A steel rod of sectional area 250 sq. mm connects two parallel walls 5 m apart. The nuts at the ends were tightened when the rod was heated to 100°C. If α_steel = 0.000012/C°, E_steel = 0.2 MN/mm², the tensile force developed at a temperature of 50°C, is

Given,
Es = 0.2 MN/mm² = 0.2x10^6 N/mm²
α = 0.000012/C°
T = 100 - 50 = 50 C°
Tensile force = E x α x T
= 0.2 x 10^6 x 0.000012 x 50 = 120 N/mm²

Its very simple:
Stress (sigma) = E × strain...........(i)
And strain due to Temp changes = alpha × (delta T)
Now delta T = 100-50 = 50°
E = 0.2 × 10^6 N/mm^2
Alpha =0.000012/c°
Equation becomes
Stress (sigma) = (0.2 ×10^6) × (0.000012 ×50°)
= 120 N/mm^2

Temperature (T1)=100°c
Temperature (T2)=50°c
Difference in temperature ∆T=100°-50°=50°c
Length(L)=5m
Area (A)=25mm2
Alpha(a)=0.000012/°c=1.2×10to the power -5
Esteel=0.2×10 to the power 6
At temperature of 50°c,strain developed is
∆L/L=alpha(a)×∆T
∆L/L=1.2×10-5×50°c
est=6×10 to the power -4
Tensile stress developed in the steel
Sigma st=Est×est
=0.2×106×6×10-4
=0.2×6×102
=120N/mm2

tharmal stress tensile/compressive = @.t.E ; where @ is the linear expansion ; t is the temprature increment ;E is the modulas of elasticity of given matarials then; given value @ = .000012/C ;E = .2MN/mm squre =.2*1000000 ; and t = 50 degree then @tE = .000012*50*.2*1000000 =120N/mm squre

it is tensile stress not tensile force

thermal stress = t*alfa*E =50 * 0.000012 * 0.2 * 10 power 6

Anyone plzz post solution for this question