 and sinkL become zero. Then, kL is equal to n (n = integer) and bucking load Pcr is expressed as shown in the following equation. Pcr = k 2 E I =

(nπ )2 E I L2

(4.8)

(6) n of Eq. (4.8) is a constant related to buckling mode, and buckling load of first mode is the value given when n is 1 as shown in the following Eq. (4.9). Pcr =

π2E I L2

(4.9)

(7) The above equations are used to compute buckling load for a column simply supported at both ends as shown in Fig. 4.11, and a general buckling load equation according to boundary condition of a long column is expressed as Eq. (4.10). The effective buckling length of a column according to boundary condition is shown in the following table. Pcr =

π2E I L 2e

(4.10)

4.1 Buckling Test of a Spaghetti Noodle

81

4.1.8 Task

L

1.5L

2L

• Compute buckling strengths of long columns of the same material (E) and the same section (A) as shown in the following.

L

L

L

Le

Le

Effec ve buckling lengths according to boundary condi ons

▲ Fixed-Fixed

▲ Fixed-Hinge

▲ Fixed-Fixed rotation

L e = 0.50 L

L e = 0.7 L

Le = L

L design = 0.65 L

L design = 0.8 L

L design = 1.2 L (continued)

82

4 Compressive Member

L

L

Le

(continued)

▲ Hinge-Hinge

▲ Fixed-Free

▲ Hinge-Fixed rotation

Le = L

L e = 2.0 L

L e = 2.0 L

L design = L

L design = 2.1 L

L design = 2.0 L

notes Le is the effective buckling length according to boundary condition Ldesign is the effective buckling length encouraged for structural design

4.2 Buckling Test of a Styrofoam Column 4.2.1 Purpose of Test • Obtaining the modulus of elasticity of a Styrofoam through buckling test assuming that the Styrofoam is a column. – The buckling load is obtained through a column buckling test, when a Styrofoam is used as a column similar to the test of Sect. 4.1. – The concept of a theoretical buckling equation is to be comprehended, and a method of obtaining modulus of elasticity of a material without a tensile test is to be understood. – The buckling load of a structural model according to each support condition will be predicted by using the Styrofoam as four columns to be discussed in following Sect. 4.3.

4.2.2 Materials and Tools (1) ½ sheet of Styrofoam with size of 600 mm × 900 mm × 5 mm (2) Box cutter, ruler, electronic scales with 3 N and 30 N capacity.

4.2 Buckling Test of a Styrofoam Column

83

4.2.3 Test Method (1) Cut the Styrofoam in 50 mm width and lengths of 600–200 mm with 100 mm spacing as depicted in Fig. 4.12. The tip of the member is processed to be knife-edge of about 45°. (2) Place the prepared Styrofoam specimen vertically on a scale, as shown in Fig. 4.13, and apply a force vertically on the specimen. (3) Take video-recording to record the maximum load during its buckling and repeat the process for each length. Fig. 4.12 Specimen preparation

5mm

Fig. 4.13 Setup and loading

84

4 Compressive Member

4.2.4 Test Results (1) Measure the Styrofoam section to obtain moment of inertia I with respect to weak axis [3, 4]. I =

bt 3 12

(4.11)

where b and t are the width and thickness of Styrofoam (mm), respectively (2) Compute modulus of elasticity E using a buckling equation for a column. Since load applying is difficult for a short column, it is hard to measure accurate value. Accordingly, a minimum of three test data obtained mainly from long members is used [2]. Pcr =

π2E I (L e )2

E=

Pcr (L e )2 π2I

(4.12)

where Pcr E I Le

is the buckling load (N), is the modulus of elasticity (MPa), is the moment of inertia (mm4 ), and is the effective buckling length (mm).

(3) For example, if Pcr of Styrofoam with L e of 300 mm is measured as 300 N, E can be calculated following. I = E=

50 × 53 bt 3 = = 520.8 mm4 12 12

300 × 3002 Pcr (L e )2 = 2 = 5252.8 MPa 2 π I π × 520.8

(4) Theoretical buckling load is computed and plotted along with test data on a graph.

4.2.5 How to Prepare a Report (1) Since it is impossible to measure accurate test value for short columns due to the difficulty in applying the load, modulus of elasticity is computed based on the data mainly of long columns. (2) The test result should be clearly shown through a table and a graph.

4.2 Buckling Test of a Styrofoam Column

85

4.2.6 Samples (1) Test data: See Table 4.3 (2) The modulus of elasticity E was computed to be 77.34 N by taking the average of five modulus of elasticity of five long columns as shown in Table 4.3. Theoretical buckling load along the column length was computed by using theoretical buckling equation π 2 EI/(L e )2 and average E of a Styrofoam section, and the result is plotted in a graph of Fig. 4.14. (3) Calculation process ➀ Record test results in Table 4.4. ➁ Compute the following I =

50 × 53 bt 3 = = 520.8 mm4 12 12

L e = 200, 390, 400, 500, 600 mm E=

Pcr-test (L e )2 ( = 2 π I

Table 4.3 Summary of test results (sample)

)×( 520.8π 2

)2

=(

) MPa

L (mm)

Pcr-test (N)

E (MPa)

Pcr-theory (N)

200

9.636

74.99

9.93

300

4.900

85.80

4.41

400

2.461

76.61

2.48

500

1.554

75.58

1.59

600

1.049

73.47

1.10

Fig. 4.14 Comparison of test results and Euler’s formula

Experiment

200

Euler's formula

Pcr (N)

150

100

50

0 0

200

400 L (mm)

600

800

86

4 Compressive Member

Table 4.4 Summary of test results

L (mm) 200

300

400

500

600

Pcr-test (N)

E (MPa)

1

9.636

74.99

2

–

–

3

–

–

1

4.900

85.80

2

–

–

3

–

–

1

2.461

76.61

2

–

–

3

–

–

1

1.554

75.58

2

–

–

3

–

–

1

1.049

73.47

2

–

–

3

–

➂ The computation of average modulus of elasticity is based on the average of those of long columns with 400–600 mm length. E =(

) MPa

➃ Test data are plotted on a Fig. 4.15. ➄ Buckling load Pcr of each tested length L e of Styrofoam is computed by using the calculated average of elasticity modulus of ➂ and Pcr = π 2 EI/(L e )2 , and the Pcr-test is marked on a Fig. 4.15. Fig. 4.15 Comparison of test results and Euler’s formula

200

Pcr (N)

150

100

50

0

0

200

400 L (mm)

600

800

4.3 Buckling Test of a Long Column (Depending on Boundary Condition)

87

4.3 Buckling Test of a Long Column (Depending on Boundary Condition) 4.3.1 Purpose of Test • Determining the buckling length and strength by the support condition of a column – The difference in strength of a long column depending on end support condition and lateral support by using a Styrofoam is to be understood. – The strength of a long column is determined by Euler’s buckling formula, and the bending shape and failure strength are governed by the support condition of the column. – The fact that there can be a difference of strength up to sixteen times, even though they are the columns of the same length, is to be understood through the tests.

4.3.2 Materials and Tools (1) (2) (3) (4)

½ sheet of Styrofoam with size of 600 mm × 900 mm × 5 mm ½ sheet of foam board with size of 600 mm × 900 mm × 5 mm Box cutter, ruler, adhesive for Styrofoam, push-pins, packing tape A4 (or letter) size papers, electronic scale.

