Solving Practical Engineering Mechanics Problems: Statics 9781681731889, 1681731886

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Series ISSN: 2573-3168

Solving Practical Engineering Mechanics Problems

BAKHTIYAROV

Synthesis Lectures on Mechanical Engineering STATICS

Sayavur I. Bakhtiyarov, New Mexico Institute of Mining and Technology

store.morganclaypool.com

MORGAN & CLAYPOOL

ABOUT SYNTHESIS This volume is a printed version of a work that appears in the Synthesis Digital Library of Engineering and Computer Science. Synthesis lectures provide concise original presentations of important research and development topics, published quickly in digital and print formats. For more information, visit our website: http://store.morganclaypool.com

SOLVING PRACTICAL ENGINEERING MECHANICS PROBLEMS: STATICS

Engineering mechanics is one of the fundamental branches of science that is important in the education of professional engineers of any major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics, acoustics, vibrations, etc. are based on engineering mechanics courses. In order to absorb the materials of engineering mechanics, it is not enough to consume just theoretical laws and theorems—a student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many problems independently. This book is a part of a four-book series designed to supplement the engineering mechanics courses. This series instructs and applies the principles required to solve practical engineering problems in the following branches of mechanics: statics, kinematics, dynamics, and advanced kinetics. Each book contains between 6 and 8 topics on its specific branch and each topic features 30 problems to be assigned as homework, tests, and/or midterm/final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided. This first book contains seven topics of statics, the branch of mechanics concerned with the analysis of forces acting on construction systems without an acceleration (a state of the static equilibrium). The book targets the undergraduate students of the sophomore/junior level majoring in science and engineering. The author welcomes all feedback/comments from the reader. Please feel free to contact him at [email protected].

Solving Practical Engineering Mechanics Problems STATICS

Sayavur I. Bakhtiyarov

Synthesis Lectures on Mechanical Engineering

Solving Practical Engineering Mechanics Problems: Statics

iii

Synthesis Lectures on Mechanical Engineering Synthesis Lectures on Mechanical Engineering series publishes 60–150 page publications pertaining to this diverse discipline of mechanical engineering. The series presents Lectures written for an audience of researchers, industry engineers, undergraduate and graduate students. Additional Synthesis series will be developed covering key areas within mechanical engineering. Solving Practical Engineering Mechanics Problems: Statics Sayavur I. Bakhtiyarov October 2017 Resistance Spot Welding: Fundamentals and Applications for the Automotive Industry Menachem Kimchi and David H. Phillips October 2017 Unmanned Aircraft Design: Review of Fundamentals Mohammad Sadraey September 2017 Introduction to Refrigeration and Air Conditioning Systems: Theory and Applications Allan Kirkpatrick September 2017 MEMS Barometers Toward Vertical Position Detection: Background Theory, System Prototyping, and Measurement Analysis Dimosthenis E. Bolanakis May 2017 Vehicle Suspension System Technology and Design Avesta Goodarzi, Amir Khajepour May 2017 Engineering Finite Element Analysis Ramana M. Pidaparti May 2017

Copyright © 2018 by Morgan & Claypool All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means—electronic, mechanical, photocopy, recording, or any other except for brief quotations in printed reviews, without the prior permission of the publisher. Solving Practical Engineering Mechanics Problems: Statics Sayavur I. Bakhtiyarov www.morganclaypool.com ISBN: 9781681731889 print ISBN: 9781681731896 ebook DOI 10.2200/S00799ED1V01Y201709MEC008 A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON MECHANICAL ENGINEERING, #8

Series ISSN: 2573-3168 Print 2573-3176 Electronic

Solving Practical Engineering Mechanics Problems: Statics Sayavur I. Bakhtiyarov

New Mexico Institute of Mining and Technology

SYNTHESIS LECTURES ON MECHANICAL ENGINEERING #8

M &C

MORGAN

& CLAYPOOL PUBLISHERS

vi

ABSTRACT

Engineering mechanics is one of the fundamental branches of science that is important in the education of professional engineers of any major. Most of the basic engineering courses, such as mechanics of materials, fluid and gas mechanics, machine design, mechatronics, acoustics, vibrations, etc. are based on engineering mechanics courses. In order to absorb the materials of engineering mechanics, it is not enough to consume just theoretical laws and theorems—a student also must develop an ability to solve practical problems. Therefore, it is necessary to solve many problems independently. This book is a part of a four-book series designed to supplement the engineering mechanics courses. This series instructs and applies the principles required to solve practical engineering problems in the following branches of mechanics: statics, kinematics, dynamics, and advanced kinetics. Each book contains between 6 and 8 topics on its specific branch and each topic features 30 problems to be assigned as homework, tests, and/or midterm/ final exams with the consent of the instructor. A solution of one similar sample problem from each topic is provided. This first book contains seven topics of statics, the branch of mechanics concerned with the analysis of forces acting on construction systems without an acceleration (a state of the static equilibrium). The book targets the undergraduate students of the sophomore/junior level majoring in science and engineering. The author welcomes all feedback/comments from the reader. Please feel free to contact him at [email protected].

KEYWORDS

force, moment, torque, gravity, equilibrium, center of gravity

vii

Contents Acknowledgments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ���������������������������� ix 1

Topic S-1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ����������������������������� 1 1.1 Determination of the Reaction Forces of Supports for Rigid Body  ������������������������������������������ 1 1.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  15 1.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  16

2

Topic S-2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 19 2.1 Application of the Method of Joints to Find Unknown Forces in a Plane (2D) Truss Structure  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  19 2.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  32 2.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  32 2.3.1 Determining Reaction Forces by Analytical Method   ������������������������������������������������ 32 2.3.2 Determining the Forces in the Truss Members  ���������������������������������������������������������� 33

3

Topic S-3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 37 3.1 Determination of Reaction Forces of Supports for Composite Stud  ���������������������������������������� 37 3.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  45 3.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  45

4

Topic S-4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 49 4.1 Determination of Reaction Forces of Supports for Composite Construction (System of Two Bodies)  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  49 4.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  62 4.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  62

5

Topic S-5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 67 5.1 Determination of Reaction Forces of Supports for Composite Construction (System of Three Bodies)  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  67 5.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  81 5.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  81

6

Topic S-6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ��������������������������� 85 6.1 Determination of reaction forces of rods supporting rectangular plate  ������������������������������������ 85 6.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  100 6.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  101

7

Topic S-7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 105 7.1 Determination of Center of Gravity  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  105 7.2 Sample Problem  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  121 7.3 Solution  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  122



Author Biography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ������������������������� 125

ix

Acknowledgments The author acknowledges that this work is essentially a translation and a revision of selected problems provided by Professor A. A. Yablonski (Collection of Problems for Course Projects in Theoretical Mechanics, 2nd ed., Vischaya Shkola Publishers, 1972, in Russian). The author intended to introduce this unique work to western academia, which is the product of material covered by him in many classes over a period of four decades in a number of universities and colleges.

1

CHAPTER 1

Topic S-1 1.1

DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

Find reaction forces in supports of the given construction systems schematically shown in Figures 1.1‒1.30. The sizes are in meters and the loads are shown in the table below. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

G 10 12 8 14 16 6 10 10 12 10 4 20 25 20 5 15

kN

P 5 8 4 6 10 6 7 6 8 4 6 10 6 4 10 5 10 4 10 8 10 7 6 14 16 4 10 6 10

M kN m 20 10 5 8 7 4 5 6 4 9 7 8 6 10 4 10 8 6 7 8 10 7 20 14 8 7 8 14

q kN m 1 4 2 3 1 2 1 2 2 1 0.5 2 1 2 2 0.5 1 0.5 0.5 1 2 1.5 0.5 1 2.5 3 1 0

α Degrees 30 60 60 30 45 60 45 60 30 30 45 45 30 45 60 45 45 30 45 45 30 30 30 30 45 30 45 30 15 30

2

1. PROBLEM S-1

q α

A B

M G

P

2

2

Figure 1.1.

C

P α

q

α D

M

A

B

G 2 Figure 1.2.

1

3

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

P

q

α

A

B

M G 2

2

Figure 1.3.

1

2

3

A

B M

q G Figure 1.4.

α

3

4

1. PROBLEM S-1

A

4 M C

B

p

2

α q D

Figure 1.5.

P

q

α 3

2

M A

Figure 1.6.

1

B

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

2 B α

P 2

q

α

M

A Figure 1.7.

4

P

1

3

α D

A Figure 1.8.

