STRENGTH OF MATERIALS [For Engineering Degree^ Diploma and A. M. I, E, Students]
By S.
RAMAMRUTHAM B.E.,
PrincipaJ,
Modern
{CMh M.l.C.B. New Delhi
Colleise of Engineering,
Auihor of Design of Reinforced Concrete Structures, Design of Steel Structuies, Theory of Structures, Applied Mechanics etc.
DHANPAT 1682, N.\I
RAI
& SONS
SARAK, DELHMiC«006
(H.O.
JULLI^OR)
Design Of Rei^niced Concrete StructnKs
Design of Steel Stmctures
Theory of Straetares Prestressed Concrete
Applied Mechanics Hydraulics. Fluid Mechanics and Fluid Machines Steel Tables
First Edition: 1962
Reprint: 1970
CONTENTS Chapter 1.
Pages
*
‘
Simple Stresses and Strains
Introduction— Definitions, stress, strain, tensile and compresstresses— Sheat stress— Plastic limit— Hooke’s law — sive poisson’s ratio— Modulus of Elasticity -Modulus of Rigidity, Bulk Modulus— Bars of varying section '—Extension of a tapering rod— Composite section -modular ratio- Bar of uniform strength- Equivalent area of composite sections— Temiterature stresses— Hoop stress— Stresses on oblique sections— State of simple shear - Relation between the Elastic constants— Volumetric Strain— Rectangular blodt subject to normal stresses— Diagonal tensile and diagonal compressive stresses— Solved problems 1 to 71 —Problems for exercise. 1
2.
Strain
- 100
Energy— Impact Leading
Strain Energy —Elastic, plastic and rigid members— Stresses due to different types of axial loading —Gradually applied loads— Suddenly applied loadj— Impact loads— Solved problems 72 to 84— Problems for exercise. 101—118 3-
Centre of Gravity and
Moment of Inertia
of Gravity— Definition— Lamina — Moment of an area— Centroid of a uniform lamina— Centroids of laminae of various shapes— Triangle, circle, semicircle, trapezium — Built-in sections— Analytical and graphical methods— Moment of Inertia of a lamina— Definition —Parallel axes theorem— Perpendicular axes theorem— Moment of Inertia of laminae of different shapes - Rectang ilar, ciicular, triangular and composite sections— Solved problems 85 to 104— Problems for Centre
continuous beams -Conception of Shear Force and Beading Moment— Sign conventions—shear force and Bending Moment diagrams for cantilevers, beams supported at ends. Beams with overhangs— Point of contrafiexure -Member subjected to couples— Members subjected to Oblique loading— Miscellaneous types of members and corresponding S.F.
and B.M. diagrams-;-Iqter-reIation between S F. and B. m! diagrams— To obtain the B.M. diagrams from S.F. diagram- Solved problems 105 to 130 -Problems for exercise.
158-227 5. Stresses in
Beaau
Definition— Pure or 8im{de bending— Theory of simple bending— Netural layer— Neutral axis— Beading Stress distribution-moment of resistance— Assomp^os in the theory of
—
(i7)
Chapter
Pages
simple binding— Practical application of bending equation modulus— Section moduli for different shapesSectioi) Rectangular, triangular, circular, I-section, T-section— Normal force on a partial area of a beam section — Moment of resistance of a partial area of a section -Fliiched beams— Equivalent section— Beams of uniform strength— Shear stress distribution on a beam section— Shear stress distribution on circular, triangular, I and T sections— Shear rectangular, stresses in bolts connecting components in laminated beams. Proportion of B VI and S F. resisted by flange and web of an [ section— Shear centre— Solved problems I3l to 199—
bmm
Problems 6.
228—337
for exercise.
Direct and Bending Stresses Stress
distribution of the section
rectangular column.
The middle
of an eccentrically loaded rule— Core or kernel of
third
—
a section -Circular section—Hollow section Structural section— Walls and pillars— Solved Problems 200 to 223 — 338—370 Problems for exercise. 7.
Masonary Dams Forces acting on a dam— Stress distribution on the base of a dam. Stability of a dam— Minimum bottom width of a dam 371—388 section. Solved Problems 224 to 228,
8.
Deflection of
beams
Member
10.
9.
bending into a circular are— slope, deflection and ladius of curvature — Derivation of formulae for slope and deflection— Cantilever— Propped cantilevers- Beams— Macaulay's Method — Beams subjected to couples— Moment area method -Mohr’s theorems— Relations between maximum bending stress and maximum deflection — Beams of varying section -Strain energy stored due to bending— Law of reciprocal deflections - Bette’s law— The first theorem of Castigliano— Impact loading on beams— Laminated SpringsConjugate beam method— Solved problems 229 to 312 Pro?-, 389—52'’ blcms for exercise.
Fixed and Continoons Beams Fixed beam -Relation between the free B.M. diagram and the fixed B.M, diagram-slope and deflection— Effect of sinking of supports— Fixed beam subjected to couple— Degree of fixing— Advantages and disadvantages of fixing beams— Continuous beam— Clapyron’s thcorm of three moments 528—581 Solved problems 313 to 324— Problems for exercise. Toraioii of Shafts
Pure Torsion— Theory of Pure Torsion —Torsional mement of resistance. Assumptions in the theory of pure Torsion— Polar modulus— H.P. transmitted by a shaft— Torsional Rigidity— Stepped shafts— Composite shafts — Keys— Couplings — Shear and Torsional resiltenoe— Shafts of non^drcular section— Close
coiled helical springs— Torsion of a tapering rod. Problems 325 to 361 —Problems for exercise. 11. Priacipal stresses
Solved
’582-632
and strains
Normal stresses— Tangential or shear streses— Principal
stresses
—Principal planes— Graphical and analytical methods— Ellipse of stress— Determination of principal stresses and strains— Obliquity— Mohr’s circle of stress— Combined bending and Torsion— Strain energy in terms of principal stresses— Equivalent bending moment and equivalent torque— Principal strains— Criterion for failure — Ellipse of strain— Solved prob633—682 lems 362 to 388. Problems for exercise. 12.
Thin Cylinders and Spheres
Thin cylinders— circumferential ond longitudinal Stresses— Riveted cylinderical boilers— Wire bound pipes. Thm spherical shells— Biaxial stresses in doubly curved walls of pressure vessels— Stresses in a conical tank. Solved probkms 342 to 683—705 349. Problems for exercise. 13.
Hiick cylinders and Spheres
Thick cylinders— Derivation of formulae— Lamme’s equations
—Hoop
413— Problems 14.
radial pressure distribution— compound spherical Shells - Solved problems 350 to 706—726 for exercise.
stresses
—Thick
cylinders
and
Colnmns and Struts
—
Introduction Axially loaded compression members— Crushing load— Buckling or critical load or crippling load— Euler’s theory of long columns— Different end conditions— Effective length of colums— Assumptions made in Euler’s theory— Limitations of Euler's formula— Empirical formulae— Rankine’s formula Straight line formula —Johnson's parabolic formula— Formula given by the I S. code— Column •abjected to eccentric loading— Euler's method— Rankinc’s method— Prof Perry’s formula— Columns with initial curvature —Laterally loaded struts— Solved problems 413 to 429. Problems for exercise. 727— 768
—
15.
Rhetcd
Jtrfats
Tjrpes of joints— Lap and butt joints— Failure of a riveted joint— Tearing strength, shearing strength, bearing strengthEfSciency of a joint— Riveted joints in structural steel work— Chain riveting and diamond riveting— Eccentric Riveted conn^ions— Resistance of a rivet against translation and rotation. Solved problems 430 to 442. Problems for -
exercise. 16.
769 —798
Wdded Couaectioiis The welding process— Advantages of welded connection — Disadvantages of welded connection —Types of weld— Minimum sizes of weld— Effective length- Minimum length— Fillet
weld applied tp the edge of a plate— Angle between fusion faces - Throat thickness -Intermittent
-
fillet
welds— Lap
joints
welds in slots or holes— End returns— Bending about fillet -Permissible stresses in single welds— Combined a 17. in welds— Eccentric welded connections. Solved stresses 799— 830 problems 443 to 462 Fillet
Analysis of Framed Stroctures Perfect frame -Deficient frame— Redundant frame— Reactions supports— Analysis of a truss -Method of joints -Method
at
of section— Graphical method. Solved problems 463 to 489.
831-914 18.
Simple Mechanical Properties of Metals Yield or flow of material— Tensile stress— Stress— Strain diagrams for Mild Steel Specimen— Limit of proportionality— Ultimate stress— Working stress— Factor of safety— Measurement of ductility -Unwin’s Method based on reduction of sectional area— Hardness— Scratch test— Indentation test— testing— Fatigue of metalsBrinnel’s method— Impact Endurance limit. Solved problems 449 to 491. 915—922
19.
Elements of reinforced Conaete General principles of design— Assumptions— Singly reinforced beams-Nctural axis-Lever arm— Moment of resistance— or critical sections— Unbalanced Balanced or economic sections— Under-reinforced and over-reinforced sections—
Doubly reinforced beams— Shear in beams- Shear stressesDiagonal tensile and diagond compressive stresses in concrete-Stirrups-Diagonal reinforcement— Bond stresses— End ancliora.e -Standard hook— Reinforcement— T and L beams- Axially Loaded Columns- Combined bending and direct stresses. Solved problems 492 to 520, Problems for
923— 998
exercises.
Appendix Index
Useful tables.
999-1035 1036—1038
1 Stresses and Strains §1.
btrodactioB
we come across may be
Materials which plastic
and
materials.
rigid
An
elastic
claailQed into elastic,
matoial
undeigoes a deformation when subjected to an external load^ such deformation disappears on the removal oftheloi$ng. plastic material undergoes a continuous deformation Ai^ng the ^riod of loading and the deformation is permanent and materitd does not regain its original dimensions on the removd' df the lAaffiag rigid material does not undergo any deformation when subjected
A
^
A
to
an external loading. In practice no material
We attribute these
is
absolutely elastic nor plastic nor rigid.
when the deformations me within certain Generally we handle a member in its elastic range. Structural limits. members are all generally designed so as to remain in the riastic properties
condition unde r the action of the working loads. §2.
Resistance to Deformation
A material when subjected to an external load system goes a deformation. Against this deformation the ateri al will offer a resistance which tends to prevent the defonnatitin. This material as long as the ntember is resistance is offered by the forced to remmn in the deformed condition.^ This resistance is offered by the material by virtue of its srrengrA. In the elastic stage, the resistance offered by the material is proportional to the deformation brought about on the materia] by the external tna/Ung The material will have the ability to offer the necessary resistance when the deformation is within a certain limit. loaded member remains in equilibrium when the resistance offered by the member against the ddbrmation and the applied load are in equifibriom. When the member is incapable of offering the necessary leristance against the external forces, the deformation wOl continue leading to the failure of the member.
m
A
§3.
Stiem
The is called
the food.
force of resistance offered by a body against the stress. The external force acting on the body
the
The load
is
in tte material of the
b^.
Tj^s of stresses. area
eppUed on the body white the Fig.
1
(a)
shows a rod of uniform tiie aids A and B.
A and subjected to axial leads F «t
is
stress is
naiw
» .
SntSNOTB OF MAtWIAlS 2
,-^„ .i«.io.yyi»niialtottc
longitaillMl
nit »f
”*'*^theiiieii>bwtalakentooiMSiit(>tl«ol)iirBC«iia by the sertioa XX. whicli it is dMded of the part C. consider the equilibrium Let us
See Fig
1
P®"
(W-
is
subjected to the
mlo
.
_ _
, , F externally
in equilibrium, the part
kli thifoS
J
C oflm a r«i8-
^t
In other words, we may against a possible „ offering the resistance stress. against the deformation is the R resistance This
SnolraUhe^Son
XX is
This part
B
Mie
say,
XX.
R
Son XJf. equal
ai>ove, *hc rwislto^ Obviously for the case mentioned offered by the section to the load. If the resistance
Mdo^ite
defonnatioD be assumed to be uniform across the^ scctioii> in^sity of the resistance per unit area of the section is called At ist&uity of stress or unit stress. In common usage the word sftoss is used to mean the intensity of stress.
iMiml tli9 difc
Intensity of stress
=p=-jR
A
P
=-
.•
A
Let doe to the application of the load the length of the dtange from / to /+
The
memha
ratio of the change in length to the original called strain.
is
•
a f
. 8train=e=-j-
member
length of the
=
3
SDfPU STRESSES AND STRAINS
of stresses. material is capable of offering the following ftrpes section of a offered by resistance the When (0 Tensile stress. the section is said to offw length, in increase an against is member a by the a tensile stress. For example, the stress offered intensity of the tensile The stress. tensile is a Fig. 1 in rod of the stress is given by
A
s^on XX
R The corresponding
_ ..