4.3.3 Test Method (1) Foam boards are used to make top plate (of 210 mm × 297 mm or 216 mm × 280 mm) and bottom plate (of 310 mm × 297 mm or 316 mm × 280 mm), as shown in Figs. 4.16 and 4.17. The dimension shown in Figures focus on A4 paper. (2) Cut the Styrofoam in 40 mm width to be used as a column and as a brace of 5 mm width and 310 mm length. Refer to Table 4.5 for the length. (3) After making one specimen [c.f., Figs. 4.16, 4.17, 4.18, 4.19, 4.20, 4.21, 4.22, 4.23 and 4.24] by each team, carry out the test to compare buckling loads. (4) Since the hinge connection of a column should be made such that it can rotate freely, the end part of a Styrofoam should be sharpened as shown in Fig. 4.23. It is then placed to be in contact with the loading plate. After applying a packing tape, it is then checked for free rotation. (5) Fixed connection is made such that the end part would not rotate and move by affixing a block on both sides of the column as shown in Fig. 4.24. The effective column length excluding the fixed part is set to be 300 mm as shown in

88

4 Compressive Member

210

Fig. 4.16 Top plate (section a-a in Fig. 4.19)

30 50

30

50

297

Brace direction

310

Fig. 4.17 Bottom plate (section b-b in Fig. 4.19)

80

50

80

50

297

Brace direction

Table 4.5 Specimens for buckling test (unit: mm) Support condition

Column

Brace

Fixing block

Styrofoam (w × t Styrofoam (w × t Foam board (w × l)/pieces × l)/pieces × t × l)/pieces (a)

Hinge-Hinge

X-bracing

40 × 5 × 300/4

5 × 5 × 310/4

(b)

Hinge-Fixed

–

40 × 5 × 310/4

–

25 × 5 × 40/8

(c)

Hinge-Fixed

X-bracing

40 × 5 × 310/4

5 × 5 × 310/4

25 × 5 × 40/8

–

(d)

Fixed-Fixed

–

40 × 5 × 320/4

–

25 × 5 × 40/16

(e)

Fixed-Fixed

X-bracing

40 × 5 × 320/4

5 × 5 × 310/4

25 × 5 × 40/16

Top plate: 210 × 297 × 5 or 216 × 280 × 5/Bottom plate: 310 × 297 × 5 or 316 × 280 × 5

4.3 Buckling Test of a Long Column (Depending on Boundary Condition) Fig. 4.18 Hinge at both ends and X-bracing

89

Weight using papers

300

210

Brace

310

Fig. 4.19 Fixed-Hinge

Weight using papers

210

a

300

a

b Fixing block

10

b

310

Figs. 4.19, 4.20, 4.21 and 4.22. For example, if two 5 mm foam boards are used as fixing blocks at fixed both ends, the total column length should be effective length of 300 mm plus 20 mm. (6) Bracing is accomplished by connecting the top and bottom of the column in X configuration with push-pins as shown in Figs. 4.23 and 4.24. (7) Take video-recording until the specimen collapses, and carefully watch and sketch the bending shape of the column to write the report.

90

4 Compressive Member

Fig. 4.20 Fixed-Hinge and X-bracing

Weight using papers

10

300

210

310

Fig. 4.21 Fixed at both ends

Weight using papers

10

300

10

210

310

(8) Loading is applied with papers as shown in Fig. 4.25. The final collapsing load is measured by weighing the loading papers and the top plate. Note that the load on a column can greatly differ depending on the constraint condition of a column and a brace. (9) Test data are recorded in Table 4.6, and plot the graph.

4.3 Buckling Test of a Long Column (Depending on Boundary Condition) Fig. 4.22 Fixed at both ends and X-bracing

91

Weight using papers

10

300

10

210

310

5

5

Fig. 4.23 Detail of hinge and brace

5 Push-pin

5

5

Packing tape

5

5

Fig. 4.24 Detail of fixed and brace

5 5

Push-pins

25

5

25

40

92

4 Compressive Member

Fig. 4.25 Buckling shape depending on boundary condition

4.3.4 Test Results (1) Test results are summarized in Table 4.6. (2) Sketch the buckling shape of column prior to collapse. (3) Plot the buckling load Pcr and effective length L e using MS Excel as shown in Fig. 4.26. Table 4.6 Summary of test results (sample) Effective length L e (mm)

Critical load Pcr-test (N)

Comparison with ➀ Theory

Test

➀ L e = 1.0L

300

–

1.00

1.00

➁ L e = 2.0L

600

–

0.25

–

➂ L e = 0.7L

210

–

2.00

–

➃ L e = 1.0L

300

–

1.00

–

➄ L e = 0.5L

150

–

4.00

–

12 10 8

Pcr

Fig. 4.26 Effective buckling length versus critical load

6 4

Pcr =

2 0

0

50

100

150

Le

π 2 EI ( Le ) 2

200

250

4.3 Buckling Test of a Long Column (Depending on Boundary Condition)

93

4.3.5 Applied Column Design Suggest a method for increasing the strength of a column using the prepared test specimen and verify it with a test. The limitation for additional material as a brace with 5 mm × 5 mm is total length of 900 mm or below

4.3.6 Buckling Characteristics of a Long Column

Le=0.5L Le=L

Le=2L

Le=L

Le=0.7L

(1) When small compressive force is loaded on a long column, the length of a member will be shortened without lateral displacement. However, when the increased force reaches the critical load, the column suddenly bends laterally. This phenomenon is called buckling [1]. (2) Buckling strength of a column is inversely proportional to squared length. In other words, if material or cross-sectional shape is the same, effective buckling length L e is a variable [2–4]. (3) Effective buckling length L e varies with support condition of the end part and installation of a brace [5, 6] (Fig. 4.27).

: Inflexion point

Fig. 4.27 Effective buckling length

94

4 Compressive Member

4.4 Competition for Buckling Strength of a Column 4.4.1 Purpose of Test • Determine buckling length depending on the support condition of a column and predict its strength. Note that the quantity of total materials is limited.

4.4.2 Materials and Tools (1) (2) (3) (4)

½ sheet of foam board with size of 600 mm × 900 mm × 5 mm ½ sheet of Styrofoam with size of 600 mm × 900 mm × 5 mm Box cutter, ruler, instant adhesive, glue gun and glue A4 (or letter) size papers, electronic scale with about 5 N capacity, weight scale.

4.4.3 Test Method (1) Use a foam board to make a top plate and a bottom plate in A4 (or letter) paper size of 210 mm × 297 mm dimension (or 216 mm × 280 mm). (2) Use a Styrofoam to make a column and a brace. (3) The effective space between top plate and bottom plate (i.e., column height) is set to be 300 mm. (4) Four columns at minimum are prepared and installed. (5) Instant adhesive or glue is used for bonding. (6) The total weight after all preparation is to be 1.4 N (140 gf) at maximum as shown in Fig. 4.28. If the total weight surpasses 1.4 N, the weight of column member should be reduced. (7) Test should be carried out with two persons per team, and sketch a drawing to explain the team’s column and arrangement of a brace. (8) Place the prepared column specimen on the scale (Fig. 4.29), and papers are gradually stacked on the top plate (Fig. 4.30). (9) Record the weight with a scale to measure the maximum load.

4.4 Competition for Buckling Strength of a Column

Fig. 4.28 Measurement of self-weight

Fig. 4.29 Test setup on weight scale

95

96

4 Compressive Member

Fig. 4.30 Stacking papers

4.4.4 Consideration of Key Point (1) Composition of a column [5] ➀ Take a choice between whether to use strong columns of less numbers or ➁ weak columns of many numbers (number of columns: 4, 6, 8, etc.) (2) Arrangement of brace [6] ➀ Which section size is to be chosen as the lateral support? ➁ In which direction is the lateral support installed? (3) Should the columns be laid out flat or be stacked on each other to be two-stories? (4) Should the arrangement of top and bottom be the same? (5) Support conditions ➀ Is the column support hinge or fixed? ➁ What is the efficiency of the material to make up fixed support?

