Question 1.15: Determine the stresses induced in steel bolt and aluminium c...

First consider the temperature stresses only.

Equilibrium condition: Since no external force is acting at the final position of the assembly, the equilibrium of internal forces in free body diagram [Fig. 1.25(d)] gives,

$F_s = F_a$ (i)

Compatibility condition: Fig. 1.25(b) shows the free expansion of the steel bolt and aluminium collar. The expansion of aluminium collar will be more than steel bolt since its coefficient of thermal expansion is higher. Thus at equilibrium aluminium collar will shorten by an amount $\Delta L_a$ and steel bolt will extend by an amount $\Delta L_s$, as shown in Fig. 1.25(c) such that,

$\Delta L_s+\Delta L_a=\left(\alpha_a-\alpha_s\right) \times L \times \Delta t$ (ii)

Substituting stress-strain relation in Eq. (ii),
$\frac{F_s \times L_s}{A_s \times E_s}+\frac{F_a \times L_a}{A_a \times E_a}=\left(\alpha_a-\alpha_s\right) \times L \times \Delta t$ (iii)

or, $\frac{F_s}{600 \times 210000}+\frac{F_a}{900 \times 70000} =(23-18) \times 10^{-6} \times 50$

or , $F_s=F_a$ = 10500 kN

Stress in steel, $\sigma_s=10500 / 600 =17.5  MPa$ (Tension)
Stress in aluminium, $\sigma_a =10500 / 900$ = 11.67  MPa (Compression)

Combined stresses: If the external loads are also acting with rise in temperature, the mechanical and thermal stress will superimpose as follows.

Stress in steel, $\sigma _s=\sigma _s {_{,mech}}+\sigma _s {_{,therm}}$ = 175 + 17.5 = 192.5  MPa (Tension)

Stress in aluminium,

$\sigma _a=\sigma _a {_{,mech}}+\sigma _a {_{,therm}}$ = 116.67 + 11.67 = 128.34 MPa (Compression)