Question 4.2: Analysis of Bolt–Tube Assembly In the assembly of the alumin...

The forces in the bolt and in the sleeve are denoted by $P_{b}$ and $P_{t},$ respectively.

Statics: The only equilibrium condition available for the free body of Figure 4.2b gives

$P_{b} = P{t}$

That is, the compressive force in the sleeve is equal to the tensile force in the bolt. The problem is therefore statically indeterminate to the first degree.

Deformations: Using Equation 4.1, we write

$\delta = \frac {PL} {AE}$    (4.1)

$\delta_b=\frac{P_b L_b}{A_b E_b}, \quad \delta_t=\frac{P_t L_t}{A_t E_t}$          (c)

where

$\delta _{b}$ is the axial extension of the bolt

$\delta _{t}$ represents the axial contraction of the tube

Geometry: The deformations of the bolt and tube must be equal to Δ = 0.002/2 = 0.001 m, and the movement of the nut on the bolt must be

$\begin{array}{r} \delta_b+\delta_t=\Delta \\ \frac{P_b L_b}{A_b+E_b}+\frac{P_t L_t}{A_t E_t}=\Delta \end{array}$         (4.6)

Setting $P_{b} = P_{t}$ and $L_{b} = L_{t},$ the preceding equation becomes

$P_b\left(\frac{1}{A_b E_b}+\frac{1}{A_t E_t}\right)=\frac{\Delta}{L_t}$            (4.7)

Introducing the given data, we have

$P_b\left[\frac{1}{600(200) 10^3}+\frac{1}{300(70) 10^3}\right]=\frac{0.001}{0.6}$

Solving, $P_{b}$ = 29.8 kN.