Question 1.16: If an external compressive load of 60 kN is applied on the s...
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.
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)