^
P
strain is called .
a
tensile st^in.
Increase in length
Tensile strains
If the bar >45 of Fig. 1 (ii) Compressive stress. to pushing axial loads as shown in Fig. 2, a resistance against a decrease in length. any section such as
be subjec^ set up by
is
XX
Fr..2
This resistance is called a compressive resistance. of the compressive resistance or stress is given by
The
intensity
Let due to the external loading the len^h of the member decrease by dl. The ratio of the decrease in length to the original length is called a compressive strain.
Compressive strain
Decrease in length Original length
SfRENGTH OP MATSRIALS
4
shows a rectangular block of height (no Shear stress. Fig. 3 unity. width and L / and length
Fig. 3
Let the bottom face of the block be fixed to a surface EF. Let a force P be applied tangentially along the top face of the Such a force acting tangentially along a surface is called a block.
hear
force.
For the equilibrium of the block, the surface EF will offer a tangential reaction P equal and opposite to the applied tangential force P. Let the block lx taken to consist of two parts G and Consider the equilibrium to which it is divided by a section XX. of the part G.
H
See Fig. right, the part
ttot
4.
In order the part a resistance
H will offer
G may
not
R along
the section
move from
left
XX
to
such
R=P.
ic)
(a) Fig. 4
Similarly, considering the equilibrium of the part H, that the part G will offer a resistance along the section that R^P.
R
The
resistance
R along
the section
XX is
we
find
XX such
called
a shear
rests-
tmee.
XX
Fig. 5 shows a failure at the section caused by the tangential loads acting on the top and bottom faces of the block. This type of ftflure is called a shear failure. In a shear failure, the two parts into which the block is separated, slide over each other. Hence if udi a shear failure should not occur, the section must be able to offer tangential resistances along the section opposing the force qt file t<^ face and the force P at the bottom face. For the
XX
P
equifi.
SIMPLH STRtSSHS
AND >rRAlNS
uz>
5
id
(b>
Hig. 5
brium of tbe system the shear resistance
R shtndd be equal
to the
tangential load P,
R^P. The
intensity of the shear resistance along
the section :
XX
is
called the shear stress.
Sheai stress - ^ ~
R A
Shear resistance Shear area
R _ P
Lx
l
"~L x
1
Shear deformation. Fig. 6 (a) shows a rectangular block subjected to shear forces on its top and bottom faces.
P
Fig. 6
When the block does not fail in shear, a shear deformation occurs as shown in Fig. 6 (6). If the bottom face of the block bo fixed, it can be realized that the block has deformed to the position A^BiCD. Or we can say, that the face ABCD has been distorted to the position A\B\CD through the angle BCBv=^^. Let us now imagine that the block consists of a number of horizontal layers. These horizontal layers have undergone horizontal displacements by dilferenl amounts with respect to the bottom face. can assume that the horizontal displacement of any horizontal layer is proportional to its distance from the lower face of the block.
We
be dh
Let the horizontal displacement of the upper face of the blodc Let the height of the block be /.
STRENGTH OF HATBRiAtS considered any other horizontal layer
shear strain. We could have which is at a distance x from the lower face. say the layer the layer XX. dx be the horizontal displacement of
XX
Let
dx Then shear Since ^
is
strain
very small,
^=tan^= j- = shear
Hence, the angular deformation the shear strain. §4.
^
in radian
strain.
measure represents
Elastic limit
A material is said to be elastic when it undergoes a deformation on the application of a loading such that the deformation disappears on the removal of the loading. When a member is subjected to an
When the its section svill olfer a resistance or stress. removed, obviously the stress will vanish and the deformaBut this is true when the deformation caused tion will also vanish. by the loading is within a certain limit. For every maieriaJ the property of assuming or regaining its previous shape and size is exhibited on the removal of the loading, when the intensity of stress If the loading is so is within a certain limit called the elastic limit. large that the intensity of stress exceeds the elastic limit, the member If after exceeding the loses to some extent its pioperty of elasticity. elastic limit the loading is removed, the member will not regain its original shape and a residual strain or permanent set remains. axial
loading,
loading
is
Hooke's law. It is observed that when a material is loaded such that the intensity of stress is within a certain limit, the ratio of the intensity of stress to the corresponding strain is a constant which is characteristic of that material. Intensi ty of stress
^constant.
strain
In the case of axial loading, the ratio of the intensity of tensile or cotnpressive stress to the corresponding strain is constant This ratio
is
called
denoted by
E
Young’s Modulus or Modulus of Elasticity* and
is
e In the case of shear loading also, the ratio of the shear stress to the corresponding shear strain is found to be a constant when the shear deformation is within a certain limit. This ratio is called Shear Modulus or Modulus of Rigidity and is denoted by C, or G.
N
fS.
Units
In this book SI andMKS units are adopted to express quantities of various magoitudes. The following nomenclature is
SiSdT
tlllPLK STRESSES
AND STRAMS
KILO
7
MILUIQT^
I0«
MEGA GIGA
MICRO 10-« NANA 10-*
10«
10»
TERRA
PICA
1012
I0-»*
In the SI units force is generally expressed in newtats. newton ikN) means 1000 newtons. In the
MKS units force is expressed in kg
was to express force
in
Stress intensity
is
kg
The
kilo-
(the earlier practice
wt).
expressed in various forms
newtonlmn^, newtonim*, kglcm\ kglnfi 1
J\r/meife2= 10-« iV/mm*
1
Njmm^ =10® N/metr^-
Problem
A rod of
!.
l
mega Newtonlmei^^
25 mm in ^tnketer and 200 cm an axial pull of 450(1 kg. Find {i) the
steel is
The rod is subjected to long. intensity of stress, (it) the strain,
and (Ui) ehngation. TUdce E—Tl X lOA
kglcm*. SoIutiOD.
Diameter of
Length of rod
PoU
rod=d=25 mm=2‘5 cm —1—200 cm =P=4500 kg
.
(i) Intensity
of stress
—
kx2’5*
=A— ntP
Area of the section
=/
P
^
'
.
«=4 909 cnr
4500
,
7 kglotfi
Strain
916'7 f “ .£’“21x10®“^
(Hi) Elongation
=d/=StramXorigina! length
(ii)
» 00004365 X 200 cm ‘0-0873 cm.
mm
in diameter tmd 2 metre long Problem 2. (SI) A steel rod 25 subjected to an axial pull of 45 kN. Find (/) the intensity of stress, (ii) the strain, tmd (Hi) elongation. Take E—2(X) GNfmetr^. is
Solution. Diameter of rod =25
Length of rod Puii
Area of rod
mm
**2 metn
P=45
=-J
jt)^=45000
N
(25)*-490-9 mnfl
=490-9 X10“« metr^ (i) Intensity
of jtresr=/—
P
STRENGTH
8
45000
MATERIALS
..
.
“490'9X 10r« ‘=9167 X /O® Nlmetr^~=9r67 Njmm^
MN/metr^
m>9J'67 (h) Strain
—d/^Strain X original length =0’0004583 x 2 metre
(h7) Elongation
=^0 0009166 metre •sO'9166 Probicin
metre long. the the
member tie
A wooden
3.
It is
is
tie is 7'
mm.
wide, 15 cm deep and I'SO of 4500 kg. The stretch of Find the Young's Modulus for
5 cm
subjected to an axial pull
found
to
be 0 0638 cm.
material.
Area of tie=j4*=7‘5x 15=112’5 cm^
SoIntioB.
=/>-4500 kg
Pull .'.
hg/cm~
Stress
Change Strain
in length
Original length
Young's Modulus
_ 0 0638 =00004253 ISO
= E— f e
40
“cF0W4253 ProUem
A
load of 400
kg/cm-.
kg has
to he raised at the end of a must not exceed 800 kgjcm^ what is the minimum diameter required ? If^ai will be the extension of 3' 50 metre length of wire ? Take E—2x /(/> kgfem^. steel wire.
4.
If the unit stress in the wire
SoIntioB.
Load on the wire= 1F=400 kg
=A=*y
Area required
Let the diameter of the wire be
ueP •••
^=
d cm
0-5
^
d-yj ^^0-7979 cm 7'979
mm
£xt0irjiMi=«d7’=Strainx original length
cm^
MMPU STRESSES AMO STRAINS
9
800
X3S0 an 2X10* =m0’l4 cm =ar4
mm
ProUen 5. (SI) A wooden tie is 60 mm wide, 120 mm deep and 1‘50 metres long. It is stdtjected to an axial pull of 30 kN. The stretch of the member is found to be 0‘625 mm. Find the Young's Modulus for the tie material. SoiathM.
Area ofthetie=X==60x 120= 7200 mm®=-7200xl0’* metr^
Puli=P=30
ifciV=30,000 JV
Stre88-/=
^=
_ Mrain-c=
72^^-6 =
Change
.
^
0‘625
in length
Qjiginal
10* Nlmetre^
len^ “ rsx 1000
= 4 167X 10-« Young's Modulus—
E=— e
4167X10*
-
,
,
W=«-
=10^ Nfmetr^ =10xl(F N/metre^
=10 GNJmetr^. ProUem 6 tensile
divisions in the O' 00 1
(SI).
load of 40 kN.
200
mm, find the
A 20 mm diameter
brass rod was subjected to a The extension of the rod was found to be 254
mm
extensometre.
If each division
is
modulus of brass.
electric
Stdathm. 7C
Area of the rod=y4=-^ (20)*
mm*
=3 14’ 16 mm* =314*16 xl0-« metre*
PuIl=P=40 r
Stress- /=•-
^ ^
itiV= 40,000
_
N
40,000 3j4*i6’x i0-«
^
= l•2732Xl08^/mefre* Lmgth of specimen=/=200 mm «d7= 254 XO’OOl =0*254 mm
Extension
Strain=e=
=0*00127
,
etpial
to
STKENGTIi OF tIAlEUAXS
10
r2732x
t
Modulus=E=
Young's
‘e
“
108
Njmetr^
0 00127
=^1002-5xmNlmetr^ =‘l00-25xl(fl Nlmetr^
= 100-25
GNImetre^.
Problem 7. A hollow steel column has to carry an axial load of 200,000 kg- If the external diameter of the column is 25 cm, nd the Take the ultimate stress for the steel column to be internal diameter. 4800 kglcm^ and allow a load factor of 4. Soiotion.
=D =25 cm
External diameter
-d
Internal diameter
Load on
the column
=
200,000 kg
Ultimate stress
=4800 kgjcm^
Factor of safety
=4
Safe stress =/=
Ultimate stress Factor of safety
4800
4
= 1200 kgjcrn^
Sectional area required
W
200,000
“l200~ ( 252
= V6-61
(7w2
-
625-d*^212'21
d2=41279 d°‘20'30 cm.
Problem 8. (SI) meted on a mild steel
The following data
refer to
(0 Diameter of the Gauge length
steel
(vi)
V
a load of 100 ktl is
elastic limit
Maximum
(v)
ivii)
Extension at
Load at
(»v)
tensile
test
coif
-30 mm
bar
(a) (itf)
a
bar.
load
Total extension
—200 mm =0139 mm =230 kN =360 kN
=56 mm
Diameter of the rod at failure
=22' 25
mm
J^'^*^^ote ia) The Young's modsdus (b) The stress at elastic
Fig. 7 shows a bar which consists of three lengths /i, h and h with sectional areas Ai, Az and Aa and subjected to an axial load P.
Even though the intensities
of stress
For
instiUDce,
will
total force on each section is the same, the be different for the three sections.
Intensity of stress for the portion
p
lAB—fi= -j-
STRENGTH OF MATERIALS
12
and. Intensity of stress for the portion
Let
E be the Young’s
and
p ^ ^3
Modulus.
—
Strain of the part
and
CD^fs^
Strain
of the
part
BC
Strain
of the
part
CD=
Change
in length
of the part AB--=dli=eiIi
Change Change
in length
of the part
in length
of the part
BC—dh=cxh CD=dh=exh
Total change in length of the bar=rf/=
Problem
9.
shows a bar consisting of three lengths. Find and the total extension of the bar for an Take E= 2xl(f> kgfcm\
Fig. h
the stres-ics in the three parts
axial pull of 4 tonnes. Solution. Intensity
Load P-4 tonnes =4000 Ag. of stress on the part A B
~565'8 kg/cm^.
SIMPLE STRESSES
AND
S111AINS
Intensity of stress
13
on the
part
BC-fz== 1274
Intensity of stress
on the
part
kg.fcnfi.
CD-fz>
=480
kg.lem*.
