Chapter 5, Problem 5.5.30P

Beam ABCDE has a moment release just right of joint B and has concentrated moment loads at D and E. In addition, a cable with tension P is attached at fand runs over a pulley at C (Fig, a). The beam is constructed using two steel plates, which arc welded to form a T cross section (see Fig. b). Consider ßexuralstresses only Find the maximum permissible value of load variable P if the allowable bending stress is 130 M Pa. Ignore the self-weight of the frame members and let length variable L = 0.75 m.

  

  

To determine

The maximum permissible load.

Answer to Problem 5.5.30P

The maximum permissible load is $675.38\text{N}$ .

Explanation of Solution

Given information:

The allowable bending stress is $130\text{MPa}$ , the length of the beam is $0.75\text{m}$ .

Write the expression for the force equilibrium in x direction.

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=0\end{array}$ .....(I)

Here, the reaction at point $A$ of the beam is $A_x$ .

Write the expression for the moment at the point of release.

  $\begin{array}{l}{\displaystyle \sum M_{rel}}=0\\ A_y\left(4L\right)+\frac{4P\left(3L\right)}{5}=0\\ A_y=-0.6P\end{array}$ .....(II)

Here, the length of the beam is $L$ , the load on the beam is $P$ .

Write the expression for the moment about point $E$ .

  $\begin{array}{l}{\displaystyle \sum M_E}=0\\ D_y=\frac{1}{4L}\left(8PL-2PL+P\left(8L\right)-\left(A_y\right)\left(16L\right)\right)\end{array}$.....(III)

Here, the vertical reaction at point $D$ is $D_y$ .

Write the expression for the force equilibrium in y direction.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ E_y=P-A_y-D_y\end{array}$ ….. (IV)

Here, the vertical reaction at point $E$ is $E_y$ .

Write the expression for the moment at $0\le x\le 4L$ .

  $M\left(x\right)=A_{y}x$ .....(V)

Here the moment is $M\left(x\right)$ .

Write the expression for the moment at $4L\le x\le 8L$ .

  $M\left(x\right)=\left[A_{y}x+\frac{4}{5}P\left(3L\right)-\frac{3}{5}P\left(x-4L\right)\right]$.....(VI)

Write the expression for the moment at $8L\le x\le 12L$ .

  $M\left(x\right)=A_{y}x-P\left(x-8L\right)$.....(VII)

Write the expression for the moment at other points.

  $M\left(x\right)=E_y\left(16L-x\right)+8PL$ .....(VIII)

Write the expression for the area of the $T$ shape.

  $A=b_f{t}_f+b_w{t}_w$ .....(IX)

Here, the width of the flange is $b_f$ , the thickness of the flange is $t_f$ , the width of the web is $b_w$ , the thickness of the flange is $t_w$ .

Write the expression for the centroid.

  $c_2=\frac{b_f{t}_f\left(b_w+\frac{t_f}{2}\right)+b_w{t}_w\left(\frac{b_w}{2}\right)}{A}$….. (X)

Here, the distance of the centroid from one end is $c_2$ .

Write the expression for the other distance of the centroid.

  $c_1=b_w+t_f-c_2$ .....(XI)

Here, the distance of the centroid from other end is $c_1$ .

Write the expression for the moment of inertia about z axis.

  $I_z=\frac{b_f{t}^3{}_f}{12}+\left(b_f{t}_f\right){\left(c_1-\frac{t_f}{2}\right)}^2+\frac{b^3{}_w{t}_w}{12}+\left(b_w{t}_w\right){\left(c_2-\frac{b_w}{2}\right)}^2$ .....(XII)

Here, the moment of inertia about z axis is $I_z$ .

Write the expression for the section modulus at the top section.

  $S_{top}=\frac{I_z}{c_1}$ .....(XIII)

Here, the section modulus at top section is $S_{top}$

Here, the section modulus of top section is $S_{top}$ .

Write the expression for the section modulus at the bottom section.

  $S_{bot}=\frac{I_z}{c_2}$ .....(XIV)

Here, the section modulus of bottom section is $S_{bot}$ .

Write the expression for the maximum permissible load.

  $P_{\max }=\frac{{\sigma }_a{S}_{bot}}{M_{\max }}$.....(XV)

Here, the allowable stress is ${\sigma }_a$ , the maximum permissible stress is $P_{\max }$ .

