Chapter 5.3, Problem 89P

Beams AB, BC, and CD have the cross section shown and are pin-connected at B and C. Knowing that the allowable normal stress is +110 MPa in tension and –150 MPa in compression, determine (a) the largest permissible value of w if beam BC is not to be overstressed, (b) the corresponding maximum distance a for which the cantilever beams AB and CD are not overstressed.

Fig. P5.89

To determine

The largest permissible value of w for the condition that the beam BC is not overstressed.

Answer to Problem 89P

The largest permissible value of w is $\underset{\_}{1.485\text{kN/m}}$ if the beam BC is not overstressed.

Explanation of Solution

Given information:

The allowable normal stress of the material in tension is ${\left({\sigma }_{all}\right)}_T=+110\text{MPa}$.

The allowable normal stress of the material in compression is ${\left({\sigma }_{all}\right)}_C=-150\text{MPa}$.

Calculation:

Show the free-body diagram of the section BC as in Figure 1.

Determine the vertical reaction at point C by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -\left(w\times 7.2\right)\times \frac{7.2}{2}+C_y\left(7.2\right)=0\\ -25.92w+7.2C_y=0\\ C_y=3.6\text{kN}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ B_y-\left(w\times 7.2\right)+C_y=0\\ B_y-7.2w+3.6w=0\\ B_y=3.6w\end{array}$

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

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-\left(w\times a\right)-B_y=0\\ A_y-wa-3.6w=0\\ A_y=wa+3.6w\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -\left(w\times a\right)\times \frac{a}{2}-3.6wa+M_A=0\\ M_A=\frac{w{a}^2}{2}+3.6wa\end{array}$

Show the free-body diagram of the section CD as in Figure 3.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ D_y-\left(w\times a\right)-C_y=0\\ D_y-wa-3.6w=0\\ D_y=wa+3.6w\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ \left(w\times a\right)\times \frac{a}{2}+3.6wa-M_D=0\\ M_D=\frac{w{a}^2}{2}+3.6wa\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=wa+3.6w$

$\begin{array}{c}{V}_B-V_A=-w\times a\\ V_B=-wa+V_A\\ =-wa+wa+3.6w\\ =3.6w\end{array}$

$\begin{array}{c}{V}_C-V_B=-w\times 7.2\\ V_C=-7.2w+3.6w\\ =-3.6w\end{array}$

$\begin{array}{c}{V}_D-V_C=-w\times a\\ V_D=-wa+V_C\\ =-wa-3.6w\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	$wa+3.6w$
B	3.6w
C	–3.6w
D	$-wa-3.6w$

Plot the shear force diagram as in Figure 4.

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 4;

Use the similar triangle concept.

$\begin{array}{l}\frac{3.6w}{x}=\frac{3.6w}{7.2-x}\\ 7.2-x=x\\ 2x=7.2\\ x=3.6\text{m}\end{array}$

The maximum bending moment occurs at a distance of $\left(a+3.6\right)\text{m}$ from left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=-\left(\frac{w{a}^2}{2}+3.6wa\right)$

$M_B=0$

$\begin{array}{c}{M}_{\max }-M_B=\frac{1}{2}\times 3.6w\times 3.6\\ M_{\max }=6.48w+M_B\\ =6.48w+0\\ =6.48w\end{array}$

$M_C=0$

$M_D=-\left(\frac{w{a}^2}{2}+3.6wa\right)$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	$-\left(\frac{w{a}^2}{2}+3.6wa\right)$
B	0
Max BM	6.48w
C	0
D	$-\left(\frac{w{a}^2}{2}+3.6wa\right)$

Plot the bending moment diagram as in Figure 5.

Show the free-body diagram of the T-section as in Figure 6.

Determine the centroid in y-axis $\left(\overline{y}\right)$ using the equation.

$\overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}$

Here, the area of the section 1 is $A_1$, the depth of the section 1 from the bottom is $y_1$, the area of the section 2 is $A_2$, and depth of the section 2 from the bottom is $y_2$.

Refer to Figure 4;

$\begin{array}{l}{A}_1=\left(200\times 12.5\right){\text{mm}}^2;y_1=156.25\text{mm}\\ A_2=\left(12.5\times 150\right){\text{mm}}^2;y_2=75\text{mm}\end{array}$

Substitute $\left(200\times 12.5\right){\text{mm}}^2$ for $A_1$, 156.25 mm for $y_1$, $\left(12.5\times 150\right){\text{mm}}^2$ for $A_2$, and 75 mm for $y_2$.

$\begin{array}{c}\overline{y}=\frac{\left(200\times 12.5\times 156.25\right)+\left(12.5\times 150\times 75\right)}{\left(200\times 12.5\right)+\left(12.5\times 150\right)}\\ =121.43\text{mm}\end{array}$

Determine the moment of inertia (I) using the equation.

$I=\frac{b_1{d}_1^3}{12}+A_1{\left(y_1-\overline{y}\right)}^2+\frac{b_2{d}_2^3}{12}+A_2{\left(y_2-\overline{y}\right)}^2$

Here, the depth of the section 1 is $d_1$, the width of the section 1 is $b_1$, the depth of the section 2 is $d_2$, and the width of the section 2 is $b_2$.

