Chapter 4.7, Problem 120P

a)

To determine

Show that the maximum compressive stresses are in the ratio 4:5:7:9.

Explanation of Solution

Given information:

The load act on the point of the bars is P.

Calculation:

At the point A:

Show the cross-sectional diagram of the square bar as in Figure 1.

Here, $a$ is sides of the square bar.

Refer to Figure 1.

The maximum compressive stress of the square bar ${\left({\sigma }_A\right)}_s$ is as follows:

$\begin{array}{c}{\left({\sigma }_A\right)}_s=-\frac{P}{A_s}-\frac{Pe{c}_A}{I_s}\\ =-\frac{P}{A_s}\left(1+\frac{A_{s}e{c}_A}{I_s}\right)\end{array}$ (1)

Here, e is the eccentricity of the load and $c_A$ is the distance between the centroid from extreme fibre.

The cross-sectional area of the square bar $\left(A_s\right)$ is $a^2$.

The eccentricity of the load (e) is $\frac{a}{2}$.

The distance between the centroid from extreme fibre $\left(c_A\right)$ is $\frac{a}{2}$.

The moment of inertia $\left(I_s\right)$ of the section of the square bar is $\frac{a^4}{12}$.

Calculate the maximum compressive stress of the square bar ${\left({\sigma }_A\right)}_s$:

Substitute $a^2$ for $A_s$, $\frac{a}{2}$ for e, $\frac{a}{2}$ for $c_A$, and $\frac{a^4}{12}$ for $I_s$ in Equation (1).

$\begin{array}{c}{\left({\sigma }_A\right)}_s=-\frac{P}{A_s}\left(1+\frac{a^2\times \frac{a}{2}\times \frac{a}{2}}{\frac{a^4}{12}}\right)\\ =-\frac{P}{A_s}-\frac{12a^{4}P}{4a^4{A}_s}\\ =-\frac{P}{A_s}-\frac{3P}{A_s}\\ =-\frac{4P}{A_s}\end{array}$

Show the cross-sectional diagram of the circular bar as in Figure 2.

Here, $c$ is radius of the circular bar.

Refer to Figure 2.

The maximum compressive stress of the circular bar ${\left({\sigma }_A\right)}_c$ is as follows:

$\begin{array}{c}{\left({\sigma }_A\right)}_c=-\frac{P}{A_c}-\frac{Pe{c}_A}{I_c}\\ =-\frac{P}{A_c}\left(1+\frac{A_{c}e{c}_A}{I_c}\right)\end{array}$ (2)

The cross-sectional area of the circular bar $\left(A_c\right)$ is as follow:

$\begin{array}{c}{A}_c=\pi c^2\\ =a^2\end{array}$

The eccentricity of the load (e) is $\frac{a}{\sqrt{\pi }}$.

The distance between the centroid from extreme fibre $\left(c_A\right)$ is $\frac{a}{\sqrt{\pi }}$.

The moment of inertia $\left(I_c\right)$ of the section of the circular bar is as follows:

$\begin{array}{c}{I}_c=\frac{\pi }{4}{c}^4\\ =\frac{\pi }{4}\times \left(\frac{a^4}{\sqrt{\pi }}\right)\\ =\frac{\pi }{4}\times \frac{a^4}{{\pi }^2}\\ =\frac{a^4}{4{\pi }^2}\end{array}$

Calculate the maximum compressive stress of the circular bar ${\left({\sigma }_A\right)}_c$:

Substitute $a^2$ for $A_c$, $\frac{a}{\sqrt{\pi }}$ for e, $\frac{a}{\sqrt{\pi }}$ for $c_A$, and $\frac{a^4}{4{\pi }^2}$ for $I_c$ in Equation (2).

