251 103 25MB
English Pages 1042
Table of contents :
Strength of Materials  1 (Pg 1  90)......Page 1
strength of materials part2 (Pg 91200)......Page 103
strength of materials part3 (Pg 201  300)......Page 213
strength of materials part 4 (Pg 301368)......Page 313
Strength of Materials Part  5 (Pg. 369454)......Page 381
Strength of Materials Part  6 (455500)......Page 467
Strength of Material  7 (Pg 501614)......Page 513
Strength of Material Part  8 (Pg 615690)......Page 627
Strength of Material Part  9 (Pg 691770)......Page 703
Strength of Material Part  10 (Pg 771 856)......Page 783
Strength of Material Part 11 (Pg 857936)......Page 869
Strength of Material  12 (Pg 9371030)......Page 949
STBEIGTl
1
or
MAIBBIALS .
( IN MKS/SI UNITS) (A Text Book for Engineering Students of all Disciplines]
U.C.JINDAL B.E. (Mech.); M. Tech. (Mech.); Ph.D.
Professor in Mech. Engg. Deptt. Delhi College of Engineering, Kashmere Gate, Delhi  110006 (India)
UMESH PUBLiCAtI6NS Publishers of Scientific, Engineering & Technical 5B, Nath Market, Nai Sarak, Delhi110006
Books ·
Strength of Mater~als U. C. Jindal
©Author All rights reserved. No part of thi6 pubiieation ma;, be ,ri!produced or transmitted, in dn}' {Nm or by any means, without the written permission of tfie autlwr and the ptibfisheYs.
1st Ed. : 1989
Published by : Umesh Publications, SB, Nath Market, Nai Sarak, Delhi110 006. Phone: 2915961
Printed at : Himdeep Printers, Pad am Nagar, Kish.an G'anJ1 Delhi110007.
Dedicated to the loving memory of my son
..
~
PANKAJ
. •.
Preface This book on Strength of Materials covers firstly the introductory course on the subject for the engineering students of all disciplines i.e. Mechanical, Production, Civil, Electrical, Electronic Engg. and Computer Sciences in the Engg. Colleges as well as in Polytechnics and secondly the advanced course on the subject for the students of Mechanical and Civil Engg. disciplines. This book will act as a faithfull companion to the students studying a course on Machine Design and computing stresses in machine members and to engineers serving in design offices of various Research and Development Organisations. The author is teaching the subject for the last 23 years and is fully conversent with the difficulties experienced by the students. Therefore, while preparing the text of the book, the point of view of the students was constantly kept in mind. The contents of the book have been designed in a manner to help all grades of the students. For the relatively mediocre students unable to attend classes regularly, there are simple examples and exercises, a thorough study of which would impart confidence and a clear understanding of the subject. For the brighter students, there are complicated problems and exercises, the understanding and solution of which will help them go a long way in securing exceptionally good marks and in assuring a place of distinction in any competitive examination. In brief, the book contains the following' features :( l) A rigorous treatment given to the subject to meet the current requirements of the students. (2) Providing a clear understanding of the basic principles of the subject through the worked examples which are more than 500. (3) Thoughtprovoking and selftesting objective type questions which are more than 200 in number. (4) Information provided about testing the mechanical properties of the materials m the laboratory. (5)
Solution of examples and problems both in MKS and SI units.
The advanced chapters on Bending of Curved Bars, Rotational Stresses, Energy Methods, Unsymmetrical Bending, Shear Centre and Torsion of Non Circular Shafts are no doubt available in many books but either the treatment given is too elementary or the examples given are insufficient. As a result, the students are apprehensive of these chapters when they appear in the examinations. Therefore, these topics have been thoroughly explained and a large number of solved examples are given so that the students can very well understand these advanced topics. Constructive suggestions for improvement of the book are always solicited. The development of this book has been strongly influenced by the author's colleagues, students and the numerous books on the subject published in India and abroad. The author is deeply indebted to the inspiration received initially from his brilliant and genius son and this book is dedicated to his loving memory.
pr, U. C. Jin~!
REFERE~'Cl:. BOOK
NOT FOR ISSUE.
Contents Chapter 1.
1· 1. 1·2. 1·3. 1·4. 1·s. 1·6. 1·1.