M

E

F

q

G B

α

C

5

6

1. PROBLEM S-1

1

1

2

q B

A M

α

α

G

P Figure 1.9.

1

1 q

P

M

B

E

A

α

C

2

1.5

G

2.5 2α

D Figure 1.10.

B 2α

M

q α A Figure 1.11.

4

3

P

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

C

B

P

G

M

α

A

Figure 1.12.

P F

D

α

q

M

α 1

C

1

1 G

Figure 1.13.

A

B F

1

E

7

1. PROBLEM S-1

q 1

3 M

1

A

2

α

G

P Figure 1.14.

1

2 q

G α

4

8

M A Figure 1.15.

P

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

q

P 2

2

2

G A

B

Figure 1.16.

4

P

1.5

2

α

B

q

G

A Figure 1.17.

α P

A Figure 1.18.

α

G

M

G

α

B

α

9

10

1. PROBLEM S-1

A

1

M q

2

2q B

3

α

P Figure 1.19.

2

q M 3

A Figure 1.20.

α

B 4

α P

C

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

M

P

α

2

4

q

4

A

B α Figure 1.21.

α

P

M

2 A

B α Figure 1.22.

2

2

q

11

12

1. PROBLEM S-1

P M B

q α

A 2 Figure 1.23.

3 M

B

P

2α 2

α

q A Figure 1.24.

C

1.1 DETERMINATION OF THE REACTION FORCES OF SUPPORTS FOR RIGID BODY

q

P α

B

2

2

2 M

A Figure 1.25.

2q

P

α

3

B 2

3

A

M 2

Figure 1.26.

2

q

13

14

1. PROBLEM S-1

Figure 1.27.

A

2

M

q B 2 P Figure 1.28.

α

2

1.2 SAMPLE PROBLEM

2 q

M

α

2

1

2α P

B α

A Figure 1.29.

P B

α

M A

α

G

Figure 1.30.

1.2

SAMPLE PROBLEM

Define the reaction forces in support A and rod CD of the construction schematically shown in Figure 1.31. G = 10 kN; P = 5 kN; M = 8 kN m; q = 0.5 kN/m; a = 30°. The sizes are in meters.

15

16

1. PROBLEM S-1

D q

P

2

1

M

C

α

A

1

B 2

G Figure 1.31.

1.3

SOLUTION

Let’s consider the equilibrium of the forces applied to the stud AB. Remove the supports: pin support at the point A, the member CD, and the string attached to point B. The actions of the support are replaced by the appropriate reaction forces (Figure 1.32). Because the direction of the reaction force of the pin A is unknown, we need to determine its components XA and YA. We also show the reaction SCD in member CD and the reaction S in the string, which is by magnitude equal to P. Evenly distributed load of the intensity q is replaced by the concentrated force Q, which is equal: Q = 2 ∙ q = 2 ∙ 0.5=1 kN and applied in the center of gravity of the distributed forces.

y

90° YA

A

C

α

XA

1 Figure 1.32.

SCD

Q

M

B

G 2

S

1

2

x

1.3 SOLUTION

17

Three equilibrium equations can be written for the plane (2D) system of forces applied to the system: ∑ MiA = 0; ‒Q ∙ 1 ‒ G ∙ 3 + SCD ∙ 4 sin 30° ‒ M + S ∙ 6 = 0

(1.1)

∑ Yi = YA ‒ Q ‒ G + SCD cos 60° + S = 0

(1.3)

∑ Xi = 0;XA ‒ SCD cos 30° = 0

(1.2)

From Equation (1.1): SCD =

Q ∙ 1 + G ∙ 3 + M ‒ S ∙ 6 1 ∙ 1 + 10 ∙ 3 + 8 ‒ 5 ∙ 6 = = 4.5 kN. 4 sin 30° 4 ∙ 0.5

From Equation (1.2): XA = SCD cos 30° = 4.5 ∙ 0.866 = 3.90 kN. From Equation (1.3): YA = Q + G ‒ SCD cos 60° ‒ S = 1 + 10 ‒ 4.5 ∙ 0.5 ‒ 5 = 3.75 kN. The values of XA, YA, and SCD are positive. It means the assumed directions of these forces coincide with their true directions.

19

CHAPTER 2

Topic S-2 2.1

APPLICATION OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

Using a method of joints find the forces in the members 1, 2, 3, 4 and 5 of the given truss structure. The truss structures are schematically shown in Figures 2.1‒2.30. The external forces (loads) acting on the trusses are given in the table below. The sizes are in meters. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

P1

P2

5 5 10 10 5 10 5 10 10 20 10 10 10 10 10 10 10 10 10 10 10 10 5 10 10 10 10 20 10 10

5 10 5 30 5 20 10 20 20 10 20 20 10 10 20 20 20 40 10 40 20 10 10 20 20 20 20 20 20 10

P3 kN 5 20 20 50 20 10 20 30 30 10 40 30 10 10 10 20 20 20 40 20 10 20 10 20 20 20 10 10 20 20

P4

P5

20 30 20 20 10 20 30 20 10 30 20 20 40 30 40 20 40 30

30 20 20 20 20 30

a m 2 2 2 2 3 3 3 3 2 2 2 2 2 -

h m 6 5 6 1.5 5 3.5 3.5 4 2.4 2.4 2.3 3 2.2 -

α degrees 30 30 30 30 45 45 45 -

20

2. PROBLEM S-2

P1 P2

3a

P3

3

2

4

1

A P4

B

5

P5

6a Figure 2.1.

P1

3 A

4

5 P2 6a

Figure 2.2.

2 P3

1

α P4

B

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

3

P2

90°

4

2

A

5

1

h

P1

B

P4

P3

6a Figure 2.3.

P1

2a

3

2

4

B

A

1 P2

P3

5

4a Figure 2.4.

21

22

2. PROBLEM S-2

P1 3

90°

P2

2

5

4

A

α

1 P3

B

P4

6a Figure 2.5.

P1

5

3

2 A

P2

P3

P4 5a

Figure 2.6.

4

1

α P5

B

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

45° 3

B

P2

P1

α P3

4

2

5

1

A

3a

Figure 2.7.

P1

3 h

90°

P2

2 A

4

1

5

P3

B

4a

2a

P1 a

A

3 4

B

2

1 P3

6a Figure 2.9.

a/2

Figure 2.8.

5

P2

23

24

2. PROBLEM S-2

B 60°

P1

3

4

α

P2

2

5 A

1 P3 a

a

P4 a

a

a

a

Figure 2.10.

B a

a

a

P1 A

3 α

a

a

2

a

5

4 P2

1

P3 Figure 2.11.

P1 3

P2

h

90° 2

5

A

4 P3

2a Figure 2.12.

B

1 P4

4a

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

P2

a

a

P3

a

a

3

5

2

α

P1

P4

4

α

4a

P5

A

B

Figure 2.13.

P3

P4

5

1

2

3

P1 4

3a

B Figure 2.14.

A

P5

3a

P2

25

26

2. PROBLEM S-2

3/2 a

3a

P3

2h

h

P1 3

4

P2

5

2h

2

1 B

A

Figure 2.15.

P3 3a P1 5 P2

3a

3 2 1

A Figure 2.16.

B

4

P4

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

P2

P1

5

3

a

2

4 P3

A 1 h

4a

B Figure 2.17.

3

5

h

3/2 h

P1

2 A

1

4 P2

6a Figure 2.18.

P3

B P4

27

28

2. PROBLEM S-2

P1 3 3/2 h

90°

4

P2

A

h

2

1

5

P3

B P4

6a Figure 2.19.

a/2

5

P1

a

3 2

A

4

1 P2

B

P3

6a Figure 2.20.

P2

P1

3

a

2

A

1

P3

B P4

6a Figure 2.21.

4

h

5

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

P2 P1

45°

3

a

2 A

5

4

P3

1

B P4

6a Figure 2.22.

6a 3

P1 2 4

1 P4

3a

P2

P5

P3 A

B

Figure 2.23.

P2 3

5 2 A

4 1

B

6a Figure 2.24.

P4

h

P1

P3

29

2. PROBLEM S-2

P1

P2 5

A

3 2

4

P3

B h

30

1

6a Figure 2.25.

5

3 2

A

h

P1

4

1

B P3

P2 6a Figure 2.26.

P2 P1

P3

a

3

2 A

1

5 B

5a Figure 2.27.

