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.

Question

Accepted Answer

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,

[latex]F_s+ F_a = 60 imes 10^{3} N[/latex] (i)

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

[latex]\Delta L_s = \Delta L_a [/latex] (ii)

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

[latex]\frac{F_s imes L_s}{A_s imes E_s}=\frac{F_a imes L_a}{A_a imes E_a} [/latex] (iii)

or, [latex] F_s=\frac{600 imes 210 imes 10^3}{900 imes 70 imes 10^3} F_a=2 F_a[/latex]

Substituting in Eq. (i)

[latex]F_a = 60 imes 10^{3}/3 = 2 imes 10^{4} [/latex] N

and [latex]F_s = 4 imes 10^{4}[/latex] N

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

[latex]\sigma_s=4 imes 10^4 / 600=66.67[/latex] MPa (Compression)
[latex]\sigma_s=2 imes 10^4 / 900=22.22[/latex] MPa (Compression)

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

[latex]\sigma_s= 192.5 - 66.67[/latex]= 125.83 MPa (Tension)

[latex]\sigma_s[/latex] = 128.34 + 22.22 = 150.56 MPa (Compression)

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

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