A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1-8a. Axial loads are applied at the positions indicated. Determine the stress in each section.
Figure 1.8a Solution:
We must first determine the axial load in each section to calculate the stresses.
Solve for the internal axial load of the bronze
The free-body diagram of the bronze section
∑ F x = 0 → − 4000 lb + P b r = 0 P b r = 4000 lb (tension) \begin{align*}
\sum_{}^{}F_x & = 0 \to \\
-4000\ \text{lb}+P_{br} & = 0 \\
P_{br} & = 4000 \ \text{lb} \ \text{(tension)}
\end{align*} ∑ F x − 4000 lb + P b r P b r = 0 → = 0 = 4000 lb (tension)
Solve for the internal axial load of the aluminum
The free-body diagram of the aluminum section
∑ F x = 0 − 4000 lb + 9000 lb − P a l = 0 P a l = 5000 lb (Compression) \begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000 \ \text{lb} + 9000 \ \text{lb} - P_{al} & = 0 \\
P_{al} & = 5000 \ \text{lb} \ \text{(Compression)}
\end{align*} ∑ F x − 4000 lb + 9000 lb − P a l P a l = 0 = 0 = 5000 lb (Compression)
Solve for the internal axial load of the aluminum
The free-body diagram of the steel section
∑ F x = 0 − 4000 lb + 9000 lb + 2000 lb − P s t = 0 P s t = 7000 lb (Compression) \begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000\ \text{lb} + 9000 \ \text{lb} + 2000\ \text{lb} - P_{st} & =0 \\
P_{st} & = 7000 \ \text{lb} \ \text{(Compression)}
\end{align*} ∑ F x − 4000 lb + 9000 lb + 2000 lb − P s t P s t = 0 = 0 = 7000 lb (Compression) We can now solve the stresses in each section.
For the bronze
σ b r = P b r A b r = 4000 lb 1.2 in 2 = 3330 psi (Tension) ( Answer ) \begin{align*}
\sigma_{br} & = \frac{P_{br}}{A_{br}} \\
& = \frac{4000\ \text{lb}}{1.2 \ \text{in}^2} \\
& = 3330 \ \text{psi}\ \text{(Tension)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} σ b r = A b r P b r = 1.2 in 2 4000 lb = 3330 psi (Tension) ( Answer ) For the aluminum
σ a l = P b r A a l = 5000 lb 1.8 in 2 = 2780 psi (Compression) ( Answer ) \begin{align*}
\sigma_{al} & = \frac{P_{br}}{A_{al}} \\
& = \frac{5000\ \text{lb}}{1.8 \ \text{in}^2} \\
& = 2780 \ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} σ a l = A a l P b r = 1.8 in 2 5000 lb = 2780 psi (Compression) ( Answer ) For the steel
σ s t = P s t A s t = 7000 lb 1.6 in 2 = 4380 psi (Compression) ( Answer ) \begin{align*}
\sigma_{st} & = \frac{P_{st}}{A_{st}} \\
& = \frac{7000\ \text{lb}}{1.6 \ \text{in}^2} \\
& = 4380\ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*} σ s t = A s t P s t = 1.6 in 2 7000 lb = 4380 psi (Compression) ( Answer )
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