4

2.1 APP OF THE METHOD OF JOINTS TO FIND UNKNOWN FORCES IN A PLANE (2D) TRUSS STRUCTURE

P1 A

P2

3

B

5 4

a

2

1

P3

4a

Figure 2.28.

P2 45°

5 4

h

P1

3 2

1

A

P3

B

P4 6a

Figure 2.29.

P1

3

5 2

B

4 P2

1 P5

P4 3a

Figure 2.30.

a

A

P3

31

32

2. PROBLEM S-2

2.2

SAMPLE PROBLEM

The truss schematically is given in Figure 2.31. Define the truss forces S1, S2, S3, S4, and S5. P1 = 58 kN; P2 = 50 kN; P3 = 85 kN.

3

a/2

4

90°

P3

a

2 5

A

1 P1

P2 6a B

Figure 2.31.

2.3

SOLUTION

2.3.1

DETERMINING REACTION FORCES BY ANALYTICAL METHOD

Let’s consider an equilibrium of the forces applied on the truss. Remove the supports at A and B by replacing them with appropriate reaction forces. Reaction force at point A is resolved into two components: XA and YA in the directions of the coordinate system axes. The reaction force at the pin support B will be shown vertically up along member BN. The force P3 is resolved into two components P ′3 and P ″3 . The magnitudes of these components: P  ′3 = P3 cos α and P ″3 = P3 sin α (Figure 2.32). Calculate sin α and cos α: DE sin α = CD = CE cos α = CD =

a 2√a2 +(2a )2 a 2√a2 +(2a )2

= 0.447; = 0.894.

Determine the magnitudes of P   ′3 and P ″3 : P  ′3 = 85 ∙ 0.894 = 76 kN; P ″3 = 85 ∙ 0.447 = 38 kN. For forces applied to the 2D truss we write three equilibrium equations: ∑ MA = 0; ‒ P1 ∙ a ‒ P2 ∙ 2a ‒ P  ′3 ∙ 5a + P ″3 ∙ a + RB ∙ 5a = 0;

(2.1)

2.3 SOLUTION

∑ Xi = 0; ‒ P ″3 + XA = 0;

(2.2)

∑ Yi = 0; YA ‒ P1 ‒ P2 ‒ P  ′3 + RB = 0.

(2.3)

33

y

P3’

a

a/2

D

E

YA A

XA

α

α

P3 P3”

C

x

N P1

P2

RB 6a

B

Figure 2.32.

From Equation (2.1): RB = P1 + P2 ∙ 2 + P  ′3 ∙ 5 ‒ P ″3 = 58 + 50 ∙ 2 + 76 ∙ 5 - 38 = 100o kN. 5 5

From Equation (2.2):

XA = P ″3 = 38 kN. From Equation (2.3): YA = P1 + P2 + P  ′3 ‒ RB = 58 + 50 + 76 ‒ 100 = 84 kN.

2.3.2

DETERMINING THE FORCES IN THE TRUSS MEMBERS

Let’s determine the forces in truss members using method of joints. In order to find the forces in the members 1, 2, and 3 (Figure 2.33) we do a section I-I and we consider an equilibrium of the forces applied to one of the parts of the truss (Figure 2.34). It is advisable to consider an equilibrium of that part of the truss where less computations are needed. Also, it is better if the selected each equilibrium equation contains only one unknown force. It will allow for the determination of each force independently from forces acting on other members. We will assume that all members are in extension. Then a negative sign in answer will indicate that the member is in compression.

34

2. PROBLEM S-2

y

4

II

I

3

III

P3’ P3

2

YA

5

P3”

1

XA

x

N P1

Figure 2.33.

P2

II

S3

I

3/2 a

B

P3’

F

P3”

G

S1

III

P3

β

S2

H

RB

I

2a

I

RB B

Figure 2.34.

To define the force S1 we will write a moment equation about the point where the lines of actions of the forces S2 and S3 are crossing: ∑ MiF = 0; ‒S1 ∙ 1.5a ‒ P  ′3 ∙ 2a ‒ P ″3 ∙ 0.5a + RB ∙ 2a.

From here:

S1 =

RB ∙ 2 ‒ P  ′3 ∙ 2 ‒ P ″3 ∙ 0.5 100 ∙ 2 ‒ 76 ∙ 2 ‒ 38 ∙ 0.5 = = 19.3 kN. 1.5 1.5

To find S2 we will project the forces on axes Ay: ∑ Yi = 0; ‒S2 cos β ‒ P  ′3 + RB = 0,

FG = where cos β = FH

1.5a

√a2 + (1.5a)2

= 0.832. From here:

2.3 SOLUTION

RB ‒ P  ′3 100 ‒ 76 = =28.8 kN. cos β 0.832

S2=

To define the force S2 we will write a moment equation about the point where the lines of actions of the forces S1 and S3 are crossing: ∑ MiH = 0; S3 ∙ 1.5a ‒ P  ′3 ∙ 3a + P ″3 ∙ a + RB ∙ 3a = 0.

From here:

S3=

P  ′3 ∙ 3a + P ″3 ‒ RB ∙ 3a 76 ∙ 3 ‒ 38 ‒ 100 ∙ 3 = = -73.3 kN. 1.5 1.5

In order to find the forces in the member 4 we do a section II-II and we consider an equilibrium of the forces applied the left side of the truss (Figure 2.35).

II

L

K

YA

S6

α V

XA P1 Figure 2.35.

S4

S7 II

An equilibrium equation: ∑ MiV = 0; ‒YA ∙ a ‒ S4 ∙ VK = 0,

where VK = VL ∙ cos α = a · 0.894. From here: S4 =

YA ∙ a 84 =‒ = ‒94 kN. 0.894a 0.894

In order to find the force S5 we do a section III-III and we consider an equilibrium of the forces applied the right side of the truss (Figure 2.36).

35

36

2. PROBLEM S-2

III

P3’

S8

α

P3 P3”

S5

M

S9 III 2a Figure 2.36.

Equilibrium equation: ∑ MiM = 0; P  ′3 ∙ 2a + S5 ∙ 2a + P ″3 ∙ 2a = 0.

From here:

S5 = ‒

P  ′3 ∙ 2 + P ″3 76 ∙ 2 + 38 =‒ = ‒ 95 kN. 2 2

37

CHAPTER 3

Topic S-3 3.1

DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD

Determine the reaction forces and the forces in the interim pins of the composite stud. The studs and acting forces are schematically shown in Figures 3.1‒3.30. All sizes are in meters. The external forces (loads) acting on the studs are given in the following table. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

P1 6 7 12 8 9 10 4 5 10 8 10 7 8 10 14 11 8 10 9 12 6 12 8 9 9 6 10 7 5 18

P2 kN 10 4 18 10 15 7 6 4 7 14 5 12 12 7 9 15 12 15 13 6 10 20 14 15 15 12 16 10 13 7

P3 5 9 14 13 6 5 8 12 16 14 -12

M1 25 20 36 30 32 18 20 16 24 26 12 15 24 30 26 40 20 35 25 32 30 30 25 28 24 26 32 26 30 16

kN m

M2 25 28 20 18 28 24 35 -

q kN/m 0.8 1 1.4 1.8 0.9 1.5 1.2 1 1.5 1.6 1.2 1 1.4 0.8 1.5 2 1 1.2 1.5 1.4 1.3 2 -

38

3. PROBLEM S-3

P1

M1

A

B

60°

C

q

P2

2.0

2.0

2.0

2.0

2.0

F

E

D 2.0

4.0

Figure 3.1.

q

P1 A

P2

B 2.0

M1

E

60° 2.0

D

C

2.0

F

2.0

2.0

45°

2.5

3.5

Figure 3.2.

q

P1 A B

45°

2.0

2.0

D

C

3.0

P2

M1 E

2.0

2.5

F

60°

2.5

2.0

Figure 3.3.

P1

P2 60°

A

45° B

2.0 Figure 3.4.

2.0

P3

M1

1.5

D

E

M2

C 2.5

2.0

1.5

2.5

2.0

F

3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 39

q

P1

A

E

B

1.5

2.0

M1

F 60°

D

C

60°

P2

3.5

2.0

2.5

3.0

1.5

Figure 3.5.

q

P1 A

P2

M1

60° B 2.0

2.0

E

D

C

2.0

2.0

3.5

F

2.0

30° 2.5

Figure 3.6.

P1

A

P2

M1

45°

M2

C

B

2.0

2.0

P3

E

F

30°

D 1.5

1.5

2.5

2.5

2.0

2.0

Figure 3.7.

