Question

When two components are initially tightened by a bolt with a pre load of 6 kN, the alongation of the bolt is 2 mm and the compression of the components together is 1 mm. When the assembly is subjected to 9 kN (tensile on the bolt) of external load, what will be the net compression in the components under this load?

• A. 0.5 mm
• B. 1.5 mm
• C. 2 mm
• D. zero

Answer 1

The correct option is D zero

$F_i\text{= 6 kN,}$

$({\delta }_i{)}_b\text{= 2mm (elongation)}$

$(\delta i{)}_c\text{= 1 mm (compression)}$

Stiffness of bolt and component

$K_b=\frac{6\times {10}^6}{2}\text{= 30 MN/m}$

$K_c=\frac{6\times {10}^6}{1}\text{= 6 MN/m}$

When external load is applied on the member, it shared by both bolt and component and both are elongated.

$P_b=\frac{K_b}{K_c+K_b}{P}_{ext},P_c=\frac{K_c}{K_c+K_b}{P}_{ext}$

$P_{ext}=9kN$

$P_b\text{and}{P}_c$ are part of external load shared by bolt and component respectively.

$P_c=\frac{6}{6+3}\times 9=6kN$

${\delta }_c^{{}^{\prime }}=\frac{6\times {10}^3}{6\times {10}^6}\times {10}^3=1mm\text{(elongation)}$

$\text{Net compression = 1 - 1 = 0}$