-
( 3 25)2
Total extension
dl‘=dh+dh-\-dlz
= 4'^i+=|-/2+^.fa ( fth +A&+/3iyi
2 x lO®
[565-8x18+1274x26
+480x16]
cm.
=0‘0255 cm. 10. A bras.i bar having a cross-sectimud area of 10 sq. subjected to axial forces shown in fig. 9. Find the total change in length of the bar. Take E> ’^^ J OSx Kfi kg.Icnfi.
ProUem
cm.
is
B 5,000kg.
\
C
8,oo
\*-60cm -»|*
2,0(mkg.
I
\toookg.
—mcwFig. 9.
Part AB. The section of the bar in this part jected to a tension of 5000 kg. SolirtioB.
.'.
Extension of
is
sub-
AB= ^rh AH 5000 x 60
idx 105x10®
=
cm. (Extension)
Part BC. The se^on of the bar cominession of 8000—5000^3000 kg. Contraction of
BC=~^^BC-
in this part is subjected to
a
STRENOIB OF MAIimiALS
14
3000X100
loxrosxio*
= CD.
Part
The
(extraction)
section of the bar in this part is subjected to
a
compression of 1000 kg. .'.
Contraction of
CD '
.4E 1000 xJ20
“ 10x105 xl0« = -— cm. =0’01 14 j .'.
Change
in length of the
“Js —CtOlM
cm. (contraction)
member
cm. (Decrease in length).
ABCD
nnbiem 11 (SI). A member is subjected to poin loads P\, P%, Pi, awif P4 as shown in Fig. 10. Calculate the force Pz necessary for equilibrium if Pi =‘120 kN, Pa=220 and P\^I60 kN.
kN
Determine
also
the net
chmge
h
length
of
the
member.
Take
£<=200 GN/m‘.
z IB* ,
Sototion.
Resolving the for^s
on the rod nlong
its &xiS;i
120+P8-2M+160
Pz=2^ kN Part
AB
Force on the cross-section-Pi=i 20
Extewdon
kN- 120,000 N (tensile)
ofAB=-+r -JjO.OOO X075 [1600 X I0~^[200 X 10®J .
“ +0 00028 metre— +0*28 mm
wc
+
8IMPLB snssnss Part
1
AND STRAINS
15
BC
= 120— 220 =» — 100 kN ==—100,000 N (compressive)
Force on the cross-section =Fi—P2
Contracuon of
100000 X
BC^-
Fo^l
=— O’OOOS metre*" —O'SO mm CD
Part •
»)rce
on the cross-section=P4 « 160^kN
= 160000 N Extension—
Net change
in length
(fensUe)
16OO0axl.2 X lO^'^’^^x 900 [
lO*]'”^*'^^
=-i-0’00107 metre»+l‘07 of the member
mm
=+02i-0‘i0+r&l ^+0‘55 mm
(exteniion).
P
A member ABCD
PraUeB
2, is subjected to point loads, Pi, 12. Pa and P4 as shown in figure 11. Calculate the force 2 necessary for equilibrium, if Pi— 4500 kg., Pz— 45000 kg.. Pi— 13000 kg. Determine the total elongation of the member, assuming the modulus of elasticity to be 2' lx IC^ kg.jcnr^.
P
M /!—4:
US
I
L
ftocm
±J
—
—mm. Fig. 11
Sohition.
Resolving the forces on the rod along
its axis,
have,
Pi+Ps-Pa+Ps Bat Pi=4500 .-. 4500
+
.'.
kg.,
Ps=45000
kg.
and
P4= 13000 kg.
45000=Pa+ 13000 Pj =36500 kg.
Part AB Force on the cros8>8ection=Pi
=4500 Extension of
AB=
kg. (tensile)
Pih
AiE 4500 XlM
'
6-25> 86400
E
(extatsion)
we
—
STRENGTH OP MATERIALS
16 Part
BC
Force on the cross-section =Pi—i*8
«450C- 36500
— Contraction of
BC=
—32000
32000 x 60
25 76800
Part
kg.
kg. (.compressive)
E
cm.
cm. (contraction)
CD
Force on the cross-section=P4 =73000 kg. (tensile) Extension of
CD= _
Pih
„ '44E AiE
13000 x 90
£
12-5
93600 —— ^
cm.
cm. ,(extension), .
Total change in length of the 'member
_
86400 ,76800 ,93600 103200
.
103200
21 X
10«
=0'049143 cm. (extension) subjected to a tensile
load
SotatloR.
Area required for the middle portion
Rg.
12.
^
SntPU
STRESSES
AND
STRAINS
17
Let the diameter of the middle portion be
nJ* ~4
y6
“
7
cms
d’=‘3'7I8
.
Let the length of the middle portion be Stress in the
d rms.
x
cms.
end portions 5200 ttx5“
^' _
1
4
-774
kg-lcm^ ,
kg/cmK
1
Total extension of the rod--0‘016 cm.
Extension of the
end portions-f Extension
ol
ti'.c
portion
n
•
middle
-0016 cm
V
(.70-.v)4Kt
-0016
r
(3(1
nv»|6x£ 056x2xlO«
.v;4
•
0
r
•
774 1(^0
14(Xiv^ 32000
8"77
Vy'i.x
i
.T-"
FfoWfm
a
tr, :
t
In
A
14.
rojh's
(t;)
thi
rca rj
Mock
pidh y from wluck
J C*K
t.
ff ^ is suyponded load H'and (A). In hath la.so. rU rapes have ki^ enfl. rnr and the value of E h
L'i)
b
imtimtons
iB ami
In it) 14'
'
•
Pt
tided
and
W
CB are
m
such a
nded from a
is
topes pined Wtvr tear both ropes
imioiifit
bind, for t th
*
downward morettu m application of the
'
ABC
ror\:
striuh by ibe same
•’*
'
cjadva'lv
I v
soio!' frir loon'vs
to a
-3:('00
14(KH-
*
23223- ^74
M
32000
had
*
and A the
^he stre.^srs In i
idbyand
the rope<
the block dm:
[
and find the the gradual
{London University)
Strength of materials 18 In
(a)
Solution,
this
T-]
tension in since the the rope is uniform Let the smooth. is pulley P kg tension in the rope be ol Hence for the equilibrium
arrangetnent the
h52m
T
the s>stcm,
2P-H^
W
/
4
2
-250 /.
4-56m.
45601
I
kg.
Intensity of stress
on
the rope section ’-P
nr
P A
-
250 -kgicmr-
=3
1
‘25
M'-i
TONNt'
r UNfie
(»•
kg lent*.
Fig. 13.
56M'52« 10'64 m.
Length of the ropc==4 56 4 4
The
increase in length of the rope
31-25
Let the
3 394
<10-64x100 cm.
cm
downward movemeni of 2S- 3 394 cm
the pulley be
cm.
«-77><>r,v2.
second arrange lunt the h>aJsshai 'i1 by the two should bc\nC\ th it thc\ estend by the same Let the stress ifiionsiry on the sections of the lopes 4B and
ih) In
ropes
AB
ftmount.
CB be pi
the
and
ard Pi
respccci\cl>\
Biquatmg the extensions of the two
m P‘i
h'0^_
J3
U
^
4‘5n ^
'
4
.
in
BqI tennon
^
r >pe
- jpi
in i4B4-tcnsion in
kg
CB
-*
W
i
wc
]ia\e.
^X SIMPLE STRESSES
AND
=
19
STRAINS
500
/»i+P2=—j,
4
500
3P8+;»2=— 7 3
^~
500 8
p2’^26 8 kgicnfl. Pi =
and
35‘7 kglcm^.
26 8
Downward movement of the
pulley
^Extension of the rope
AB or CB
^V662 cm. Problem 15. A steel subjected to a pull of St,
tie
rod 4 cm.
in diameter
add 2
long is
To what
length the bar should be bored centrally so that the total extension will increase by 20% under the same pull, the bore being 2 cm. diameter.
Take E^2000
tjern^.
et
at Fig. 14.
ilutioo.
*
4
«
— (4)2«47c
cm^.
«
4n
/ 'E
‘
're
X 2000
x200=
Extension after the bore
is
l_
5«
made=r2x
J'5J5
6 Lei the bar be bored to a length cl reduced section A' "=4ji
—" 4
I
metres.
Area at the
(2)®
4nr-ic=>3n
enfi.
Extension of the rod
imttrn
I
pi. i<
STRi
20
- 1) (2 lOn 2 -/
^
25n
_i_ 25si
15Jt
^
2/
,
^
3nx2U00
'“itx2d00^^
NCIH OP MATERIALS
15 “25 To 30-15/+20/-'36 "^
5/- 6
/=»r2 m.
Problem 16 A rectanr gular base plate is fixed at each of its four corners by a 20
mm.
and
nut
diameter bolt as shown in
Fig. 16.
The
I
ig.
16
between nut and plate are of 22
mm.
plate
rests
on
washers of 22 mm. inter-nal diameter and SO mm. external diameter. Upper washers which are placed internal diameter and 44 mm.
external diameter.
which the
/
If the base plate carries a load of 12 t {including self weight is equally distributed at the four corners) calculate the stress on er washers before the nuts are tightened-
What would be the stress in the upper and lower washers when the nuts are tightened so as to produce a tension of 0’5t on each bolt ?
(AMIE May SoIntioD.
Area of the lower washer
Load
(5®— 22*) cm.®
4 =
15-83 cm.®
transmitted fo one lower washer =
4 Stress intensity in the lower
1971)
washer*
3/
= 3000 kg.
kg.jcm.*
= 189'4 kg.lcm.^ When
the nuts are tightened the compressive irasner •tension the bolt =0-5/ =500 kg.
m
Area of the upper washer
Stress intensity in the lower
load in the upper
""X ^4‘^*“2'2*) cm®. = 1140 cm*. washer* 1 1
40
kg.fcrrfi.
.43’8S kg.lcm
*
—
SIMPLE STRESSES
AND STRMNS
Now the compressive
21
load on the lower washer
-^000+500- ^500 washer
Stress intensity in the lower
3500 15‘83
= 22 /
kg.Jcnfi.
Fig. 17 shows a rigid bar ABC hinged at A and ProbleoDi 17. suspended at two points B and C by two bars BD and CE made of aluminium and steel respectively^ The bar carries a load of 2000 kg. midway between B and C. The cross-sectional area of the aluminium bar BD is 3 sq. mm. and that of the steel bar CE is 2 sq. mm. Deter^ mine the load taken by the two bars BD and CE. {A.M.LE. Nov., 1965) Modulus of Elasticity for aluminium Eai^7000 kg Modulus of Elasticity for steel Ef^^ 20^000 kglmm.^
Solution.
STEEL
A
rod
rigid
will
not bend. It will remain straight. But the aluminium and steel rods will extend due to their elasticity.
HOD
Fig. 17 shows the position of the rigid bar after the aluminium and steel rods have undergone their extensions. Fig, 17.
Let the tensions in the aluminium and\steel rods be
S
kg. respectively. Since the rigid bar remains straight, the extensions are proportional to their distances from A.
BD
Let the extension of Extension of
3D
be
33% = 8
C£ = CCi =
— xS=25 2
—
But extension of £Z?== 8 = 7 A ah a
and extension of
.
/
R of
kg.
and
CE and
STRENGTH OF MATERIALS
22
Now
consider the equilibrium of the rigid bar.
on the
laking moments of the forces A, we have,
rigid bar,
about the end
/?x1+5x2 =2000x1-5 J?+25 =3000 But
...if)
U 5+25=3000 ‘^„'
s - 3,(100
^^
3000
x80.
^ j^ 26 kg.
(tensile)
181
80 §7.
X 1326=348 kg.
(tensile)
Extension of a tapering rod
l ilt, i:/
Fig. 19
shows a bar uniformly tapering from a diameter di
one end to a diameter d?
at the
at
other end.
Let the member be subjected to an axial tensile load P.
dx of the bar at a distance x
Let *
•
) d'^di—kx. Cross-sectional area at distance
^
5=s >1'
—
--^Wt-Arx)®
4 Intensity of stress
x from the
^
on the section ‘P
=P
.
AP x(di-A:x)®
larger
end
SIMPLE STRESSES
AND
23
STRAINS
P
Strain
'
E
AP
=
ntHdi — kxi~
Extension of the elemental length
d'c
^edx 4P ~~KE(di-kx,^
dx
Total extension of the bar /
=8.
_4P
f
dx
^E
1
{
0 /
4P
J
k=-
But
fcjf I rfi— d\
4P nEk I di-kl di - d2
r/if
r
8-.
-
’
\
r:E{di--d‘i)\di~ d^-hdz
(1 ~
=_
nFXdx—dzyx. dz
jvf’frfi
—
U
dif
‘
'l
di)
K£rf,t/3
For the particular case when the dx
and for
r«>d is
of unifonn diameter,
=d2 ~d
this case
s
4P/
^~\EtP tapers xmiformly rom 30 mrn. to ] 5 mm. If tin odhe subjected to
ProMem
18.
