UNIT-01. SIMPLE STRESSES &
STRAINS
Lecture Number - 06
Prof. S.C.SADDU
MECHANICAL DEPARTMENT
SIT LONAVALA
Strength of Materials
Agenda
• To study temperature stresses and temperature
strain
Strength of Materials
Thermal Stresses
• A temperature change results in a change in length or
thermal strain. There is no stress associated with the
thermal strain unless the elongation is restrained by
the supports.
• Treat the additional support as redundant and
apply the principle of superposition.
PL
 T   T L
P 
AE
  thermal expansion coef.
• The thermal deformation and the deformation from
the redundant support must be compatible.
  T   P  0
  T   P  0
P   AE T 
 T L 

PL
0
AE
Strength of Materials
P
  E T 
A
Temperature stresses
• Extension in bar due to raise in temperature is,
L=  *T *L
• Temperature Strain is, =  *T
• Temperature Stress developed is,
=  *E*T
Strength of Materials
Example: 1 A steel tube having an external diameter of
36 mm and an internal diameter of 30 mm has a
brass rod of 20 mm diameter inside it, the two
materials being joined rigidly at their ends when the
ambient temperature is 18 0C. Determine the
stresses in the two materials: (a) when the
temperature is raised to 68 0C (b) when a
compressive load of 20 kN is applied at the increased
temperature.
• For brass: Modulus of elasticity = 80 GN/m2; Coefficient of
expansion = 17 x 10 -6 /0C
• For steel: Modulus of elasticity = 210 GN/m2; Coefficient of
expansion = 11 x 10 -6 /0C
Strength of Materials
30
Brass rod
20
Steel tube
Area of brass rod (Ab) =
Area of steel tube (A s) =
 x 202
 314.16 mm2
4
 x (362  302 )
 31102
. mm2
4
As E s  311.02 x 10 m x 210 x 10 N / m 2  0.653142 x 108 N
6
2
9
1
 153106
.
x 10 8
As E s
Strength of Materials
36
Stress in steel tube =
5444.71N
 17.51N / mm2  17.51 MN / m2 (Tension)
2
311.02 mm
Stress in brass rod =
5444.71N
2
2

17
.
33
N
/
mm

17
.
33
MN
/
m
(Compression)
2
314.16 mm
(b) Stresses due to compression force, F’ of 20 kN
F ' Es
20 x 103 N x 210 x 109 N / m2
2
s 


46
.
44
MN
/
m
(Compression)
8
E s As  Eb Ab
0.653142  0.251327 x 10
F ' Eb
20 x 103 N x 80 x 109 N / m2
b 

 17.69 MN / m2 (Compression)
8
E s As  Eb Ab
0.653142  0.251327 x 10
Resultant stress in steel tube = - 46.44 + 17.51 = 28.93 MN/m2 (Compression)
Resultant stress in brass rod = -17.69 - 17.33 = 35.02 MN/m2 (Compression)
Strength of Materials
Example: 2 A railway is laid so that there is no stress in
rail at 10º C. If rails are 30 m long Calculate,
1. The stress in rails at 60 º C if there is no allowance
for expansion.
2. The stress in the rails at 60 º C if there is an
expansion allowance of 10 mm per rail.
3. The expansion allowance if the stress in the rail is to
be zero when temperature is 60 º C.
4. The maximum temp. to have no stress in the rails if
the expansion allowance is 13 mm/rail.Take  = 12 x
Strength of Materials
-6
10 per 1ºC
E= 2 x 10 5 N/mm 2
Solution:
1. Rise in temp. = 60 º - 10 º = 50 ºC
so stress =  t E =12 x 10 -6 x50x 2 x 10 5
= 120 MPa
2. tp x L/E =
 = (L t -10)
= (30000 x 12 x 10 -6 x50-10)
= 18 -10 = 8 mm
tp =E /L = 8x 2 x 10 5 /30000
= 53.3 MPa
Strength of Materials
3. If stresses are zero ,
Expansion allowed =(L t )
= (30000 x 12 x 10 -6 x50)
=18 mm
4. If stresses are zero
tp =E /L*(L t -13)=0
L t=13
so t=13/ (30000 x 12 x 10 -6 )=360 C
allowable temp.=10+36=460c. Strength of Materials
Example: 3 A composite bar made up of aluminum and
steel is held between two supports. The bars are
stress free at 400c. What will be the stresses in the
bars when the temp. drops to 200C, if
(a) the supports are unyielding
(b)the supports come nearer to each other by 0.1 mm.
Take E al =0.7*105 N/mm2 ;al =23.4*10-6 /0C
ES=2.1*105 N/mm2
s =11.7*10-6 /0C
Aal=3 cm2 As=2 cm2
Strength of Materials
Free contraction =Ls s t+ LALAlt
Since contraction is checked tensile stresses will
be set up. Force being same in both
As s= Aal al
contraction of steel bar s = (s/Es)*Ls
contra.of aluminum bar al = (al/Eal)*Lal
(a) When supports are unyielding
s + al =  (free contraction)
(b) Supports are yielding
s + al = ( - 0.1mm)
Strength of Materials
Example: 4 A steel bolt of length L passes through a
copper tube of the same length, and the nut at the
end is turned up just snug at room temp.
Subsequently the nut is turned by 1/4 turn and the
entire assembly is raised by temp 550C. Calculate the
stress in bolt if L=500mm,pitch of nut is 2mm, area of
copper
tube
=500sq.mm,area
of
steel
bolt=400sq.mm
• Es=2 * 105 N/mm2 ;s =12*10-6 /0C
• Ec=1 * 105 N/mm2 ;c= 17.5*10-6 /0C
Strength of Materials
Solution:
Two effects
(i) tightening of nut
(ii)raising temp.
tensile stress in steel = compressive force in copper
[Total extension of bolt +Total compression of tube]
=Movement of Nut
[s+  c] = np
( where p = pitch of nut)
• PL/AsEs + s L t) +(PL/AcEc- c L t)=np
Strength of Materials