Temperature stresses or Thermal Stresses

The stresses induced in a body due to change in temperature are known as thermal stresses.

The materials have a tendency to expand or contract when they exposed to temperature changes.

If the length of the bar AB is L, the Coefficient of linear thermal expansion is α 
T = Rise or fall of temperature (t2 - t1)
E = Young’s modulus

The change in length dL due to temperature change is given by 

 ∆t = α.T.L

Thermal strain, e = ∆t/L = α . T 

If the temperature deformations are not allowed freely, the stresses are induced in the body. These stresses are called thermal or temperature stresses.

Thermal stress,  stp= e.E = α.T.E 

Yielding of Supports

Suppose if the support yield by an amount d, thermal stress 

stp = (D- d).E/L = (Lat - d).E/L

Example:1

Q.1 A railway is laid so that there is no stress in rail at 10º C. If rails are 30 m long Calculate, 

a. The stress in rails at 60ºC if there is no allowance for expansion. 

b. The stress in the rails at 60ºC if there is an expansion allowance of 10 mm per rail. 

c. The expansion allowance if the stress in the rail is to be zero when temperature is 60ºC. 

d. The maximum temp. to have no stress in the rails if the expansion allowance is 13 mm/rail. 

Take a = 12x10^-6 per 1ºC E= 2 x10⁵ N/mm² 

Solution:

a) Length of rail = 30m

Rise in temp.  t = 60º - 10º = 50ºC

 stress in rail = a t E = (12x10^-6).50.(2x105)                           stp =  120 MPa

b) stp = (D- d).E/L = (Lat - d).E/L

           stp x L/E   = (La t -10)                                                              = (30000 x 12x10^-6 x50 -10)                                     = 18 -10 = 8 mm

        stp =DE /L  = 8x 2 x10⁵ /30000

                           = 53.3 MPa

c) If stresses are zero , 

 Expansion allowed =(La t )                                                                    = (30000 x 12x10^-6 x50)                                           =18 mm

d. If stresses are zero,when d = 13

stp =E /Lx(La t -13)=0

La t =13

so t =13/ (30000 x 12x10^-6 )= 36ºC

allowable temp.=10 + 36 = 46ºC 

Thermal Stresses in Composite section

Consider a composite section having two different materials A and B.The material A has low coefficient of thermal expansion than material B. 

Due to rise in temperature the two bars expand freely as shown in condition (2). Material B expands more, hence it has high coefficient of thermal expansion and material A has low coefficient of thermal expansion.

If the differential expansion of the bars are prevented in composite sections, it will reach for equilibrium condition(3) 

In equilibrium condition, tensile force exerted by material B on A is equal to compressive force exerted by A on B.

Therefore to achieve equilibrium condition, the material A is exerted by tensile force and material B is exerted by compressive force.

The change in length in bar A = ∆ta

∆ta = [Elongation due to temperature + elongation due to tensile force]A =[αtL+ (PL/AE)]A

The change in length in bar B = ∆tb

∆tb=  [Elongation due to temperature + elongation due to compressive force]B =[αtL– (PL/AE)]B

But ∆ta =∆tb

Example 2:

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 =500 sq.mm, area of steel bolt = 400 sq.mm.,

Es=2x10⁵ N/mm² ;and α_s =12x10^-6 /⁰C ,

Ec=1 x10⁵ N/mm² ;  α_c= 17.5x10^-6 /⁰C 

Solution:

In this problem we have two effects

 (i) tightening of nut and  (ii) raising temp.

Note:In case of nut and bolt used on a tube with washers, the tensile load on the bolt is equal to the compressive load on the tube.

tensile stress in steel bolt = compressive force in copper tube

[Total extension of bolt

+Total compression of tube] = Movement of Nut

[Ds+ Dc] = np     ( where p = pitch of nut)

(PL/A_sE_s + a_s L t) +(PL/A_cE_c- a_c L t)= np

P (1/A_sE_s +1/A_cE_c) = t(a_c - a_s)+np/L

P[1/(400*2*10⁵) + 1/(500x1x10⁵) ]

               =(17.5-12)x10^-6 +(1/4)x2/500

P  =  40000N

Stress in steel bolt s_s = 40000/400                                                           = 100 MPa(tensile)

The stress in copper tube s_c= 40000/500                        =80 MPa(compressive)

Assignment:

1.A brass rod 2m long is fixed at both its ends. If the thermal stress is not to exceed 76.5 MPa, calculate the temperature through which the rod should be heated .

Take the value of α = 17x10-6 /C and E= 90 GPa.

 (Ans: t = 50 c)

2. A steel bar, fixed at its both ends, is heated through 15 C. Calculate the stress developed in the bar, if modulus of elasticity and coefficient of linear expansion for the bar material is 200 GPa and 12x10-6 /C respectively.

(Ans:stress 36 Mpa,)

3. An alloy bar 2m long is held between two supports. FInd the stress developed in the bar, when it is heated through 30 C, if both the ends (i) do not yield (ii) yield by 1 mm. Take the value of E and coefficient of thermal expansion for the alloy as 120 GPa and 24x10-6 /K .

(Stress= 86.4 MPa, 26.4 Mpa)

4. An alloy bar 6m long is held between two supports by a steel rod 25 mm diameter passing through metal plates and nuts at each end.. The rod is held at temp. of 100 C. Determine the stress in the rod, when the temperature falls down to 60 C, if

(i) do not yield   (ii) yield by 1 mm. 

Take the value of young's modulus for the alloy as 200 GPa and α = 12x10-6 /c.

(Stress= 96 MPa, 62.6MPa)

5. Two parallel walls 6m apart are stayed together by a steel rod 2.5 cm diameter at a temperature of 80 0 c passing through washers and nuts at each end. Calculate the pull exerted by the rod when it has cooled to 220 C if,

(i) do not yield (ii) yield by 1.5 mm. 

Take the value of E and α for the alloy as 200 GPa and 11x10-6 /c .

(Stress= 127.6 kN, 77.6m/mm2)

6. A steel bar of 30 mm diameter is heated to 800 C and then clamped at the ends. It is then allowed to cool down to 300 c. During cooling, only 1 mm contraction was allowed. Calculate the temperature stress developed and reaction at the clamps. 

Take the value of E and α for the alloy as 200 GPa and 12x10-6/C.Take length of bar 10 meters.

(F=70.68 KN)