A rod of 30
centimeter'^.
£
kg-Ictrfi.
Solution.
The
extension of the rod S=.
•nEdxdi
is
given by,
w
STRENGTH OP MATERIALS
24 /’=--600 kg.
fn our case,
/^-30 an. {li --
r/2
mm. -3 cm,
30
— 15 mm
^1*5 cm.
E^ixnf kg.lcm^.
and
4x600x30
.
75x2x108x3x1*5 :=.0'()02547
Problem
from error inwheil
Modulus
tension
19.
If a
cm.
diam
I*!
ushv*
’tcr
the
to
cm
test bar is found to taper uniformly {D{-a) cm. diameter prove that the
mean diameter
to
the
Young's
{A.M.LE. May, 1965}
percent.
is
calculate
j
tad^di^iDA- a) ^di — (D—a)
Diameter at the larger
Solution.
Diameicr
ai the smaller
end
Let the length of the bar be
/
Let the Young’s Modulus be Let the extension of the
E
member be
S
JFl /:
API
-
^d\d%^
But
di - D~\~a and
d%—D-’a
API If the
mean
Youngs Modulus.
dian
Dbvi.msly
be adopted let E' be the computed E is erroneous.
4/V Ttn^r
4Pl tzD'H vji
J
error in
the
Modulus, when the mean diameter
coraputatioD is
E-E' (^ £
of the
Youngs
adopted
}
JPL
API
nZ)‘iS
API
X 1 00 percen
SIMPLE STRESSES
25
AND STRAINS
°
^
xlOO
percent
— ^X 100 percent 2
10 a v-
(
percent.
D
)
Problem 20. A bar of steel /v of lenf'th I and is of uniform The width of the bar laries uniformly from u at one end thickness Find the extenshm of the rod when it carries an to b at the other end. axial pull
jP.
.Solution.
Fig. 20
shows the tapering rod.
Consider any section
X— X distant
Width of the
= a—
section
a*
from the bigger end.
b
-
y-^A*
where
Thickness at the
section ^
k
t
Area of the section = f (a ~ Ax) o
Stress
/.
^
on the
•
F
~
Extension of an elcmenlaJ length dx
t
(a
-kx)
E
^
^
—
STRENGTH OF MATERIAU
26
dx tE
a—kx
J 0,
(a-kx) ]
_a
P
a-Kl
iLk
a-b
A-
But
1
I
PI
R ®
a
,
b'
£/ (a-fc)
Probiem 21. A siraif^iii bar of itecl rectangular in section is SOO cms. long and is of uiiijonn thickness I 5 cm. The width of the rod varies uniformly from cms. at one end to 4 cms. at the other. If the rod is subjecleJ to an axial tensile load of 3000 kg.,JinJthe extension of the rod Take Ta --2x I O’’ kg-jcm'^. Solation.
pi Extension of the rod
,
^ ^
Lt \a—o)
In our case
log,
0
kg. /
= 5(H)
1-15 a
~
I
ms.
cms.
10 cms.
6=4 cms and
£=
i0« Aif./cm*
2
X
3000X100 X rs (10-4)
!!)«
,
10
4
^^H)x300x0'9163 2 x io*irr‘5x6
—0‘045815 cm.
AND STRAINS
tSIMPLB STRESSES §8.
27
Bars of Composite Sections
Suppose the cross-section of a member consists of different
memW will be shared by
the various
on
load applied
the
materials,
components
P
the the
of section. For instance suppose a column consists of an outer tube of area Ai and Young’s Modulus El and an inner tube of area A2
and Young’s Modulus
£2
Let
.
the length of the column be /. Suppose a load P be applied on the column. Let the unit stresses on the outer and inner tube sections be />] and p 2 .
Load on outer tube-f-Ioad on the inner tube
~ Total
load on the column
P\<\+P%42—P
...(i)
Fig.
Let dl be the decrease
in length
2l
of the column.
Strain of each tube
But
From computed.
equations
(i)
From Eqn.
£i__
Pz
E\
Ez
and
(it)
pi=
{ii\
...(«)
the stresses pi
Et p^*.
-
•£2
Problem 22. A internal diameter and
i
compound tube conusn of a steel tube IS cms. cm. thickness and an outer brass tube 17 cm.
and / cm. thickness. The two tubes are of the The compound tube carries an axial load of WO tonnes. .stre.sses and the load carried by each tube and the amount it Length of each tube is 15 cms. Take E$=2xl(P ks tern * ^‘
internal diameter
same
length.
Find the shortens.
and E.== lx
m
SoIntioB
kg/cm^.
Area of steel tubc=/l.= -—(17*— 15*) cm.*
= 50-27 ^
The ratio
^
is callc
outer and inner tubes.
i
cm
*
the modular rauo between the materials of thj
STRENGTH OF MATERIALS
28
Area of brass tubc™>4i»— ~(I9‘^— 17^) cm,^
= :6 55c/w2 Let the stresses in steel and brass be p»
and ph kg.jcm.^ respec-
lively.
Strain in brass
Strain in steel
Eb
Es
Pb
ps-~^2pi>
jL.h
Load on stecl+Ioad on copper = Total load p,A,+pt>Ab—P
i-e.,
2pi-x50 27+p(,x56-55= l00,00() kg. 1
57 09 p(,= 100,000
_ 100 000*^'^^'”“
„
,
,
,
157 09
pi,
=636' 5 kg-lcm.^ 2
.
.
X 636
5 = 1273
kg /cm.^
Load on the brass tube=Pi,=pi,4i,=636 5x56-55
=35990 Load on
kg.
the steel tube
127.,
= 64010 Decrease in length of the
x 50 27
kg.
compound tube
=Decrease
in length
-Decrease
in length
of either of the tubes of brass tube
I
E.
-
«36 5
bt section.
=^0-009548 cm.
xT6«
1
7^
cm.XJO cm.
The cohmn carries a load of 18 tonnes and the steel bars Take E ~>i t tm ?^ ^
stresses in concrete f^S-Irm.^ and Ec^014xl(fi
kg./cm.*
SohtioB.
Area of i4.
steel
bars
= 8x3 l42=25 siy 25 14
136
c/w.2
cm2
Actual area of concrete
"'^‘="3‘’*-25 14=874-86 cm* and steel be p. and p,
, i-ct the stresses in concrete
respectively.
SIMn£ STRBSSES AND STKAINS Strain in concrete— Strain in steel
PL. "E.
E,
P.-E,- P»
21 pt~\Spt Load on
steel
s=Tot8l load i.e., I
.„(/)
+ load on concrete on the column.
F4R,
ptAt-^ptAc—P Spc X 25- 14+p. X 874-86 =P== 18000
22
kg.
1251-96^,-18000 ^' =
18000
,
,
,
l25r96*^/^-
— 14' 37 kg.tcm.^ p$ — l5x 14' 37 kg.fcmJ^
/.
=215 55 kg Icm^ ProUein 24. A load of 30 kg. is applied on a short concrete column 25 cm. 25 cm. The column is reinforced by steel bars of total area 56 cm.^ If the modulus of elasticity for steel is 15 times that of concrete find the stresses in concrete
and steel.
If the stress in concrete should not exceed 40 kg.fcm.^^ find the area of steel required so that the column may support a load of
60ft00 kg.
When
SolotioD.
the column carries
a load of 30M0 kg.
=*^#=56 cm.* 25*— 56= 569 cm.* Area of concrete Let the stresses in steel and concrete be p» and pe Area of steel
respectively.
= Strain in concrete
Strain in steel JL-
P-L
En. E
ps^e/p^ pt=l5pt load on steel+load on concrete
=Load on piAi‘\~pcA.=P 15 p»
X 56-\-p* X 569
column.
...(0
STRENGTH OF MATERIALS
30 ^-TO.OOO kg.
30,000 30,000
1409
pc=2r29 p,= When
the
\5
,
,
,
kg-lcm.^
^21 29= 3I9’35 kg-lcm.^
column carries a load of 60,000 kg.
Let the area of steel bars be An cm.^
Area of concretc=^r=(625-^.) cm.^ Strain in concrete
Strain in steel
f’-P''E. Ec
•
Pf=p-pc p.-^-l5pr.
But
Pf
.
^ 40
kg. Jem ®
P..--I5X
Load on
steel 4- load
40-600
kg.lcm.^
on concrete
=Load on column peAc
~P
600 /l,+40 (625-.4.S)
= 60,000 560 /!.- 35000
35000
.
2
Problem 25. A compound tube is made by shrinking a thin steel tube on a thin brass tube. A$ and Ab are the sectional areas of the steel and brass tubes, and £% and Eh are the corresponding values of Young^s Modulus. Show that for any tensile load the extension of the compound is equal to that of a single tube of the same length and total cross^
tube
sectional area^ but having a Youfigs
Modulus of As'^-Ab
Solution.
Area of
Let the load on the compound tube be
steel
P
tube
Aica of brass tube Let the stresses in
^Ab
steel
JP*
Es
and brass be
=r.
^Eb
ErP'>
and pb respectively,
AND STR MNS
SIMPLE STRESSES
Load on
steel
31
+ load on brass = Total load on the compound
tube
psAs-{-pbAit=P
i.e.,
E»
pbA»+pbAi>’=‘F
Eb
]=/> rEiiA$'\~Ei^Eo
,
"]
'•'‘[fiAfBx)’ Extension of the
compound tube =df ==ExteRSion of steel or brass tuoc tb
dh
P-
4\E,A,+EbAb
\i }
E be
Young’s Modulus of a tube of area the same load and undergoing the same extension. Let
" *-( {Aa+At,)E jP/
carrying
)
PI
(E^Ab^ Eo Ah)
(Aa-\-Ab)E
F<^A.^EuAh
Problem 26. A tube of aluminium 4 cm, external diameter and 2 cm, internal diameter is snugly JiUtd on to a solid steel rad of 1 cm, diameter. The composite bar is loaded in compression by an axial load P. find the stress in aluminium when the load is such that the stress in steel is 700 kg.lcm.'^ What is the xalue ofP ? £*> «»2 x /O® kg.fcm\
(AMLE. May
Ea^7xl(fi kg./cm.^
42^22
“
SolutioB.
( i4s~“(2)‘^=^7c cni.^
4 j 1 Modular
ratio
=i?i
£• = -zr= 20
3jc
cm
2
1969)
SIRENGTH OF MATERIALS
-
fa
X 700
P
Total load
lA
245 kg.jcm,^
kg
i
-245 x3Ji-f700X7t« 1435 n -4/OeS
kg.
Air.
Vrohhm 27/ A rompouiui bar consists of a central steel strip placed between two strips of brass 25 cm wide and 0 64 cm. each 2 5 cm wide and 1 cm thick. The strips are firmly fixed together 2’ 5 cm. wide and to form a compound bar td rectangular section {2t+0 64) cm thick. Determine <
on the com-
Let the load
Solution.
pound bar he P
•
tonnes.
Let the stresses in bra^s and
pb and pt tonnesjan
^
sem
BRA^S
be
steel
?
respectively.
Strain in steel
2
o^cm
L
strain in biass
iim
SttAtfS
x_
p» .
Eb
Id
Hut load on
steel
"DTO
P’
Eb'
114)
P'
1-816
load on brass = Total load
]
Pf^An+pbAh1*816 p//(2*5x *64)
Fig. 23
E.
p' \r
P x2rx2’5
P-
-
P
tonnes.
pdim+5t)
tonnes.
Area of the composite section
-
.4
2*5x0*64 f
2*5
x2r cm.^
r60 r5/) cm.^ Apparent Young’s Modulus i
E" Strain
1
570 tonnesjem!^
^
AE />d2*9054-5/)
(L6U+5t)1570 This must be equal to the strain of brass or
steel
-
SiliPLE STRESSES
AND
/>fc^2'90S4~S
•
••
33
STRAINS
0
ll-60+5<)l 570
pi
“1140
2*905+51 1*60+51
1570 1140
1*377
2*905+ 5/-2*203+6*»8M 1*8851=0*702
=
1
0 702
cm
0*j72
1*885
Since ps= 1*816 pb and since tbe stress in eitbcr biMS or steel should not exceed 1*570 tomelcml^
pt—VSlQ tomefcm.^
Let
/>6=-|rg^j “0*865 laww/cm.* .*.
Load on
the
bar=P=^.^.+p6^>
= 1 *57 X 2 5 X 0 64+0*865 x 2*5 x 2 X0*372 umne$
=2*512+ 1*608 ='^*/2
tmmes.