$\begin{array}{c}{\left({\sigma }_A\right)}_c=-\frac{P}{A_c}\left(1+\frac{a^2\times \frac{a}{\sqrt{\pi }}\times \frac{a}{\sqrt{\pi }}}{\frac{a^4}{4{\pi }^2}}\right)\\ =-\frac{P}{A_c}-\frac{4{\pi }^2{a}^{4}P}{a^4{\pi }^2{A}_c}\\ =-\frac{P}{A_c}-\frac{4P}{A_c}\\ =-\frac{5P}{A_c}\end{array}$

Show the cross-sectional diagram of the diamond shape bar as in Figure 3.

Here, $a$ is side of the diamond shape bar.

Refer to Figure 3.

The maximum compressive stress of the diamond shape bar ${\left({\sigma }_A\right)}_d$ is as follows:

$\begin{array}{c}{\left({\sigma }_A\right)}_d=-\frac{P}{A_d}-\frac{Pe{c}_A}{I_d}\\ =-\frac{P}{A_d}\left(1+\frac{A_{d}e{c}_A}{I_d}\right)\end{array}$ (3)

The cross-sectional area of the diamond shape bar $\left(A_d\right)$ is as follow:

$A_d=a^2$

The eccentricity of the load (e) is $\frac{\sqrt{2}a}{2}$.

The distance between the centroid from extreme fibre $\left(c_A\right)$ is $\frac{\sqrt{2}a}{2}$.

The moment of inertia $\left(I_d\right)$ of the section of the diamond shape bar is as follows:

$I_d=\frac{a^4}{12}$

Calculate the maximum compressive stress of the diamond shape bar ${\left({\sigma }_A\right)}_d$:

Substitute $a^2$ for $A_d$, $\frac{\sqrt{2}a}{2}$ for e, $\frac{\sqrt{2}a}{2}$ for $c_A$, and $\frac{a^4}{12}$ for $I_d$ in Equation (3).

$\begin{array}{c}{\left({\sigma }_A\right)}_d=-\frac{P}{A_d}\left(1+\frac{a^2\times \frac{\sqrt{2}a}{2}\times \frac{\sqrt{2}a}{2}}{\frac{a^4}{12}}\right)\\ =-\frac{P}{A_d}-\frac{12\times 2a^{4}P}{4a^4{A}_d}\\ =-\frac{P}{A_d}-\frac{6P}{A_d}\\ =-\frac{7P}{A_d}\end{array}$

Show the cross-sectional diagram of the triangular bar as in Figure 4.

Here, $s$ is side of the triangular bar.

Refer to Figure 4.

The maximum compressive stress of the triangular bar ${\left({\sigma }_A\right)}_t$ is as follows:

$\begin{array}{c}{\left({\sigma }_A\right)}_t=-\frac{P}{A_t}-\frac{Pe{c}_A}{I_t}\\ =-\frac{P}{A_t}\left(1+\frac{A_{t}e{c}_A}{I_t}\right)\end{array}$ (4)

The cross-sectional area of the triangular bar $\left(A_t\right)$ is as follow:

$\begin{array}{c}{A}_t=\frac{1}{2}\times s\times \frac{\sqrt{3}s}{2}\\ =\frac{\sqrt{3}{s}^2}{4}\end{array}$

The distance between the centroid from extreme fibre $\left(c_A\right)$ is as follows:

$\begin{array}{c}{c}_A=\frac{2}{3}\times \frac{\sqrt{3}s}{2}\\ =\frac{\sqrt{3}s}{3}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{3s}{3\sqrt{3}}\\ =\frac{s}{\sqrt{3}}\end{array}$

The eccentricity of the load (e) is $\frac{s}{\sqrt{3}}$.

