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Chapter 1

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.

Step-by-Step

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)

Screenshot 2022-10-07 011005