PfoUen 28. Two vertical wires are suspended at a ^stance of 50 an. apart as shown in Fig. 24. Their upper ends are firmly secured and their lower ends support a rigid horizontal bar which carries a load fV. The left hand wire has a diametir of l'6 mm. and is made of copper and the right hand wire has a diameter of 0 9 mm. and is made of steel. Both wires initially are 4 5 m. long (a)
Wf both
Determine the position of the wires extend by the
line
of action of
W
if due
to
same amount
(h/) Determine the slope of the rigid bar if a load of 20 kg. is hung at the centre of the bar. Neglect the weight of the bar.
Take £i=*2*l XIO* kg./cm.^ and £«= Sidntioa.
x
1*3
|0* kg.jem.* {A.M.I.E. Nov., 1968)
^•=0*00636 cm.^
^*=0*02010
c/n*.
=1*615
fi^mft
^
f '*
i4«+fnil«
mW
.
Ar^mA, I-6I5X('*0063« 0*0201
+
1
*615
*||
X00063
J
1
STRENGTH OF MATERIALS 34
W
r*=^0-338
Taking moments about copper
WxSO^lVx
0-331
^>0
x—16'900 cm.
Oae(b) When 20 Load
i
is al
kg. load
COPPFR U6nim
mid spin
0!A.
each wire=10 kg.
I
10x450 *‘“00201 xr^ix
10*
10X450 00636x2x10*
'0
ca
>0"1722 cm.
0‘3b3^ cm.
J
Let * be the inclination of the rigid
^
,
OtA
W
30
Fig. 24
=0 003632
® JO*®
e=(r
n\ Two
Problem 29.
rods one of steel and the other of and are 50 cm apart. .md length of each rod are ^ cm and 4 metres respectively. Diameter A cross fixed to the rods at the lower ends carries a load of 500 kg such tlMt the cross bar remains horizontal even after loading. ^ Find the stress in each rod and the position the copper are
e^h
vertical
rigidly fixed at the top
of
Take
E,=2xi0«
kg.jcm.^
and E,
load on the bar. x /OO ^^,^2
-
(A.M.I.E. Winter 1978) Solution.
Area of steel rod t'
-d*-^-(2)*=7t
Modular
=
jc
cm"^
OtA
\copm te
400
cm
ratio
Et
2 cm
Ksrm
Area of copper rod =* /4 1
2 cm OtA
c/m2
1
X 10*
^
is the cross bar
remainVhoSntal ?h!
^onsofthe
steel
tods are equal
Sinr^Tk
hnvethe same
orSS
strami of these rods*are
1500 kg ks ^ the
e^Sd
— a:—} Fiib.
*
25
— soms SrRflSBS AND sntAINS
35
Strain in steel^Strain in copper
et—e»
f±^A Em
Ee
f,==mfe,hM
Let
7*«
/•= 2 /. and Te be the tensions *
in the steel
Ti^fgAt^TlftAi^Tfe T»=^fcA»=ft
m^2 and copper rods.
ft
ie
T*=‘2T*
r.+r4-50OA* 2r.+r,-5oo kg.
But
r.=
500
, _ j kg. and
Let the SOO kg load be at a distance
^ r.=
1000
x from
Considering the equilibrium of the mcnnents about the right end,
Ay-
the copper rod.
cross bar,
and taking
500*=r.x50 500 X.
1000 " 3
X50
x=33'33 an. 30. Two vertical rods are each the upper end at a distance of
Problem fastened at
63 arts, apart. Each rod is 300 cms. long and 12 nun. in diameter. A horizontal rigid cross bar cormects the lower ends of the rods and on it is placed a load of 450 kg. so that the cross bar renuuns horizonttd. Find the position of the load on the cross bar and die stresses in each rod. One rod is of steel for which snei E=r96 X 104 kg.jcm.* and the other of bronze ejuf for which E-^0‘63x 104 kg.fcm.* SehrtiM.
BRO/Of
Area of eadi bar
=1131 cm.* Let the stresses in sted and bronze be pt aadpv kg-fan.* respecdvdy.
1
Since the rigid bar remains horizontal, the catensions of tiw steel and bronze bars are equal.
Ps 450i^
FIs.
26
SnCBNGTH
OTMAmii^
36 Strain in steel
» Strain in brass.
p± Et
Et Et Ih
Ei
1% 0 63 ph.
Load on
f load on br«izc-fot|ll load
steel
i4.=.4»=l'311
But 3'111
/>» X
1*311 +ifex
cm*
1*311—450
4*650p»=450
=9^-80 kg./trrt*. jp,=3*lll x96*8— 30/*l kf./on.* Cdnader the equilibrium of the rimd rodt see Fig. 26. load be api^ed at x ems. from the steel bar. Taking moments about the .*.
left
Let
tiie
aid we have
Ax60*=450xjc 96*8 X 1*131 x60=450:s 96*»x 1*131X60 ctns.
X’‘t4'6
.*•
cm.
Hence the load must be If 14' 6 cats, from the steel bar.
on the
a
diatahee
bar 50 cm. long and 7 cm. Omiaer 7 5 cms. inside diameter and The ahminium cylinder is (7015 cm. axial load of 60,000 kg. is mpHedt^e cover plates as shotm in Fig. 27.
Load OB steel+Ioad on alumimam =totaI load on the poflopoate ptAf\-p»A.^P (>-0015) 44000 x 38*4H-14000»X 3^36=6000011: i
SIMNCTH OP IMIBUAU 16938+48I 04*=60+25-4 2174048-85-4
/.
85-4
«
*- 21^045=0 03928 an. /».=(S-0015) 44000 kg.Icm.*
-(0 03928-0 015) 44000 kg./an.*
~ 1068-32 kg.lem.^ 14000 5
and
= 14000 X 0 03928 kgjcn^. =549 92 kg.fcnt,^ §9.
EquWaleot Area of a Compomid Sectioa
compound Suppose a column consists of a concrete oolumn reinforced with steel
•
•
bars.
Let the gross area of the
oolumn be
^
>4.
Let the area of
steel
^ ^ ^O^RALL / AR£Ag^A
ai ARfACf
be
Si sna^A^
A$.
Actual area of con-
mte-^AcMA—As).
^
^ '
Let the stresses in conCrete and steel be pc and p,
~ '
p.
it^)ectively.
Strain in
Concrete— Strain
^
in Steel
P*^mpc
...(#)
where
E, ‘s
called the
modular ratio between
OQilcrete.
Load on
steel -{-load
on concrete =load on column
P*At-\-pcAt^P
pd.A»~i~mAt)
=P
P.^
iL-_
steel
and
SIMPLE STRESSES
AND
STRAINS
A,"‘(A—A$)
But
_
P A — At-f-mAt
P ^‘’°A+{m-l)A. Suppose in place of the composite section, a plain concrete column of area A-\-{m— l)A, had been provided, the stress in concrete
P A-\-(m—ljAt
Hence for determining the stress in concrete, we may consider that the given reinforced concrete column is equivalent to a plain concrete column whose sectional area
=At^A-{-(m — This area >(<=/4-H(m—l)
is
lyA*
called
the equivalent concrete
area.
Stress in
concrete= „ Equivalent concrete area
Stress in steel
The above
.
,
= modular ratio x stress in concrete
principle, for instance,
can be applied to problem
23, page 28.
column=30
In the problem, Size of the
Modular .•.
ratio
=m=
p-
=15
XSc
Equivalent concrete area
=i4»=i4+(m — 1) At
=30*+(15-l)x25T4
cm.®
=1251-96 cm.® Stress in concrete
_ ~P‘'^
_
Load on the column Equivalent concrete area
1800CL
"l25r96
Stress in
kg.lcm.^
—14‘37 kg.fcm.* steel=p»=mp« — 15 x 14*37 kg.jcm.^ =s215'55 kg.jcm.^
Problem 32. A steel strip of cross-section 40.'' mm. X 10 mm. is two copper strips one on either side, each of cross-sectUm 40 mm. X 7 5 mm. to transfer the load. There are two bolts on the line of the pull. Show that neglecting frictim and the deformation of the bolted to
STKBNGTH OF MATERIAU
bolts
of 3
the ratio a pull applied to the pint will be shared by the bolts in
Assume
to 4.
E for steel is twice
of copper.
that
Solntioa.
members
Figure 30 shows the details of connection of the in the problem.
mentioned
P
kg., i.e., the load Let the load applied on the connection be the load transLet plate. the steel of end of P kg. is applied at the the load ferred to the bolt AbtPi kg. Hence between the two bolts to transferred will be load This kg. will be (P-Pi) plate in the steel
the bolt B.
The load Pi transferred to the bolt A two copper plates between the two bolts. between the (wo bolts. Load on the
plates
^Pi
^
in copper.
/
1
(1
^
Vi yim
ySTE(i
•
A
\
i
>J
\
jJ
.
L
J
—
IOtbw ,
IJ
the plates
/Ctms
n
mmm ^
tratisferred to the
kg.
* Strain
Strain in steel
n
be
Now consider
— (P— Pi) kg.
steel plate
Load on the copper
will
Fig. 30
P-Pi_ A$E» ArEr
Pi
.
Pi AkEc At
Ec
“ A, 2x40x75
P-Pi’‘A.E. Pi
P-Pi“ 40X10 -h.
P-Pi“ But load on the andJhe load on
bolt
E, \_ •
2
24
A =Pi P«*P— Pi
the bolt
Ratio of the loads shared by the bolts
A and B=3 -.4. Two copper rods and a steel rod, together support a rigid uniform beam weighing P kg as shown in Fig. SL The stKsm ProUem
respectively.
supported.
33.
Find the magnitude of the load Young s Modulus for steel is twice
P
that
Area of copper component
-/4.=2f4x 41 -32 cm*.
that
can be safelv ^ ^
of cooner.
ample SntBSSES A^D STRAINS
41
Area of steel component
=/!»= 5 X S'* 25 cm*. Length of copper component=f«= 15 an. Length of steel component =1*== 25 an. Decrease in length of coppcr=Decreaae in lengdi
of
steel
=8
6=gc/«=e»/» •
^5
“25 =
If.
Ce
Stress Stress in
in
0-6
steel
copper=p«=ec£ii
£l.==IL po
€e
=0-6x2=r2 Et
p$’^Vlpe :. when p, reaches 600 kg.jcm^. p, wiU reach r2X600720 kg /an*, whidi is less than its permissible value. P^peAfi-ptAt =(600 X 32)+(720 x 25) ’-‘37200 kg.
kg.
SnUBNGTH or materials:
42
FroUen 34. support a load the rods.
Two copper rods and one
of 25000 kg. as shown
rod together Find the stresses Us
steel
in Fig. 32.
Take Et^2xl(fi kg./cm.^ and Ec=1^10^ kg.fcm.* Sdotim. Each rod will be compressed by the same amount. Let the decrease in length of each rod be * cm. Let the strain in copper and
.
steel
be
and
respectively.
le
.
7~
€$
»
e.Let the stresses in
Y
€e
.
e.
and copper be p, and pc respectively. pt=e,Ec ind p,—eeEc steel
^‘==— Pc
-
•
Ec
Cc
P»
p,-j p» Load on steel+Load on copper=Total load applied i.e.,
p»/4«'4"Pe^«=7*
*'**-^^
4
y p«X 16+p< X(2x 10)— 2SOOO kg. 64
-yPc+20 p«-2SpOO kg.
SOiPU STKBSSES AND STRAINS 124
-3- /».==25000 kg.
25000 x3"
kg.lcm.*
”124 *
/.
pt=‘604’84kglcm.* p.
= j X 604 M=S06 45 kg.lcm.*
PraUm 35. Two
and one steel rod together losses ^^PPff • f and 1200 kg-lcm-
copper rods
su^rt a
load as shown in steel are not to exceed 600
Fig. 33. kg.lcnt.*
If the
P Kg
ROD’-s.
smi fiO0]
COPPER ROD
<«c»x4
5C1flX3C1ll
I
3CIIIX3CW 12
cm I
kW/gr
/////777?y a’cR
%}////m//77P Fig. 33.
load that cat be supported.
Young's Modulus for steel
is
twice that
of capper.
be compressed to the. same extent. Let the decrease in length of eiach rod be * cm. e# respectivdy. Let the strain in steel and copper be e» and 3— Cl. beetle
Each rod
will
•
L
et
12
= 0-6
20 e» in steel and copper be p» and p# pB^enE* and pt^eeEo /»
Let the stresses
•
pt
et
respectively.
