Chapter 4.5, Problem 55P

4.55 and 4.56 Five metal strips, each 0.5 × 1.5-in. cross section, are bonded together to form the composite beam shown. The modulus of elasticity is 30 × 10⁶ psi for the steel, 15 × 10⁶ psi for the brass, and 10 × 10⁶ psi for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 12 kip∙in., determine (a) the maximum stress in each of the three metals, (b) the radius of curvature of the composite beam.

Fig. P4.55

To determine

Find the maximum stress in Aluminum, Brass, and Steel.

Answer to Problem 55P

The maximum stress in Aluminum  is $\underset{\_}{6.86\text{ksi}}$.

The maximum stress in Brass is $\underset{\_}{6.17\text{ksi}}$.

The maximum stress in Steel is $\underset{\_}{4.11\text{ksi}}$.

Explanation of Solution

Given information:

The dimension of each strip is $0.5\times 1.5-\text{in}\text{.}$

The modulus of elasticity of aluminum is $E_a=10\times {10}^6\text{psi}$.

The modulus of elasticity of brass is $E_b=15\times {10}^6\text{psi}$.

The modulus of elasticity of steel is $E_s=30\times {10}^6\text{psi}$.

The beam is bent about a horizontal axis by a couple of moment $M=12\text{kip}\cdot \text{in}\text{.}$

Calculation:

Consider aluminum as the reference material.

Calculate the modular ratio $\left(n\right)$ as shown below.

$n=\frac{E_{\text{base}}}{E_{\text{reference}}}$ (1)

For steel.

Substitute $30\times {10}^6\text{psi}$ for $E_s$ and $10\times {10}^6\text{psi}$ for $E_a$ in Equation (1).

$\begin{array}{c}{n}_s=\frac{30\times {10}^6}{10\times {10}^6}\\ =3\end{array}$

For brass.

Substitute $15\times {10}^6\text{psi}$ for $E_b$ and $10\times {10}^6\text{psi}$ for $E_a$ in Equation (1).

$\begin{array}{c}{n}_b=\frac{15\times {10}^6}{10\times {10}^6}\\ =1.5\end{array}$

For aluminum.

Substitute $10\times {10}^6\text{psi}$ for $E_a$ and $10\times {10}^6\text{psi}$ for $E_a$ in Equation (1).

$\begin{array}{c}{n}_a=\frac{10\times {10}^6}{10\times {10}^6}\\ =1\end{array}$

Sketch the transformed cross section as shown in Figure 1.

Refer to Figure 1.

The moment of inertia of the cross sections $I_1=I_5$ and $I_2=I_4$.

Calculate the moment of inertia for each section as shown below.

For section (1).

$I_1=I_5=n_a\left(\frac{b_1{h}_1^3}{12}+b_1{h}_1{d}_1^2\right)$

Substitute $1$ for $n_a$, $1.5\text{in}\text{.}$ for $b_1$, $0.5\text{in}\text{.}$ for $h_1$, and $\left(1.25-\frac{0.5}{2}\right)\text{in}\text{.}$ for $d_1$.

$\begin{array}{c}{I}_1=1\times \left(\frac{1.5\times {0.5}^3}{12}+1.5\times 0.5{\left(1.25-\frac{0.5}{2}\right)}^2\right)\\ =0.7656\text{in}{\text{.}}^4\end{array}$

Hence, $I_5=0.7656\text{in}{\text{.}}^4$.

For section (2).

$I_2=I_4=n_b\left(\frac{b_2{h}_2^3}{12}+b_2{h}_2{d}_2^2\right)$

Substitute $1.5$ for $n_b$, $1.5\text{in}\text{.}$ for $b_2$, $0.5\text{in}\text{.}$ for $h_2$, and $\left(0.75-\frac{0.5}{2}\right)\text{in}\text{.}$ for $d_1$.

$\begin{array}{c}{I}_2=1.5\times \left(\frac{1.5\times {0.5}^3}{12}+1.5\times 0.5{\left(0.75-\frac{0.5}{2}\right)}^2\right)\\ =0.3047\text{in}{\text{.}}^4\end{array}$

Hence, $I_4=0.3047\text{in}{\text{.}}^4$

For section (3).

$I_3=n_s\frac{b_3{h}_3^3}{12}$

Substitute $3$ for $n_s$, $1.5\text{in}\text{.}$ for $b_3$, and $0.5\text{in}\text{.}$ for $h_3$.