References

97

References 1. AISC. Steel Construction Manual, 15th edn. American Institute of Steel Construction (AISC) (2017) 2. Chen, W.F., Lui, E.M.: Structural Stability—Theory and Implementation. Elsevier Science Publishing Co., Inc (1987) 3. Gere, J.M.: Mechanics of Materials, 5th edn. Nelson Thornes Ltd (2001) 4. Hibbeler, R.C.: Structural Analysis, 10th edn. Pearson (2018) 5. Salmon, C.G., Johnson, J.E., Malhas, F.A.: Steel Structures—Design and Behavior (Emphasizing Load and Resistance Factor Design), 5th edn. Pearson Education, Inc (2009) 6. Salvadori, M., Heller, R.: Structure in Architecture, 2nd edn (1962)

Chapter 5

Truss

5.1 Determination of Tensile and Compressive Members in a Truss 5.1.1 Purpose of Test • Inspecting visually the tensile or compressive member in the given shape and loading condition of truss. – Since a truss is connected between members with a pin, not flexural moment but only axial force (tension and compression) takes place, making it an effective structure. – While triangle configuration of a truss makes a stable structure, rectangular or pentagonal configuration of connected members accelerates deformation, making it an unstable structure. – Tensile or compressive status of each truss member connected with a pin can be examined through this test. If a member is removed after making a truss, the removed part becomes rectangular or rotates around one point, making it an unstable structure. When excessive deformation takes place after removing a member, tension or compression of the member can be figured out by observing the difference in the distance between the joints. – In case of actual structure, because all the joints are not pin and upper and lower chord members are continuous, moment occurs at the end part of a member. However, if the section size is relatively smaller than the length of a member, the resistance against moment is very small in comparison with the resistance in axial force. Accordingly, there are many instances of ignoring the moment value during design of a truss [1, 6].

© Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_5

99

100

5 Truss

Fig. 5.1 Bots and nuts

5.1.2 Materials and Tools (1) ½ sheet of foam board of 600 mm × 900 mm × 2 mm (or 3 mm) size (2) Twenty bolts and nuts for pin connection (φ2 mm and 25 mm length, refer to Fig. 5.1) (3) Box cutter, ruler, awl

5.1.3 Test Method (1) Specimen preparation ➀ Specimens are prepared as shown in Figs. 5.2, 5.3 and 5.4. ➁ The dimensions in the drawing of Fig. 5.2 represent the distance between centers of members. For example, in member (a)–(c), 10 mm are added to the center length of 80 mm in both sides to make the specimen of 100 mm length as shown in Fig. 5.4.

(c)

Detail A (d)

(e)

(f)

(g)

(j)

(k)

(l)

(b)

60

(a)

(h)

80

(i)

80

80

80 480

Fig. 5.2 Truss shape and dimensions (unit: mm)

80

80

5.1 Determination of Tensile and Compressive Members in a Truss Fig. 5.3 Assembly of detail A

101

Upper chord

Vertical member Diagonal member

Bolt & nut

Upper chord

Upper chord

Di m e agon mb al er

20

10

10

10

60

80

100

10

10

Fig. 5.4 Members (unit: mm)

10

Vertical member

➂ Use an awl to bore a hole at the place of connecting members and apply a bolt to connect each member at the joint. The connecting bolt should not be too tightened for easy rotation.

102

5 Truss

Fig. 5.5 Loading and determination

(2) Test method ➀ Install a truss in simple support condition as shown in Fig. 5.5, and apply a load after removing a member to determine tension or compression status. If the distance between connected members becomes further, then it is a tension member. If the distance becomes closer, it is determined to be a compression member [3, 4]. ➁ When upper or lower chord member is determined, loosen a nut of only one joint to find the pivot point. The pivot point is useful as a section method in structural mechanics. ➂ Diagonal brace is removed by loosening the nut, and tensile or compressive influence is determined after observing the deformation of the quadrilateral. ➃ Draw tensile (+) or compressive (–) influence for members (f)–(j) and (e)–(k) in Fig. 5.2. ➄ Carry out a test to find zero-force member as shown in Fig. 5.6. A zeroforce member refers to a member not influenced by tensile or compressive force. For example, if a load is vertically applied to joint (j), member (e)– (j) becomes zero-force member. The truss is in stable condition without it [member (e)–(j)].

Fig. 5.6 Determination of zero-force member

5.1 Determination of Tensile and Compressive Members in a Truss

103

Fig. 5.7 Truss of cantilever type

➅ Perform a test on members (f)–(j) marked with a solid line, and carry out second test for the member (e)–(k) marked with a dashed line by changing the direction of diagonal member only. (3) Analysis of test result and applied test ➀ Results • After determining tensile or compressive influence on each case of members (f)–(j) and (e)–(k), sort out advantageous arrangement of a truss system to enhance the strength and explain the reason. • The strength of a tensile member is not affected even if it becomes longer. On the contrary, the strength of compressive member is reduced due to buckling, if it becomes longer [7]. • Be able to determine what is a tensile member, a compressive member, or a zero-force member by looking at existing actual truss structure. ➁ Applied test • Determine if members of the truss, which were fabricated similar to that of Fig. 5.7, are a tensile member or a compressive member against external force by changing support condition (cantilever or vertical truss). • Make a different truss type by each team, and determine if the members are a tensile member, a compressive member, or a zero-force member. (4) Write a Report

5.2 Strength Test of a Truss Made of Spaghetti Noodles 5.2.1 Purpose of Test • Verifying the theoretical collapse strength of a truss through a test – Design a simple truss and make it with paghetti sections, and obtain collapse strength through a test.

104

5 Truss

– The specimen should be prepared with a single layer as slender as possible in order to compare the test result with theoretical result. – Member forces of the tested truss specimen are computed by method of joints or method of sections, and then, it is compared with test results. The closer the theoretical value and the test value are, the better evaluation is given. – A truss design method is to be comprehended through test and analysis, and the ability to predict collapse strength of a truss is improved.

5.2.2 Materials and Tools (1) One bag of spaghetti noodles of around 2 mm diameter (2) A sheet of foam board in 600 mm × 900 mm × 3 mm dimension (3) Box cutter, ruler, glue gun & glue, electronic scale with 60 N capacity

5.2.3 Test Method (1) Specimens are prepared in the specification shown in Figs. 5.8, 5.9 and 5.10, and the completed truss is shown in Fig. 5.11. Additional 15 mm length is required in order to ensure the support condition without dropping. (2) The joints are bonded with a glue, and minimum amount of glue is used to have the joint behave like a hinge [5–7]. (3) Only one spaghetti noodle as a member between two joints should be used for comparing theoretical and test results. Fig. 5.8 King post truss (unit: mm)

86 .6 2

86.67

86

.6 2

3 0°

15

300

15

75

75

Fig. 5.9 Howe truss (unit: mm)

15

300

15

5.2 Strength Test of a Truss Made of Spaghetti Noodles

105

75

75

15

300

15

Fig. 5.10 Pratt truss (unit: mm)

Fig. 5.11 Completed King post truss

(4) The length of the member between joints should be 50 mm at the minimum. (5) Specific device to prevent out-of-plane failure of the specimen as shown in Figs. 5.12 and 5.13 is made of transparent acrylic with 5 mm thickness. Fig. 5.12 Test setup and loading

106

5 Truss 10

400 Top part

150

Top part

20

10

Bottom part

60

Bottom part

50

300

35

50

Side View

35

350

14

50

300

65

50

35

130

Front View

150

70

Assembly using bolt and nut

65

Top part Assembly using bolt and nut

4 50

Bottom part

60

15

35

60

10

60

Top part

Top View

Details of parts

Fig. 5.13 Details of the device for preventing out-of-plane deformation (unit: mm)

(6) Truss specimen is inserted between top parts so that only in-plane deformation can take place. Specimen is supported with span length of 300 mm, and centerpoint loading is applied on it. A foam board with 3 mm thickness is used as a loading stick to reduce the friction between top parts. (7) During loading, the scale is video-recorded in order to report maximum load of the truss and the failure mode. Play back the recorded video to check failure mode and maximum load. (8) Member forces are calculated by method of sections or method of joints [3, 4], and the failure load of truss system is computed using buckling formula [1, 2]. When calculating member forces and buckling load, symbols of load P and length L are substituted into the equation for variable analysis. Modulus of elasticity E of a spaghetti section is obtained through buckling test aforementioned in Sect. 4.1. (9) Compare theoretical failure load to the failure load obtained from the test.

5.2 Strength Test of a Truss Made of Spaghetti Noodles

107

5.2.4 How to Prepare a Report (1) Since compressive strength of a truss member is smaller than tensile strength, given the same section and length, the behavior of a truss is governed by strength of a compressive member. Thus, behavior of a truss system is predicted, assuming the strength of a system is governed by the failure of the compressive member after truss analysis. (2) The computation result of theoretical analysis is documented by comparing with test result.