1·s. 1·9. " 1· 10. 1· 11. 1· 12. 1· 13. 1·14. 1· 1s.
2.
Page Nos.
Simple Stresses and Strains
1
Normal and shear forces Normal strain Bars of varying cross section Tapered bars Bar subjected to various forces Extension under self weight Bar of uniform strength Shear stress and shear strain Volumetric stress and volumetric strain Tensile test on mild steel Strain energy and resilience Sudden load Impact loads Stress concentration in members under tensile force Factor of safety Summary Multiple choice questions Exercises
!.
4 8 11
15 19 20 22 26 27 35 39 41 43 45 66 68 70
Composite Bars and Temperature Stresses
75
2 ·1. 2·2. 2.3. 2·4.
Stresses in a composite bar Composite bar with more than 2 bars of different materials Composite systems Bars of different lengths subjected to loads . , 2·s. Bolt and tube assembly tightened with a nut ., 2·6. Temperature stresses in a single bar " 2·1. Temperature stresses in a composite bar Summary Multiple choice questions Exercises · 3.
75 78 80 84 86 89 90 115 116 118
Principal Stresses and Strains
123
, 3·1. Stresses on an inclined plane 3·2. Principal stresses , 3·3. Graphical soiutioµ
123 129 p•
l~J
(ii)
Chapter
Page Nos.
3·4, Ellipse of stresses 3·5, Strain components 3'6. Strain components an inclined plane 3·1. Mohr's strain circle 3·s. Principal strains jn terms of principal stresses 3·9, M odified modulus of elasticity Summary Multiple choice questions Exercises 4.
137 139 140 142 J4S 146 163 164 165
Relations Between Elastic. Constants
4· 1. Young's modulus of elasticity and Poisson's ratio 4' 2. Modulus of rigidity 4'3. Relation between E and G 4'4. Relation between modulus of elasticity and bulk modulus Summary Multiple choice questions Exercises
169
171 172
174 183 183 185
5. Thin Cylindrical and Spherica] Shells
187
5' I. Stresses in thin cylinders 5 ·2. Thin spherical shells 5'3. Cylindrical shell with hemispherical ends 5·4. Wire winding of thin cylinders Summary Multiple choice questions Exercises
6.
187 192 194 196 215 216
.218
Thick Cylinders
6"1.
6'2. 6"3. 6'4. 6"5.
221
Lame's equations Thick cylinder subjected to internal fluid pressure Compound cylinders Shrinkage allowance Hub and shaft assembly Summary Multiple choice questions Exercises
Shear Force and Bending Moment Diagrams 1· 1. Various types of beams 7·2. Shear force diagram of simply supported b~a m sqbje.< 10 5 = 17'044 x 106 cm: 1 '
Exercise 1'61. A circular steel ·bar. di~metet_i10 tµ)ll aµd length 1500 mm is fixed at its upper end. If the weight density of steel ~s 0'0078 kg/chi 5 aetermine : ' (a) maximum stress dcvelope.d in bar, ·,
l
;•.
J
(b) elongation of bar under its own 't'trlght. Oiv~n E = 2 x 106 kg/cm2. \ [Ans.; far 1·17 kg/cin2 (b) tN1875 X10 3 mm]
1'7.
BAR OF UNIFORM STRENGTH
·., A bar of varying section is shown in Fig. 1·19 such that the· stress developed in the bar at. any section is the same. Say w is the weight per unit volume of the bar. The area of cross section aj: the bottom edge is A1 and area of cross sectioJ.?. at top fixed edge is A 2 and length of the bar is H. j
 I
:•
i
•
• i
J
= A '= A + dA 1
= wA . dy
I
H·,
f
Weight of~'tr1p·,
'
I'
\ '
Let i us consider a small elementary strip ABCD rof length •dy at a distance of y .; from the .hottoi;n : edge . Say the stress developed in tQ.e i J:jar ~brqug~ µt its length . f. ! ' • ! ! lS I i i , , ~ t ; !.,.. ~ t Say area at (:!J =A Area at AB
I
,.,
I
,
J '.
