Chapter 6, Problem 6.2.13P

A simply supported wooden I-beam with a 12-ft span supports a distributed load of intensity q = 90 lb/ft over its length (see figure part a). The beam is constructed with a web of Douglas-fir plywood and flanges of pine glued to the web, as shown in the figure part b. The plywood is 3/8 in. thick: the flanges are 2 in, × 2 in, (actual size). The modulus of elasticity for the plywood is 1,600,000 psi and for the pine is 1,200,000 psL

1. Calculate the maximum bending stresses in the pine flanges and in the plywood web.

What is q, if allowable stresses are 1600 psi in the flanges and 1200 psi in the web?

  

  

To determine

The bending stress that is maximum in the plywood web and pine flanges

Answer to Problem 6.2.13P

Bending Stress Maximum for plywood $=1130.629psi$

Bending Stress Maximum for pine $=969.11psi$

Explanation of Solution

Given:

The given figure

  

The I beam that is wooden has the span of 12 ft over its length supporting a distributed intensity load q=90lb/ft. The beam is made of Douglas fir plywood web and pine flanges with the plywood having a thickness of 3/8 in and that of flanges 2in.*2in. The elasticity modulus of the plywood is 1,600,000psi and for pine it is 1,200,000psi.

Concept Used:

Bending Stress Maximum for plywood,

  ${\sigma }_{pw}=\frac{M_{\max }\left(\frac{h_1}{2}\right)\left(E_{pw}\right)}{E_{pw}{I}_1+E_p{I}_2}$

Where,

  $\begin{array}{l}{M}_{Max}=Maximumbendingmoment\\ I_1=PlywoodInertiamoment\\ I_2=PineInertiamoment\\ E_{PW}=Plywood\mathrm{mod}ulusElasticity\\ E_P=Pine\mathrm{mod}ulusElasticity\\ h_1=height\end{array}$

Calculation:

Bending moment maximum is given as

  $M_{\max }=\frac{q{L}^2}{8}$

Substituting the values we have,

  $M_{\max }=\frac{90*{12}^2}{8}$

  $\begin{array}{l}=\frac{90*{12}^2}{8}\\ =\frac{90*144}{8}\\ =\frac{12960}{8}\\ =1620lb.ft\end{array}$

Plywood moment of inertia is given as,

  $I_1=\frac{t*h_1{}^3}{12}$

  $\begin{array}{l}=\frac{\left(\frac{3}{8}\right)*7^3}{12}\\ =\frac{\left(\frac{3}{8}\right)*343}{12}\\ =\frac{128.625}{12}\\ =10.719i{n}^4\end{array}$

Pine moment of inertia is given as,

  $I_2=2\left(\frac{b{a}^3}{12}+ba{\left(\frac{h_1+a}{2}\right)}^2+\frac{(b-t){(h_2-a)}^3}{12}\right)+(b-t)(h_2-a){\left(\frac{h_1}{2}-\frac{h_2-a}{2}\right)}^2$

  $=2\left(\frac{2*{\left(\frac{1}{2}\right)}^3}{12}+2*{\left(\frac{1}{2}\right)}^2{\left(\frac{7+\frac{1}{2}}{2}\right)}^2+\frac{\left(2-\frac{3}{8}\right){\left(2-\frac{1}{2}\right)}^3}{12}\right)+\left(2-\frac{3}{8}\right)\left(2-\frac{1}{2}\right){\left(\frac{7}{2}-\frac{\left(2-\frac{3}{8}\right)}{2}\right)}^2$

  $=2\left(\frac{2*{\left(\frac{1}{2}\right)}^3}{12}+2*\left(\frac{1}{4}\right){\left(3.75\right)}^2+0.457\right)+\left(\frac{13}{8}\right)\left(\frac{3}{2}\right){\left(\frac{7}{2}-\frac{\left(\frac{13}{8}\right)}{2}\right)}^2$

  $=2\left(0.02083+14.0625+0.457+18.4336\right)$

  $=65.948i{n}^4$

  $E_{pw}{I}_1+E_p{I}_2=(1600000)(10.718)+(1200000)(65.948)$

  $\begin{array}{l}=(1600000)(10.718)+(1200000)(65.948)\\ =17148800000+79137600000\\ =96286217i{n}^4\end{array}$

Bending Stress Maximum for plywood,

  ${\sigma }_{pw}=\frac{M_{\max }\left(\frac{h_1}{2}\right)\left(E_{pw}\right)}{E_{pw}{I}_1+E_p{I}_2}$