5.2.5 Samples (1) Theoretical computation of a truss (Figs. 5.14, 15.15 and 15.16). (2) Loading test (Figs. 5.17, 15.18 and 5.19, Table 5.1) P

Fig. 5.14 Member forces in King post truss

P

P 0

P

0

P

0

0.87 P

0.87 P

P/2

0.5 P

0

P

P

0 0.5 P

7P 0.

0.5 P

P 7P 0.

0

P

P

0. 7P

0

0. 7P

Fig. 5.15 Member forces in Howe truss

P/2

0.5 P

P/2

P/2

0.5P

0.5P 0.5P P

7P 0.

0 P/2

P

P

0. 7

0.5P 7P 0.

0.5P

P

P

P

0.5P

0. 7

Fig. 5.16 Member forces in Pratt truss

0

0.5P

0 P/2

0.5P

108

Fig. 5.17 Before and after loading in King post truss

Fig. 5.18 Before and after loading in Howe truss

Fig. 5.19 Before and after loading in Pratt truss

5 Truss

5.2 Strength Test of a Truss Made of Spaghetti Noodles Table 5.1 Summary of test results (sample)

109

Theoretical load Tested load (N) Theory/test (N) Specimen#1 4.0

6.095

0.66

Specimen#2 5.5

11.388

0.48

Specimen#3 2.5

8.576

0.29

Note The reason that the test result value is much greater than theoretical value is because the joints of the truss are assumed to be a pin in theoretical analysis while they are fixed with a glue in actual test

Fig. 5.20 Member forces by test in King post truss

Fig. 5.21 Member forces by test in Howe truss

Fig. 5.22 Member forces by test in Pratt truss

(3) Draw member forces with tested load using method of joints or method of sections, assuming center-point load is applied to the truss (Figs. 5.20, 5.21 and 5.22).

110

5 Truss

5.2.6 Task • Compute all member forces of the following truss system. 3m

3m

5 kN 30°

4m

4m

4m

5.3 Strength Competition of a Spaghetti Truss 5.3.1 Purpose of Test • Propose the truss system which can manifest maximum strength by using limited amount of spaghetti noodles. – Design and fabricate a truss, which can manifest maximum strength, by using spaghetti noodle, and then carry out strength contest of the fabricated truss. – Use what you have learned through the test of previous Section and carry out strength contest experiment. A rational design method for the manifestation of maximum truss strength is proposed under limitation of the given spaghetti quantity, truss dimensions, and total weight of the truss system. – The truss model, which manifests the maximum strength, attains the highest score. In case of the same strength, the truss with less weight is evaluated to possess better strength.

5.3.2 Materials and Tools (1) (2) (3) (4)

Four bags of spaghetti noodles of about 2 mm diameter and 260 mm length Glue gun & glue, U-bolt (Fig. 5.23) Loading board of 100 mm × 50 mm × 6 mm dimension Loading device, weight (or steel plate) with 5 N and 10 N.

5.3 Strength Competition of a Spaghetti Truss

111

6

40

U bolt

100

Loading board

Fig. 5.23 Installation of loading plate and U-bolt (unit: mm)

5.3.3 Test Method (1) Specimen preparation ➀ Each truss member can be glued to attach several spaghetti noodles with glue for making the section area bigger. ➁ Using only the material given, a spaghetti section bridge of the length over 500 mm is built. The distance between support points is 450 mm as shown in Fig. 5.24. ➂ The width and height of the truss bridge are to be 70 mm or above and 300 mm or below, respectively. ➃ The weight of the truss bridge including U-bolt and loading board is to be 5 N or below. U-bolt and loading board are given because all team should have the same weight. ➄ The bonding of spaghetti noodles is to be performed with given glue only. Do not use any other adhesives such as an epoxy. ➅ U-bolt and loading board are ready installed at the bottom of the bridge during its fabrication. Fig. 5.24 Test setup

m 70 m

Height

300 mm

th Wid

Loading

Span length = 450 mm Total length

500 mm

112

5 Truss

Fig. 5.25 Loading

(2) Test method ➀ One member of each team increases the weight at loading device (cf. Figure 5.25). ➁ Loading is to be carried out with weights of 5 N or 10 N, and the team members would confer themselves to decide on additional weight. ➂ The addition of weight is to be carried out at each step with the minimum interval of ten seconds. ➃ The weight, at which the specimen sustained for the last ten seconds before its collapse, is regarded as official maximum weight. ➄ Video-recording is carried out during the test, and then, it is replayed to examine the failure mode. Problems in specimen fabrication are examined through the video observation, and a method to increase the strength is investigated. ➅ The contest is ranked by the official maximum weight on the specimen.

5.3 Strength Competition of a Spaghetti Truss

113

5.3.4 How to Prepare a Report (1) Establish a truss design concept, prepare design drawing, and record failure mode. (2) Document possible solutions to the problems in preparing a truss system and the room for improvement on the fabrication of a truss system in the future [1, 6, 7].

5.4 Competition for Effective Reinforcement of a Truss 5.4.1 Purpose of Test • A truss structure with 400 mm span is fabricated with limited weight (0.16 N or 16 gf) of foam board to carry out first test. The maximum strength of the first test is given the score of 50 points. After reinforcing the first tested specimen with an additional member of 0.08 N, carry out the strength contest with additional 50 score points of strength increase through the second test – Use foam board to fabricate truss and carry out strength contest. Suggest a way of reinforcement after the first test in order to increase the strength of the truss specimen with limited amount of materials. – Observe failure mode through the first test and find out problematic issues. Strengthen first tested truss specimen with a suggested method, and then, verify the strength increase through second test. – Observe various failure modes of the truss structure, and investigate rational ways of manifesting maximum strength of the structure.

5.4.2 Materials and Tools (1) ½ sheet of foam board of 600 mm × 900 mm × 5 mm dimension (2) Glue gun & glue, box cutter, ruler, electronic scale of around 1 N capacity

5.4.3 Test Method (1) Fabricate the specimen in consideration of such methods as truss, arch, and lateral support with lattice structure. The test should not end with one fabrication of the specimen but continue with reinforcement of the specimen through 3–4 tests to examine the effect of reinforcement.

114

5 Truss

Fig. 5.26 Weight for the first test

(2) Specimen preparation ➀ Measure the weight of the specimen to see if it complies with the limit weight of 0.16 N (16 gf) for the first test as shown in Fig. 5.26 and if the maximum weight of the specimen after strengthening it is 0.24 N (24 gf) for the second test as shown in Fig. 5.27. ➁ Fine members and coarse members are rationally arranged in consideration of weight limitation of the material. For example, coarse member is set to be compressive, and fine member is to be tensile and brace elements. ➂ Prevent lateral buckling with the placement of supporting member perpendicularly connecting main trusses [1, 2]. ➃ Span length is to be 400 mm, but the total length of actual truss is to be 450 mm. (3) First test method ➀ Support jig is fabricated according to loading span as shown in Fig. 5.28, and it is placed on a scale. Center-point loading is applied as shown in Fig. 5.29. Fig. 5.27 Weight for the second test

5.4 Competition for Effective Reinforcement of a Truss

115

Fig. 5.28 Support jig

Fig. 5.29 Loading

➁ Loading is to continue even if the truss is collapsing to a side without giving an effort to prevent it from collapsing. ➂ Examine failure modes of the member and investigate what problem caused the failure. ➃ Record the maximum load of the first test (Fig. 5.30). (4) Second test method ➀ Reviewing the failure of first test, the failure part is partially reinforced with additional material of 0.08 N (8 gf). 6) The net weight after the reinforcement should not exceed 0.24 N. ➂ Look for the optimized place of additional reinforcement to increase the maximum strength through 3–4 preparatory tests (Caution: Do not make the specimen again). ➃ Record the maximum load of the second test and compute the ratio of load to weight (Fig. 5.31).