Substitute 12.5 mm for $d_1$, 200 mm for $b_1$, $\left(200\times 12.5\right){\text{mm}}^2$ for $A_1$, 156.25 mm for $y_1$, 121.43 mm for $\overline{y}$, 150 mm for $d_2$, 12.5 mm for $b_2$, $\left(12.5\times 150\right){\text{mm}}^2$ for $A_2$, and 75 mm for $y_2$.

$\begin{array}{c}I=\frac{200\times {12.5}^3}{12}+200\times 12.5{\left(156.25-121.43\right)}^2+\frac{12.5\times {150}^3}{12}+12.5\times 150{\left(75-121.43\right)}^2\\ =32,552.08+3,031,081+3,515,625+4,042,021.69\\ =10.621\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^4\\ =10.621\times {10}^{-6}{\text{m}}^4\end{array}$

Refer to Figure 4;

$\begin{array}{l}{y}_{bottom}=-121.43\text{mm}\\ y_{top}=41.07\text{mm}\end{array}$

Tension at Points B and D:

Refer to Figure 5;

Determine the moment at points B and D using the relation.

$M_B\text{and}{M}_D=-\frac{{\left({\sigma }_{all}\right)}_{T}I}{y_{top}}$

Substitute 110 MPa for ${\left({\sigma }_{all}\right)}_T$, $10.621\times {10}^{-6}{\text{m}}^4$ for I, and 41.07 mm for $y_{top}$.

$\begin{array}{c}{M}_B\text{and}{M}_D=-\frac{110\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 10.621\times {10}^{-6}}{41.07\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ =-28.45\text{kN-m}\end{array}$

Compression at Points B and C:

Refer to Figure 5;

Determine the moment at points B and D using the relation.

$M_B\text{and}{M}_D=-\frac{{\left({\sigma }_{all}\right)}_{C}I}{y_{bottom}}$

Substitute –150 MPa for ${\left({\sigma }_{all}\right)}_C$, $10.621\times {10}^{-6}{\text{m}}^4$ for I, and –121.43 mm for $y_{bottom}$.

$\begin{array}{c}{M}_B\text{and}{M}_D=-\frac{-150\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 10.621\times {10}^{-6}}{-121.43\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ =-13.12\text{kN-m}\end{array}$

Tension at maximum bending moment:

Refer to Figure 5;

Determine the maximum moment using the relation.

$M_{\max }=-\frac{{\left({\sigma }_{all}\right)}_{T}I}{y_{bottom}}$

Substitute 110 MPa for ${\left({\sigma }_{all}\right)}_T$, $10.621\times {10}^{-6}{\text{m}}^4$ for I, and –121.43 mm for $y_{bottom}$.

$\begin{array}{c}{M}_{\max }=-\frac{110\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 10.621\times {10}^{-6}}{-121.43\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ =9.62\text{kN-m}\end{array}$

Compression at maximum bending moment:

Refer to Figure 5;

Determine the maximum moment using the relation.

$M_{\max }=-\frac{{\left({\sigma }_{all}\right)}_{C}I}{y_{top}}$

Substitute –150 MPa for ${\left({\sigma }_{all}\right)}_C$, $10.621\times {10}^{-6}{\text{m}}^4$ for I, and 41.07 mm for $y_{top}$.

$\begin{array}{c}{M}_{\max }=-\frac{-150\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 10.621\times {10}^{-6}}{41.07\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ =38.79\text{kN-m}\end{array}$

Refer to the calculated distribution loads; the smallest value controls the design.

Refer to Figure 5;

Equate the maximum bending moment calculated and the maximum bending moment in the tension side.

$\begin{array}{l}6.48w=9.62\text{kN-m}\\ w=1.485\text{kN/m}\end{array}$

Therefore, the largest permissible value of w for the condition that the beam BC is not overstressed is $\underset{\_}{1.485\text{kN/m}}$.

To determine

The maximum distance a for the condition that the beams AB and CD are not overstressed.

Answer to Problem 89P

The maximum distance a for the condition that the beams AB and CD are not overstressed is $\underset{\_}{1.935\text{m}}$.

Explanation of Solution

Refer to Part (a), Figure 4;

The maximum bending moment in the beams AB and CD occurs at the ends A and D.

The calculated maximum bending moment at the points A and D is as follows:

$M_A\text{and}{M}_D=-\left(\frac{w{a}^2}{2}+3.6wa\right)$

The maximum allowable compression moment at the points A and D is as follows:

$M_B\text{and}{M}_D=-13.12\text{kN-m}$

Equate the values;

$-\left(\frac{w{a}^2}{2}+3.6wa\right)=-13.12\text{kN-m}$

Refer to the answer of the part (a); $w=1.485\text{kN/m}$.

Substitute $1.485\text{kN/m}$ for w.

$\begin{array}{l}-\left(\frac{1.485a^2}{2}+3.6\times 1.485a\right)=-13.12\text{kN-m}\\ 0.7425a^2+5.346a-13.12=0\\ a=1.935\text{m}\end{array}$

Therefore, the maximum distance a for the condition that the beams AB and CD are not overstressed is $\underset{\_}{1.935\text{m}}$.