  $\begin{array}{l}=\frac{\left(1620*12\right)\left(\frac{7}{2}\right)\left(1.6*{10}^6\right)}{96286217}\\ =\frac{\left(19440\right)\left(\frac{7}{2}\right)\left(1.6*{10}^6\right)}{96286217}\\ =1130.629psi\end{array}$

Bending Stress Maximum for pine,

  ${\sigma }_{pw}=\frac{M_{\max }\left(\frac{h_1}{2}+a\right)\left(E_P\right)}{E_{pw}{I}_1+E_p{I}_2}$

  $\begin{array}{l}=\frac{(1620*12)\left(\frac{7}{2}+\frac{1}{2}\right)\left(1200000\right)}{96286217}\\ =\frac{(19440)\left(4\right)\left(1200000\right)}{96286217}\\ =969.11psi\end{array}$

Conclusion:

Thus, the bending stress that is maximum in the plywood web and pine flanges is given by equating the maximum bending movement

To determine

The q_max with 1600psi maximum stress in flanges and for web being 1200psi.

Answer to Problem 6.2.13P

  $q_{\max }=95.5lb.ft$

Explanation of Solution

Given:

The given figure:

  

The I beam that is wooden has the span of 12 ft over its length supporting a distributed intensity load q=90lb/ft. The beam is made of Douglas fir plywood web and pine flanges with the plywood having a thickness of 3/8 in and that of flanges 2in.*2in. The elasticity modulus of the plywood is 1,600,000psi and for pine it is 1,200,000psi.

Concept Used:

Web maximum stress,

  ${\sigma }_{pw}=\frac{M_{pw}\left(\frac{h_1}{2}\right)\left(E_{pw}\right)}{E_{pw}{I}_1+E_p{I}_2}$

Where,

  $\begin{array}{l}{M}_{pw}=Maximum\text{ allowable }stress\\ I_1=PlywoodInertiamoment\\ I_2=PineInertiamoment\\ E_{PW}=Plywood\mathrm{mod}ulusElasticity\\ E_P=Pine\mathrm{mod}ulusElasticity\\ h_1=height\end{array}$

Calculation:

Maximum stress Maximum for Web,

  ${\sigma }_{pw}=\frac{M_{pw}\left(\frac{h_1}{2}\right)\left(E_{pw}\right)}{E_{pw}{I}_1+E_p{I}_2}$

Substituting the values we have,

  $1200=\frac{M_{pw}\left(\frac{7}{2}\right)\left(1600000\right)}{96286217}$

  $\begin{array}{l}1200*96286217=M_{pw}\left(\frac{7}{2}\right)\left(1600000\right)\\ M_{pw}=\frac{1200*96286217}{\left(\frac{7}{2}\right)\left(1600000\right)}\end{array}$

  $\begin{array}{l}=\frac{1200*96286217}{\left(\frac{7}{2}\right)\left(1600000\right)}\\ =\frac{72214.663}{\left(\frac{7}{2}\right)}\\ =20632.76lb.in\Rightarrow 1719.397lb.ft\end{array}$

Maximum stress Maximum for flange,

  ${\sigma }_{pine}=\frac{M_p\left(\frac{h_1}{2}+a\right)\left(E_p\right)}{E_{pw}{I}_1+E_p{I}_2}$

Substituting the values we have,

  $\begin{array}{l}1600=\frac{M_p\left(\frac{7}{2}+\frac{1}{2}\right)\left(1200000\right)}{96286217}\\ 1600*96286217=M_p\left(4\right)\left(1200000\right)\\ M_p=\frac{1600*96286217}{\left(4\right)\left(1200000\right)}\end{array}$

  $=32095.406lb.in\Rightarrow 2674.62lb.ft$

Bending moment that is maximum,

  $M_{all}=\min (M_{pw},M_p)$

  $\Rightarrow 1719.40lb.ft$

Maximum uniform load distributed is given as,

  $M_{all}=\frac{q_{\max }{L}^2}{8}\Rightarrow q_{\max }=\frac{8M_{all}}{L^2}$

  $\begin{array}{l}=\frac{8*1719.397}{{12}^2}\\ =\frac{8*1719.397}{144}\\ =95.522lb.ft\end{array}$

Conclusion:

Thus, the q_max with 1600psi maximum stress in flanges and for web being 1200psi is given by the equation of maximum load.