^-=0*6x2=r2 £•
p,^\1p,. Suppose
steel is
permitted to reach
kg.lcm.* the oonrespoiKhng stress in
its
safe stress
of 120^
cof^ wiBhr^^ 'WO®
STRBNOIH OP MjllBtlALS
44
kg.lcm.^ which exceeds, the safe stress of 600 kg-lan.* for copper. Therefore let copper be allowed to reach its safe stress of 600 kg/cm.*
Corresponding Total
T2— 720 kg.fcnt^^
wdl be 600 x
stress in steel
load—'P==load on
steel
t-load
on copper
~p$At-i-peAc
,
=720 X 16 + 600 x 2 x 9 ^22320 kg.
kg.
Three vertical rods equal in length and each 12 and together diameter are equispaced in a vertical support a load of 1000 kg. the rods being so adjusted as to share the load equally. If now an additional load of 1000 kg. be added determine the stress in each rod- The middle rod is of copper and the 2 X 10^ kg.jcm.* Take outer rods are of steel. Et= 1 X 10^
Problem 36.
mm.
pl^
in
a^
kg-lcm
*
SolntioB.
(/)
Stresses due to the initial load
Area of each bar the load
Initially
4
of 1,000 kg.
cm.^ = T 13 cm.*
xV2^
on each bar 1000,
The
each bar
initial stress in
1000
po— 294 '9 (a)
kg./cm.*
Stresses due to additional load stresses in
...
,
copper and
of 1000 kg.
steel
additional load of 1000 kg.
= Strain in
Strain in copper
Et
steel
steel.
Eg
E*
Load on
^
Ee
^
Pc—2pc
+ load on copper = Total load.
P$ /4»+pe Ac'=P
2;p.(2xi-I3)+p.xil3
+ 000 %. 5-65 p.
=
1000 kg.
p. = ^kg.lcm.* ft
and
be pe and
—
m
kg./cm.*
p.-2x 177= 354
/>»
due to the
SmPIB STEBS8ES AMD Fiaal
45
STRAINS
stms
294*9+ 177— 47/ P kgjcnfl.
in copper
=sp<»+ps=:294'9+354=^^5'P kgjcnfi*
Final stress in steel
A steel rod 18 mm.
in diameter passes centrally through a steel tu^ 25 mm. in internal diameter and 30 mm. in’ extenusd diameter. The tube is 75 cm. long and is closed by rigid wadiers of negligible thickness which are fastened by nuts threaded on the rod. The nuts are tightened until the compressive load on the tube is 2 tonnes. Calculate the stresses in the tube and the rod.
Pioble^ 37.
find the increase
in these str^s$es
me quarter of a turn relative to cm. Take
when one nut
the other.
is
tightened
by
There are 4 threads per
E^2000 tonnesjcm^. When the nuts
arc tightened the tube will be compressed and the rod will be elongated. Since HP external forces have been applied, the compressive load on the tnbe must be equal to the tensile load on the rod. SointipB.
f—
U
0 m
scM
^
r™ Fig. 34
(32-2-5*)-
Area of the tube
4
=Ar=~
Area of the rod
81
x(l-8)*--,«\ K
crn^.
Let the stresses in the rod and tube be p. and pt tonnesicm.*' reflectively.
Tensile load
on the rod— compressive load on the tube prAt ^pi-A *
At
UK
100
275
^“'324^ (0 Vnten
the eompressive load an the tdbe is
2
tonnes.
2
.'..Stfess.m flie
tobe—p*^ ^l^romes/em*.
—
toimeslcn/^
—lt926 totmeslen^ {eomprasMlTtey
STRBNGTH 275 Stress in the
rod=/?r^
^0'7S6 (ii)
When one nut
is
XO 926
W MATIUAU
twinesjcm^
tonneslcm^, (tensile)
tightened by one quarter of a turn.
Lct/t and/r be the stresses due ty tightening of the nut quarter of a turn.
Obviously ft
is
a compressive stress and ft
is tensile,
by one
and
324^' Reduction
in the length oi the
.
E
tube
/
_/,x75 2000
cm.
Extension of the rod
-•I' .Ax 75
^
cm.
275
/fX75
‘324
2000
68-66/.
cm.
2000
But contraction of tAc tube'\-extension of the rod’^axial advanca fif the nut.
75/
,
63-66/
“* ,
2ot)0
2000
(
~ ) = 0 0625 cm.
75/+63-66/,=.125 138-66/ = 125
fi==0'90 tlcm^, (compressive)
275 f'‘°
^
324
0 9 = 0-754 tjem^. {temiU) *
•netal
bar of diameter
D and
and E the Yomg s Modulus of the materud efthebm. (A.M.I.E. May, 1975)
SBIPLB CniESSeS
rod
AND STKAINS
47
Consider any sectton SohitiOB. distant x from the lower end.
XX
Weight of the rod below the sectioa
of the
XX
where 41— sectional area of the rod. Stress at the section
XX^f^
«»YJf
Consider an elenMntal loigth dx of the rod from XX.
Extennon of the elemental length of rod ’=‘~dx
FIS.3S
STRENGTH OF UATEUAIA
48
Suppose the area of the upper and lower ends be Ai aad Aa respectivdy.
end. end.
Let the area of the section be /I at a distance x from the lower Let the area be 4 at a distance x+dx from die lower
Let the weight per unit volume of the
Ginsider the equilibrium of the strip
upwards— Total
Total force acting
Let
member be w
EFGH force
acting
downwards
p be the uniform stress intensity.
p{A-\-dA)—pA+wA dx p dA~wA dx j —dA = w dx A p Integrating, we get .
.
—
log.
where
Cl ^constant of integration
At
jr=0, log*
• •
Aa~ Cl
—-x+iog. A%
log.
log.
'
Ai
w X p wx
wx A=Aaf p
...(/>
w/
^1=^2/
Obviously
Eq-
(i)
may
2'3k^o
..Hi)
also be written as
At
p
X
Vthe w tfthe bw IM ^n£ii
the tie is to
€*liRtltwe^8tms.leHL^
...mo metres hng.
uJdof ^
^
material
aMPLE
STRESSES
AND STRAINS Area of the
Solatioa.
Intensity
of
49 at the bottom
tie
=/»=
stress
Wt. per unit volume =M’ = Tfce area at
=A2—5 cnfi.
— =14000 —^000 10
,
lUUJ
kg.jcrr^.
kg-lcm^.
any distance x from the bottom end
2'31ogio
is
given by
^
j Ai
p
Let the area at the upper end be A\
23Iogr„-^ =
J/ 8 X 180 0
^
1000x14000 9 8'/50
logic
-4^ =rooi A2
Ai^V00lx5^5W5 cm\ shows a Problem 40. Fig, rigid square platform of negligible
P
weight and of side I supported by four identical elastic pillars each of height h If a load P be applied at
a point
distant a and h from the adjacent sides and find the pressure on each pillar and the depression of the centre of the plat-
AB
AD
form, Soiotion.
Let the pressures
on the legs AA\^ BBi, CCi, DDi be Pa, Pb, P and Pd respectively. For the equilibrium of the platform.
Pa+PK+P.+Pd
=P
Taking moments about AB, we have, Pel-^-Pal^^^Pa
P.+P4 =
Pa I
m
STRENGTH OF MATERIALS
50
Taking moments about AD, we have i>»/+/»c/*=Ph ...(/«)
Let
Sa>
adO
CCi and DDi.' These pressures
on the
be the depressions of the pillars AAj, BBi, depressions are proportional to the respective
pillars.
Where AT is constant
8a— AlPa.
SaoiF,
h=KPi, dc=--KPe
h=KPi — Average of depressions of A O
Depression of
and
C
—Average of depressions of B and
D
KiPa+Po) "
__8.+8.
-
2
2
“
“
2
K(P^+P4 2
P.+Pt^Po+Pi Rewriting the above equations.
P.+P.+/>.+Pi-P
...(0
Pc+Pd=^
...(iV)
Pi+Pc=~
...(f/f)
P«+Po=P»4-P
From
equations
(/)
and
Pa'f'Po
Subtracting eq.
(if)
(jv),
we
get
— Pt-f-Pd — -^
from
>1 «
eq.
(ff/),
1
we
get
..(v)
SaiFU SntESSES and strains
SI
Substituting the value of Pd in equation (v/)
<»-«)}
P-=
1-2 (b-a)[
Substituting the value of Pa in equation
P‘=
(//)
-4y|l-2
P«=-j^|2 (a+b)— Substituting the value of P» in equation (>y)
J*.-P.+Pd-P.= -|—
2(a+b)~/|
/-2(a+b)| Thus the pressures on the
pillars are.
P.= -J|3f-2
(u+b)|
P*= £-{/+2 (*-<»)}
P.=-^|2 (o+*)-/)|
Let the Young's modnius of the material of pillan be Depression of the centre O of the platform
=8=J
where
A
is
pillar.
P«+P«=“-^
ProHea 41. in
(Depression of the pillar AAi^ depression of the pillar CCi)
the sectional area of each
But
shown
E
Fig. 39.
BD
CD
and support a tkaiPas AD, If the three ropes are of the same sectioned ana.
Three ropes
STRENGTH
52 (Ike
middk rope be
and the other ropes beat
vertical
JM the load carried by each
(Of
UATERIAIS
6 with the vertical^
rope.
Fig.J9
Let the length of /
CD
k. be
n y
D
/t+2ec6s6-i*
y=l(PInraease in length of
AD
Increase in length of
BD
R) sec 8
..(0
But
we
Differentiating,
get,
2/i
h
~dl
-cos
h AE _=cos8 R AE‘ I
cos* 6
-j^cos 0
Q^R cos* But
e
- J?)
Rcos*
2R
6=
P-R 2
.(fl)
sece
sec®
cos* 0=(j»-/()
J? (1+2 cos*
6)-?
1+2 cos*
I
..(i)
SIMFU SIKBSSES AND STKAOV
53
^„_Pcos*0 r+2cos*'8 Problem 42*. A rigid horizontal beam of kngth 21 is earied by three wires each of length I hut of sectional areas Ait A%, As as shown be placed at a distance ,Klfrom the Uft endt in Fig. 41. If a load find the tensions in the three wires neglecting the weight of the beam.
W
Solotioa. in the wire
Let the tension
lit
AA' he P,
and tension
r*
wire
in
the
in
the wire
1
^:/j/////////////M////////////,//,C
CCheQ and tension SB' he R.
<+«
I
@
5^
0
R
P
For the equilibrium, we
^
g
have
/>4-e+«=»' left
-(o
—
Taking moments about the end of the beam, we have
’I"*
n
RI+Q2I^Wia
R+2Q^WK
Fig 41.
P^JV-Q-R ==lV-i(fVK—R)~R
^W-lfVK~ V (2-a:)-
...m Since the
beam
Extension of
will
always remain straight
BB'
=mean
Let the extensions of AA’,
Viiii/or siiideuls
may
of the extensions of A A' and CCf.
CC and BB' be Sx, Sg and 9^
leave this numerical.
STKEMOTH
54 C»
«8=
Since
AzE “
—
“f" Sj
-
we have
lAiE
2
MATOtlAtS
CH>
^
Q-
l ‘
AiE
--^1 At Substituting for
P and R
4
A3
_
.
AAi
1
4Ag
a. '
2Ai
Ai
y<2
2-£ -LI -_E| 4
Aa _
I
R A2
Ai
[
2-K Ax
_
R being known the tensions equations (ii) and (//7).
] )
^
W
Aa
rC.-U. ^ ^ 4^2
4,4i
1 J
._WK~R
K>izA
f
WK-R
+
2>Ji
Aa
[_L ^ -J_
1
Q, we have, W(2 K)-R f
F and Q can
4
K Aa
be determined from
Problem 43. A uniform rope of length I units hangs verticallyFind the extension of the first a units of length of the rope from the top due to the weight of the rope itself Find also the total extension of the rope. Solalion. Consider an elemental length dx of the rope at a distance x from the botto n of tne rope.
Let the weight per unit volume of the rope be
p.
Let the cross- secti,)jal area of the rope be A*• Force on tae cross-section of the elemental
— P Ax .
.
Stress
on the section of the elemental part
Extension of the elemental part
/ •*.
Total extension
~S —
£
1^“*^
0
part
— SMPLB
SrRGSSBS
AXD STRAINS
55
PP 2E Exteosiou of the top a units of length
“Total extension— extension of the bottona (/— fl) units of length.
=
pp
p (/-fl)*
2E
2E
2El
P-il-df
P
§ 11.
2 /a— o*
" 2E
^
II
aUll
] ^
—d)
Temperature Stresses
When the temperature of a material changes there will be corresponding change in the dimension. When a member is free ta expand or contract due to rise or fall of temperature, no stresses will be induced in the member. But, if the natural change in length due to rise or fall of temperature be prevented, stresses will be offered.
