Chapter 15.2, Problem 1FP

Determine the minimum dimension a to the nearest mm of the beam’s cross section to safely support the load. The wood has an allowable normal stress of σ_allow = 10 MPa and an allowable shear stress of τ_allow = 1 MPa.

To determine

Find the minimum dimension a to the nearest beam cross section.

Answer to Problem 1FP

The minimum dimension a to the nearest beam cross section is $\underset{\_}{139.2\text{mm}}$.

Explanation of Solution

Given information:

The allowable bending stress is 10 MPa.

The allowable shear stress is 1 MPa.

The width of the cross section is a.

The depth of the cross section is 2a.

Calculation:

Sketch the free body diagram of beam as shown in Figure 1.

Consider a section, $V_A$ is the shear force and load acting downward direction.

Apply the equilibrium condition along y –direction.

$\sum F_y=0$

$\begin{array}{c}{V}_A=6+6\\ =12\text{kN}\end{array}$

Determine the bending moment at point A.

$\begin{array}{c}{M}_A=\left(6\times 2\right)+\left(6\times 1\right)\\ =12+6\\ =18\text{kN}\cdot \text{m}\end{array}$

Calculate the moment of inertia (I) using the relation.

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

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

Substitute a for b and 2a for d.

$\begin{array}{c}I=\frac{a\times 2a^3}{12}\\ =\frac{2}{3}{a}^4\end{array}$

Calculate the dimension (a) value using the relation:

${\sigma }_{\text{allow}}=\frac{M_{\max }c}{I}$

Here, $M_{\max }$ is the maximum moment, I is the moment of inertia, and c is the centroid from the neutral axis.

Substitute $\frac{2}{3}{a}^4$ for I, $18\text{kN}\cdot \text{m}$ for $M_{\max }$, 10 MPa for ${\sigma }_{\text{allow}}$, and a for c.

$\begin{array}{l}10\times {10}^6=\frac{18\times {10}^3\left(a\right)}{\frac{2}{3}{a}^4}\\ a=0.1392\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =139.2\text{mm}\end{array}$

Use, $a=140\text{mm}$

Hence, the minimum dimension a to the nearest beam cross section is $\underset{\_}{139.2\text{mm}}$.

Find the value of I using the relation:

$I=\frac{2}{3}{a}^4$

Substitute 140 mm for a.

$\begin{array}{c}I=\frac{2}{3}{\left(140\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =\frac{2}{3}{\left(0.14\right)}^4\\ =0.256\times {10}^{-3}{\text{m}}^{\text{4}}\end{array}$

Find the value of $Q_{\max }$ using the relation:

$Q_{\max }=Ay$

Here, A is the area of the cross section and y is the centroid of area.

Substitute $\left(0.14\times 0.14\right){\text{m}}^2$ for A and $0.07\text{m}$ for y.

$\begin{array}{c}{Q}_{\max }=0.07\times 0.14\times 0.14\\ =1.372\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Check:

Calculate the $\left({\tau }_{\max }\right)$ using the relation:

$\left({\tau }_{\max }\right)=\frac{V_{\max }{Q}_{\max }}{It}$

Here $V_{\max }$ is the maximum shear force, I is the moment of inertia, and t is the thickness, and $\left({\tau }_{\max }\right)$ is the shear stress.

Substitute $12\text{kN}$ for $V_{\max }$, $1.372\times {10}^{-3}{\text{m}}^{\text{3}}$ for $Q_{\max }$, $0.256\times {10}^{-3}{\text{m}}^{\text{4}}$ for I, and 0.14 m. for a.

$\begin{array}{c}\left({\tau }_{\max }\right)=\frac{12\times {10}^3\left(1.372\left({10}^{-3}\right)\right)}{\left[0.2561\left({10}^{-3}\left(0.14\right)\right)\right]}\\ =\frac{16.46}{3.58\times {10}^{-5}}\\ =0.459\text{MPa<}{\tau }_{\text{allow}}=1\text{MPa}\end{array}$

Hence it is OK.