P1

A

60°

q

P2

M1

D

C

F E

B 1.5 Figure 3.8.

2.0

2.5

2.0

2.5

2.0

30°

3.5

40

3. PROBLEM S-3

P1

q

P2 60°

A

P3

M1

B D

C

2.0

2.0

2.0

2.0

2.0

E

45°

1.5

2.0

F

2.5

Figure 3.9.

P1

M1

A

q

P2 F

30° C

B

45°

2.5

2.5

D

1.5

2.0

E

2.5

2.0

3.0

Figure 3.10.

P1

A

P2

M1

60°

C B

2.5

3.5

1.5

P3

M2

E D

3.0

1.5

30°

2.0

2.0

Figure 3.11.

P1 A

P2

q D 60°

45° B

2.0 Figure 3.12.

2.0

M1

C

4.0

4.0

2.0

2.0

E

3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 41

P1

M1

45° A

E 60°

C

B

2.0

1.5

P3

q

P2 D

2.5

1.5

4.0

2.5

2.0

Figure 3.13.

P1

A

P2 30°

q

M1

E

B

2.0

2.5

D

C

2.5

2.0

30°

5.0

2.0

Figure 3.14.

P1

M1

A

P2

60°

D

B

3.5

M2

P3 45°

C

2.0

2.0

E

2.5

2.0

1.5

2.5

Figure 3.15.

P1

P2

q C

A B

45° 3.0 Figure 3.16.

M1

2.0

60° D

3.5

3.5

E 2.0

2.0

42

3. PROBLEM S-3

P1

P2

M1

A

B

M2

45° C

2.0

3.5

2.5

D 2.0

E

2.5

3.5

Figure 3.17.

P1

P2

M1

30°

C

60°

B

A

2.5

P3

2.0

D

1.5

3.5

E

2.5

2.0

2.0

Figure 3.18.

P2 M1 P1

A

q

B

60°

C

30° 3.0

2.0

2.0

2.5

D

E 5.0

1.5

Figure 3.19.

P2

P1 45° A

M2

q B D

C

2.0 Figure 3.20.

2.0

3.0

2.0

4.0

E

3.0

3.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE STUD 43

P1

q

M1

60° A

P2 E

B D

C

2.0

4.0

2.0

F

3.0

30°

1.5

2.0

1.5

Figure 3.21.

P1 A

45°

M1

q

B

P2

M2

D

F E

C

2.0

2.5

2.5

2.0

3.5

2.0

1.5

Figure 3.22.

P1

q

M1

A

60° F

B

30°

2.0

P2

3.0

D

C

2.5

2.5

E

3.0

1.5

1.5

Figure 3.23.

q

P1 C

A

30°

D

Figure 3.24.

F E

B

4.0

P2

M1

2.0

3.0

1.5

2.0

1.5

45°

2.0

44

1. PROBLEM S-1

P1

P2 60°

A

C

45°

D

B

1.5

P3

M2

M1

F

E

2.5

1.5

2.0

2.0

2.5

1.5

2.5

Figure 3.25.

q

P1

P3

P2

A

M1

C

45°

60°

D E

B 2.0

2.0

2.5

F

1.5

2.5

2.0

1.5

2.0

Figure 3.26.

q

M1

A

E

C

30°

F

D

B

4.0

P2

P1

2.0

2.5

2.0

1.5

2.0

2.0

Figure 3.27.

P2

P1

q

A

D B 2.0

Figure 3.28.

1.5

E

C

60° 2.0

M1

P3

3.5

2.0

3.0

2.0

3.3 SOLUTION

q C A

60°

E

D

B 5.0

P2

P1

M1

2.5

2.5

45

45°

2.0

2.0

2.0

Figure 3.29.

P2

P1 45°

M1

B

P1

60°

E

C

30° 2.5

1.5

3.0

D 2.0

2.0

3.5

Figure 3.30.

3.2

SAMPLE PROBLEM

A composite stud schematically shown in Figure 3.31. Determine the reaction forces and the forces in the interim pins if P1 = 12 kN; P2 = 20 kN; M = 50 kNm; and q = 2 kN/m.

P1 A

B

M

60° C

4.0

4.0

4.0

P2

q

E

D 4.0

2.0

2.0

Figure 3.31.

3.3

SOLUTION

A composite stud consists of a system of the simple studs connected via pins. Therefore, we will consider the system of the forces in equilibrium applied to each simple stud. The forces in the pins connecting these studs will be taking into the account. First, we will consider the stud DE (Figure 3.32), as a number of unknown forces (RE, XD, and YD) applied to this stud is equal to the number of equilibrium equations: ∑MiD = 0; RE ∙ 4 ‒ P2 ∙ 2 cos 30° ‒ Q1 ∙ 1 = 0,

where Q1 = q ∙ 2 = 2 ∙ 2 = 4 kN.

(3.1)

46

1. PROBLEM S-1

∑Xi = 0; XD ‒ P2 cos 60° = 0,

(3.2)

∑Yi = 0; YD ‒ Q1 ‒ P2 cos 30° + RE = 0.

From Equation (3.1): RE =

(3.3)

P2 ∙ 2 cos 30° + Q1 ∙ 1 20 ∙ 2 ∙ √ 2 + 4 ∙ 1 = = 9.66 kN. 4 4 3

From Equation (3.2): XD = P2 cos 60° = 20 ∙ 0.5 = 10 kN. From Equation (3.3): YD = Q1 + P2 cos 30° ‒ RE = 4 + 20 ∙ √32 ‒ 9.66 = 11.6 kN.

P2

YD

60°

XD D

Q1

1.0 1.0

RE E

2.0

Figure 3.32.

For the stud BD (Figure 3.33): ∑ MiB = 0; ‒M + RC ∙ 4 ‒ Q2 ∙ 6 ‒ YD′ ∙ 8 = 0,

(3.4)

∑ Xi = 0; XB ‒ XD′ = 0;

(3.5)

where Q2 = q ∙ 4 = 2 ∙ 4 = 8 kN.

∑ Yi = 0; YB + RC ‒ Q2 - YD′ = 0.

From Equation (3.4): RC =

M + Q2 ∙ 6 + YD′ ∙ 8 50 + 8 ∙ 6 + 11.6 ∙ 8 = = 47.7 kN. 4 4

From Equation (3.5): XB = XD′ = 10 kN. From Equation (3.6): YB = RC + Q2 + YD′ = 47.7 + 8 + 11.6 = 28.1 kN.

(3.6)

47

YB XB

M

B

RC X D’ C

4.0

Q2

D

2.0

2.0

YD’

Figure 3.33.

For the stud AB (Figure 3.34): ∑ MiA = 0; MA ‒ P1 ∙ 4 ‒ YB′ ∙ 4 = 0;

(3.7)

∑ Yi = 0; YA ‒ P1 ‒ YB′ = 0.

(3.9)

∑ Xi = 0; XA ‒ XB′ = 0;

YA

A MA

(3.8)

P1

XA

B

4.0

4.0

XB’

YB’

Figure 3.34.

From Equation (3.7): MA = P1 ∙ 4 + YB′ ∙ 8 = 12 · 4 ‒ 28.1 · 8 = -177 kNm. From Equation (3.8): XA = XB′ = 10 kN. From Equation (3.9): YA = P1 + YB′ = 12 ‒ 28.1 = -16.1 kN. To verify the obtained results we need to make sure that any equation of equilibrium written for the forces applied to the whole construction is valid. For example: ∑Yi = 0; YA ‒ P1 + RC ‒ Q2 ‒ Q1 ‒ P2 cos 30° + RE = 0.

49

CHAPTER 4

Topic S-4 4.1

DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (SYSTEM OF TWO BODIES)

Determine the reaction forces and the forces in the interim pin of the composite construction. The constructions and the acting forces are schematically shown in Figures 4.1‒4.30. All sizes are in meters. The external forces (loads) acting on the constructions are given in the table below. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

P1 6 5 8 10 12 14 16 12 14 8 15 15 7 5 6 8 9 7 6 7 8 5 14 10 11 15 11 12 10 9

kN

P2 8 10 12 8 6 10 8 6 10 11 15 16 18 16 17 6 10 13 10 15 14 12 9 10

M kNm 25 26 33 25 27 18 20 28 26 29 28 15 30 24 31 26 27 35 32 30 34 36 28 33 18 36 30 35 29

q kN/m 0.8 1.1 1.3 1 0.9 1.4 1 1.4 0.9 1 1.5 1.1 0.9 1.5 0.8 1.1 0.8 1.1 0.8 1.2 1.5 1.2 1.3 1 1.4 1.5 1.1 1.3 1.5

50

4. PROBLEM S-4

90°

P1

C

3.5

q

2

M A

B 2

4 Figure 4.1.

1.5

C

P1

P2

M

B

A 2 Figure 4.2.