$\begin{array}{c}{I}_3=3\times \frac{1.5\times {0.5}^3}{12}\\ =0.0469\text{in}{\text{.}}^4\end{array}$

Calculate the moment of inertia $\left(I\right)$ as shown below.

$I=I_1+I_2+I_3+I_4+I_5$

Substitute $0.7656\text{in}{\text{.}}^4$ for $I_1$, $0.3047\text{in}{\text{.}}^4$ for $I_2$, $0.0469\text{in}{\text{.}}^4$ for $I_3$, $0.3047\text{in}{\text{.}}^4$ for $I_4$, and $0.7656\text{in}{\text{.}}^4$ for $I_5$.

$\begin{array}{c}I=0.7656+0.3047+0.0469+0.3047+0.7656\\ =2.1875\text{in}{\text{.}}^4\end{array}$

Calculate the maximum stress $\left(\sigma \right)$ as shown below.

$\begin{array}{l}\left|\sigma \right|=\left|-\frac{Mny}{I}\right|\\ \sigma =\frac{nMy}{I}\end{array}$ (2)

For steel.

Substitute $12\text{kip}\cdot \text{in}\text{.}$ for M, $2.1875\text{in}{\text{.}}^4$ for I, $0.25\text{in}\text{.}$ for y, and $3$ for n in equation (2).

$\begin{array}{c}\sigma =\frac{3\times 12\times 0.25}{2.1875}\\ =4.11\text{ksi}\end{array}$

Hence, maximum stress in steel is $\underset{\_}{4.11\text{ksi}}$.

For brass.

Substitute $12\text{kip}\cdot \text{in}\text{.}$ for M, $2.1875\text{in}{\text{.}}^4$ for I, $0.75\text{in}\text{.}$ for y, and $1.5$ for n in equation (2).

$\begin{array}{c}\sigma =\frac{1.5\times 12\times 0.75}{2.1875}\\ =6.17\text{ksi}\end{array}$

Hence, maximum stress in brass is $\underset{\_}{6.17\text{ksi}}$.

For aluminum.

Substitute $12\text{kip}\cdot \text{in}\text{.}$ for M, $2.1875\text{in}{\text{.}}^4$ for I, $1.25\text{in}\text{.}$ for y, and $1$ for n in equation (2).

$\begin{array}{c}\sigma =\frac{1\times 12\times 1.25}{2.1875}\\ =6.86\text{ksi}\end{array}$

Therefore, maximum stress in aluminum is $\underset{\_}{6.86\text{ksi}}$.

To determine

The radius of curvature of the composite beam.

Answer to Problem 55P

The radius of curvature of the composite beam is $\underset{\_}{151.9\text{ft}}$.

Explanation of Solution

Given information:

The dimension of each strip is $0.5\times 1.5-\text{in}\text{.}$

The modulus of elasticity of aluminum is $E_a=10\times {10}^6\text{psi}$.

The modulus of elasticity of brass is $E_b=15\times {10}^6\text{psi}$.

The modulus of elasticity of steel is $E_s=30\times {10}^6\text{psi}$.

The beam is bent about a horizontal axis by a couple of moment $M=12\text{kip}\cdot \text{in}\text{.}$

Calculation:

Refer to part (a).

The moment of inertia of the beam is $I=2.1875\text{in}{\text{.}}^4$.

Calculate the radius of curvature $\left(\rho \right)$ as shown below.

$\frac{1}{\rho }=\frac{M}{E_{a}I}$

Substitute $12\text{kip}\cdot \text{in}\text{.}$ for M, $2.1875\text{in}{\text{.}}^4$ for I, and $10\times {10}^6\text{psi}$ for $E_a$.

$\begin{array}{c}\frac{1}{\rho }=\frac{12\text{kip}\cdot \text{in}\text{.}\times \frac{1,000\text{lb}}{1\text{kip}}}{10\times {10}^6\text{psi}\times \text{2}\text{.1875}\text{in}{\text{.}}^4}\\ \rho =\frac{1}{5.4857\times {10}^{-4}}\\ =1,823\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =151.9\text{ft}\end{array}$

Therefore, the radius of curvature of the composite beam is $\underset{\_}{151.9\text{ft}}$.