Strength of Materials Problem 101 – Stress in each section of a composite bar

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.

Strength of Materials by Andrew Pytel and Ferdinand Singer Problem 101
Figure 1.8a

Solution:

We must first determine the axial load in each section to calculate the stresses. The free-body diagrams have been drawn by isolating the portion of the bar lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered.

Solve for the internal axial load of the bronze

Free body diagram for the internal axial load of the bronze section for Problem 101 of Strength of Materials by Ferdinand Singer and Andrew Pytel
The free-body diagram of the bronze section
∑Fx=0→−4000 lb+Pbr=0Pbr=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*}

Solve for the internal axial load of the aluminum

Free-body diagram of the aluminum section for problem 101 of Strength of materials by Andrew Pytel and Ferdinand Singer
The free-body diagram of the aluminum section
∑Fx=0−4000 lb+9000 lb−Pal=0Pal=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*}

Solve for the internal axial load of the aluminum

The free-body diagram of the steel section
∑Fx=0−4000 lb+9000 lb+2000 lb−Pst=0Pst=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*}

We can now solve the stresses in each section.

For the bronze

σbr=PbrAbr=4000 lb1.2 in2=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*}

For the aluminum

σal=PbrAal=5000 lb1.8 in2=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*}

For the steel

σst=PstAst=7000 lb1.6 in2=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*}
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