# 4.16 Stresses due to Change in Temperature—Thermal Stresses

Whenever there is some increase or decrease in the temperature of a body, it causes the body to expand or contract. A little consideration will show that if the body is allowed to expand or contract freely, with the rise or fall of the temperature, no stresses are induced in the body. But, if the deformation

of the body is prevented, some stresses are induced in the body. Such stresses are known as thermal stresses.

Let l = Original length of the body,

t = Rise or fall of temperature, and

α = Coefficient of thermal expansion,

∴ Increase or decrease in length,

δl = l. α.t

If the ends of the body are fixed to rigid supports, so that its expansion is prevented, then

compressive strain induced in the body,

εc =

. .

.

l l t

t

l l

δ = α = α

∴ Thermal stress, σth = εc.E = α.t.E

Notes : 1. When a body is composed of two or different materials having different coefficient of thermal expansions, then due to the rise in temperature, the material with higher coefficient of thermal expansion will be

subjected to compressive stress whereas the material with low coefficient of expansion will be subjected to tensile stress.

2. When a thin tyre is shrunk on to a wheel of diameter D, its internal diameter d is a little less than the wheel diameter. When the tyre is heated, its circumferance π d will increase to π D. In this condition, it is slipped on to the wheel. When it cools, it wants to return to its original circumference π d, but the wheel if it is assumed to be rigid, prevents it from doing so.

∴ Strain, ε =

D d D d

d d

π − π = −

π

This strain is known as circumferential or hoop strain.

∴ Circumferential or hoop stress,

σ = E.ε =

E (D d)

d

Example 4.13. A thin steel tyre is shrunk on to a locomotive wheel of 1.2 m diameter. Find the internal diameter of the tyre if after shrinking on, the hoop stress in the tyre is 100 MPa. Assume E = 200 kN/mm2. Find also the least temperature to which the tyre must be heated above that of the wheel before it could be slipped on. The coefficient of linear expansion for the tyre is 6.5 × 10–6 per °C.

Solution. Given : D = 1.2 m = 1200 mm ; σ = 100 MPa = 100 N/mm2 ; E = 200 kN/mm2

= 200 × 103 N/mm2 ; α = 6.5 × 10–6 per °C

Internal diameter of the tyre Let d = Internal diameter of the tyre.

We know that hoop stress (σ),

100 =

E (D d) 200 103 (D d)

d d

− = × −

∴ 3 3

100 1

200 10 2 10

D d

d

− = =

× ×

…(i)

3

1

1 1.0005

2 10

= + =

×

D

d

∴ d =

1200

1199.4 mm 1.1994 m

1.0005 1.0005

= = = D

Ans.

Least temperature to which the tyre must be heated Let t = Least temperature to which the tyre must be heated.

We know that

π D = π d + π d . α.t = π d (1 + α.t)

α.t = 3

1

1

2 10

D D d

d d

π − = − =

π ×

…[From equation (i)]

∴ t = 3 6 3

1 1

77 C

2 10 6.5 10− 2 10 = = °

α × × × × ×

Ans.

Example 4.14. A composite bar made of aluminium and steel is held between the supports as shown in Fig. 4.16. The bars are stress free at a temperature of 37°C. What will be the stress in the two bars when the temperature is 20°C, if (a) the supports are unyielding; and (b) the supports yield and come nearer to each other by 0.10 mm?

It can be assumed that the change of temperature is uniform all along the length of the bar.

Take Es = 210 GPa ; Ea = 74 GPa ; αs = 11.7 × 10–6 / °C ; and αa = 23.4 × 10–6 / °C.

Solution. Given : t1 = 37°C ; t2 = 20°C ; Es = 210 GPa = 210 × 109 N/m2 ; Ea = 74 GPa

= 74 × 109 N/m2 ; αs = 11.7 × 10–6 / °C ; αa = 23.4 × 10–6 / °C , ds = 50 mm = 0.05 m ; da = 25 mm

= 0.025 m ; ls = 600 mm = 0.6 m ; la = 300 mm = 0.3 m

Let us assume that the right support at B is removed and the bar is allowed to contract freely due to the fall in temperature. We know that the fall in temperature,

t = t1 – t2 = 37 – 20 = 17°C

∴ Contraction in steel bar

= αs . ls . t = 11.7 × 10–6 × 600 × 17 = 0.12 mm

and contraction in aluminium bar

= αa . la . t = 23.4 × 10–6 × 300 × 17 = 0.12 mm

Total contraction = 0.12 + 0.12 = 0.24 mm = 0.24 × 10–3 m

It may be noted that even after this contraction (i.e. 0.24 mm) in length, the bar is still stress free as the right hand end was assumed free.

Let an axial force P is applied to the right end till this end is brought in contact with the right hand support at B, as shown in Fig. 4.17.

