Chapter 11.3, Problem 40P

To determine

Find the strain energy of the prismatic beam AB.

Answer to Problem 40P

The strain energy of the prismatic beam AB is $\underset{\_}{U=\frac{2M_0^{2}L}{Eb{d}^3}\left\{1+\frac{3E{d}^2}{10G{L}^2}\right\}}$.

Explanation of Solution

Given information:

Taking into account the effect of both normal and shearing stresses.

Calculation:

Calculate the moment of inertia (I) for the rectangular cross section as shown below.

$I=\frac{b{d}^3}{12}$

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the cross section (A) as shown below.

$A=bd$

Calculate the centroid (c) as shown below.

$c=\frac{d}{2}$

Calculate the reactions as shown below.

Take moment about B is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ R_{A}L+M_0=0\\ R_{A}L=-M_0\\ R_A=-\frac{M_0}{L}\end{array}$

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_B-\frac{M_0}{L}=0\\ R_B=\frac{M_0}{L}\end{array}$

Calculate the shear force as shown below.

Shear force at A is $S{F}_{@A}=-\frac{M_0}{L}$.

Shear force at B is $\begin{array}{c}S{F}_{@B}=-\frac{M_0}{L}+\frac{M_0}{L}\\ =0\end{array}$

Calculate the bending moment as shown below.

Bending moment at A is $B{M}_{@A}=M_0$.

Bending moment at B is $\begin{array}{c}B{M}_{@B}=M_0-\frac{M_0}{L}\times L\\ =0\end{array}$

Sketch the shear force and bending moment diagram as shown in Figure 1.

Refer to Figure 1.

Maximum shear force $V=-\frac{M_0}{L}$.

Maximum bending moment $M_{\max }=M_0$.

Bending moment at a distance $v$ from B $M=\frac{M_0}{L}v$

Calculate the strain energy due to bending $\left(U_1\right)$ as shown below.

$U_1={\displaystyle {\int }_a^b\frac{M^2}{2EI}dv}$

Substitute $M=\frac{M_0}{L}v$ for M and apply the limits.

$\begin{array}{c}{U}_1={\displaystyle {\int }_0^L\frac{{\left(\frac{M_0}{L}v\right)}^2}{2EI}dv}\\ =\frac{1}{2EI}{\displaystyle {\int }_0^L\frac{M_0^2{v}^2}{L^2}dv}\\ =\frac{M_0^2}{2EI{L}^2}{\left(\frac{v^3}{3}\right)}_0^L\\ =\frac{M_0^2}{6EI{L}^2}\times L^3\end{array}$

$=\frac{M_0^{2}L}{6EI}$

Substitute $\frac{b{d}^3}{12}$ for I.

$\begin{array}{c}{U}_1=\frac{M_0^{2}L}{6E\left(\frac{b{d}^3}{12}\right)}\\ =\frac{2M_0^{2}L}{b{d}^{3}E}\end{array}$

Calculate the shear stress $\left({\tau }_{xy}\right)$ as shown below.

${\tau }_{xy}=\frac{3}{2}\frac{V}{A}\left(1-\frac{y^2}{c^2}\right)$

Calculate the strain energy density $\left(u\right)$ as shown below.

$u=\frac{{\tau }_{xy}^2}{2G}$

Here, G is the modulus of rigidity.

Substitute $\frac{3}{2}\frac{V}{A}\left(1-\frac{y^2}{c^2}\right)$ for ${\tau }_{xy}$.

$\begin{array}{c}u=\frac{{\left(\frac{3}{2}\frac{V}{A}\left(1-\frac{y^2}{c^2}\right)\right)}^2}{2G}\\ =\frac{9V^2}{8G{A}^2}\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)\end{array}$

Substitute $bd$ for A.

$\begin{array}{c}u=\frac{9V^2}{8G{\left(bd\right)}^2}\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)\\ =\frac{9V^2}{8G{b}^2{d}^2}\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)\end{array}$

The value of $v=bdy$.

Differentiate both sides of the Equation as shown below.

$dv=bdydx$

Calculate the strain energy due to shear as shown below.

$U_2={\displaystyle \int udv}$

Substitute $\frac{9V^2}{8G{b}^2{d}^2}\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)$ for u, $bdydx$ for $dv$, and apply the limits.

$\begin{array}{c}{U}_2={\displaystyle \underset{0}{\overset{L}{\int }}{\displaystyle {\int }_{-c}^c\frac{9V^2}{8G{b}^2{d}^2}\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)bdydx}}\\ =\frac{9V^2}{8Gb{d}^2}{\displaystyle \underset{0}{\overset{L}{\int }}{\displaystyle {\int }_{-c}^c\left(1+\frac{y^4}{c^4}-2\frac{y^2}{c^2}\right)dydx}}\\ =\frac{9V^2}{8Gb{d}^2}{\displaystyle \underset{0}{\overset{L}{\int }}{\left(y+\frac{y^5}{5c^4}-2\frac{y^3}{3c^2}\right)}_{-c}^{c}dx}\\ =\frac{9V^2}{8Gb{d}^2}\left\{\left(c+\frac{c^5}{5c^4}-2\frac{c^3}{3c^2}\right)-\left(\left(-c\right)+\frac{{\left(-c\right)}^5}{5c^4}-2\frac{{\left(-c\right)}^3}{3c^2}\right)\right\}{\left(x\right)}_0^L\end{array}$

$\begin{array}{c}=\frac{9V^2}{8Gb{d}^2}\left\{c+\frac{c}{5}-\frac{2c}{3}+c+\frac{c}{5}-\frac{2c}{3}\right\}L\\ =\frac{9V^{2}cL}{8Gb{d}^2}\times \frac{\left(15+3-10+15+3-10\right)}{15}\\ =\frac{9V^{2}cL}{8Gb{d}^2}\times \frac{16}{15}\\ =\frac{6V^{2}cL}{5Gb{d}^2}\end{array}$

Substitute $-\frac{M_0}{L}$ for V and $\frac{d}{2}$ for c.

$\begin{array}{c}{U}_2=\frac{6{\left(-\frac{M_0}{L}\right)}^2\left(\frac{d}{2}\right)L}{5Gb{d}^2}\\ =\frac{3M_0^2}{5GbdL}\end{array}$

Calculate the total strain energy as shown below.

$U=U_1+U_2$

Substitute $\frac{2M_0^{2}L}{b{d}^{3}E}$ for $U_1$ and $\frac{3M_0^2}{5GbdL}$ for $U_2$.

$\begin{array}{c}U=\frac{2M_0^{2}L}{b{d}^{3}E}+\frac{3M_0^2}{5GbdL}\\ =\frac{2M_0^{2}L}{Eb{d}^3}\left(1+\frac{3E{d}^2}{10G{L}^2}\right)\end{array}$

Hence, the strain energy of the prismatic beam AB is $\underset{\_}{U=\frac{2M_0^{2}L}{Eb{d}^3}\left\{1+\frac{3E{d}^2}{10G{L}^2}\right\}}$.