2

4

1.5

90°

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

P1

M

C

1

30° P2

2

3.0

q A

B 5

Figure 4.3.

M 3

q P1

B

A

1 Figure 4.4.

C

1

4

q M

P1

B

30°

C

4

4

A Figure 4.5.

D

4

4

51

52

4. PROBLEM S-4

q

P2

P1

C

45°

B 45°

2

1

2

2

2 A Figure 4.6.

1

P1

3 C M

q

2

D

P2 60°

A 3

B 3

Figure 4.7.

3

3 q 1

C 30°

M

2

P1

A

B 3

Figure 4.8.

P2

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

q

P1 45°

C

2.5

M A

B 3

2

2

Figure 4.9.

1

M

C

q

P1

3

90°

A

B 2

Figure 4.10.

1.5

1.5

53

54

4. PROBLEM S-4

1

2

1 q

P1

D 30°

2

C

1

2

45° P2

M

B

A

Figure 4.11.

P1 60°

2

1

2

M

1

C

P2

2

4

q

B

Figure 4.12.

A

3

D

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

q

P2

B

60°

C M

4

P1 90°

A

D

2

2

2

3

2

Figure 4.13.

C

M

5.0

B q

A Figure 4.14.

4

90°

55

56

4. PROBLEM S-4

2 C

2

q

M

B

60°

A

30°

3

P1

45° P2

3

3

Figure 4.15.

A P2

P1

90°

2

q 60°

D

C

1 Figure 4.16.

4

M

B

1

3

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

3

P2

3

P1

45°

2

C

A

M

2

q B Figure 4.17.

2

2

M 30°

C

2 2

B

q

2

P1

A

60° P2 Figure 4.18.

57

58

4. PROBLEM S-4

A P2

q

2

30°

2 2

P1

M

C 60°

2

3

B

Figure 4.19.

D P1

2

2

C

B

2

P2

A

60°

4 Figure 4.20.

q

3

M

4

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

D P1

2

C

2

3

M

q

2

P2

B

A

60°

4

4

Figure 4.21.

3 M

P1

P2

q

C

45°

A

D 1

2

B 1

30°

1

Figure 4.22.

P1

45°

A

M

B 1

Figure 4.23.

q

P2

2

1

D

C 1.5

2

1

59

60

4. PROBLEM S-4

D

P1

4

q

P2

M

60°

C 4

B

2

5

4

2.5

2.5

A Figure 4.24.

q 2

B

P2

3 P1

1

M

90°

A

D 45°

1

C

1.5

1.5

Figure 4.25.

P1 30°

F 2 Figure 4.26.

M A

4

P2

C

B 2

D

60°

2

2

q

30°

4.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (2 BODIES)

D 4 q

B F

M

A C

60° P1

2

4

2

Figure 4.27.

C

M

A

45°

3

2

P1

q

B

D

60°

2

3

P2

Figure 4.28.

q P1

A

B 4

Figure 4.29.

30°

M

C 2

2

P2 45°

D 2

2

P2

61

62

4. PROBLEM S-4

2

P2

4 P1

A

60° 30°

q

C

M

B 2

2

Figure 4.30.

4.2

SAMPLE PROBLEM

A composite structure is schematically shown in Figure 4.31. Determine the reaction forces and the forces in the interim pin if P1 = 10 kN; P2 = 12 kN; M = 25 kNm; q = 2 kN/m; and α = 60°.

P1

α

C M P2

4

q

A 3

B 3

2

2

Figure 4.31.

4.3

SOLUTION

First, we will consider the system of the forces in equilibrium applied to the entire construction (Figure 4.32). To simplify a calculation of the moments of the force P1 we will resolve it into two components P1′ and P1″ as: P 1′ = P1 cos α = 10 ∙ 0.5 = 5 kN,

4.3 SOLUTION

63

P 1″ = P1 sin α = 10 ∙ 0.866 = 8.66 kN. The equilibrium equations will be written as: ∑ MiA = 0; P1′ ∙ 4 + P1″ ∙ 3 ‒ Q ∙ 2 ‒ M ‒ P2 ∙ 5 + YB ∙ 7 = 0,

(4.1)

∑ Yi = 0; ‒P1″ + YA ‒ P2 + YB = 0,

(4.2)

∑ Xi = 0; XA + XB ‒ P 1′ + Q = 0.

(4.3)

where = q ∙ 4 = 2 ∙ 4 = 8 kN.

From Equation (4.1): YB =

‒P 1′ ∙ 4 ‒ P 1″ ∙ 3 + Q ∙ 2 + M + P2 ∙ 5 ‒5 ∙ 4 ‒ 8.66 ∙ 3 + 8 ∙ 2 + 25 + 12 ∙ 5     = = 7.86 kN. 7 7

From Equation (4.2): YA = P 1″ + P2 ‒ YB = 8.66 + 12 ‒ 7.86 = 12.8 kN. Equation (4.3) contains two unknowns. It is impossible to define those unknowns. We can only determine the relationship between them.

y P1”

P1

α

C

P’

2

M P2

YA

A 3

2

Q

YB

XA

3

2

2

B

XB

x

Figure 4.32.

Let’s consider an equilibrium of the forces applied to the right side of the construction (Figure 4.33): ∑ MiC = 0; ‒M ‒ P2 ∙ 2 + XB ∙ 4 + YB ∙ 4 = 0,

(4.4)

64

4. PROBLEM S-4

∑ Xi = 0; XB + XC = 0,

(4.5)

∑Yi = 0; YC ‒ P2 + YB = 0.

(4.6)

From Equation (4.4): XB =

M + P2 ∙ 2 ‒ YB ∙ 4 25 + 12 ∙ 2 ‒ 7.86 ∙ 4 = = 4.39 kN. 4 4

From Equation (4.5): XC = XB = ‒4.39 kN. From Equation (4.6): YC = P2 ‒ YB = 12 ‒ 7.86 = 4.14 kN. From Equation (4.3): XA = ‒XB + P 1′ ‒ Q = ‒4.39 + 5 ‒ 8 = ‒7.39 kN.

YC

XC

C

M

4

P2

YB XB

2

2

B

x

Figure 4.33.

To verify the obtained results we need to make sure that any equation of equilibrium written for the forces applied to the entire construction is valid (Figure 4.31). For example: ∑ MiB = 0; P 1′ ∙ 4 + P 1″ ∙ 10 ‒ Q ∙ 2 ‒ YA ∙ 7 ‒ M + P2 ∙ 2 = 5 ∙ 4 + 8.66 ∙ 10 ‒ 8 ∙ 2 ‒ 12.8 ∙ 7 ‒ 25 + 12 ∙ 2 = 130.6 ‒ 130.6 = 0. The results of simulations are shown in the following table.

4.3 SOLUTION

XA -7.39

YA 12.8

Forcers, kN XB YB 4.39 7.86

XC -4.39

YC 4.14

65

67

CHAPTER 5

Topic S-5 5.1

DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (SYSTEM OF THREE BODIES)

Determine the reaction forces and the forces in the interim pin of the composite construction. The constructions and the acting forces are schematically shown in Figures 5.1‒5.30. All sizes are in meters. The external forces (loads) acting on the constructions are given in the following table. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

P1 6 11 9 10 8 10 16 13 11 12 8 12 15 10 12 13 7 9 12 11 8 14 13 10 15 12 9 6 8 15

kN

P2

M1

8 12 14 15 17 14 16 8 7 9 14 -

25 34 20 30 22 28 36 25 29 34 28 36 30 35 32 26 23 29 33 38 36 28 32 35 40 37 29 25 30 34

kNm

M2 30 34 37 21 34 -

q kN/m 0.8 1 1.1 1.4 1 1.2 0.9 1.3 1.5 1 1.4 1.2 1.5 1.1 0.8 1 1.2

68

5. PROBLEM S-5

B 90°

P1

C 3.5

M1

2

q

A 4

D

2

Figure 5.1.

1.5

C

P1

P2

M1 A

Figure 5.2.

