The composite bar consisting of Steel and Aluminium, shown in the figure is connected to two grips at a temperature of $60^oC$ - Take $E_s = 2 × 10^5 N/mm^2 , E_A = 0.70 × 10^5 N/mm^2$

$α_s = 1.17 × 10^{-5} per^oC , α_A = 2.34 × 10^{-5} per^oC $

Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: Dec 2013

Draw the same diagram

When there is a drop in the temperature of the grips, steel and Aluminium, both try to contract, which cause a tensile stress in each of the bars.

Steel:

$E_s = 2 × 10^5 N/mm^2$

$α_s = 1.17 × 10^{-5}/^oC $

$A_s = 250mm^2$

$L_s = 800mm$

Aluminium:

$E_{Al} = 0.7 × 10^5 N/mm^2$

$α_{Al} = 2.34 × 10^{-5}$

$A_{Al} = 375mm^2$

$L_{Al} = 400mm$

By the load sharing relationship,

Forced induced in steel = Force induced in Aluminium

$σ_s × A_s = σ_{Al} × A_{Al}$

Putting in the data, we have,

$250σ_s = 375σ_{Al}$

By the strain relationship,

${δ_L}_{Steel} = {δ_L}_{Aluminium}$

$[Free Contraction - Induced Expansion]_{Steel} + [Free Contraction - Induced Expansion]_{Aluminium} = Contraction Allowed$

$[αTL - \frac{PL}{AE}]_{Steel} - [αTL - \frac{PL}{AE}]_{Aluminium} = Contraction Allowed$

Putting in the values, we have:

$[0.3744 - (4 × 10^{-3})σ_s] + [0.3744 - (5.71 × 10^{-3})σ_Al] = Contraction Allowed$

$0.7488 - [(4 × 10^{-3})σ_s + 5.71 × 10^{-3})σ_{Al}] = Contraction Allowed....(ii)$

i When the ends do not yield

Contraction Allowed = 0

From (II),

$0.7488 - [(4 × 10^{-3})σ_s + 5.71 × 10^{-3})σ_{Al}] = 0 .....(iii)$

From (I) and (III), we get

$σ_s = 95.918 N/mm^2$

$σ_{Al} = 63.945 N/mm^2$

This is the value of the stresses, when the ends do not yield.

ii When the ends yield by 0.25mm

Contraction allowed = 0.25mm

From (II),

$0.7488 - [(4 × 10^{-3})σ_s + 5.71 × 10^{-3})σ_{Al}] = 0.25 .....(iv)$

From (I) and (IV), we get

$σ_s = 63.894 N/mm^2$

$σ_{Al} = 42.596 N/mm^2$

This is the value of the stresses, when the ends yield by 0.25mm.