Question 4.3: Thermal Stresses in a Bolt–Tube Assembly Determine the axial...
Thermal Stresses in a Bolt–Tube Assembly
Determine the axial forces in the assembly of bolt and tube (Figure 4.2a), after a temperature rise of ΔT.
Given: ΔT = 100°C, \alpha _{b} = 11.7 × 10^{–6}/°C, and \alpha _{t} = 23.2 × 10^{–6}/°C
Assumptions: The data presented in the preceding example remain the same.
Only force–deformation relations, Equation (c), change from Example 4.2. Now the expressions for the extension of the bolt and the contraction of the sleeve are
\delta_b=\frac{P_b L_b}{A_b E_b}, \quad \delta_t=\frac{P_t L_t}{A_t E_t} (c)
\begin{array}{l} \delta_b=\frac{P_b L_b}{A_b E_b}+\alpha_b(\Delta T) L_b \\ \delta_t=\frac{P_t L_t}{A_t E_t}-\alpha_t(\Delta T) L_t \end{array} (d)
Note that, in the foregoing, the minus sign indicates a decrease in tube contraction due to the temperature rise.
We have L_{b} = L_{t} and P_{b} = P_{t}. These, carried into \delta _{b} + \delta _{t} = \Delta, give
P_b\left(\frac{1}{A_b E_b}+\frac{1}{A_t E_t}\right)+\left(\alpha_b-\alpha_t\right) \Delta T=\frac{\Delta}{L_t} (4.8)
where, as before, Δ is the movement of the nut on the bolt. Substituting the numerical values into Equation 4.8, we obtain
P_b\left[\frac{1}{600(200) 10^3}+\frac{1}{300(70) 10^3}\right]+(11.7-23.2) 10^{-6}(100)=\frac{0.001}{0.6}
This yields P_{b} = 50.3 kN.
Comment: The final elongation of the bolt and the contraction of the tube can be calculated by substituting the axial force of 50.3 kN into Equation (d). Interestingly, when the bolt and tube are made of the same material (\alpha _{b} = \alpha _{t}), the temperature change does not affect the assembly. That is, the forces obtained in Example 4.2 still hold.