2

1.5

90°

B

D 2

4

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

P1

M1

B

1.5

1

1

C

30° P2

2

3.0

q A

D

2.5 Figure 5.3.

P2

1

M A

B

2

90°

2

1.5

D

P1

E

60°

C

2

2

3

Figure 5.4.

P1

1

B

60°

D

3

1.5 3.0

3.0

q M1 A

Figure 5.5.

1.5

P2

C

E

69

70

5. PROBLEM S-5

2

1

45°

2

M1

E

D

2

1.5

P1

C

B

1.5

60°

P2

2

A Figure 5.6.

P1

M1

C

D

2

3.5

B

2

2

45°

1

E

1

M2 A Figure 5.7.

P2

1

1

2

E 3 Figure 5.8.

M1

C

M2

2.5

A

B

45°

2

2 F 3

D

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

2

2

2

B

M2

1.5

M1

E

D

1

P1

2

C

A Figure 5.9.

P1

q M1 1.5

1.5

3.0

2.5

B

A Figure 5.10.

C

60°

D

E 1

1.5

71

72

5. PROBLEM S-5

P1 60°

C 1

1

1

2

D M1

2

3.0

q

A

B

E

1.5

1.5

Figure 5.11.

P1

P2 60°

A

B 1

E 1

2

1

3.0

1

C Figure 5.12.

F

M1 D

1

1

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

B M1

1.5

P1

2

1.5

2

M2 1.5

D

2.5

4.5

C

E

A Figure 5.13.

P2

60°

1.5

C

1.5 D

A

B 2

Figure 5.14.

M1

2

1.5

P1

E 3

F 2

73

74

5. PROBLEM S-5

q B

E 45°

P1

3.5

2

90°

P1

M1

C

A 1

1

D 3.5

2.5

F 2.5

2.5

Figure 5.15.

C

2

M1 B

4 3.0

q A

Figure 5.16.

1.5

P1

D

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

M1

A

B D

P1

3

2

3.0

M2

C

2

Figure 5.17.

M1

C 1

B

P1

q 4

60°

5 A

D

Figure 5.18.

P2

M1

B 1

Figure 5.19.

C

60°

1

2

90°

P1

A

E

D 1.5

1.5

1.5

1.5

75

76

5. PROBLEM S-5

3

M1

B

P1

3

2

45°

D

q

2

C

A

E

Figure 5.20.

A

E C

B

2

3

2

D 2

Figure 5.21.

2

q

M1

60°

P1

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

2

C 60°

P2

D

P1

45°

2

1.5 A

M1

3

3

B

2 Figure 5.22.

C 90°

B

45°

3.0

q

A

Figure 5.23.

3

D

1

45°

q

3

P1

2

M1

E

77

78

5. PROBLEM S-5

A q

P1

2

M1

2

B 1

E 1

2

1

45°

2

2

D

C Figure 5.24.

2.5 D 3

1

M1

2

P1

q

A 2

2

45°

B Figure 5.25.

E

C

30°

5.1 DETERMINATION OF REACTION FORCES OF SUPPORTS FOR COMPOSITE CONSTRUCTION (3 BODIES)

q

E 1

D

C 1.5

2

2

3.0

1

P1

1.5

A

M1 B Figure 5.26.

A 2

q

M1 D 3

E

45°

2

3

1

B

2

C Figure 5.27.

P1

79

80

5. PROBLEM S-5

E

2

45°

2

P1

1

A

2

3

M1

C

q

2

B D

Figure 5.28.

P1

B 1.5

A

M1

1

60°

D

1.5

2

P2

1 C

3 Figure 5.29.

E

1

5.3 SOLUTION

81

P1

A

2

60°

B 1

2 60°

1

C

2

E 2

D

M1

q

Figure 5.30.

5.2

SAMPLE PROBLEM

A composite structure is schematically shown in Figure 5.31. Determine the reaction forces and the forces in the interim pins if P1 = 10 kN; P2 = 15 kN; M = 40 kNm; and q = 1.6 kN/m.

2

B

P1

C

2 M 4

A

60°

2

q

P2

2

2 D 2

Figure 5.31.

5.3

SOLUTION

First, we will consider the system of the forces in equilibrium applied to the stud BC (Figure 5.32). The equations of the equilibrium are: ∑ MiB = 0; ‒P2 ∙ 2 sin 60° + YC ∙ 4 = 0,

(5.1)

82

5. PROBLEM S-5

∑ MiC = 0; ‒YB ∙ 4 + P2 ∙ 2 sin 60° = 0,

(5.2)

∑ Xi = 0; XB ‒ P2 ∙ cos 60° ‒ XC = 0.

(5.3)

From Equation (5.1): YC =

P2 ∙ 2 sin 60° 15 ∙ 2 ∙ √32 = 6.5 kN. = 4 4

From Equation (5.2): YB =

P2 ∙ 2 sin 60° = YC = 6.5 kN. 4

P2 XB

B

60°

2

C

XC

2

YB

YC

Figure 5.32.

The components XB and XC can be determined from the equations of equilibrium of the forces applied to the element CD. Let’s consider the equilibrium of the forces applied to the element CD (Figure 5.33): ∑ Yi = 0; YD ‒ YC′ = 0,

(5.4)

∑ Xi = 0; XC′ + XD = 0.

(5.6)

∑ MiC = 0; M + XD ∙ 4 + YD ∙ 2 = 0,

(5.5)

From Equation (5.4): YD = YC′ = 6.5 kN.

From Equation (5.5): XD=

M + YD ∙ 2 40 + 6.5 ∙ 2 = = ‒13.3 kN. 4 4

From Equation (5.6): XC′ = ‒XD = 13.3 kN. From Equation (5.3): XB = P2 ∙ cos 60° + XC = 15 ∙ 0.5 + 13.3 = 20.8 kN.

.

5.3 SOLUTION

83

2 XC’

C YC ’

4

M

YD XD

D Figure 5.33.

Finally, we will consider an equilibrium of the forces applied to the element AB (Figure 5.34): ∑ Xi = 0; XA + Q ‒ XB′ = 0,

(5.7)

∑ Yi = 0; YA ‒ P1 ‒ Y B′ = 0,

(5.8)

where Q = q ∙ 2 = 1.6 ∙ 2 = 3.2 kN.

∑ MiB = 0; MA + XA ∙ 2 ‒ YA ∙ 4 + P1 ∙ 2 + Q ∙ 1 = 0.

From Equation (5.7):

XA = Q + XB′ = ‒3.2 + 20.8 = 17.6 kN. From Equation (5.8): YA = P1 + YB′ = 10 + 6.5 = 16.5 kN. From Equation (5.9): MA = ‒XA ∙ 2 + YA ∙ 4 ‒ P1 ∙ 2 ‒ Q ∙ 1 = ‒17.6 ∙ 2 + 16.5 ∙ 4 ‒ 10 ∙ 2 ‒ 3.2 ∙ 1 = 7.6 kNm.

(5.9)

5. PROBLEM S-5

YB’

4

XB’

2

B

YA XA

A

Q

1

P1

1

84

MA Figure 5.34.

To verify the obtained results we need to make sure that any equation of equilibrium written for the forces applied to the entire construction is valid (Figure 5.31). For example: ∑ Xi = 0; XA + Q ‒ P2 ∙ cos 60° + XD = 0.

17.6 + 3.2 ‒ 15 ∙ 0.5 ‒ 13.3 = 20.8 ‒ 20.8 = 0. ∑ Yi = 0; YA ‒ P1 ‒ P2 ∙ cos 30° + YD =0.

16.5 ‒ 10 ‒ 15

√3

2

+ 6.5 = 23 ‒ 23 = 0.