116

5 Truss

Fig. 5.30 Failure at first test

Fig. 5.31 Loading at second test

5.4.4 Test Results (1) After recording the result of the first test, compute the load per unit weight of truss (Fig. 5.32). (2) Likewise, record the load and compute the load per unit weight of truss through second test. (3) Suggest an opinion on how to make a truss to be more efficient in increasing the maximum load (Table 5.2).

Fig. 5.32 Buckling of compressive elements through first test

5.4 Competition for Effective Reinforcement of a Truss

117

Table 5.2 Summary of test results Weight (N)

Failure load (N)

Failure load/weight

Increase (%)

1st test 2nd test

5.4.5 Efficiency of Material in a Truss (1) Compressive members and tensile members should have large section and small section, respectively [3, 4]. (2) Connecting materials are appropriately used to prevent lateral buckling [5]. (3) Since failure of the joint results in premature failure of a structural model, care should be taken to prevent it [6, 7].

References 1. AISC. Steel Construction Manual, 15th edn. American Institute of Steel Construction (AISC) (2017) 2. Chen, W.F., Lui, E.M.: Structural Stability—Theory and Implementation. Elsevier Science Publishing Co., Inc. (1987) 3. Gere, J.M.: Mechanics of Materials, 5th edn. Nelson Thornes Ltd. (2001) 4. Hibbeler, R.C., Structural Analysis, 10th edn. Pearson (2018) 5. Kulak, G.L., Fisher, J.W., Struik, J.H.A.: Guide to Design Criteria for Bolted and Riveted Joints, 2nd edn. Wiley (1987) 6. Salmon, C.G., Johnson, J.E., Malhas, F.A.: Steel Structures—Design and Behavior (Emphasizing Load and Resistance Factor Design), 5th edn. Pearson Education, Inc. (2009) 7. Salvadori, M., Heller, R.: Structure in Architecture, 2nd edn (1962)

Chapter 6

Other Tests

6.1 Vibration Test (Dynamics) 6.1.1 Purpose of Test • Computing natural period by using rubber bands (spring) and coins (mass) – The concept of spring stiffness with rubber bands and coins is to be comprehended. – Since rubber band shows inelastic behavior, find the elastic range of band. Then, measure the frequency and period using coins as mass and compare them with theoretical values. – Compute the frequency and period according to mass and spring stiffness, and then, understand the basics of dynamics by verifying it with actual measurement.

6.1.2 Materials and Tools (1) Rubber band, thumbtack, plastic bag, 600 mm ruler. (2) Electronic scale of 10 N or below capacity. (3) Twenty coins (e.g., quarter in USA, ten pence in UK, e0.2–1 in Europe, and 100 won in South Korea).

6.1.3 Test Method (1) Ten coins are placed on a scale to measure the weight and then divided by number of coins to obtain the coin unit weight. For example, © Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7_6

119

120

6 Other Tests

(a) Connection of bands

(b) Fixity of bands

(c) Put 5 coins in a bag

(d) Put additional 5 coins in a bag

(e) Deflection measurement

Fig. 6.1 Test procedure

Weight of ten coins 0.54 N/10 coins = 0.054 N/coin (2) Five rubber bands are serially connected as shown in Fig. 6.1a, and the increased length with weight is measured to compute the spring constant of the rubber bands. After obtaining spring constant, the frequency and period are measured through vibration test to be compared with theoretical values. The following equation is used. F = kx, k =

F x

(6.1)

where F is the increased weight (N), x is the increased length (mm), and k is the spring constant (N/mm). (3) Press a thumbtack on a book rack (Fig. 6.1b), and then, hook a rubber band on top and connect the bottom to a plastic bag. (4) The rubber band length, when five coins are put in the plastic bag, is set to be the initial value. At this time, the initial length is measured from the start point of the rubber band to the end point of the band (Fig. 6.1c). (5) Five coins are additionally put in the plastic bag, and the changed length of the rubber bands is measured (Fig. 6.1d and e). (6) Add five coins at a time to the plastic bag until the total reaches 20 coins. Each time, when five coins are added, the length is measured and recorded as shown in Table 6.1. Weight increase F and length increase x are computed according to Eq. (6.1) to obtain spring constant k. (7) Modulus of elasticity of rubber band is scattered between 7.5 and 4.7 MPa. In other words, as load increases, it decreases while showing nonlinear behavior.

6.1 Vibration Test (Dynamics)

121

Table 6.1 Summary of test results (sample) Step

Weight (N)

Measured length (mm)

Increased weight Increased F (N) length x (mm)

Spring constant k = F//x (N/m)

Step 1–5 coins

0.27

412

–

–

–

Step 2–10 coins

0.54

448

0.27

36

7.5

Step 3–15 coins

0.81

496

0.27

48

5.6

Step 4–20 coins

1.08

553

0.27

57

4.7

6.1.4 Comparison of Test Frequency and Theoretical Frequency (1) Test frequency ➀ Seventeen coins are placed in the plastic bag, and the time to reach 10 cycles is measured. Mass of coins m = 17 coins × 5.4 g = 91.8 g = 0.0918 kg ➁ Measuring method is to pull the coin bag downwards, and then let go of the hands to bring about top-to-bottom free vibration to measure the time taken for ten vibrations. Use stopwatch feature of a smart cell phone. If the measured time is 7.91 s, the natural period is 7.91 s/10 cycle = 0.791 s/cycle. In other words, T = 0.791 s and f = 1/T = 1.26 Hz. (2) Theoretical frequency ➀ Pay attention to dimension of unit mass, which is kg = N·s2 /m, and make all units congruent.  

4.7 = 7.16 rad/s ➁ ω = 2π f = mk = 0.0918   1 k 1 4.7 ➂ f = 2π = 2π = 1.13 Hz m 0.0918

(3) Comparison ➀ The difference in measured frequency and theoretical frequency was 1.26/1.13 = 1.12, indicating about 12% of error. ➁ Modulus of elasticity of rubber band shows nonlinear behavior, because it is a material of large damping constant. Larger damping makes longer period and reduced frequency.

122

6 Other Tests

6.1.5 Applied Vibration Test • The number of rubber bands can be increased or reduced, or the test can be carried out by connecting them in parallel.

6.1.6 Vibration Theory (1) Undamped free vibration in a single-degree-of-freedom (SDF) system ➀ The motion of SDF systems such as a mass-spring system or one-story frame as shown in Fig. 6.2, subjected to zero external force, has the following motion equation [2, 5]. m u(t) ¨ + ku(t) = 0

(6.2)

where m is the mass (kg or N·s2 /m), u(t) is the displacement of mass (m), ü(t) is the acceleration of mass (m/s2 ), and k is the stiffness (N/m). ➁ Free vibration is initiated by disturbing the system from static equilibrium position by imparting the mass some displacement u(0) and velocity u(0) ˙ at time zero. Subject to these initial conditions, the solution to the homogeneous differential equation of Eq. (6.2) is obtained by standard methods as shown in the following. u(t) = u(0) cos(ωn t) +

u(0) ˙ sin(ωn t) ωn

(6.3a)

 2 [u(0)]2 + [u(0)/ω ˙ n ] cos(ωn t − φ)

or u(t) =

u(t) m k

u(t) m

(a) Mass-spring system Fig. 6.2 Single-degree-of-freedom

k/2

k/2

(b) One-story frame

(6.3b)

6.1 Vibration Test (Dynamics)

123

where u(0) is the initial displacement (mm), u(0) ˙ is the initial velocity (mm/s), is the natural circular frequency of vibration (rad/s), and ωn φ = tan

−1



u(0)/ω ˙ n u(0)



➂ The time required for the undamped system to complete one cycle of free vibration is the natural period T n in units of seconds. It is related to the natural circular frequency of vibration ωn in unit of radians per second Eq. (6.4). The natural cyclic frequency of vibration is denoted by Eq. (6.5) (Fig. 6.3). 2π s ωn

(6.4)

1 ωn Hz = Tn 2π

(6.5)

Tn = fn =

(2) Undamped free vibration in multi-degree-of-freedom (MDF) system ➀ The motion of MDF systems of three-story frame as shown in Fig. 6.4, subjected to zero external force, can be expressed in the following motion equations. Fig. 6.3 Free vibration of a system without damping

u(t)