21
SIMPLE STRESSES AND STRAINS
For equilibrium f. A + w. A dy= f(A + dA) wA dy=f . dA
or Integrating both the sides A0
H
Jdy
dA w y=y
J
A,
O . A2,
llti A~=;
or
·H
A1
or
or or at any distance, Example t·71, A vertical circular bar 150 cm high is. subjected to a uniform stress of 2 6 kg/cm2 throughout its length. The diameter at the bottom edge is 6 cm, determine the diameter at the top edge if it is fixed in the ceiling. Given the weight density =0'0078 kg/cm3 • s,,Iution.
Arca of cross section at the bottom edge, A 1 =
;
62 = 911 cm2
density, w=·0078 kg/cm3; Height, ll= l50 cm; Uniform stress,/= 2'6 kg/cm 2 • Area of cross section at the top fixed edge wH
A2 =A1
el ·0078 X 150
=9n e ~  = 9i; x e·45 = 9ft X l '5683= ~ D'i.3 4
Diameter at the top edge, D~=
r 9n X 1"56~3 X4 *
. rv ·
= 'V56'4588 = 7'514 cm.
.,
Weight
'STRENGTH OF MATERIALS
22
Exercise 1·11. A copper bar of uniform strength, 2 metres long is suspended vertically with its top edge fixed in the ceiling. The uniform stress developed in the bar is 17·8 kg/cm2 • If the diameter at the bottom edge is 8 mm, what will be the diameter at the [Ans. s·tt mm] top edge. Given weight density of copper= S·9 X 10 3 kg/cm3 •
t·s. SHEAR STRESS AND SHEAR STRAIN Fig. 1· 1 (c) shows a rectangular block distorted under the actiI
.c.
( b)
Fig. I ·20
Fig. 1·20 shows a block of length L, breadth Band height H subjected to a force P at the top surface while the bottom surface DCEF is fixed. Under the action of this force, the block is distorted to a new shape A'B'G'H ' DCEF. · shear force, P Shear stress = area  of ·· the plane, ABGH
=
q
P Bx L
(generally denoted by this letter)
,1, . displacement AA' Shear stram = AD = tan .,, .
But the angle ¢, is very very small within the elastic limit of the material. Shear strain tan if>!"" sin ef> ,,=,, ¢,.
=
So the shear strain is given by the angle of displacement
ef>.
Jf the force Pis gradually increased, the shear stress q also gradually increases and the angle of displacement if, changes with q. Fig. 1·20 (b) shows the variation of ¢, with respect to q. Upto a particular l imit shear stress q is directly proportional to !lthear strain r/>. This limit is called the elastic limit. Within the elastic limit, if the shear force is removed from the block, the block returns to its original shape and original dimensions. But if the shear force is
23
SIMPLE STRESSES AND STRAINS
removed after the elastic limit stress (q.), there will be permanent deformation or distortion left in the block. Within the elastic limit q ex: ¢,
q= G · where G is the proportionality constant
or
G shear stress, q  shear strain, ¢ ·
Ratio of {  is defined as the Mo dulus of rigidity, G. Complementary shear stt'ess. Fig. 1·21 shows a b lock subjected to shear stress q at the top surface, and a shear stress q (due to the reaction) at the bottom surface. The shear
_rq~r?·" qxlxB C l ockwise co uple
Anticlockwis e couple ·
Fig. 1·21
, j,·
angle ¢, is very small (and not so large as shown in the fi gure) the length D C, breadth A'H' and height A'D' can be considered as n egligibly changed fr om their original dimensions L, B and H respectively. +
Shear force at the t op surface
= q XL X B +
Shear force at the bottom surface= q XL XB. These two forces constitute a couple of arm H , tending to rotate the body in the clockwise direction. Moment of the couple
= qXL X B x H;)
For equilibrium this applied couple has to be balanced by the internal resistance developed in the body. Say the resisting shear stress on the vertical faces is q' as shown in the figure. Shear force on the surface CEG'B' = q' x Bx Ht Shear force on the surface DFH'A'=q'xBx H ,I.. These two forces constitute an anticlockwise couple of arm L, resisting the applied couple. Moment of the resisting couple = q' x B x H x L. ') For equilibrium q' x BHL= q X LBH
or
q'= q.