The results of simulations are shown in the table below. Moment, kNm MA 7.6

XA 17.6

YA 16.5

XB 20.8

Forcers, kN YB XC 6.5 13.3

YC 6.5

XD -13.3

YD 6.5

85

CHAPTER 6

Topic S-6 6.1

DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

Determine the reaction forces in rods supporting a thin horizontal rectangular plate of weight G under action of force P applied along the side AB. The constructions and the acting forces are schematically shown in Figures 6.1‒6.30. All sizes (in meters) and the external forces (loads) acting on the plate are given in the table below. Problem # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

G 20 28 18 28 32 22 20 30 36 20 8 33 24 33 35 14 20 33 20 33 28 16 20 28 30 20 33 30 30 20

kN

P

a

b

25 30 20 30 35 25 25 35 40 25 10 35 25 35 40 15 25 35 25 35 30 20 25 30 35 25 35 25 35 25

8 5.5 4 7 8 9 4 5.5 6 8.5 3 6 9.5 6 7 4 8 6 4 6 7 5.5 4 7 6 8 6 6 5.5 4

2.5 5 4.5 4 4 2.5 5 5.5 6 2.5 2.5 5.5 2.5 5.5 5 3.5 2.5 5.5 5 5.5 4 3 5 4 5 2.5 5.5 5 5 5

m

c

d

3.5 3.5 3.5 4 4 4.5 3 4 4 3.5 1.5 3 3.5 4 4 3 3.5 4 4 4 4 3 3 4 4 3.5 3 3 4 4

1 1 1.5 2 1.5 1 1 1.5 1 1.5 -

86

6. PROBLEM S-6

d

6 B

b

c

3

P

A

5

4

1 2

Figure 6.1.

P

b

6

5

B 2

1

Figure 6.2.

c

A 3 4

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

b B

P

A c

5 6

4

2

3

1

d

Figure 6.3.

6 c

P A

B 4

5

1

b Figure 6.4.

3

2

87

88

6. PROBLEM S-6

b P B

5

6

3

1

c

A

4

2

Figure 6.5.

P

B A

6

5 c

2

1

3

b Figure 6.6.

4

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

6

d

B

5

4

P

3

c

A 1

2

b Figure 6.7.

P A

c

B 4

6

5

1 b Figure 6.8.

3

2

89

90

6. PROBLEM S-6

b

P A B

4

6

1

3

c

5

2

Figure 6.9.

b P A d

6

c

5

1

2 3

Figure 6.10.

B

4

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

P

A

c

B

6

1

4

5

3

2 b

Figure 6.11.

B

d

5

A P

3

c

4

6 b Figure 6.12.

1

2

91

92

6. PROBLEM S-6

b P

B

A

5 4

c

6 1 3

2

Figure 6.13.

b

B A

5 1

Figure 6.14.

6

4

2

c

P

3

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

b B

5

4

6

2

P

c

A

3

1

Figure 6.15.

b P

B

A

3

5 6

1 2

Figure 6.16.

c

4

93

6. PROBLEM S-6

6

B

d

94

4

P

A

c

5

3 1 2 b

Figure 6.17.

b P

B 5

4

3

c

6 1

Figure 6.18.

A

2

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

P

B

c

A

4

5

2

6

3

1

b Figure 6.19.

P

b A B

c

5 1

Figure 6.20.

6

4

3 2

95

96

6. PROBLEM S-6

d

6 B 4

5

c

3 A

2

P 1

b Figure 6.21.

b

B 4 1

2

Figure 6.22.

3

6

c

P

5

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

A

P

c

d

6

5

4

1

B

3

2

b Figure 6.23.

P

B

3 5

1

6

b Figure 6.24.

2

4

c

A

97

98

6. PROBLEM S-6

5

b B A P

c

4

2

6

1

Figure 6.25.

d

6

P

B

A 5

1 2 b Figure 6.26.

4

c

3

3

6.1 DETERMINATION OF REACTION FORCES OF RODS SUPPORTING RECTANGULAR PLATE

b

P A B

5

c

6

1

3

4

2

Figure 6.27.

B A 4

P

3 5

2

1 b Figure 6.28.

c

d

6

99

100

6. PROBLEM S-6

b

2

A

3

5

c

B

P

4

1 6

Figure 6.29.

b B

P A

c

5

6

3

1

2

4

Figure 6.30.

6.2

SAMPLE PROBLEM

A construction is schematically shown in Figure 6.31. Determine the reaction forces in the supporting rods 1, 2, 3, 4, 5, and 6 if G = 6 kN; P = 9 kN; a = 5 m; b = 5 m; c = 4 m; and d = 2 m.

d

6.3 SOLUTION 101

6

B

P

c

A 5

4

3

2 1

b

Figure 6.31.

6.3

SOLUTION

Let’s consider the system of the forces in equilibrium applied to the rectangular plate (Figure 6.32). The given forces are the gravity force G applied in the center of the gravity of the plate, and the force P. Assuming that all rods are in extension, their reaction forces S1, S2, S3, S4, S5, and S6 are directed from the joints. We select the Cartesian coordinate system Oxyz, so its axes will be crossing as much rods as possible. In this case the equilibrium equations will be simple. We draw right parallelepiped and introduce the angles η, ψ, θ, and φ. The force S6 is resolved into three components along the axes (Figure 6.32): S6′ = S6 sin θ sin η; S6″ = S6 sin θ cos η; S6‴ = S6 cos θ.

102

6. PROBLEM S-6



θ

z

z1

B

A

η

P

S6

φ

S4

S6’’’

G

S2

S6’

S6 ” S5

D

S3

x

S1

c

d

b

ψ

Figure 6.32.

Next, we write six equations of equilibrium: b

∑ Mix = 0; G ∙ 2 + S4 ∙ b + S5 sin φ ∙ b = 0,

(6.1)

∑ Miz = 0; ‒P ∙ b + S5 cos φ ∙ b ‒ S6″ ∙ a = 0,

(6.3)

∑ Yi = 0; ‒S3 cos ψ ‒ S6″ = 0,

(6.5)

∑ Miy = 0; G ∙ a2 + S6‴ ∙ a = 0,

∑ Xi = 0; ‒P + S1 cos φ + S5 cos φ ‒ S6′ = 0,

∑ Zi =0; ‒G ‒ S1 sin ϕ ‒ S2 ‒ S3 sin ψ ‒ S4 ‒ S5 sin φ + S6‴ = 0.

Through the given sizes we can determine the trigonometric functions of the angles: sin φ =

c 4 = = 4 ; √a2 + c2 √52 + 42 √41

cos φ =

a = 5 ; √a2 + c2 √41

sin ψ =

c 4 = = 4 ; √b2 + c2 √52 + 42 √41

(6.2)

(6.4)

(6.6)

6.3 SOLUTION 103

b = 5 ; cos ψ = √b2 + c2 √41 sin η = cos η = sin θ = cos θ =

a 5 1 = = ; √a2 + b2 √52 + 52 √2 b

√a2 + b2

=

1

;

√2

√a2 + b2

√a2 + b2 + d 2

=

√52 + 52

√52 + 52 + 22

=

5√2 5 = ; √54 3√3

d 2 = . √a2 + b2 + d 2 √54

From Equation (6.2): S6″ = 12 ∙ G = 12 ∙ 6 = 3 kN. As: S6″ = S6 cos θ, From Equation (6.3): S5 =

then:

S6 =

S6″ 3√54 = = 11.03 kN. cos θ 2

P ∙ b + S6″ ∙ a P S ∙ a sin θ cos η 9√41 1.5 ∙ √54 ∙ 5 ∙ = + 6 = + 5 ·5 b cos φ cos φ b cos φ 5

5√2 1 54 ∙ √2 ∙

√41

= 3.3 ∙ √41 = 21.1 kN.

From Equation (6.1):

4 S4 = ‒ 12 ∙ G ‒ S5 sin φ = ‒ 12 ∙ 6 ‒ 3.3∙√41 ∙ = ‒3 ‒ 13.2 = ‒16.2 kN. √41

From Equation (6.5):

‒S ″ S sin θ sin η ‒1.5 ∙ √54 ∙ 5√2 ∙ 1 ∙ √41 √54 √2 S3 = cos6ψ = ‒ 6 cos ψ = = ‒1.5 ∙ √41 = ‒9.62 kN. 5

From Equation (6.4):

1 P S′ S sin θ sin η 9√41 P 1.5 ∙ √54 ∙ 5√2 √54 ∙ √2 ∙√41 = (1.8 ‒ 3.3 + S1 = cos φ ‒ S5 + cos6 φ = cos φ ‒ S5 + 6 cos φ = ‒ 3.3√41 + 5 5 1.5) ∙ √41 = 0.

From Equation (6.6): S2 = ‒G ‒ S1 sin φ ‒ S3 sin ψ ‒ S4 ‒ S5 sin φ + S 6‴ = ‒G ‒ (S1 + S5) sin φ ‒ S3 sin ψ ‒ S4 + S6 cos θ = ‒6 4 4 2 ‒ (0 + 3.3 ∙ √41) ∙ ‒ (‒1.5 ∙ √41) ∙ ‒ (‒16.2) + 1.5 ∙ √54 · = 6 kN. √41 √41 √54 The results of simulations are shown in the following table. Rod number Reaction forces, kN

1 0

2 6

3 -9.62

4 -16.2

5 21.1

6 11.03

As seen from the table, the rods 2, 5, and 6 are in extension, the rods 3 and 4 are in compression, and the rod 1 has zero reaction force.