AB
Suppose a rod of length / be fixed at the rads A and £. Let the temperature rise by T. If the member was free to expand, the free expansion of the member would be
^ 1
BB'=»Tl where a is the coefficient of linear expansion. If the member is allowed to freely expand no stresses will be induced. But if the member is prevented from expanding, compressive stresses will be induced.
(h>
f — s.
(t)
N,
=5 =:) B
a'
IZ±f 1
This can be realized as follows. Let the rod B be of length t Let its ends A and B be fixed. Suppose there is a rise of temperature. The rod tends to expand
+
by
B
B is
B'
—
IJ,
(W)
Fig.
44
«77.
Suppose the fixture at the end removed so that the rod freely expands by aJI so that BB'^^a-Tl. Let now an external load be applied at B so that the rod is decreased in its length from (/-|-«77) to /.
Let a be the sectional area of the rod.
Compressive 8tress=p^
P a
SIRENGTH OF MATERIALS
56
Strain
z+ar/ Stress ,
/
’E (Young’s modulus)
Strain
Hence when the rod
is
prevented from
expanding temperature
^olTE.
stress
The
thrust
on the rod section
~P~pa ~~(xT£a In generah the temperature strain
_
Expansio n or contracti on prevented original length
Suppose a rod of length L when subjected ro a rise of temperature is permitted to expand only by 5, the temperature strain
^^JExpansion preverUed Original length
1
Temperature
stress
^F^Ee E(oTl-l) /
Problem 44. A rod is 2 metres long at pansion of the rod when the temperature is raised
Find the exto
pansion is prevented, find the stress in the material. Take kg./cm.^ and « = 0-0000/2 per °C (AM/E, Solution.
Rise in temperature^ T= 80— 10
Free expansion
=a
If this ex-
80*C.
E=-IxJO^
May
1974)
= 70°C.
T/
.=0-000012 x 70 X 2x100 cm. —O' 168 cm. Temperature stress
=a r£ =0 000012 X 70 X 1 X 10« =840
ture
of
ture me
is IS
2^^ rtdsed to raised
is
6rc.
kg./cm.^
kg. I cm. ^
Fmd the
20 metres long at a tempera^hen the tempera-
temperature stress produced (/)
when
E5
SIMPLE STRESSES
AND STRAINS
57
the expansion of the rod is prevented (ii) when the rod is permitted expand by 5'S mm. Take ol^12x 10~^ per *C and £= 200 GNjm.^
to
Solttilon.
Free expansioo of the rod~a77
-12X10 --^-JO'8
When
1
(65-- 20)20
0108 metre
=^-0
(')
«
mm.
the expansion is fully prevented
emperature
stress
—ar£
- 12 X 10-^’ X (65- 20) X 200 X 10»
Njmi
~ 1 08 X ] 0*' Njmetre^ ^108 MNImetre^ When
(a)
the
rod
is
permitted to expand by
In this case, expansion prevented /.
5' 8
mm.
= 10’8 — 5*8 — 5 mm.
^Expansion prevented
Strain
Original length 5
1
20xldb0“~4000 Temperature stress ^^Strmnx 1
4000
X 200 X 10»
'-=^50X10^ N/mctre-
—50
MN metre^ !
Problem 46. (S.I.). A 15 mm. diameter steel rod passes centrally Through a copper tube 50 mm. external diameter and 40 mm. internal diameter. The tube is closed at each end by rigid plates of negligible thickness. The nuts are tightened lightly home on the projecting parts of the rod. If the temperature of the assembly is raised by 6(fC. Calculate the stresses developed in copper
Take
and
steel.
Eb^ 210 GNImetre^ £c-^I05 ONImetre^ a, = 72 X 70-6 per 'C a, - 1 7' X 70~^ per ;
;
;
‘C.
Solution.
Area of the
steel
rod
=
/4.s
—
(1 5)^
mm.^
—56*25 n mm.^ Area of the copper tubc«v4c— ~*(502— 40^) mm.^
—225 w mml^ Free expansioo of steel jFree
—v-bTI
expansion of copper— «cT/
STRENGTH OP MATERIALS
58
Let the actual expausion of each component be 8
aLcTl>S>asTI steel is in tension
and copper
is
in compression.
and copper.
Let /• and fc be the stresses in steel
For the equilibrium of the system Tension in stccl= Compression in Copper. f,A,-=fcA<
/-4/. Actual expansions of steel
Actual expansion of copper
-
«.r+A =a.r- jEr £,s
But/«-4/r and substituting for 12
X 10“® XbO+r—
ats,
= i7s X
720 X 103
we
Es and Ec,
get
y fio
^ ,
ac,
j^,3_
/•
3/ 105
=330 X 103 jLy /
— U 'S^x
rneirc^
Njmetre^ 55 MNjmetr^
Z=4/.==4 X ends
pa^^t^ueh
exter^TdiLetTr^
Fit
rt TJ
1
\
i(fi
SS=46 2 MNImetr^.
A
screwed at the
SO mm.
internal
and
of the whole assembly is raised 126°C and the iuitt nn screwed lightly home on the «»* of the tube Fnd common temperamre Iw falleFtol6^''a^ to
Coefficient Coefficient
of expansion for
steel = 1 2x KT* per'C of expansion for gun metal ’=20 x 10~* per *C
SlfMPLB STAfiSSBS
AND
59
SIRAINS =2'1
Modulus of Elasticity for steel Modtdus of Elasticity for gun metal Solatioii.
Area of
steel
x 70* kg-lcm^.
=0'94 X Kfi kg.Icnfi. IE., May 1966} (A
M
tube
32-2-5® Jem®.
Area of the gun metal
rod=^i/=—
= 1*21
x2*2® cm^.
n cm^.
Let the length of the rod and tube between the nuts be
/
cms.
two members had been free to contract, 77 of the gun metal rod— =«• Tl contraction of the steel tube
If the
free contraction free
Since a,/ is greater than ol* the free contration of the gun metal rod is greater than the free contraction of the ^teel tube. But, since 1-
^ZZZZZZZZZZZZZZZZZZl
6UN metal 90D
Fig.
45
the ends of the rod have been provided with nuts the two members are not free to contract fully, each of the members will contract by Let S cm, be the final contraction of each rod. the same amount. The free contraction of the gun metal rod is greater than 5, while the Hence the steel tute free contraction of the steel tube is less than S. will be subjected to compressive stress while the gun metal rod will and pt, be the stresses in steel be subjected to tensile stress. Let and gun metal.
For the equilibrium of the whole system. Total compressive force in steel
—Total tension .*•
in
gun metal.
psAn'=p(iAn
r21n P"
'16 />««=1*76 pg
STRENGTH OF kUTHMAU
60 Final contraction of steel
= Final
contraction of
gun metal
J,TI+
a,T+
P.t
Ei
T^m - 16 =V/ 0’C 'Jv
12 X U?
2’i
:
1
X lie
Pa 0-94 Xl0«
20x I0-«XI10
lUM-
X i0« X
I-
10-
i
’r+ok)
'•(
2
Pf^46?'7 kg.icfn^ {tensile) i 76 A 462*7 kgjcnfips
**.
‘-8I4‘J kg.lcm-. (compressive)
Problem 48 A steel bar placed between two copper bars each having the same area and length as the steel bar at A5®C. At this stage they are rigidly connected together at both the ends. When the temperature iv raised to 315' C, the length of the bars increases by 0*15 cm. Determine the original length and the final stresses in the bars.
Take
£1
-**
2* I
---0
x
/O® kg. /cm?
000012 per
Er-=^ I
:
a.--
;
X
I(fi
kgfemi^
;
0 0000175 per
{AM IE, Summer
COPPER COPPCA.
h
mi
t
•0‘1S
cm Fig.
46
Solatioii.
Let the sectional area of the
steel
Sectional area of the copper
component be A cm.®
component=2y4 cm.*
Free expansion of steel component =a< 77 Free expansion of copper component=«cr/ Let 8 be the
at^l
expansion «.ri
each bar.
<8
1978)
SIMPLE SrRBSSES Steel
AND STRAINS is
6t
and copper
in tension
Let /« and fc be the stresses in
is
in
compression.
and copper
steel
respectively.
For the equilibrium of the system Tension in
compression
steel
in copper.
fA^f(2A) Actual expansion of
steel
— Actual
a,r/+-^ /-a.2Y-4
expansion of copper.
/
^
a,7’+-4-=- a'7’-
£
r-ais-is-soo'c 0 000012 x 300
12x300+^1 1-9524
---
f X
10 ®
1650 tS'^5* 1 1
f—2 X 845
kgjcm.^ (compressive) ] 1
= 1690 22
kg.jcm,'^ (tensile)'
consider say the steel bar.
Actual expansion of the
steel
=
bar
15
^+2-f x*^j 06 1
/•
x 300-
17-5x300-/,
/ft
Now
-0-0000175
2n"xio«
690*22
12X300
150000
3600 H- 804*87 /- 150000 4404*87 /- 150000
1—24 05
/.
cm.
Problem 49. A 12 mm- diameter steel rod passes centrally through a copper tube 48 mm. external and 36 mm. internal diameter and 2 50 metres long. The tube is closed at each end by 24 mm. thick The nuts are tightened until steel plates which are secured by nuts. The whole ossemthe copper tube is reduced in length by O' 508 mm. bly is then raised in temperature by 60*C. Calculate the stress in after the rise of temperature^ assuming copper and steel before diat the thickness of the plates remains unchanged.
md
Take
kg.jcmf,
u.^I2y.lO-^rC
and
Ec^E05x l(fi kg-jcmK n.^I7
5xl0^rc
STRENGTH OF MATERIALS
62 Solution.
Area of the
steel
rod=.4.“ ^r2)*cw2.==0-36«cm2.
Area of the copper tube -3’62)
Caseii). Stresses the tightening When the nuts nuts. are tightened the steel rod will be subjected
Me
cm.*=2’52n
arfi.
*
to
K
Fig. 47 the and me tensile stress ana to tcnsiic pper tube will be subjected to compressive stress.
the stresses in copper and
Let pe and p$ be
steel.
Total compression in coppcr=Total tension in steel.
pC^O^^psAs
2 52n 0 36k
Ar ps-
Ai
Pi^lpe c* Strain
„
copper = er
in
change
=
.
f-
in -length -
original length
OOSOg e«==
250 Stress in
copper=pc=ec Ec 0*0508 "
Xl05xl0«*g./rm.2
250
=213' 4 kgjcm,^ (compressive) Stress in steel
=pc=7/;a = 7x213*4
- 1493 Case
(a).
If the
S kg.lcm:^
Stresses due to rise
two members had been
Free expansion of steel
==
of temperature.
free to
expand.
uTL
Free expansion of copper^ x/r/n Since ac is greater than the free expansion of copper is grratcr than the frw expansion of steel. But since the ends of the rod are provided with washers and nuts the members are not free to expand fully Final expansion of each of the members
will be the Let this final expansion be 5. The free expansion of copper IS greater than 8 while the free expansion of steel is less thanl. Hencethestee rod will be subjected to a tensile stress while the cop^r tube will be subjected to a compressive stress. Let f, and f. be the stresses in steel and copper. For the equilibrium of Ac whole "uuw system,
same.
SIMPLE STRESSES
AND
STRAINS
63
Total tension in steel=Total compression in copper
ftAt—feAe r
f
•
17-^'
Final expansion of steel
=Final expansion of copper a,Tl.+
^.
U^olcTL- §-/. iLc
£0$
T=(ffC,
But
/.=250+4-8 = 254-8 cnw. /«— 250 cms.
and
12xl0-«X60 x 254-8+^-^.J‘^^^Jg*
=
17
5>cl0-«x60 x 250-
/cx250 105x~10«
= 72‘64 kg.lcm? (compressive) f, = 7 xn-64 kg./cm.^ fc.
Final stresses
perature
f,=S08’4S kg Im.^ (tensile) due to tightening the nuts and
rise
of tem-
:
Stress in copper
•^pc+fc=2l3'4+72'64=286'04 kg./cm,*
Stress in steel
=p.+/s = 1493’8+508'41 ’=2002' 28
(compressive)
kg.jcm.' (tensile)
Problem. 50. A steel rod 20 mm. diameter and 6 metre long is connected to two grips one at each end at a temperature of 120‘C. Find the pull exerted when the temperature falls to 40*C (/) if the ends x l(fi kg.lcm.^ do not yield (it) if the ends yield by 0 I J cm. Take
E=2
md«.^l2>tlO-^
Case
Length of the rod=/=600 cms
temperaturc=7’= 120— 40=«80°C
(/)
When
the ends do
mt yield.
Temperature stress=*7!E -
Pull in the rod
l2xl0-«x80x2xl0«A:g./cm.2
= 1920 kg.lcm.^ = Stress x area
(tensile)
= 1920x~x2*fcg ^6033
kg.