24
STRENG:rH OF MA'HiRJA:LS
This resisting stii:ess q' is called .the ·complementary sh.ear str~ss a:nd always acts at an angle of 90° to the applied shear stress . Moreover if the applied stre,s is positive i.e .., tending to rotate the body in the clockwise d irection, then the compl~!llentary shear stress will be negative i.e., tending to rotate the body in the anticlockwise direction. The use of the concept of complementary shear stress 'Will be made in the chapter 3 on principal stresses.
Example 1·8l. Fig. 1 ·22 shows a rivet joining two plates of thickness t= 1·2 cm and width= 6 cm. The plates are subjected to force F = 1200 kg. If the diameter of t"'1 ~ rivet is
diameter of rivet
Fig. 1·22
15 mm, determine the shear stress developed in the rivet. The ultimate shear strength of the material of the rivet is 2·s tonnes/cm2. How much maximum load the plates can carry. Solution. The plates carry the force F which acts as a shear force on the circular plane aa' of the rivet.
This plane aa' is along the contact surface between the two plates. Force F = qOO kg Area of cross section under shear force,
A= Shear stress in the rivet, q=
TC
4
(d):!=
TC
4 ( I ' 5)2=
~ = J1.;~~
I '767 cm2
= 679' I 2 kg/cm 2.
If one section of the rivet is subjected to shear force, the rivet is said to be in single shear.
N0w the ultimat,e shear stress of the materiiil of the rivet q .. u= 2800 kg/crn 2 Maximum shear force,
= qu11 x A = 2800 x 1·767 = 4947'6 kg.
Un
of proportionality or
Ill Ill.
or
E .,.
Volum~tric str a i n
Bulk Fig. 1·26
27
SIMPLE STRESSES AND STRAINS
Example 1'91. A spherical ball of a material 100 mm in diameter goes d own to a depth of 500 metres in sea water. If the weight density of sea water = I040 kg/m 3 and the Bulk modulus of the material is 16 x 10s kg/cm 2, determine the change in the volume of the ball. Solution. Weight density, w= 1040 kg/m3
= 1040 x 10
6
kg/cm a = '00104 kg/cma Depth of water upto the ball, h= 500 metres = 500 X 100 cm Hydrostatic pressure or volumetric stress,
p = wh Bulk modulus,
= '00104 x 500 x 100 = 52 kg/cm 2 K = l6 x 10s kg/cm~
Volumetric strain (reduction in volume), 52  3'25 10s 16 x 10sX
p
f.o= K
IW f..,= V or Original volume,
av, change in
volume= f.u .
V
V = nD3  = n X (I 00)3 = 0'5236 X 106 mm a 6 6 oV=V. f.o= 0'5236 X 106X 3 25 x 10 5 = 17'017 mm3= '017 cma.
Exercise 1·91. A cube of 200 mm side of a mater ial is subjected to a volumetric stress of 50 N/ mm2 • What will be the change in its volume if the Bulk modulus of the m aterial [ Ans. 4000 mm 3 ] is lOO X 103 N/mm 2 •
1'10. TENSILE TEST ON MILD STEEL . !'vfikl steel is the m1tterial most commonly used in m achi ne members and in structural ~pplicat10ns. A specimen of circular section and of the shap e shown in F ig. 1·27 is cla mped m the fixtures of a testing machine. Collars are provided at both the ends so that the specimen
~~ Fracture
Fig. 1'27
28 is firmly fixed in the fixtures of the machine. t he cc11trai portibn, where section is uniform is called the guage length. Then t he spt cuncn is gradually extended a nd the hiternal resi1;tailce ofl'er'ed by the specimen (i.e., tensile load) gradually increases. During the init ial stage i.e., when p oc 8/, extensometer is used to measure very small changes in length. AfLer this stage, vernier scale on the machine is used to measure cxteuj;ion. Load and extension are simul . taneously recorded t ill the specimen breaks into two pieces. P and 8/ are now plotted on a graph taking suita ble scales. 0 to A is a straight line ; stress at A is called the limit of prl!)portionality. The material obeys Hooke's law i.e., p oc. 8/ P 81 or A ocT stress, f oc
E,
strain
/ = EE
wher e Eis the Young's modulus of elasticity.
