Chapter 6, Problem 6.3.9P

A simple beam thai is IS ft long supports a uni¬form load of intensity a. The beam is constructed of two angle sections, each L (1 × 4 × 1/2, on either side of a 2 in. x 8 in. (actual dimensions! wood beam (see the cross section shown in the figure part a]. The modulus of elasticity of the s I eel is 10 limes that of the wood,

(a) If the allowable stresses in the steel and wood are 12,000 psi and 900 psi. respectively, what is the allow atile load a t. A olc. Disregard the weight of the beam, and see Table F-5(a) of Appendix I ' for I lie dimensions and properties of the angles.

(b) Repeal partial if a I in. 10 in. wood Hange tactual dimensions) is added i see figure pallhi b).

  

To determine

The allowed load $q_{allow}$ .

Answer to Problem 6.3.9P

The allowed load $q_{allow}$ is $264.54\text{}\text{lb/ft}$

Explanation of Solution

Given:

We have the following data for calculation.

Length of the beam, $L=\text{}\text{18ft}$

Load with intensity of as q

Two angle sections used for construction of the beam, $L6\times 4\times \frac{1}{2}\text{ and 2 in}\times \text{8}\text{}\text{in}$

Allowed stress of steel, ${\sigma }_s=12,000\text{}\text{psi}$

Allowed stress of wood, ${\sigma }_w=900\text{}\text{psi}$

Concept Used:

First, Area, centroids and moment of inertia for wood and steel sections is determined.

This step is followed by further calculation involving moment for wood and steel section and maximum moment.

Calculation:

We have $n=20$ .

Let us first calculate for wood.

1. Centroidal distance,

  $\begin{array}{l}{\overline{y}}_1=\frac{h}{2}-h_1\\ {\overline{y}}_1=\frac{8}{2}-h_1\\ {\overline{y}}_1=4-h_1\end{array}$

2. Area of the wood beam,
$\begin{array}{l}A=b\times h\\ A=2\times 8\\ A=16\end{array}$

Now, calculation for two sections of steels.

  $\begin{array}{r}{\overline{y}}_2=h_1-d\\ \text{For section L6}\times \text{4}\times \frac{1}{2}\\ I_Z=17.3\text{}{\text{in}}^4\\ d=1.98\text{}\text{in}\\ {\text{A}}_S=4.75\text{}{\text{in}}^2\\ h_s=6in\\ So,{\overline{y}}_2=h_1-1.98\end{array}$

So, further calculating,

   $\begin{array}{l}{A}_2{\overline{y}}_2=-2n{A}_s{\overline{y}}_2\\ A_2{\overline{y}}_2=-2\left(20\right)\left(4.75\right)\left(h_1-1.98\right)\\ A_2{\overline{y}}_2=-190\left(h_1-1.98\right)\\ So,\\ {\displaystyle \sum \overline{y}dA}=0\\ 16\left(4-h_1\right)-190\left(h_1-1.98\right)=0\\ 64-16h_1-190h_1+376.2=0\\ -206h_1+440.2=0\\ 206h_1=440.2\\ h_1=2.1369\text{}\text{in}\end{array}$

Now, performing transformation of moment if inertia below :

   $\begin{array}{l}{I}_{\gamma }=\left[\frac{b{h}^3}{12}+bh{\left(\frac{h}{2}-h_1\right)}^2\right]+2h\left[I_Z+A_s{\left(h_1-d\right)}^2\right]\\ I_{\gamma }=\left[\frac{2\times 8^3}{12}+\left(2\times 8\right){\left(\frac{8}{2}-2.1369\right)}^2\right]+2\left(20\right)\left[17.3+4.75{\left(2.1369-1.78\right)}^2\right]\\ I_{\gamma }=85.33+55.538+696.667\\ I_{\gamma }=837.545\text{}{\text{in}}^4\end{array}$

We are determining moment for wood.

   $\begin{array}{l}{M}_1=\frac{{\sigma }_w{I}_{\gamma }}{h-h_1}\\ M_1=\frac{\left(900\right)\left(837.545\right)}{8-2.1369}\\ M_1=128565.2319\text{}\text{lb}\cdot \text{in}\\ M_1=128.565\text{}\text{k}\cdot \text{in}\end{array}$

Moment for steel sections,

   $\begin{array}{l}{M}_2=\frac{{\sigma }_s{I}_{\gamma }}{n\left(h_s-h_1\right)}\\ M_1=\frac{\left(12000\right)\left(837.545\right)}{20\left(6-2.1369\right)}\\ M_1=130083.87\text{}\text{lb}\cdot \text{in}\\ M_1=130.083\text{}\text{k}\cdot \text{in}\end{array}$