STRENGTH OF MATERIALS
64
When
Case (ay
the ends yield
by O il cm.
Contraction prevented
Temperature
=
strain
Origin^TkiigVh aTt — ^
80x600— OTl
12X10
600
0'^ 600 Temperature stress =Strain x Young’s Modulus
^ 0'466 60^
_ / n 0* kg.lcm.^ X 2 X 1,nA,
-1553 kx-lcm.^ Pull in the rod
—Sttess •'area
= 1553 X Problem
A
X 2®
kg.
= 4878
kg.
cm. external diameter and 3
mm.
thick encloses centrally a solid copper bar of 3 cm, diameter. bar and the tube are rigidly connei (ed together at the ends at a
The
51.
steel tube
temmetal when heated ta
Find the stress in each perature of 30"C. Also find the increase in length if the original length of the I80*C, assembly is 30 cm Coefficients of expansion for steel and copper and 1 7xp)'^ rc\pectively per degree centigrade^ are J OSxIO
E^2' I XlO^ kg^janr for
sit'd
and
I' 1
x iO^ kg.jcm'^ for copper.
Solution.
Area of
steel
tube
Area of copper bar -Jr-
(4
^
^
52— 3 ’9^)-= 3*959
cm-.
Since the coefficient of expansion for copper is greater than that of steeL the free expansion of the copper bar is greater than the free expansion of the steel tube. Since the two components are rigidly
connected together at the ends,
actual expansion of steel --^-actual expansion of copper. 8 Let be the actual expansion of each component.
Obviously, 8 is greater than the free expansion of the steel tube than the free expansion of copper. Hence steel is in tension and the copper bar ms in compression. For the equiBbrium of the ^ system.
Total tension in steel— Total compression in copper Let the stresses in steel and copper be/# and
ft respectively*
miKB SnUBSn AND SIXAINS
C5
f*A*
7069 ^ 3-959
A,-'"
1-785/;
Actual expansion of steels Actual expansion of copper
a,Tl+
^/=«.
».T+
n—^l Jbc
JDt
/.
.acT-^
In
jt,(t
r=180-30=150"C /.
rOixiO-6xl50+2Tj47ol“l'^’‘*<^**» 1620+
1-785/.
21
-=2550-
^
-2550-0-9091 1-7591 /.=930
1620+0-85
/.
/« ““
—flMOf
/,
=^52S’8 kg.lcm.* p930 75^
iam^resatfe}
f'‘‘l7i5xS2gr»-~943-9kg.lem» component
(UHsOey
Increase in length of either
=«.
( =[
n+ ^
/
“•^+6-)' 1-08x10-6x150+
30
c»r.
*0'062 cm.
A weight of 20 tomes is supported by three short 52. each -5 cm.^ in section. The central is of steel and the The pillars are so adjusted that at a temouter ones are of copper. perature of J5°C each carries equal load. The temperature is then raised to 115'C. Estimate the stress in each piUar at JTC and IIS^C. Take El ’=‘2 y 10^ kg Icm.^ ; Et=^0'8xl0^ kgJem.*;tu>^l2xfO-AfC and »e=18 5 x W'^jC. {.A.M.LE. May, 1965) Solution. Area of each pillar— .4— 5 cm.* ProUem
pt^
pillars,
Initial stresses
At 15®C each
pillar
carries-^ tonnes.
Stress in each pillar—
toimeslem*.
-I333’33kg.lcm.*
I
STRENGTH OF MATERIALS
66
of temperature alone. Let the stresses due of temperature aloue be pc kg Icm.^ (compressive) in copper
Stresses due to rise
to rise
and
steel. p* kg-tern.^ (tensile) in
Lit the change in length of each
member be
I^oicTl—^-
«.r/+-§'
,,T+^^c,cT-Pf PL Ec P*
•
•
.
.£i«r(«.-«.)=(ll5-15)(18-5-12)10-« Er
P'
I
looxesxio'^
‘lx 10*"*'0'8xl0« I’Ll jPl ==650 8 2
^0
2/>.+5/>.=2600 pc At
Also
...(i)
= pc At
p.X5 -prXlO p.=2p. Substituting in Eqn<' (0,
2
...(«)
we
get,
X 2pc-l-5pc =2600 P(-==--^-=288’89
and
^rg./cm.*
(compressive)
Pc=2x288-89 = 577-78 kg.lcm.^
(tensOe)
Final stress in copper
= 1333 33+288 89= /622-2.? kg.Jcm.* Final stress in steel=l333
33- 57778 = 755-55 %/cm.*
ProMem 53. A flat bar of aluminium alloy 24 mm, wide and 6 mm. thick is placed between two steel bars each 24 mm, wide and 9 mm. thick to form a composite bar 24 mm. x 24 mm. as shown in The three bars are fastened together at their ends when the temperature is 1(PC. Find the stress in each of the materials when the temperature of the whole assembly is raised to SOTC.
Fig. 48.
If at the new temperature a tensiie load of 2000 kg. is applied to the composite bar what are the final stresses in steel and allov ? Take **
m,-m 24 xl(r*perC.
X IC^ per “C and
sun PLB STRESSES
Solution*
AND STRAINS
67
Area of aluminium
^a=2-4 X 0-6 - 1-44 cm.2 Area of steel in«=2 X 2*4 X 0*9=4*32
cnfi
Stresses due to rise of ternIf the two members had perature. been free to expand, (i)
free expansion of steel ^dsTl free expansion of aluminium --OiaTl
But since the members arc fasFig. 48 tened to each other at the ends final expansion of each member would be the same. Let this expansion be The free expansion of aluminium is greater than 8 while the free expansion of steel is less than Hence steel is subjected to tensile stress while aluminium is subjected to compressive stress. Let ps and Pa be the stresses in steel and aluminium. For the equilibrium of the whole system Total tension in steel=Tota] compression in aluminium /?»
Final increase
PiA$"^paAa X4*32=«/?aX 1*44
in
length of steel=Final increase in length of
aluminium.
But
7'-=50-I0=40’C
12X10-«X40+ ,-fi73-24xl0-*x40-
,
^
— X10«
480+^=.960—|-p. ^+’Y/>«='480 1920
^+3ip.-19» Pk’^S76 kg./cm,* (jeompressive)
x
SIMMOTH
to enternal load
Let the stresses due nod alumiaium.
of 2000 kg.
to the external loading
be
and
/ in sleel
alaminittm. Strain in 8teel=8traio in
^L
.
£•
fi
E.
.
But load on steel+Load on aluininiom>»' Total load
V44fa+Anf,^7m r-M/.+4 32x3/.-2000 14'4(>/.-2000 f
.8
2000.
/•“144 **•/<»»•
..
’^69'44 kg.fcm.* (compressive)
/.*3x 69-44 Final stresses due to Stress in
JIrg./an*
^208.32 hg.fcm.^ (compressive) rise of temperature and loading
Two steel rods one of 8 cm. diameter and the other end by means of a turn budUe. The other end of each rod is rigidly fixed and there is inltialty a small tension in the rods. If the effective length of each rod is 4 Problea. S4.
^6 cm. diameter are joined end to
metres, find the increase in this tension when the turn buckle is turned by one-quarter of a turn. On the rod of bigger diameter there are I’S meads per centimetre while there are 2 threads per centimetre on the other rod. Neglect the extension of the turn buckle, also what rise in temperature
Selatien.
would nullify the increase in tension. Take E=2 J0» kg./cm.* and e.^J2'X.l(r^fC.
Cross-sectional area of the stnalif- r
4
“M-2S
X 6® cm.*
cwi.*
bar
mVLB STRBSns AND STRAINS
69
Fl|.
49.
Cross-sectional area of the bigger bar
X8* cm.*
4
-50-27 cm.*
When
the turn buckle
is
Extension of the smaller
turned by one quarter of a tsrn»
bar=-^
Extension of the bigger bar
^ cm.
•
—
cm.
Total extension of the two
rods»
o
O
cm.
7 -24 Let the tension in each bar be
T kg.
Total extension in the two bars
400P f
‘2xl0«V
“
+ ,
\ 1 50-27 I
400 7iS5P 2x10®'' 28-28 x 50-27
P-26380 In order this tension must be total
1
28-28
kg. nullified
by
expansion of the two rods mustbeequal to
Let the
.-.
rise
of temperature be TX!
-®xrx 800 -;5
I 2 xl 0
7_ 24
-
24X12X800 -J0-58*C.
rise
7
of tempomture
— STRENGTH OP MATERIALS
ProUem 55. A
steel
rod 32
mm.
in diarneter is fixed co^entria^
^ ^
outside ai^ inpde diameters of 48 nm. in a brass tube which has tube are 40 cms. long 34 mm. respectively. Both the rod and the wAicA is held between two stops rod compound The level. their ends are raised then is bar the temperature of the are exactly 40 cms. apart and by 60rC. distance between the stresses in the rod and tube if the (o)
the stops
Find (0 remains constant
(il) is
0 025 cm.
increased by
stops if the Find the increase in the distance between the 8000 kg. force exerted between them is Ei.’=0'9 x 10* kg.fcm.* kg.jcm.^ rake E>=2x ib)
m
«,= 12 X 70 Solution,
(a) (i)
*
and
per°C
the distance between
When
the stops remains
constant.
m^ctsTEa
Stress in steel
Stress in brass
= 12 X 10"® X 60 X 2 X 10® kg.fcm.* = 1440 kg./cm.* (compressive) — a»r£» =21 X
10"*
X 60 X 0*9 X 10® kg.jcm.^
= 1134 kg./cm.* (compressive) (a)
When
the distance between the stops is increased
by 0’025 cm.
—
expansion prevented
,
Strain in steel
-r ,, original length
“
a.r/-8
~
/
12X10-^x60 x 40-' 0 025 40
=0-000095 Stress in steel
=£«e»=2 x
10®
x 0-000095 kg jcm.*
= 190 kg. Icm.*
(compressive)
prevented
Strain in brass
onginal length
" “
•bTl^ /
21xlO-®x60x4n-0-02S 40
=0000635 Stress in brass
=£^6 =0-9 X 10* xO-000635 kg.lcm.* —571’5 kg.lcm.* (compressive)
— SIMPU
SIHeSSBS
(b)
When
AND
8
71'
SPRAINS
the force exerted between the stops is
8000 kg.
Let the expansion of the composite member be 8 cm. c* •• Strain
m steel *
1
—
— j
“
12xl(Hx 60 x 40-8 40
=.(720xl0-«— ’“pt=Etei
Stress in steel
=2x 10* ^720X10^--;^^ kg.lcm.^ )440_^*j kg.lcm.* Similarly strain in brass
=e»—
— ——
«* Tl
j
21
”
Xl0~«x 60 X 40-8 40
=.(l260xl0-«--^) •^pt^Eb
Stress in brass
e»
=0'9 X I0«( 1260 X
kg-tcm.*
)
=( 1 134-^ X 10« 8 ) kg.lcm* Area of Steel
=^.„ix3'2* cm * 4
«=8'04I cw*.
Area of brass
=^46= ^^4'8*— 3‘4*
Load on
on brass
steel +load
cm.® =9*016 cm*. j
Total load between the stops
ptAt+pi,Ab—P (
1440- -2^-)8 04H1
1
(
1
134-— x 10« 8^ 9 016 =8000 kg.
580+10220-605000 »
8 -8000
13800 605000'
i(jr
02281 an.
SIMNOTH OP MATBUAU
72 HtacfSiMPi.
112.
SO diows a thin steel tyje of ioternal diameter d. Such a tfn can be dirank on to a wheel of slightly bigger diameter D. The la this stage the glial -tyie is heated so that its diameter exceeds D. now the tyre be cooled it it gtee» tyre Is slaved on to the whed. pnventoi fh>m assuming its original diameter d. Hence it will grip ‘
Fig.
U
the wheel. tyre.
Hence a pensile stress u induced circumferentially along the Sodi a sireas it called a hoop stress. Temperatttre strain ^
amtraction prevented ^
original length
icD—vd
~
ltd
D-d
~
d
Hoop stress due
to fall of temperature
It is Prahlem 56. rigid wheel is 3 metres in diameter. derired to shrink on to the wheel a thin steei tyre. Find the internal diameter of the tyre if qfier fitting the hoop stress in the tyre is 900 kg.lem.*. find also the least temperature to which the tyre must be luated dbofe diat of the wheel.
rigid wheel 1'25 metre in Udometer is to be If the stress in the steel tyre is not to excifid 140 MNImetr^,find the minimum diameter of the tyre. Find also the minimum temperature to which the tyre is to be raised so that it can befitted over the wheel.
