Question 4.2: Analysis of Bolt–Tube Assembly In the assembly of the alumin...
Analysis of Bolt–Tube Assembly
In the assembly of the aluminum tube (cross sectional area A_{t}, modulus of elasticity E_{t}, length L_{t}) and steel bolt (cross-sectional area A_{b}, modulus of elasticity E_{b}) shown in Figure 4.2a, the bolt is single threaded, with a 2 mm pitch. If the nut is tightened one-half turn after it has been fitted snugly, calculate the axial forces in the bolt and tubular sleeve.
Given: A_{t} = 300 mm², E_{t} = 70 GPa, L_{t} = 0.6 m, A_{b} = 600 mm², and E_{b} = 200 GPa
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.