Chapter 6, Problem 6.2.16P

A wood beam in a historic theater is reinforced with two angle sections at the outside lower corners (see figure). If the allowable stress in the wood is 12 M Pa and that in the steel is 140 M Pa, what is ratio of the maximum permissible moments for the beam before and after reinforcement with the angle sections? See Appendix F Table F-5(b) for angle section properties. Assume that e_w= 12 GPa and E₃=210 GPa.

  

To determine

The ratio of maximum permissible moment of a beam before and after reinforcement.

Answer to Problem 6.2.16P

The ratio of the maximum permissible moment before and after reinforcement is$0.755$ .

Explanation of Solution

Given: .

  $Area,A_L=1090m{m}^2$.

Moment of Inertia,$I_{11}=1.14\times {10}^{6}m{m}^4.$.

Distance from the neutral axis, d=31mm..

  $E_s$ =12GPa..

  $E_w$ =21GPa..

Width, b=240mm..

Height, h=480mm..

Calculation: .

The maximum moment if no angle section used is calculated as:.

  $M_{1\max }={\sigma }_{wa}\left(\frac{b{h}^2}{6}\right)$.

  ${\sigma }_{wa}$ - Allowable stress..

b- The width of section..

h- Height of section..

  $\begin{array}{l}{M}_{1\max }=12\times \left(\frac{240\times {480}^2}{6}\right)\\ M_{1\max }=110.592\times {10}^{6}N.mm\times \left(\frac{1KN}{1000N}\right)\times \left(\frac{1m}{1000mm}\right)\\ M_{1\max }=110.592kN.m\end{array}$.

The location of the neutral axis from the lower end is stated as:.

  $h_2=\frac{E_w\left(bh\right)\left(\frac{h}{2}+6.4\right)+E_s\left(2A_L\right)d}{E_w\left(bh\right)+E_s\left(2A_L\right)}.$.

Young modulli of steel and wood are$E_w$ and$E_s$.

Substitute$E_s$ =12GPa,$E_w$ =21GPa, b=240mm, d=31,$A_l$ =1090mm, h=480mm..

  $h_2=\left\{\frac{12\times {10}^3\left(240\times 480\right)\left(\frac{480}{2}+604\right)+\left\{210\times {10}^3\left(2\times 1090\right)31\right\}}{\left\{12\times {10}^3\left(240\times 480\right)\right\}+\left\{210\times {10}^3\left(2\times 1090\right)\right\}}\right\}$.

  $h_2=\frac{\left\{3.4062\times {10}^{11}\right\}+\left\{0.141918\times {10}^{11}\right\}}{\left\{13.824\times {10}^8\right\}+\left\{4.578\times {10}^8\right\}}$.

  $\begin{array}{l}{h}_2=0.192813\times {10}^{3}mm\\ h_2=192.813mm\end{array}$.

The distance of the neural axis from the top..

  $h_1=h+6.4-h_2$.

Substitute h=480mm and$h_2$ =192.813mm.

  $\begin{array}{l}{h}_1=480+6.4-192.813\\ h_1=293.587mm\end{array}$.

Calculate the moment of Inertia for wood section..

  $I_{{}_w}=\frac{b{h}^3}{12}+\left(bh\right){\left(\frac{h}{2}-h_1\right)}^2$.

Substitute b=240mm, h=480mm and$h_1$ =293.587mm.

  $I_{{}_w}=\frac{240\times {480}^3}{12}+\left(240\times 480\right){\left(\frac{480}{2}-293.587\right)}^2$.

  $\begin{array}{l}{I}_{{}_w}=22.1184\times {10}^8+3.3080\times {10}^8\\ I_{{}_w}=25.43\times {10}^{8}m{m}^4\times \left(\frac{1m^4}{1000m{m}^4}\right)\\ I_{{}_w}=2.543\times {10}^3{m}^4\end{array}$.

The moment of Inertia of the steel section..

  $I_s=2I_{11}+2A_L{\left(h_2-d\right)}^2$.