104

6. PROBLEM S-6

To verify the obtained results we will write an equilibrium equation, for example, a moment equation about axis z1: ∑ Miz1 = ‒S1 cos φ ∙ b + S3 cos ψ ∙ b + S6′ ∙ b = ‒S1 cos φ ∙ b + S3 cos ψ ∙ b + S6 sin θ sin η ∙ b = 0 + (‒1.5√41) 5 5√2 1 ∙ ∙ 5 + 1.5 ∙ √54 ∙ ∙ ∙ 5 = ‒37.5 + 37.5 = 0. √41 √54 √2

105

CHAPTER 7

Topic S-7 7.1

DETERMINATION OF CENTER OF GRAVITY

Determine the center of gravity of the construction consisting of the homogeneous rods of the same weights per unit of their lengths (Figures 7.1‒7.6), the plane objects (Figures 7.7‒7.18 and 7.24‒7.30) or the 3D objects (Figures 7.19‒7.23) (Figures 7.1‒7.30). In Figures 7.1-7.6 the sizes are in meters, in Figures 7.7‒7.30 in centimeters.

2.5

2.5

y

x

O 4 Figure 7.1.

106

7. PROBLEM S-7

2.5

5

y

45°

x

O Figure 7.2.

y

30° 45°

15° 30°

O Figure 7.3.

x

7.1 DETERMINATION OF CENTER OF GRAVITY 107

y 2

1.5

45°

2 45°

x

O Figure 7.4.

y

90°

60°

30°

O

3.5 Figure 7.5.

3.5

x

108

7. PROBLEM S-7

y

90°

60° 60°

60°

O

2 Figure 7.6.

2

x

7.1 DETERMINATION OF CENTER OF GRAVITY 109

4

y

x

O 60

5

Figure 7.7.

x

O

30

Figure 7.8.

110

7. PROBLEM S-7

y

10

4

2

8

2

2

2

2

x

O

Figure 7.9.

12

8

4

15

y

O 8 2 20 Figure 7.10.

x

7.1 DETERMINATION OF CENTER OF GRAVITY 111

y 60°

60°

x

O

30 Figure 7.11.

y

18

R2

2

2

9

x

16 Figure 7.12.

112

7. PROBLEM S-7

y 4

x

4

O 2 2 16

Figure 7.13.

y

7.5

7.5

7

20

2

7

O

7 Figure 7.14.

x

7.1 DETERMINATION OF CENTER OF GRAVITY 113

y

10

7

10

O

30 Figure 7.15.

x

114

7. PROBLEM S-7

y 6

4

3

6

20

O

x

10 Figure 7.16.

45°

Figure 7.17.

O

20

10

5

y

x

7.1 DETERMINATION OF CENTER OF GRAVITY 115

y

15 x

O

Figure 7.18.

25

z

x Figure 7.19.

10

15

O

25

y 10

116

7. PROBLEM S-7

17

z

15

O

10

y

40

x Figure 7.20.

z 10

10

10

25

O

20

30

x Figure 7.21.

y

7.1 DETERMINATION OF CENTER OF GRAVITY 117

z

10

15

O

5

13

15

y

25

x Figure 7.22.

10

20

z

O

y

15

x

35 10

Figure 7.23.

15

118

7. PROBLEM S-7

40

y

x

O 15

40

20

Figure 7.24.

y

15

15

25

10

x

O 15 Figure 7.25.

40

7.1 DETERMINATION OF CENTER OF GRAVITY 119

40

R15

y

x

O 65 Figure 7.26.

y

40

20

x

O 70 Figure 7.27.

120

7. PROBLEM S-7

y

7

15

25

35

x

O

40

60

Figure 7.28.

y

5 Figure 7.29.

40

O

30

5

4 4

x

7.2 SAMPLE PROBLEM 121

y

10

30

10

20

x

O 50 Figure 7.30.

7.2

SAMPLE PROBLEM

Determine the coordinates of the gravity center of the plane object shown in Figure 7.31.

y 30

R20

80 Figure 7.31.

x

122

7. PROBLEM S-7

7.3

SOLUTION

The coordinates of the gravity center are determined by formulas: xc =

∑ Fi xi ; F

yC=

∑ Fi xi . F

(7.1)

In order to use these formulas, we need to cut the area into small objects with known centers of gravity. In this problem these objects are a rectangle, a triangle, and a semicircle (Figure 7.32). The area of the semicircle will be cut from the rectangle area, so we will consider it with the negative sign. The area of the rectangle: F1 = 40 ∙ 30 = 1,200 cm2. The area of the triangle: F2 =

40 ∙ 50 = 1,000 cm2. 2

The area of the semicircle: F3 =

π ∙ 202 = 200π = 628 cm2. 2

y

40

30

C1 C

C2

yC

C3

50

x xC Figure 7.32.

The gravity centers of selected elements will have the following coordinates. 1.  Rectangle: x1 = 15 cm; y1 = 20 cm.

7.3 SOLUTION 123

2.  Triangle:

50 40 x2 = 30 + 3 = 46.7 cm; y2 = 3 = 13.3 cm.

3.  Semicircle:

x3 =

4R 4 ∙ 20 = = 8.5 cm; y3 =20 cm. 3π 3π

To calculate the coordinates of the gravity center of the entire plane object, we provide the following table. Element # 1 2 3 Total

Fi, cm2 1,200 1,000 -628 1,572

Xi, cm 15 46.7 8.5 -

Yi, cm 20 13.3 20 -

Siy = Fi xi, cm3 Six=Fi yi, cm3 18,000 24,000 46,700 13,300 -5,338 -12,560 59,362 24,700

The coordinates of the gravity center of the given plane object will be calculated by Equations (7.1): xC =

59,362 24,700 = 37.8 cm; yC = = 15.7 cm. 1,572 1,572

The center of gravity is shown in Figure 7.32. Note: The areas and the gravity centers’ coordinates of some plane objects are given in the following table. Plane object Triangle

Area

y

1 F= 2 aha

B

ha

C

yC

Gravity Centers’ Coordinates 1 xC = 3 (x1 + x2 + x3), where x1, x2, and x3 are the coordinates of the vertices 0, A, and B; 1 yC = 3 h a

A

0

x

xC a

Circular Sector

F= R

a a

xC

C

x

αR2

2R sin α R2b = xC = 3α 3F

124

7. PROBLEM S-7

π α=2

F=

πR2 2

xC =

4R 3π

π α= 6

F=

πR2 6

xC =

2R π

(semicircle)

Circular Segment

R

a a

C

xC

x

1 F = 2  R2 ∙ (2α ‒ sin 2α)

xC =

b3 4R sin3 α = 3(2α ‒ sin 2α) 12F

125

Author Biography Dr. Sayavur I. Bakhtiyarov is a Professor at New Mexico Institute of Mining and Technology (Socorro, NM, USA) and a Fulbright Fellow. Dr. Bakhtiyarov obtained his Ph.D. from the Russian Academy of Sciences in 1978, and in 1992 a D.Sc. from the Azerbaijan National Academy of Sciences. His areas of expertise are: multiphase flows, nanotechnology, nonlinear fluid mechanics, tribology, rheology, and self-healing composites. Dr. Bakhtiyarov taught engineering courses over four decades in several countries (Azerbaijan, Russia, China, UK, Turkey, U.S.). Dr. Bakhtiyarov authored 350+ scientific publications in refereed scholarly journals, books, international conferences and symposia proceedings, and 14 patents. Dr. Bakhtiyarov was elected as a foreign member of Russian Academy of Natural Sciences and International Ecoenergy Academy. He served as a Program director of US DOE and NASA research projects, INSRP US DOD coordinator for NASA’s Mars Science Lab mission. Dr. Bakhtiyarov is a lead organizer of the ASME annual symposia and forums, Editor in Chief of two international journals, Mechanics and Solids (IJM&S) and Manufacturing Science and Technology (IJMS&T), and an Editorial Board Member of i-manager's Journal on Engineering and Technology (IJET), Mathematics Applied in Science and Technology (MAST), International Journal of Applied Engineering Research (IJAER), International Journal of Dynamics of Fluids (IJDF), and Far-East Journal of Mathematics (FEJM).