. u(0)

u(0) t T

Fig. 6.4 Three-story shear building

u3(t) m3 k3/2 k3/2 m2 k2/2 k2/2 m1 k1/2 k1/2

u2(t) u1(t)

124

6 Other Tests

¨ + [K ][u(t)] = [0] [M][u(t)]

(6.6)

⎡

⎤ m1 0 0 [M] = ⎣ 0 m 2 0 ⎦ 0 0 m3 ⎤ ⎡ 0 k1 + k2 −k2 [K ] = ⎣ −k2 k2 + k3 −k3 ⎦ 0 −k3 k3 ⎤ ⎡ u 1 (t) [u(t)] = ⎣ u 2 (t) ⎦ u 3 (t)

(6.7)

(6.8)

(6.9)

where [M] is the mass matrix, [K] is the stiffness matrix, and [u(t)] is the displacement vector (3) Mode shape and vibration frequency in MDF ➀ The solution to motion equation of Eq. (6.6) in frame of Fig. 6.4 is similar in form to the solution of a SDF system and can be expressed as follows. u i (t) = φi cos(ωi t − θi ) . . . (i = 1, 2, 3)

(6.10a)

or {u(t)} = {φ} cos(ωt − θ )

(6.10b)

➁ The task of computing eigenvalue ω2 and eigenvector {φ} from the motion equation of Eq. (6.6) to satisfy Eq. (6.11) is an eigenvalue problem.

 [K ] − ω2 [M] [φ] = 0

(6.11)

➂ When {φ} = {0}, the Eq. (6.11) is always valid but useless from the perspective of engineering. Therefore, the eigenvalue problem considers the following Eq. (6.12) from the perspective of engineering [3, 4]. 

det [K ] − ω2 [M] = 0 ➃ Example of two-story shear building (Fig. 6.5)

m1 0 [M] = 0 m2



10 = 03

(6.12)

6.1 Vibration Test (Dynamics)

125

Fig. 6.5 Two-story shear building

u2(t) m2=3 k2=2

u1(t)

m1=1 k1=1





3 −2 = −2 2 





10 3 −2 =0 − ω2 det [K ] − ω2 [M] = det 03 −2 2

k + k2 −k2 [K ] = 1 −k2 k2

ω1 = 0.4380, ω2 = 1.8641 • Substituting ω1 and ω2 to Eq. (6.11) to compute {φ} makes the first eigenvector and the second eigenvector, which are the basic of mode shape. First of all, computing ω1 , 





 3 −2 0 φ11 2 1 0 = − 0.4380 φ12 −2 2 03 0





−2 φ11 0 3 − 0.43802 = −2 2 − 3 × 0.43802 φ12 0

• If substituting φ 11 = 1, [φ1 ] =

φ11 φ12

=

1.0000 1.4040

• With respect to ω2 ,

−2 3 − 1.86412 −2 2 − 3 × 1.86412



1.0000 φ21 = [φ2 ] = φ22 −0.2374



φ21 φ22

=

0 0

• The following equation shows eigenvector matrix composed of the first and second eigenvectors computed above. Mode shape is represented by the matrix as shown in Fig. 6.6. 

[φ] = φ1 φ2





φ21 = φ22

126

6 Other Tests

Fig. 6.6 Mode shape

12=1.4040

22=

_

11=1

21=1

[ 2]

[ 1]

=

1.0000 1.0000 1.4040 −0.2374

0.2374

6.1.7 How to Prepare a Report (1) Measurement of the weight of coins Weight of 10 coins is (· · ·) N/10 coins = (· · ·) N/coin (2) After five coins are added into the plastic bag at each test step, measurements are recorded in Table 6.2. Weight increase F and length increase x are computed according to Eq. (6.1) to obtain spring constant k. (3) Test frequency ➀ Seventeen coins are placed in the plastic bag, and the time to reach 10 cycles is measured. Table 6.2 Summary of test results Step

Weight (N)

Measured length (mm)

Increased weight Increased F (N) length x (mm)

Spring constant k =  F/x (N/m)

Step 1–5 coins

–

–

–

–

–

Step 2–10 coins

–

–

–

–

–

Step 3–15 coins

–

–

–

–

–

Step 4–20 coins

–

–

–

–

–

6.1 Vibration Test (Dynamics)

127

Mass of coins m = 17 coins × (. . .) g = (. . .) g = (. . .) kg ➁ Measuring method is to pull the coin bag downwards, and then let go of the hands to bring about top-to-bottom free vibration to measure the time taken for ten vibrations. Use stopwatch feature of a smart cell phone. If the measured time is ( ) s, the natural period is ( ) s/10 cycle =( ) s/cycle. In other words, T = ( ) sec and f = 1/T = ( ) Hz. (4) Theoretical frequency ➀ Pay attention to the dimension of unit mass, which is kg = N·s2 /m, and make all unitscongruent. 

➁ ω = 2π f = mk =   ( 1 k 1 ➂ f = 2π = m 2π (

( (

) )

) )

=(

=(

) rad/s

) Hz

(3) The difference in measured frequency and theoretical frequency is ( = ( ), indicating about ( )% of error.

)/(

)

6.2 Arch and Dome Structures 6.2.1 Purpose of Catenary Test • Making of a catenary form with Styrofoam and clips – Understand a concept of catenary hanging by its own weight with Styrofoam and clips. – A catenary is a structure with a perfect balance of tension through clips and gravity pulling downward. Understand the arch structure is an inverted catenary structure.

6.2.2 Materials and Tools (1) ½ sheet of Styrofoam of 600 mm × 900 mm × 5 mm dimension. (2) Box cutter, ruler, clip, packing tape, glue gun & glue.

6.2.3 Test Method (1) Cut Styrofoam of 160 mm diameter to make a bottom plate. A total of five and seven clips are affixed on four places of circular bottom plate with a packing tape as shown in Fig. 6.7.

128

6 Other Tests

Fig. 6.7 Affixing plate and linked clips

Fig. 6.8 Overturning plate and gluing

(2) Link the clips as shown in Fig. 6.8, and overturn the bottom plate to make an arch due to the weight of clips. Glue is used to fasten the drooping parts of clip links. (3) The center, where the clips are gathered, is connected through a loop as shown in Fig. 6.9. (4) After the glue is hardened, the catenary structure in this test is overturned to manifest an arch structure as shown in Fig. 6.10.

6.2 Arch and Dome Structures

Fig. 6.9 Connection at center

Fig. 6.10 Completed structural model

129

130

6 Other Tests

6.2.4 Catenary Theory (1) A catenary is a curved form exhibited when the ends of a material with the same density are held so that the center is hanging under gravity field as shown in Fig. 6.11. The following Eq. (6.3) is the theoretical equation for a catenary [1]. y = a cosh

x 

(6.3)

a

where a is the scaling of curve (2) The form of a catenary exhibits a balanced shape as shown in Fig. 6.12, and the tension on the holding ends and gravity acting on it are in perfect balance. The curved surface, where average curvature is 0 at all points, is a catenary surface only [6]. (3) When a catenary is turned upside down, the inside tension becomes compressive force. However, compressive forces also apply in parallel to the catenary line as the same as tensile force. Therefore, the force inside the arch to deform the arch does not take place owing to this balance of forces. As a result, the shape of an arch can be maintained with the weight of arch alone, and such an architectural structure can be built with minimum amount of members [3, 4]. Many structures have been built based on this principle as illustrated in Fig. 6.13.

Fig. 6.11 Catenary

Catenary

6.2 Arch and Dome Structures

Fig. 6.12 Actual catenary tested by Antoni P. Gaudi

Fig. 6.13 Structures with catenary action

131

132

6 Other Tests

6.3 Creating a Load Cell 6.3.1 Purpose of Test • Creating of a load cell using strain gauge s and cylinder steel – Understand the principles of strain gauges and how to convert strain values into loads. – Design of optimal cross-sectional area of a load cell using stress–strain relationship according to the target measuring capacity of the load cell. – Know the difference between the full-bridge connection of the Wheatstone bridge and the quarter-bridge connection and compute the coefficient values to convert strain to load.