The moment of inertia $\left(I_t\right)$ of the section of the triangular bar is as follows:

$\begin{array}{c}{I}_t=\frac{1}{36}\times s\times {\left(\frac{\sqrt{3}s}{2}\right)}^3\\ =\frac{s}{36}\times \frac{3\sqrt{3}{s}^3}{8}\\ =\frac{\sqrt{3}{s}^4}{96}\end{array}$

Calculate the maximum compressive stress of the triangular bar ${\left({\sigma }_A\right)}_t$:

Substitute $\frac{\sqrt{3}{s}^2}{4}$ for $A_t$, $\frac{s}{\sqrt{3}}$ for e, $\frac{s}{\sqrt{3}}$ for $c_A$, and $\frac{\sqrt{3}{s}^4}{96}$ for $I_t$ in Equation (4).

$\begin{array}{c}{\left({\sigma }_A\right)}_t=-\frac{P}{A_t}\left(1+\frac{\frac{\sqrt{3}{s}^2}{4}\times \frac{s}{\sqrt{3}}\times \frac{s}{\sqrt{3}}}{\frac{\sqrt{3}{s}^4}{96}}\right)\\ =-\frac{P}{A_t}-\frac{96s^{4}P}{3\times 4s^4{A}_t}\\ =-\frac{P}{A_t}-\frac{8P}{A_t}\\ =-\frac{9P}{A_t}\end{array}$

Calculate the maximum compressive stresses are in the ratio:

$A={\left({\sigma }_A\right)}_s:{\left({\sigma }_A\right)}_c:{\left({\sigma }_A\right)}_d:{\left({\sigma }_A\right)}_t$

Substitute $\left(-\frac{4P}{A_s}\right)$ for ${\left({\sigma }_A\right)}_s$, $\left(-\frac{5P}{A_c}\right)$ for ${\left({\sigma }_A\right)}_c$, $\left(-\frac{7P}{A_d}\right)$ for ${\left({\sigma }_A\right)}_d$, and $\left(-\frac{9P}{A_t}\right)$ for ${\left({\sigma }_A\right)}_t$.

$\begin{array}{c}A=-\frac{4P}{A_s}:-\frac{5P}{A_c}:-\frac{7P}{A_d}:-\frac{9P}{A_t}\\ =\frac{4}{A_s}:\frac{5}{A_c}:\frac{7}{A_d}:\frac{9}{A_t}\end{array}$

The four bars shown have the same cross-sectional area.

$A=4:5:7:9$

Hence the maximum compressive stresses are in the ratio 4:5:7:9 is proved.

b)

To determine

Show that the maximum tensile stresses are in the ratio 2:3:5:3.

Explanation of Solution

Given information:

The load act on the point of the bars is P.

Calculation:

At the point B:

Refer to Figure 1.

The maximum tensile stress of the square bar ${\left({\sigma }_B\right)}_s$ is as follows:

$\begin{array}{c}{\left({\sigma }_B\right)}_s=-\frac{P}{A_s}+\frac{Pe{c}_B}{I_s}\\ =\frac{P}{A_s}\left(\frac{A_{s}e{c}_B}{I_s}-1\right)\end{array}$ (5)

Here, the e is the eccentricity of the load and $c_B$ is the distance between the centroid from extreme fibre.

The cross-sectional area of the square bar $\left(A_s\right)$ is $a^2$.

The eccentricity of the load (e) is $\frac{a}{2}$.

The distance between the centroid from extreme fibre $\left(c_B\right)$ is $\frac{a}{2}$.

The moment of inertia $\left(I_s\right)$ of the section of the square bar is $\frac{a^4}{12}$.

Calculate the maximum tensile stress of the square bar ${\left({\sigma }_B\right)}_s$:

Substitute $a^2$ for $A_s$, $\frac{a}{2}$ for e, $\frac{a}{2}$ for $c_B$, and $\frac{a^4}{12}$ for $I_s$ in Equation (5).

$\begin{array}{c}{\left({\sigma }_B\right)}_s=\frac{P}{A_s}\left(\frac{a^2\times \frac{a}{2}\times \frac{a}{2}}{\frac{a^4}{12}}-1\right)\\ =\frac{12a^{4}P}{4a^4{A}_s}-\frac{P}{A_s}\\ =\frac{3P}{A_s}-\frac{P}{A_s}\\ =\frac{2P}{A_s}\end{array}$

Refer to Figure 2.