B is the elast ic limit, i.e., if the load is removed at thi s stage, strain will also return to zero When the material is loaded beyond this stage, plastic deformation will occur in the material, i.e., after the removal of the load, strain is not fully recovered and the r esidual strain (0r deformation) remains in the materi al. From O to B, there is elastic stage and from B to the point of fracture is the plastic stage. Beyond B i.e. , at the point C, t here is considerable extension with decrease in internal resistance. This is called the upper yield point and the stress at this point is called the upper yield strength . At D again the internal resist aJ1ce of the material increases upto the point E, i.e., the maximum load point. At this p oint, necking takes place in the specimen and further extension t akes place in the vicinity of this neck. This point is also called the point of plastic instability. The stress at t he maximum load i.e., P.,,,.,/A is called ultimate tensile strength of the material. At the p oint F , the test piece breaks making a cup and cone type of fracture as sh own in the figure which is a typical fracture for a ductile material. The two pieces can be joined together t o find out the diameter at the neck where the specimen has broken, say t4c area at. the neck is a. N ominal breaking strength Load at fracture /  A (original area of cross section) . Load at fracture Actual breakmg strength= ( f k) a area o nee I
Percentage reduction in areas
Aa
=  A X lOO
81 =   X 100 1 where 8/ is the elongation upto fracture. Percentage elongation
The specimen of mild steel suffers a considerable increase in length till it breaks. T his type of material is called a ductile material. On the contrary, a brittle material like cast iron fails after very small elongation .
In some ductile materials, which d o not exhibit a definite yield point, a proof stress (at 0'2% strain) is determined to fi nd the onset of yielding (as shown in the figure). Percentage elongat ion and percentage reduction in area give estimate about the ductility of the material.
SIMPLE STRESSES
AND
29
STRAINS
Now for differ ent gauge lengtl\S and different areas of crosssection the percentage elongation will be different and to draw a comparison between the ductility of various materials becomes difficult. To avoid this difficulty, similar test pieces must be tested for all the materials if the test pieces of same gauge length l and a1'ea A cannot be tested. Barba has sho wn th.at extensio n upto the maxi11rnm load i.e. , 8/1 is proportional to the length of the specimen and extension beyond maximum load a nd upto the point of fracture ie., M2 is proportional to VA, where A is the area ofcross section
8/1 ex: / = bl
or
8/2 cc
./A
= c1 7r 8i=ol1 +812= ·bl+ cif A
T otal elongatio n,
Similar test pieces will h ave the sa me ratio of ...,~ = D{~,
25
DI
or
rt
ratio
will be
the same for similar test pieces . The constants b and c ar e call~d the Barba's co nstants.
Example 1'10·1. An aluminium a lloy specimen 12·5 mm dia meter and 5 cm gauge length was tested under tension. During the first part of the test, following readings were recorded
Load (kg) Extension (mm)
0 0
750 '0327
I
1000
I
'0450
I
I
1250
1500
I
1750
2000
 ·~  . . '0568
'0756
I
· 120
'216
Plot the load extension diagram and determing the following values : 1. Young's modulus of elasticity ; 2. Limit of proportionality ; 3. o·1% proof stress.
Solution.
Diameter~of the specimen = 12'5 mm= 1'25 cm
Area of cross section ,
A=;
(1·25)2= 1'227 cm2 •
Graph is plotted between load in kg and extension in mm (taking suitable scales for beth). Fro m the graph OA is a straight line and beyond the point A loadextension curve is not a straight line. So the point A represents the limit for proportion ali~. Taking a point C along the straight line Load at C = 500 kg p 500 2 Stress = A= 1.227 kg/cfn
81, extension at C
= '022 mm
ao
STRENGTH OF M.ATEilIALS
0 022
Strain
=so
Young's modulus, E
2 _ 2QQ_ J'227 X 2Q_ ·022 _ 9'26 X 105 k g / cm .
_ Limit of proportionality 
1250 _ . . 1018 75 kg/cm11 . 1 227
Guage length
= 50mm
o· 1 percent extension

(See the graph)
50 x 0· 1 = 0'05 mm. 100
At 0·05 mm extension draw a line EB parallel to the straight line OA , intersecting the load extension curve at the point B. Load at point B So
= 1838 kg
o· 1% proof stress

(from graph of Fig. l '28)
~
. 1498 kg/ cm a. 1 227
2000
 ·
1S38
1750
roo 1250
I
Limit of
I
I
I
I
I
,B
I I I
0>
>