6.3.2 Materials and Tools (1) (2) (3) (4)

One cylindrical made of steel with the yield strength of 400 MPa or higher. Four strain gauges (R = 120  or 350 ) for steel material. CN bond, M-Coat A coating, sandpaper, permanent pen, acetone, cotton. Strain meter, laptop, universal testing machine (UTM), lead wire.

6.3.3 How to Make a Load Cell (1) Calculate the cross section of the load cell ➀ Calculate the cross-sectional area (A) of the cylindrical steel according to the target measurement capacity of the load cell as shown in Fig. 6.14, taking D d

L

Fig. 6.14 Size and shape of load cell

6.3 Creating a Load Cell Fig. 6.15 Stress–strain curve of steel material

133

Stress (MPa)

Fy

E

ε y = 2,000

Strain (με)

into consideration the strain (ε) to be 1000 με and the 50% safety factor of the yield strain (εy ) with reference to the stress–strain curve of Fig. 6.15. ➁ Equation (6.13a, b, c) is an equation for calculating the required crosssectional area of the load cell. σ = Eε

(6.13a)

P/A = Eε

(6.13b)

A = P/Eε

(6.13c)

where σ is the stress (MPa), E is the modulus of elasticity (210,000 MPa), ε is the strain, P is the target load of load cell (kN), and A is the cross-sectional area required for load cell (mm2 ) ➂ Equation (6.14a, b) shows the relationship between the change of resistance and the axial strain. Strain gauge coefficient (S g ) can be used to calculate the axial strain due to resistance change. The coefficient of general strain gauge for steel is about 2.0. (2.11 for the case of FLA-5-11 of Tokyo Sokki Kenkyujo company) R/R = Sg εa εa =

1 R Sg R

where εa is the normal strain along axial direction of gauge, S g is the gauge factor,

(6.14a) (6.14b)

134

6 Other Tests

B

Fig. 6.16 Wheatstone bridge circuit

+

R1

R2

A

C R3

R4

E _

D _

+ V

R is the resistance changed due to the axial strain, and R is the resistance of gauges ➃ Figure 6.16 shows the configuration of the Wheatstone bridge circuit. The potential difference (E) due to the change of resistance R1, R2, R3, and R4 can be calculated according to Eq. (6.15a, b). ➄ Firstly, verify that the operating states of the four resistors and their respective resistance values are measured similarly to the theoretical values through a quarter-bridge connection that activates only one resistor. ➅ Secondly, the load is measured by a full-bridge connection that activates all four resistors. At this time, the resistance value should be equal to the sum of the results of the quarter-bridge connection as performed in previous step ➄ above. VAB =

R1 V R1 + R2

(6.15a)

VAD =

R4 V R3 + R4

(6.15b)

E = VBD = VAB − VAD

(6.15c)

(2) Test methods and procedures ➀ Attach the strain gauges vertically (R1, R3) and horizontally (R2, R4) to the center of the cylindrical steel in the four directions as shown in Figs. 6.17 and 6.18. Here, the vertical direction gauges measure tension and compression, and the horizontal direction gauges enhance the sensitivity of the load cell and correct for the temperature and eccentric load [3, 4]. ➁ Apply the M-coat A coating to the strain gauges to prevent their damage caused by moisture after attaching them as shown in Fig. 6.18b, and wrap them with protective tape as shown in Fig. 6.18c.

6.3 Creating a Load Cell

135

(Vertical)

R3

R3

R2

L/2

d

R1

L/2

D

(Horizontal)

L

R2

R4

(Horizontal)

R4

R1

(Vertical)

(a) Top view

(b) Perspective view

Fig. 6.17 Position and direction of gauge attachment

(a) Attachment of strain gauge

(b) Application of M-coat A

(c) Wrapping with protective tape

Fig. 6.18 Procedure of gauge attachment

➂ After wrapping the strain gauges with protective tape, check that there are no damage to the four strain gauges by using an insulation resistance meter (or ohm m), and connect each strain gauge to the strain meter with a one gauge3 wire quarter-bridge method as shown in Fig. 6.19. This is the procedure to verify the operation of each resistor R1, R2, R3, R4 before full-bridge connection. ➃ UTM is used to perform the compression experiment according to the capacity of the load cell. Make sure that the strains of the gauges R1 and R3 do not exceed 1000 με. Verify that the strain of the R2 and R4 gauges

R1

Fig. 6.19 Connection of quarter bridge using one gauge-3 wire

A B(H) C D(L) E

136

6 Other Tests

A B(H) C D(L) E

R1 R2 R3 R4 Fig. 6.20 Connection of full bridge using four gauges

Table 6.3 Connection layout

No.

Color of lead wire

Lead function

Strain gauge

1

Red

+Excitation

A C

2

Black

−Excitation

3

Green

+Signal

B

4

White

−Signal

D

attached perpendicularly to the load direction is close to the experimental value multiplied by Poisson’s ratio (ν = 0.3) [see Table 6.4 and Fig. 6.24]. ➄ If the result is similar to the theoretical value, connect four strain gauges to the connection line with a four-gauge full-bridge method as shown in Fig. 6.20. Refer to Table 6.3 for cable connection method. ➅ After four-gauge connection, perform compression test again. Plot the applied load–strain graph with the measured data as shown in Fig. 6.25. Find the coefficient that converts the strain into the load by using measured data. ➆ Knowing the coefficient value of created load cell, it is possible to measure the compressive force by using the created load cell.

6.3.4 Calculating Potential Difference of Wheatstone Bridge (Example) • Assuming that the increased resistances of R1 and R3 strain gauges in the created load cell under compression are +1, the potential difference according to the state of R2 and R4 strain gauges attached perpendicularly to the load direction is theoretically calculated. And then, the sensitivity of load cell according to the potential difference is examined. (1) When the R2 and R4 are activated (tension −0.3): see Fig. 6.21 VAB =

(120 + 1) R1 V = V R1 + R2 (120 + 1) + (120 − 0.3)

= 0.5027V VAD =

(120 − 0.3) R4 V = V R3 + R4 (120 + 1) + (120 − 0.3)

6.3 Creating a Load Cell

137

B

Fig. 6.21 Wheatstone bridge circuit when activating R2 and R4

R1 (120+1)

R2 (120-0.3)

A

C

E

R3 (120+1)

R4 (120-0.3) D _

+ V

= 0.4973V E = VBD = VAB − VAD = 0.0054V (2) When the R2 and R4 are not activated (tension 0): see Fig. 6.22 R1 V = 0.5021V R1 + R2 R4 V = 0.4979V VAD = R3 + R4 E = VBD = VAB − VAD = 0.0042V VAB =

(3) Sensitivity according to potential difference Fig. 6.22 Wheatstone bridge circuit when not activating R2 and R4

B R1 (120+1)

R2 (120)

A

C R3 (120+1)

R4 (120) D _

+ V

E

138

6 Other Tests

• When the R2 and R4 gauges attached perpendicularly to the loading direction are activated and the tension of −0.3, which is generated simultaneously with compression +1, is calibrated, the potential difference becomes larger. It indicates that the larger the potential difference is, the greater the sensitivity of the load cell is.