The maximum tensile stress of the circular bar ${\left({\sigma }_B\right)}_c$ is as follows:

$\begin{array}{c}{\left({\sigma }_B\right)}_c=-\frac{P}{A_c}+\frac{Pe{c}_B}{I_c}\\ =\frac{P}{A_c}\left(\frac{A_{c}e{c}_B}{I_c}-1\right)\end{array}$ (6)

The cross-sectional area of the circular bar $\left(A_c\right)$ is as follow:

$\begin{array}{c}{A}_c=\pi c^2\\ =a^2\end{array}$

The eccentricity of the load (e) is $\frac{a}{\sqrt{\pi }}$.

The distance between the centroid from extreme fibre $\left(c_B\right)$ is $\frac{a}{\sqrt{\pi }}$.

The moment of inertia $\left(I_c\right)$ of the section of the circular bar is as follows:

$\begin{array}{c}{I}_c=\frac{\pi }{4}{c}^4\\ =\frac{\pi }{4}\times \left(\frac{a^4}{\sqrt{\pi }}\right)\\ =\frac{\pi }{4}\times \frac{a^4}{{\pi }^2}\\ =\frac{a^4}{4{\pi }^2}\end{array}$

Calculate the maximum tensile stress of the circular bar ${\left({\sigma }_B\right)}_c$:

Substitute $a^2$ for $A_c$, $\frac{a}{\sqrt{\pi }}$ for e, $\frac{a}{\sqrt{\pi }}$ for $c_B$, and $\frac{a^4}{4{\pi }^2}$ for $I_c$ in Equation (6).

$\begin{array}{c}{\left({\sigma }_B\right)}_c=\frac{P}{A_c}\left(\frac{a^2\times \frac{a}{\sqrt{\pi }}\times \frac{a}{\sqrt{\pi }}}{\frac{a^4}{4{\pi }^2}}-1\right)\\ =\frac{4{\pi }^2{a}^{4}P}{a^4{\pi }^2{A}_c}-\frac{P}{A_c}\\ =\frac{4P}{A_c}-\frac{P}{A_c}\\ =\frac{3P}{A_c}\end{array}$

Refer to Figure 3.

The maximum tensile stress of the diamond shape bar ${\left({\sigma }_B\right)}_d$ is as follows:

$\begin{array}{c}{\left({\sigma }_B\right)}_d=-\frac{P}{A_d}+\frac{Pe{c}_B}{I_d}\\ =\frac{P}{A_d}\left(\frac{A_{d}e{c}_B}{I_d}-1\right)\end{array}$ (7)

The cross-sectional area of the diamond shape bar $\left(A_d\right)$ is as follow:

$A_d=a^2$

The eccentricity of the load (e) is $\frac{\sqrt{2}a}{2}$.

The distance between the centroid from extreme fibre $\left(c_B\right)$ is $\frac{\sqrt{2}a}{2}$.

The moment of inertia $\left(I_d\right)$ of the section of the diamond shape bar is as follows:

$I_d=\frac{a^4}{12}$

Calculate the maximum tensile stress of the diamond shape bar ${\left({\sigma }_B\right)}_d$:

Substitute $a^2$ for $A_d$, $\frac{\sqrt{2}a}{2}$ for e, $\frac{\sqrt{2}a}{2}$ for $c_B$, and $\frac{a^4}{12}$ for $I_d$ in Equation (7).

$\begin{array}{c}{\left({\sigma }_B\right)}_d=\left(\frac{a^2\times \frac{\sqrt{2}a}{2}\times \frac{\sqrt{2}a}{2}}{\frac{a^4}{12}}-1\right)\\ =\frac{12\times 2a^{4}P}{4a^4{A}_d}-\frac{P}{A_d}\\ =\frac{6P}{A_d}-\frac{P}{A_d}\\ =\frac{5P}{A_d}\end{array}$

Refer to Figure 4.

