Chapter 5.5, Problem 145P

Two cover plates, each 7.5 mm thick, are welded to a W460 × 74 beam as shown. Knowing that σ_all = 150 MPa for both the beam and the plates, determine the required value of (a) the length of the plates, (b) the width of the plates.

Fig. P5.144 and P5.145

To determine

Find the length of the plates.

Answer to Problem 145P

The length of the plates is $\underset{\_}{4.49\text{m}}$.

Explanation of Solution

The length of the cover plate is $l=5\text{m}$.

The thickness of the cover plate is $t=7.5\text{mm}$.

The width of the cover plate is $b=200\text{mm}$.

The allowable normal stress in the beam is ${\sigma }_{all}=150\text{MPa}$.

Show the free-body diagram of the prismatic beam as in Figure 1.

Determine the vertical reaction at point B by taking moment at point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -\left(40\times 8\right)\times 4+B_y\left(8\right)=0\\ B_y=160\text{kN}\end{array}$

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(40\times 8\right)+B_y=0\\ A_y-320+160=0\\ A_y=160\text{kN}\end{array}$

Show the free-body diagram of the section AD as in Figure 2.

Determine the moment at point D by taking moment about point D.

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ M_D-160\left(4-\frac{l}{2}\right)+40\times \left(4-\frac{l}{2}\right)\times \frac{\left(4-\frac{l}{2}\right)}{2}=0\\ M_D-160\left(4-\frac{l}{2}\right)+20{\left(4-\frac{l}{2}\right)}^2=0\\ M_D=160\left(4-\frac{l}{2}\right)-20{\left(4-\frac{l}{2}\right)}^2\end{array}$

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The section modulus of the $\text{W}460\times 74$ section is $S=1460\times {10}^3{\text{mm}}^3$.

Determine the moment at point D $\left(M_D\right)$ of the beam using the relation.

${\sigma }_{all}=\frac{M_D}{S}$

Here, the allowable stress of the beam is ${\sigma }_{all}$.

Substitute 150 MPa for ${\sigma }_{all}$, $160\left(4-\frac{l}{2}\right)-20{\left(4-\frac{l}{2}\right)}^2$ for $M_D$, and $1460\times {10}^3{\text{mm}}^3$ for S.

$\begin{array}{c}150\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=\frac{160\left(4-\frac{l}{2}\right)-20{\left(4-\frac{l}{2}\right)}^2\text{kN-m}\times \frac{\text{1000}\text{N}}{\text{1}\text{kN}}}{1460\times {10}^3{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^3}\\ 160\left(4-\frac{l}{2}\right)-20{\left(4-\frac{l}{2}\right)}^2=219000\end{array}$

Find the length of the plate using trial and error method.

$l=4.49\text{m}$

Therefore, the length of the plates is $\underset{\_}{4.49\text{m}}$.

To determine

Find the width of the plates.

Answer to Problem 145P

The width of the plates is $\underset{\_}{211\text{mm}}$.

Explanation of Solution

The length of the cover plate is $l=5\text{m}$.

The thickness of the cover plate is $t=7.5\text{mm}$.

The width of the cover plate is $b=200\text{mm}$.

The allowable normal stress in the beam is ${\sigma }_{all}=150\text{MPa}$.

Show the free-body diagram of the prismatic beam as in Figure 3.

Determine the vertical reaction at point B by taking moment at point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -\left(40\times 8\right)\times 4+B_y\left(8\right)=0\\ B_y=160\text{kN}\end{array}$

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(40\times 8\right)+B_y=0\\ A_y-320+160=0\\ A_y=160\text{kN}\end{array}$

Show the free-body diagram of the section AC as in Figure 4.

Determine the moment at point C by taking moment about point C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ M_C-160\left(4\right)+\left(40\times 4\right)\times \frac{4}{2}=0\\ M_C-640+320=0\\ M_C=320\text{kN-m}\end{array}$

Show the cross section of the beam as in Figure 5.

Determine the section modulus (S) of the beam using the relation.

$S=\frac{M_C}{{\sigma }_{all}}$

Here, the allowable normal stress in the beam is ${\sigma }_{all}$.

Substitute $320\text{kN-m}$ for $M_C$ and 150 MPa for S.

$\begin{array}{c}S=\frac{320\text{kN-m}}{150\text{MPa}\times \frac{{\text{10}}^{\text{3}}{\text{kN/m}}^{\text{2}}}{\text{1}\text{MPa}}}\\ =2133.33\times {10}^{-6}{\text{m}}^3\times {\left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)}^3\\ =2133.33\times {10}^3{\text{mm}}^3\end{array}$

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The depth of the $\text{W}460\times 74$ section is $d=457\text{mm}$.

The moment of inertia of the $\text{W}460\times 74$ section is $I_{beam}=333\times {10}^6{\text{mm}}^4$.

Determine the moment of inertia (I) of the beam using the relation;

$\begin{array}{c}I=I_{beam}+2I_{plates}\\ =I_{beam}+2\left(\frac{b{t}^3}{12}+A{\left(\overline{y}\right)}^2\right)\end{array}$ (1)

Here, the moment of inertia of the beam is $I_{beam}$, the width of the plate is b, the thickness of the plate is t, the cross sectional area of the plate is A, and the distance between the neutral axis and the center of the plates is $\overline{y}$.

Refer to the cross section of the beam;

$A=\left(b\times 7.5\right){\text{mm}}^2;\overline{y}=\frac{457}{2}+\frac{7.5}{2}=232.5\text{mm}$

Substitute $333\times {10}^6{\text{mm}}^4$ for $I_{beam}$, 7.5 mm for t, $\left(b\times 7.5\right){\text{mm}}^2$ for A, and 232.5 mm for $\overline{y}$ in Equation (1).

$I=333\times {10}^6+2\left(\frac{b\times {7.5}^3}{12}+b\times 7.5\times {232.5}^2\right)$

Determine the distance from the neutral axis (c) to outer most fibre as follows;

$c=\frac{457}{2}+7.5=236\text{mm}$

Determine the width of the plates (b) using the relation.

$S=\frac{I}{c}$

Substitute $2133.33\times {10}^3{\text{mm}}^3$ for S, $333\times {10}^6+2\left(\frac{b\times {7.5}^3}{12}+b\times 7.5\times {232.5}^2\right)$ for I, and 236 mm for c.

$\begin{array}{c}2133.33\times {10}^3=\frac{333\times {10}^6+2\left(\frac{b\times {7.5}^3}{12}+b\times 7.5\times {232.5}^2\right)}{236\text{mm}}\\ 333\times {10}^6+2\left(\frac{b\times {7.5}^3}{12}+b\times 7.5\times {232.5}^2\right)=503465880\\ b=211\text{mm}\end{array}$

Therefore, the width of the plates is $\underset{\_}{211\text{mm}}$.