TEMPERATURE –INDUCED STRESSES
It is common knowledge that materials expand when
heated and contract when cooled. By wholly or
partially restricting such an expansion or contraction, a
stress will be induced in the material
In a bar of length L and coefficient of linear expansion 𝛼,
the magnitude of the temperature-induced stress can
be determined by first considering the bar to be free to
expand (thermal expansion ∆L = αL ∆T due to rise in
temperature from 𝑇1 𝑡𝑜 𝑇2 , and then using the
subsequent mechanical compression 𝛿 to calculate the
induced stress 𝑥 =
PL
EA
=
σL
E
N.B:
When the bar is wholly constrained
Free expansion (temperature expansion = Mechanical
Compression)
𝑥 = ∆L
σL
= αL ∆T
E
EXAMPLE 1:
A brace spacer of diameter 75 𝑚𝑚 and length 350 𝑚𝑚
fits snugly between two faces at a temperature of 15 ℃.
Calculate the compressive force on the spacer at a
temperature of 50℃
(1.1). When the expansion is completely prevented
(1.2). when the two faces yield by 0.15 𝑚𝑚
E = 90 GPa and α = 16 × 10−10 ℃
COMPOSITE BARS
Composite bars can be in series or parallel arrangement
(1). Series arrangement:
For series arrangement, both materials will be in tension
or both in compression.
For equilibrium,
Compressive force F1 = Compressive force F2
𝜎1 𝐴1 = 𝜎2 𝐴2
(1)
If brought back to the original length,
Sum of mechanical change = Sum of the thermal change
𝑥1 + 𝑥2 = ∆𝐿1 + ∆𝐿2
𝜎1 𝐿 1
𝐸1
+
𝜎2 𝐿 2
𝐸2
= 𝛼1 𝐿1 ∆𝑇 + 𝛼2 𝐿2 ∆𝑇
(2)
The stresses 𝜎1 𝑎𝑛𝑑 𝜎2 is obtained by solving equation 1
and 2 simultaneously.
Example 2:
The composite bar shown just fits between the rigid
end-stops. For a temperature rise of 70℃, calculate:
(2.1). the stresses induced in the steel and the brass
(2.2). the final distance of the shoulder AB from the lefthand end-stop
𝐹𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 ∶ 𝐸 = 200 𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼 = 11 × 10−6 ℃ ;
𝐹𝑜𝑟 𝑏𝑟𝑎𝑠𝑠 ∶ 𝐸 = 100 𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼 = 18 × 10−6 ℃
(2). Parallel arrangement
Two materials to be rigidily connected together at their
ends so that under all conditions they are of equal
lenght. The final position will be somewhere between
the two free positions, thus inducing a tensil stress in
one material nd compressive stress in the other.
EXAMPLE 3:
A rectangular strip of aluminum 50 mm wide and 15 mm
thick is clad on both the 50 mm sides by brass strip
50 𝑚𝑚 wide and 5 mm thick. The three strips are
bonded together so that under all conditions they have
the same length. At 20℃ the lenght of the composite
bar is 600 mm. If the temperature rises to 100 ℃,
calculate:
(3.1). the stresses set up in the two materials
(3.2). the final lenght of the composite bar
𝐹𝑜𝑟 𝑎𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝐸 = 70 𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼 = 22 × 10−6 ℃
𝐹𝑜𝑟 𝑏𝑟𝑎𝑠𝑠 𝐸 = 105 𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼 = 17 × 10−6 ℃
EXAMPLE 5:
A copper bar 50 mm in diameter is placed within a steel tube 75
mm external diameter and 50 mm internal diameter of exactly the
same length. The two pieces are rigidly fixed together by two pins
18 mm in diameter, one at each end passing through the bar and
tube. Calculate the stress induced in the copper bar, steel tube and
pins if the temperature of the combination is raised by 50°C.
𝑇𝑎𝑘𝑒 𝐸𝑠 = 210 𝐺𝑁/𝑚2 ; 𝐸𝐶 = 105 𝐺𝑁/𝑚2 𝛼𝑠 = 11.5 ×
10– 6/°𝐶 𝑎𝑛𝑑
𝛼𝑐 = 17 × 10– 6/°𝐶.
QUESTION 6:
A steel tube is bushed with loosely fitting bronze tube of the same
length and they are rigidly connected together at their ends. The
thickness of the steel tube is 7.5 𝑚𝑚. The external and internal
diameters are 𝑑2 and 60 𝑚𝑚 for the steel, and 60 𝑚𝑚 and 50 𝑚𝑚
for the bronze respectively. The initial length of the composite tube
is 300 𝑚𝑚. For temperature rise of 20 ℃, calculate:
(6.1). the stress set up in the steel and the bronze
(6.2). the force 𝑃𝑠 on the steel and the force 𝑃𝑏 on the bronze
(6.3). the increase in length of the composite tube
𝐹𝑜𝑟 𝑠𝑡𝑒𝑒𝑙: 𝐸𝑠 = 210 𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼𝑠 = 11 × 10−6 ℃ ;
𝐹𝑜𝑟 𝑏𝑟𝑜𝑛𝑧𝑒: 𝐸𝑏 = 97𝐺𝑃𝑎 𝑎𝑛𝑑 𝛼𝑏 = 19 × 10−6 ℃