The maximum tensile stress of the triangular bar ${\left({\sigma }_B\right)}_t$ is as follows:

$\begin{array}{c}{\left({\sigma }_B\right)}_t=-\frac{P}{A_t}+\frac{Pe{c}_B}{I_t}\\ =\frac{P}{A_t}\left(\frac{A_{t}e{c}_B}{I_t}-1\right)\end{array}$ (8)

The cross-sectional area of the triangular bar $\left(A_t\right)$ is as follows:

$\begin{array}{c}{A}_t=\frac{1}{2}\times s\times \frac{\sqrt{3}s}{2}\\ =\frac{\sqrt{3}{s}^2}{4}\end{array}$

The distance between the centroid from extreme fibre $\left(c_B\right)$ is as follows:

$\begin{array}{c}{c}_B=\frac{1}{3}\times \frac{\sqrt{3}s}{2}\\ =\frac{\sqrt{3}s}{3\times 2}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{3s}{3\times 2\sqrt{3}}\\ =\frac{s}{2\sqrt{3}}\end{array}$

The eccentricity of the load (e) is $\frac{s}{\sqrt{3}}$.

The moment of inertia $\left(I_t\right)$ of the section of the triangular bar is as follows:

$\begin{array}{c}{I}_t=\frac{1}{36}\times s\times {\left(\frac{\sqrt{3}s}{2}\right)}^3\\ =\frac{s}{36}\times \frac{3\sqrt{3}{s}^3}{8}\\ =\frac{\sqrt{3}{s}^4}{96}\end{array}$

Calculate the maximum tensile stress of the triangular bar ${\left({\sigma }_B\right)}_t$:

Substitute $\frac{\sqrt{3}{s}^2}{4}$ for $A_t$, $\frac{s}{\sqrt{3}}$ for e, $\frac{s}{2\sqrt{3}}$ for $c_B$, and $\frac{\sqrt{3}{s}^4}{96}$ for $I_t$ in Equation (8).

$\begin{array}{c}{\left({\sigma }_B\right)}_t=\frac{P}{A_t}\left(\frac{\frac{\sqrt{3}{s}^2}{4}\times \frac{s}{\sqrt{3}}\times \frac{s}{2\sqrt{3}}}{\frac{\sqrt{3}{s}^4}{96}}-1\right)\\ =\frac{96s^{4}P}{3\times 4\times 2s^4{A}_t}-\frac{P}{A_t}\\ =\frac{4P}{A_t}-\frac{P}{A_t}\\ =\frac{3P}{A_t}\end{array}$

Calculate the maximum tensile stresses are in the ratio:

$A={\left({\sigma }_A\right)}_s:{\left({\sigma }_A\right)}_c:{\left({\sigma }_A\right)}_d:{\left({\sigma }_A\right)}_t$

Substitute $\left(\frac{2P}{A_s}\right)$ for ${\left({\sigma }_A\right)}_s$, $\left(\frac{3P}{A_c}\right)$ for ${\left({\sigma }_A\right)}_c$, $\left(\frac{5P}{A_d}\right)$ for ${\left({\sigma }_A\right)}_d$, and $\left(\frac{3P}{A_t}\right)$ for ${\left({\sigma }_A\right)}_t$.

$\begin{array}{c}B=\frac{2P}{A_s}:\frac{3P}{A_c}:\frac{5P}{A_d}:\frac{3P}{A_t}\\ =\frac{2}{A_s}:\frac{3}{A_c}:\frac{5}{A_d}:\frac{3}{A_t}\end{array}$

The four bars shown have the same cross-sectional area.

$B=2:3:5:3$

Hence the maximum tensile stresses are in the ratio 2:3:5:3 is proved.