Calculation:

Substitute $-0.6P$ for $A_y$ in Equation (III)

  $\begin{array}{c}{D}_y=\frac{1}{4L}\left(8PL-2PL+P\left(8L\right)-\left(-0.6P\right)\left(16L\right)\right)\\ =\frac{1}{4L}\left(23.6PL\right)\\ =5.9P\end{array}$

Substitute $-0.6P$ for $A_y$ , and $5.9D_y$ in Equation (IV)

  $\begin{array}{c}{E}_y=P-\left(-0.6P\right)-5.9P\\ =P+\left(0.6P\right)-5.9P\\ =-4.3P\end{array}$

Substitute $150\text{mm}$ for $b_f$ , $18\text{mm}$ for $t_f$ , $100\text{mm}$ for $b_w$ , $12\text{mm}$ for $t_w$ in Equation (IX)

  $\begin{array}{c}A=\left(\left(150\text{mm}\times 18\text{mm}\right)+\left(100\text{mm}\times 12\text{mm}\right)\right)\\ =2700{\text{mm}}^{\text{2}}+1200{\text{mm}}^{\text{2}}\\ =3.9\times {10}^3{\text{mm}}^2\end{array}$

Substitute $150\text{mm}$ for $b_f$ , $18\text{mm}$ for $t_f$ , $100\text{mm}$ for $b_w$ , $12\text{mm}$ for $t_w$ , and $3.9\times {10}^3{\text{mm}}^2$ for $A$ in Equation (X)

  $\begin{array}{c}{c}_2=\left[\frac{\left(150\text{mm}\times 18\text{mm}\right)\left(100\text{mm}+\frac{18}{2}\text{mm}\right)+\left(100\text{mm}\times 12\text{mm}\right)\left(100\text{mm}+\frac{100}{2}\text{mm}\right)}{3.9\times {10}^3{\text{mm}}^2}\right]\\ =\left[\frac{294300{\text{mm}}^{\text{3}}+180000{\text{mm}}^{\text{3}}}{3.9\times {10}^3{\text{mm}}^2}\right]\\ =90.846\text{mm}\end{array}$

Substitute $18\text{mm}$ for $t_f$ , $100\text{mm}$ for $b_w$ , $90.846\text{mm}$ for $c_2$ in Equation (XI)

  $\begin{array}{c}{c}_1=100\text{mm}+18\text{mm}-90.846\text{mm}\\ \text{=118}\text{mm-90}\text{.846}\text{mm}\\ \text{=27}\text{.154}\text{mm}\end{array}$

Substitute $150\text{mm}$ for $b_f$ , $18\text{mm}$ for $t_f$ , $100\text{mm}$ for $b_w$ , $12\text{mm}$ for $t_w$ in Equation (XII)

  $\begin{array}{c}{I}_z=\left[\begin{array}{l}\left(\frac{150\text{mm}\times {\left(18\text{mm}\right)}^3}{12}\right)+\left(150\text{mm}\times 18\text{mm}\right){\left(27.154\text{mm}-\frac{18\text{mm}}{2}\right)}^2\\ +\left(\frac{{\left(100\text{mm}\right)}^2\times 12\text{mm}}{12}\right)+\left(100\text{mm}\times 12\text{mm}\right)\left(90.846\text{mm}-{\left(\frac{100\text{mm}}{2}\right)}^2\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(72900{\text{mm}}^4\right)+\left(150\text{mm}\times 18\text{mm}\right){\left(27.154\text{mm}-\frac{18\text{mm}}{2}\right)}^2\\ +\left(10000{\text{mm}}^4\right)+\left(100\text{mm}\times 12\text{mm}\right)\left(90.846\text{mm}-{\left(\frac{100\text{mm}}{2}\right)}^2\right)\end{array}\right]\\ =3.965\times {10}^6{\text{mm}}^4\end{array}$

Substitute $3.965\times {10}^6{\text{mm}}^4$ for $I_z$ , $27.154\text{mm}$ for $c_1$ , in Equation (XIII)

  $\begin{array}{c}{S}_{top}=\frac{3.965\times {10}^6{\text{mm}}^4}{27.154\text{mm}}\\ =1.46\times {10}^5{\text{mm}}^3\end{array}$

Substitute $3.965\times {10}^6{\text{mm}}^4$ for $I_z$ , $90.846\text{mm}$ for $c_2$ , in Equation (XIV)

  $\begin{array}{c}{S}_{bot}=\frac{3.965\times {10}^6{\text{mm}}^4}{90.846\text{mm}}\\ =4.364\times {10}^4{\text{mm}}^3\end{array}$

Substitute $130\text{MPa}$ for ${\sigma }_a$ , $4.364\times {10}^4{\text{mm}}^3$ for $S_{bot}$ , $0.75\text{mm}$ for $L$ , $11.2\text{N}\cdot \text{m}$ for $M_{\max }$ in Equation (XV)

  $\begin{array}{c}{P}_{\max }=\frac{130\text{MPa}\times \left(4.364\times {10}^4{\text{mm}}^3\right)}{11.2\times 0.75}\\ =\frac{\left(130\text{MPa}\right)\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\times \left(4.364\times {10}^4{\text{mm}}^4\right){\left(\frac{1\text{m}}{1000\text{mm}}\right)}^4}{\left(11.2\text{N}\cdot \text{m}\right)\times \left(0.75\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =\frac{\left(130\times {10}^6\text{Pa}\right)\left(\frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)\times \left(4.364\times {10}^4\times {10}^{-12}{\text{m}}^3\right)}{\left(11.2\text{N}\cdot \text{m}\right)\times \left(0.75\times {10}^{-3}\text{m}\right)}\\ =675.38\text{N}\end{array}$

Conclusion:

The maximum permissible load is $675.38\text{N}$ .