Substitute$I_{11}$ =$1.14\times {10}^{6}m{m}^4$ ,$h_2$ =192.813mm,$A_L$ =1090mm , d=31mm..

  $I_s=2\left(1.14\times {10}^{6}m{m}^4\right)+2\left(1090\right){\left(192.813-31\right)}^2$.

  $\begin{array}{l}{I}_s=0.0228\times {10}^8+0.570799\times {10}^8\\ I_s=0.5936\times {10}^{8}m{m}^4\times \left(\frac{1m^4}{{1000}^{4}m{m}^4}\right)\end{array}$.

  $I_s=5.936\times {10}^{-5}{m}^4$.

The maximum moment based on allowable stress in steel..

  $M_{\max }={\sigma }_{sa}\left[\frac{E_s{I}_s+E_w{I}_w}{h_2{E}_s}\right]$.

${\sigma }_{sa}$ - the allowable stress in steel

Substitute${\sigma }_{sa}$ -140MPa,$E_s$ -210GPa ,$I_s$ -$5.936\times {10}^{-5}$.

$E_w$ -12GPa ,$I_w$ -$2.543\times {10}^{-3}$.

  $M_{\max }=140\times {10}^6\left(\frac{\left(210\times {10}^9\times 5.936\times {10}^{-5}\right)+\left(12\times {10}^9\times 2.543\times {10}^{-3}\right)}{192.813\times {10}^{-3}\times 210\times {10}^9}\right)$.

  $\begin{array}{l}{M}_{\max }=148.597\times {10}^{3}N.m\times \left(\frac{1kN}{1000N}\right)\\ M_{\max }=148.597KN.m\end{array}$.

$$.

The maximum moment based on the allowable stress on top..

  $M_{w\max Top}={\sigma }_{wa}\left[\frac{E_s{I}_s+E_w{I}_w}{h_1{E}_w}\right]$.

  $M_{w\max Top}=12\times {10}^6\left(\frac{\left(210\times {10}^9\times 5.936\times {10}^{-5}\right)+\left(12\times {10}^9\times 2.543\times {10}^{-3}\right)}{293.587\times {10}^{-3}\times 12\times {10}^9}\right)$.

  $\begin{array}{l}{M}_{w\max Top}=146.387\times {10}^{3}N.m\times \left(\frac{1KN}{1000N}\right)\\ M_{w\max Top}=146.387KN.m\end{array}$.

The maximum moment based on the allowable stress in wood on bottom..

  $M_{w\max Bot}={\sigma }_{wa}\left[\frac{E_s{I}_s+E_w{I}_w}{\left(h_2-6.4\right)E_w}\right]$.

Substitute the value

Substitute${\sigma }_{sa}$ -140MPa,$E_s$ -210GPa,$I_s$ -$5.936\times {10}^{-5}$.

$E_w$ -12GPa ,$I_w$ -$2.543\times {10}^{-3}$.

  $M_{w\max Bop}=12\times {10}^6\left(\frac{\left(210\times {10}^9\times 5.936\times {10}^{-5}\right)+\left(12\times {10}^9\times 2.543\times {10}^{-3}\right)}{\left(192.813-6.4\right)\times {10}^{-3}\times 12\times {10}^9}\right)$.

  $\begin{array}{l}{M}_{w\max Bot}=230.548\times {10}^{3}N.m\times \left(\frac{1KN}{1000N}\right)\\ M_{w\max bot}=230.548KN.m\end{array}$.

The lower magnitude of the moment among the three values for the cross section is to be safe. Hence, the maximum allowable moment is$M_{w\max Top}=146.387KN.m.$.

Therefore the ratio of the maximum permissible moment before and after reinforcement is

  $ratio=\frac{M_{i\max }}{M_{w\max Top}}$.

Substitute the value$M_{i\max }$ =110.592KN and$M_{w\max Top}$ =146.387KN

  $\begin{array}{l}ratio=\frac{110.592}{146.387}\\ ratio=0.755\end{array}$.

Therefore, the ratio of the maximum permissible moment before and after reinforcement is$0.755$ .