Fig. 4.17

We know that cross-sectional area of the steel bar,

As = ( )2 (0.05)2 1.964 10 3 m2

4 4 ds π = π = × −

and cross-sectional area of the aluminium bar,

Aa = ( )2 (0.025)2 0.491 10 3 m2

4 4 da π = π = × −

We know that elongation of the steel bar,

δls = 3 9 6

0.6 0.6

m

1.964 10 210 10 412.44 10

s

s s

P l P P

A E −

× × = =

× × × × ×

= 1.455 × 10–9 P m

and elongation of the aluminium bar,

δla = 3 9 6

0.3 0.3

m

0.491 10− 74 10 36.334 10

× × = =

× × × × ×

a

a a

P l P P

A E

= 8.257 × 10–9 P m

∴ Total elongation, δl = δls + δla

= 1.455 × 10–9 P + 8.257 × 10–9P = 9.712 × 10–9 P m

Let σs = Stress in the steel bar, and

σa = Stress in the aluminium bar.

(a) When the supports are unyielding

When the supports are unyielding, the total contraction is equated to the total elongation,i.e.

0.24 × 10–3 = 9.712 × 10–9P or P = 24 712 N

∴ Stress in the steel bar,

σs = P/As = 24 712 / (1.964 × 10–3) = 12 582 × 103 N/m2

= 12.582 MPa Ans.

and stress in the aluminium bar,

σa = P/Aa = 24 712 / (0.491 × 10–3) = 50 328 × 103 N/m2

= 50.328 MPa Ans.

(b) When the supports yield by 0.1 mm

When the supports yield and come nearer to each other by 0.10 mm, the net contraction in

length

= 0.24 – 0.1 = 0.14 mm = 0.14 × 10–3 m

Equating this net contraction to the total elongation, we have

0.14 × 10–3 = 9.712 × 10–9 P or P = 14 415 N

∴ Stress in the steel bar,

σs = P/As = 14 415 / (1.964 × 10–3) = 7340 × 103 N/m2

= 7.34 MPa Ans.

and stress in the aluminium bar,

σa = P/Aa = 14 415 / (0.491 × 10–3 ) = 29 360 × 103 N/m2

= 29.36 MPa Ans.

Example 4.15. 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.

Solution. Given: dc = 50 mm ; dse = 75 mm ; dsi = 50 mm ; dp = 18 mm = 0.018 m ;

t = 50°C; Es = 210 GN/m2 = 210 × 109 N/m2 ; Ec = 105 GN/m2 = 105 × 109 N/m2 ;

αs = 11.5 × 10–6/°C ; αc = 17 × 10–6/°C

The copper bar in a steel tube is shown in Fig. 4.18.

We know that cross-sectional area of the copper bar,

Ac = ( )2 (50)2 1964 mm2 1964 10 6 m2

4 4 dc π = π = = × −

and cross-sectional area of the steel tube,

As = ( )2 ( )2 (75)2 (50)2 2455 mm2

4 4

π ⎡ − ⎤ = π ⎡ − ⎤ = ⎣ dse dsi ⎦ ⎣ ⎦

= 2455 × 10–6 m2

Let l = Length of the copper bar and steel tube.

We know that free expansion of copper bar

= αc . l . t = 17 × 10–6 × l × 50 = 850 × 10–6 l

and free expansion of steel tube

= αs . l . t = 11.5 × 10–6 × l × 50 = 575 × 10–6 l

∴ Difference in free expansion

= 850 × 10–6 l – 575 × 10–6 l = 275 × 10–6 l …(i)

Since the free expansion of the copper bar is more than the free expansion of the steel tube, therefore the copper bar is subjected to a *compressive stress, while the steel tube is subjected to a tensile stress.

Let a compressive force P newton on the copper bar opposes the extra expansion of the copper bar and an equal tensile force P on the steel tube pulls the steel tube so that the net effect of reduction in length of copper bar and the increase in length of steel tube equalises the difference in free expansion of the

two.

∴ Reduction in length of copper

bar due to force P

=

.

c . c

P l

A E

6 9

.

1964 10− 105 10 =

× × ×

P l

= 6

.

m

206.22 × 10

P l

and increase in length of steel bar due to force P

= –6 9

. .

s . s 2455 10 210 10

P l P l

A E

=

× × ×

= 6

.

m

515.55 10

P l

×

∴Net effect in length = 6 6

. .

206.22 10 515.55 10

P l P l +

× ×

= 4.85 × 10–9 P.l + 1.94 × 10–9 P.l = 6.79 × 10–9 P.l

Equating this net effect in length to the difference in free expansion, we have

6.79 × 10–9 P.l = 275 × 10–6 l or P = 40 500 N

Stress induced in the copper bar, steel tube and pins

We know that stress induced in the copper bar,

σc = P / Ac = 40 500 / (1964 × 10–6) = 20.62 × 106 N/m2 = 20.62 MPa Ans.

Stress induced in the steel tube,

σs = P / As = 40 500 / (2455 × 10–6) = 16.5 × 106 N / m2 = 16.5 MPa Ans.

Note : This picture is given as additional information and is

not a direct example of the current chapter.

* In other words, we can also say that since the coefficient of thermal expansion for copper (αc) is more than

the coefficient of thermal expansion for steel (αs), therefore the copper bar will be subjected to compressive

stress and the steel tube will be subjected to tensile stress.

and shear stress induced in the pins,

τp = 6 2

2

40500

79.57 10 N/m

2 2 (0.018)

4

p

P

A

= = ×

× π

= 79.57 MPa Ans.

…(Q The pin is in double shear )