Now, the maximum moment would be:

   $\begin{array}{l}{M}_{\max }=\min \left(M_1{M}_2\right)\\ M_{\max }=128.565\end{array}$

But the equation for maximum moment is:

   $\begin{array}{l}{M}_{\max }=\frac{q{L}^2}{8}\\ \frac{\left({10}^3\right)\left(128.565\right)}{12}=\frac{q{\left(18\right)}^2}{8}\\ q=\frac{128.565\times {10}^3\times 8}{{\left(18\right)}^2\times 12}\\ q=264.54\text{}\text{lb/ft}\end{array}$

Conclusion:

The allowed load $q_{allow}$ is calculated using the given information and moment equations.

To determine

The allowed load $q_{allow}$ .

Answer to Problem 6.3.9P

The allowed load $q_{allow}$ is $280\text{}\text{lb/ft}$.

Explanation of Solution

Given:

We have the following data for calculation.

Length of the beam, $L=\text{}\text{18ft}$

Load with intensity of as q

Two angle sections used for construction of the beam, $L6\times 4\times \frac{1}{2}\text{ and 2 in}\times \text{8}\text{}\text{in}$

Allowed stress of steel, ${\sigma }_s=12,000\text{}\text{psi}$

Allowed stress of wood, ${\sigma }_w=900\text{}\text{psi}$

Concept Used:

First, Area, centroids and moment of inertia for wood and steel sections is determined.

This step is followed by further calculation involving moment for wood and steel section and maximum moment.

Calculation:

So, the width and height of flange would be, $b_f=1\text{}{\text{in and h}}_f=10\text{ in}$

Now for transformed sections,

   $\begin{array}{l}{\displaystyle \sum \overline{y}dA}=0\\ \left(bh+2n{A}_S\right)\left(h_1+h_f-h_b\right)-h_f{h}_f\left(h_b-\frac{b_f}{2}\right)=0\\ \left(2\times 8+2\left(20\right)\left(4.75\right)\right)\left(2.1369+10-h_b\right)-1\times 10\left(h_b-\frac{1}{2}\right)=0\\ h_b=3.015\text{ in}\end{array}$

Now performing transformation of moment if inertia as below,

  $\begin{array}{l}{\left(I_{\gamma }\right)}_b=\left[I_{\gamma }+\left(bh+2n{A}_S\right){\left(h_1+b_f-h_b\right)}^2\right]+\left[\frac{b_f{}^3{h}_f}{12}+b_f{h}_f{\left(h_b-\frac{b_f}{2}\right)}^2\right]\\ {\left(I_{\gamma }\right)}_b=\left[838+\left(2\times 8+2\left(20\right)\left(4.75\right)\right)\left(2.1369+10-{3.015}^2\right)\right]+\left[\frac{1^3\times 10}{12}+1\times 10{\left(3.05-\frac{1}{2}\right)}^2\right]\\ {\left(I_{\gamma }\right)}_b=905\text{}{\text{in}}^4\end{array}$

We are determining moment for wood.

   $\begin{array}{l}{M}_1=\frac{{\sigma }_w{\left(I_{\gamma }\right)}_b}{h_0+h_f-h_b}\\ M_1=\frac{\left(900\right)\left(905\right)}{8+1-3.015}\\ M_1=136090.2256\text{}\text{lb}\cdot \text{in}\\ M_1=136.09\text{}\text{k}\cdot \text{in}\end{array}$

Moment for steel sections,

   $\begin{array}{l}{M}_2=\frac{{\sigma }_s{\left(I_{\gamma }\right)}_b}{\left(h+b_f-h_b\right)n}\\ M_2=\frac{\left(12000\right)\left(905\right)}{\left(6+1-3.015\right)20}\\ M_2=136260.98\text{lb}\cdot \text{in}\\ M_2=136.26\text{}\text{k}\cdot \text{in}\end{array}$

Now, the maximum moment would be,

   $\begin{array}{l}{M}_{\max }=\min \left(M_1{M}_2\right)\\ M_{\max }=136.09\text{}\text{k}\cdot \text{in}\end{array}$

But, the equation for maximum moment is:

   $\begin{array}{l}{M}_{\max }=\frac{q{L}^2}{8}\\ q_{all}=\frac{8M_{\max }}{L^2}\\ q_{all}=\frac{8\times 136\times 1000}{{\left(18\right)}^2\times 12}\\ q_{all}=280\text{}\text{lb/ft}\end{array}$

Conclusion:

The allowed load is calculated by the moment equations and given information.