6.3.5 Creating a Load Cell (Example) (1) When the strain of R1 and R3 becomes 1000 με, design the maximum capacity of the load cell (Fig. 6.23). P = Assuming that the capacity of load cell is P = 135 kN, A = Eε 135,000 N 2 = 642 mm is required by Eq. (6.15c). When the D 210,000 N/mm2 ×1000μ and d are chosen to be 40 and 23 mm, respectively, the cross-sectional area A is 841.16 mm2 . The length L was chosen to be 50 mm in this case. 2) Quarter-bridge connection: After connecting each of the strain gauges R1, R2, R3, and R4 to the data logger by one gauge-3 wire method as shown in Fig. 6.19, perform compressive strength test according to the designed maximum capacity of load cell. The experimental results are shown in Table 6.4 and Fig. 6.24. It is confirmed that when the average value of R1 and R3 is multiplied by 0.3 times (Poisson’s ratio, ν), the values were similar to the values of R2 and R4. (3) Full-bridge connection: Refer to Table 6.4 and connect each resistor by fourgauge method. And then, perform the compression test again. It is to confirm that the results are similar to the sum of the R1, R2, R3, and R4 strain values of the P

L

Fig. 6.23 Size of load cell (example)

d D

Table 6.4 Strains at quarter bridge when P is 135 kN Gauges

Vertical gauges R1

Measured stain (με)

−690

Average strain (με)

−603.5

Limit strain (με) (Max)

±1000

Horizontal gauges R3 −517

R2 169 177.5 ±300

R4 186

6.3 Creating a Load Cell

139

Applied load at UTM (kN)

150 120

R2

R3

R1

R4

90 60 30 0 -800

-600

-400

-200

0

200

400

Measured strain at quarter-bridge (με)

Fig. 6.24 Measured load versus strain at quarter bridge

160 y = 0.091x - 0.3493

120 80 40

Loadcell 1 Trend line

0

0

500

1000

1500

Applied load at UTM (kN)

Applied load at UTM (kN)

previous quarter-bridge test results. Figure 6.25 shows the applied load–strain relationship of four created load cells. (4) Convert strain to load: The strain can be converted to the load using the slope of the load–strain graph. The coefficients of four load cells are 0.091, 0.0902, 0.0914, and 0.0927, respectively (see Fig. 6.25). Know the coefficient value of

2000

160 y = 0.0902x - 0.4818

120 80 40

Loadcell 2 Trend line

0

0

y = 0.0914x - 0.9773

120 80 40

Loadcell 3 Trend line

0

0

500

1000

1500

2000

Measured strain at full-bridge (με)

Applied load at UTM (kN)

Applied load at UTM (kN)

160

500

1000

1500

2000

Measured strain at full-bridge (με)

Measured strain at full-bridge (με) 160

y = 0.0927x + 0.031

120 80 40

Loadcell 4 Trend line

0

0

500

1000

1500

2000

Measured strain at full-bridge (με)

Fig. 6.25 Measured load versus strain of full bridge at four load cell

140

6 Other Tests

Applied load at UTM (kN)

160 120 80

Loadcell 1 Loadcell 2

40

Loadcell 3 Loadcell 4

0

0

40

80

120

160

Measured load at load cell (kN) Fig. 6.26 Comparison of applied and measured loads after calibration with coefficient

Fig. 6.27 Measurement of bolt tension using a created load cell

the load cell, which is converted into the load, and use it as an input value when measuring the tension of the bolt (Fig. 6.26). (5) Application example of the created load cell: The created load cell can be used for bolt tension measurement as shown in Fig. 6.27 [6].

References 1. Chen, W.F., Lui, E.M.: Structural Stability—Theory and Implementation. Elsevier Science Publishing Co., Inc (1987) 2. Chopra, A.K.: Dynamics of Structures—Theroy and Applications to Earthquake Engineering, 2nd edn. Prentice Hall (2001) 3. Gere, J.M.: Mechanics of Materials, 5th edn. Nelson Thornes Ltd (2001) 4. Hibbeler, R.C.: Structural Analysis, 10th edn. Pearson (2018) 5. Paz, M., Leigh, W.: Structural Dynamics—Theory and Computation, 5th edn. Kluwer Academic Publishers (2004) 6. Salvadori, M., Heller, R.: Structure in Architecture, 2nd edn (1962)

Index

A Accelerometer, xiii Allowable stress, 22, 43, 49 Arch structure, 127 Axial force, 29, 99

B Beam, 2, 42, 51, 57 Bending, 8 Bending moment, 30, 66 Bending stress, 40 Bending test, 10, 38 Boundary condition, 80 Buckling, 51, 71, 77, 82, 92 Buckling equation, 71, 82, 84 Buckling length, 87, 94 Buckling load, 73, 78, 82, 92 Build up, 33

C Cantilever, 46 Castiliano’s theorem, 49 Catenary, 127 Centroid, 13, 16, 59 Channel section, 57 Clinometer, xiii Collapse, 3 Column, 4, 71, 73, 76, 82, 94 Composite, 33 Compression, 8, 99 Compression member, 102 Compressive force, 4, 36, 68, 76 Compressive member, 76, 77, 103 Compressive stress, 13, 24, 37 Compressive test, 10

Concentrated load, 36, 66 Conjugated beam method, 49 Critical load, 78, 93

D Damping, 121 Deflection, 44, 46, 49, 59 Dial gauge, viii Dial indicator, viii Ductile, 53 Dynamics, 119

E Eccentric force, 77 Eccentricity, 57 Effective buckling length, 73, 84, 93 Eigenvalue, 124 Eigenvalue problem, 124 Eigenvector, 124 Elastic, 119 Elastic curve method, 49 Elastic load method, 49 Euler’s buckling equation, 73 Euler’s buckling formula, 87 Extensometer, 25

F Failure load, 20, 34, 38 Fixed end, 57 Flange, 51 Flexural capacity, 1, 51 Flexural deformation, 16, 36 Flexural member, 1, 13, 16, 33, 36, 38, 40, 44, 49

© Springer Nature Singapore Pte Ltd. 2021 K.-J. Shin and S.-H. Lee, Experiment-Based Structural Mechanics, https://doi.org/10.1007/978-981-15-8311-7

141

142 Flexural resistance, 1, 68 Flexural strength, 1, 51 Flexural stress, 13, 38, 41 Fracture stress, 20, 24, 37, 38 Free end, 57 Free vibration, 121 Frequency, 119 Full bridge, 132 G Geometric moment of area, 16, 60 Goniometer, xiii H Hinge, 87 I Inelastic behavior, 119 In-plane, 51, 106 L Lateral buckling, 114 Lateral support, 51, 87, 113 Lattice structure, 113 Linear Variable Differential Transformer (LVDT), viii Load cell, xii, 132 Long column, 77, 87, 93 M Mass, 119 Mass-spring system, 122 Modulus of elasticity, 71, 82, 84 Moment, 36 Moment of inertia, 13, 16, 60, 73, 78, 84 Motion equation, 122 Multi-Degree-of Freedom (MDF), 123 N Natural circular frequency, 123 Natural period, 121, 123 Neutral axis, 13, 36, 68 Nominal stress, 30 Non-linear behavior, 121 Normal stress, 24 O Out-of-plane, 51, 106

Index P Period, 119 Plastic moment, 51, 55 Pressure cell, xiii

Q Quarter bridge connection, 132

R RC beam, 68 Reinforced concrete, 63 Reinforcement, 113 Ruler, vii

S Safety factor, 19, 22 Safety ratio, 38, 41 Scale, xi Section modulus, 13, 40 Shear center, 57, 59 Shear connector, 33 Shear flow, 60 Shear force, 33 Shear stress, 59 Short column, 77, 84 Sideway, 51 Simple beam, 42 Simply supported beam, 66 Single-Degree-of Freedom (SDF), 122 Slenderness ratio, 77 Span, 45 Spring, 119 Spring stiffness, 119 Statically determinate structure, 49 Strain gauge, ix, 25, 132 Stress block, 68 Stress concentration, 26, 29, 34 Strong axis, 77 Support, 53, 54 Support condition, 44, 82, 87, 94

T Tensile, 8 Tensile force, 26, 36, 68 Tensile member, 26, 103 Tensile strength, 13, 24, 26, 27, 37 Tensile stress, 13, 24, 26, 37 Tensile test, 11, 19, 29 Tension, 99

Index Tension member, 19, 102 Theorem of moment area, 49 Torsion, 30, 57 Torsional deformation, 57 Truss, 99, 103, 106, 110, 113 U Ultimate, 1 Unbraced length, 51, 53, 54 Uniformly distributed load, 36, 44, 46 V Vernier calipers, vii Vertical member, 4

143 Virtual work method, 49

W Weak axis, 78 Web, 51 Wheatstone bridge, 132, 134

Y Yield strength, 26

Z Zero-force member, 102, 103