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## Q. 1.16

If an external compressive load of 60 kN is applied on the same bolt and aluminium collar assembly of Example 1.15, calculate the stresses in steel bolt and aluminium collar.

## Verified Solution

First consider the external load alone. Equilibrium condition: Consider the free body diagram and assuming the direction of force as shown in Fig. 1.26,

$F_s+ F_a = 60 \times 10^{3} N$ (i)

Compatibility condition: The displacement of steel bolt and aluminium collar will be equal under external load.

$\Delta L_s = \Delta L_a$ (ii)

Substituting stress—strain relation in Eq. (ii),

$\frac{F_s \times L_s}{A_s \times E_s}=\frac{F_a \times L_a}{A_a \times E_a}$ (iii)

or,  $F_s=\frac{600 \times 210 \times 10^3}{900 \times 70 \times 10^3} F_a=2 F_a$

Substituting in Eq. (i)

$F_a = 60 \times 10^{3}/3 = 2 \times 10^{4}$ N

and $F_s = 4 \times 10^{4}$  N

Therefore, stresses due to external load of 60  kN acting alone,

$\sigma_s=4 \times 10^4 / 600=66.67$  MPa  (Compression)
$\sigma_s=2 \times 10^4 / 900=22.22$  MPa  (Compression)

Total stresses due to tightening of nut + rise in temperature + external load,

$\sigma_s= 192.5 – 66.67$= 125.83  MPa  (Tension)

$\sigma_s$ = 128.34 + 22.22 = 150.56  MPa (Compression)