Chapter 6.2, Problem 20P

A timber beam AB of Length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio τ_m/σ_m of the maximum values of the shearing and normal stresses in the beam is equal to 2h/L, where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L = 5 m, w = 8 kN/m, τ_m = 1.08 MPa, and σ_m = 12 MPa.

Fig. P6.20

To determine

To show that: The ratio $\frac{{\tau }_m}{{\sigma }_m}$ of the maximum values of the shearing and normal stresses in the beam is equal to $\frac{2h}{L}$.

Answer to Problem 20P

The ratio $\frac{{\tau }_m}{{\sigma }_m}$ of the maximum values of the shearing and normal stresses in the beam is equal to $\frac{2h}{L}$ is proved.

Explanation of Solution

Given information:

The length of the beam AB is L.

The depth of the beam is h.

Calculation:

Calculate the area of the cross section as shown below.

$A=bh$

Here, b is the width of the beam and h is the depth of the beam.

Calculate the section modulus of the cross section as shown below.

$S=\frac{1}{6}b{h}^2$

Due to the symmetry of the beam reaction at supports C and D are Equal.

$R_C=R_D=\frac{wL}{2}$

Calculate the shear force as shown below.

Shear force at A, $V_A=0$.

Shear force at A right, $V_{A,\text{right}}=-\frac{wL}{4}$

Shear force at C, $\begin{array}{c}{V}_C=\frac{wL}{2}-\frac{wL}{4}\\ =\frac{wL}{4}\end{array}$

Shear force at D, $\begin{array}{c}{V}_D=-\frac{3wL}{4}+\frac{wL}{2}\\ =-\frac{wL}{4}\end{array}$

Shear force at B, $\begin{array}{c}{V}_B=-wL+\frac{wL}{2}+\frac{wL}{2}\\ =0\end{array}$

Calculate the bending moment as shown below.

BM at A, $M_A=0$

BM at C, $\begin{array}{c}{M}_C=-\frac{wL}{4}\times \frac{L}{8}\\ =-\frac{w{L}^2}{32}\end{array}$

BM at D, $\begin{array}{c}{M}_D=-\frac{3wL}{4}\times \frac{3L}{8}+\frac{wL}{2}\times \frac{L}{2}\\ =-\frac{9w{L}^2}{32}+\frac{w{L}^2}{4}\\ =\left(\frac{-9+8}{32}\right)w{L}^2\\ =-\frac{w{L}^2}{32}\end{array}$

BM at B, $\begin{array}{c}{M}_B=-wL\times \frac{L}{2}+\frac{wL}{2}\times \frac{3L}{4}+\frac{wL}{2}\times \frac{L}{4}\\ =-\frac{w{L}^2}{2}+\frac{3w{L}^2}{8}+\frac{w{L}^2}{8}\\ =\left(\frac{-4+3+1}{8}\right)w{L}^2\\ =0\end{array}$

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

Calculate the maximum shear stress as shown below.

${\tau }_m=\frac{3}{2}\frac{V_{\max }}{A}$

Substitute $\frac{wL}{4}$ for $V_{\max }$ and $bh$ for A.

$\begin{array}{c}{\tau }_m=\frac{3}{2}\frac{\left(\frac{wL}{4}\right)}{bh}\\ =\frac{3wL}{8bh}\end{array}$ (1)

Calculate the maximum normal stress as shown below.

${\sigma }_m=\frac{M_{\max }}{S}$

Substitute $\frac{w{L}^2}{32}$ for $M_{\max }$ and $\frac{1}{6}b{h}^2$ for S.

$\begin{array}{c}{\sigma }_m=\frac{\frac{w{L}^2}{32}}{\frac{1}{6}b{h}^2}\\ =\frac{w{L}^2}{32}\times \frac{6}{b{h}^2}\\ =\frac{3w{L}^2}{16b{h}^2}\end{array}$

Calculate the ratio $\frac{{\tau }_m}{{\sigma }_m}$ of the maximum values of the shearing and normal stresses as shown below.

$\begin{array}{c}\frac{{\tau }_m}{{\sigma }_m}=\frac{\frac{3wL}{8bh}}{\frac{3w{L}^2}{16b{h}^2}}\\ =\frac{3wL}{8bh}\times \frac{16b{h}^2}{3w{L}^2}\\ =\frac{2h}{L}\end{array}$

Therefore, the ratio $\frac{{\tau }_m}{{\sigma }_m}$ of the maximum values of the shearing and normal stresses in the beam is equal to $\frac{2h}{L}$ is proved.

To determine

The depth and width of the beam.

Answer to Problem 20P

The depth of the beam is $\underset{\_}{225\text{mm}}$.

The width of the beam is $\underset{\_}{61.7\text{mm}}$.

Explanation of Solution

Given information:

The length (L) of the beam is $5\text{m}$.

The load is $8\text{kN}/\text{m}$.

The maximum shear stress is ${\tau }_m=1.08\text{MPa}$.

The maximum normal stress is ${\sigma }_m=12\text{MPa}$.

Calculation:

Refer to part (a).

$\frac{{\tau }_m}{{\sigma }_m}=\frac{2h}{L}$ (2)

Calculate the depth of the beam as shown below.

Substitute $5\text{m}$ for L, $1.08\text{MPa}$ for ${\tau }_m$, and $12\text{MPa}$ for ${\sigma }_m$ in Equation (2).

$\begin{array}{l}\frac{1.08}{12}=\frac{2\times h}{5}\\ 2h=0.45\\ h=225\times {10}^{-3}\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ h=225\text{mm}\end{array}$

Hence, the depth of the beam is $\underset{\_}{225\text{mm}}$.

Calculate the width of the beam as shown below.

Substitute $8\text{kN}/\text{m}$ for w, $1.08\text{MPa}$ for ${\tau }_m$, $5\text{m}$ for L and $225\text{mm}$ for h in Equation (1).

$\begin{array}{l}1.08\text{MPa}\times \frac{{10}^3\text{kN}/{\text{m}}^2}{1\text{MPa}}=\frac{3\times 8\text{kN}/\text{m}\times 5\text{m}}{8b\times 225\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}\\ 1944b=120\\ b=0.0617\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ b=61.7\text{mm}\end{array}$

Therefore, the width of the beam is $\underset{\_}{61.7\text{mm}}$.