Chapter 5.3, Problem 86P

Solve Prob. 5.85, assuming that the cross section of the beam is inverted, with the flange of the beam resting on the supports at B and C.

5.85 Determine the largest permissible distributed load w for the beam shown, knowing that the allowable normal stress is +80 MPa in tension and –130 MPa in compression.

Fig. P5.85

To determine

The largest permissible distributed load w.

Answer to Problem 86P

The largest permissible distributed load (w) is $\underset{\_}{108.8\text{kN/m}}$.

Explanation of Solution

Given information:

The allowable normal stress of the material in tension is ${\left({\sigma }_{all}\right)}_T=+80\text{MPa}$.

The allowable normal stress of the material in compression is ${\left({\sigma }_{all}\right)}_C=-130\text{MPa}$

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \left(w\times 0.2\right)\times \frac{0.2}{2}+C_y\left(0.5\right)-\left(w\times 0.7\right)\times \frac{0.7}{2}=0\\ 0.02w+0.5C_y-0.245w=0\\ C_y=0.45w\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ B_y-\left(w\times 0.9\right)+C_y=0\\ B_y-0.9w+0.45w=0\\ B_y=0.45w\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=0$

$\begin{array}{c}{V}_B-V_A=-w\times 0.2\\ V_B^{Left}=-0.2w+V_A\\ =-0.2w+0\\ =-0.2w\end{array}$

$\begin{array}{c}{V}_B^{Right}=-0.2w+0.45w\\ =0.25w\end{array}$

$\begin{array}{c}{V}_C-V_B=-w\times 0.5\\ V_C^{Left}=-0.5w+V_B\\ =-0.5+0.25w\\ =-0.25w\end{array}$

$\begin{array}{c}{V}_C^{Right}=-0.25w+0.45w\\ =0.20w\end{array}$

$\begin{array}{c}{V}_D-V_C=-w\times 0.2\\ V_D=0+V_C\\ =-0.2w+0.2w\\ =0\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	0
B (Left)	–0.2w
B (Right)	0.25w
C (Left)	–0.25w
C (Right)	0.2w
D	0

Plot the shear force diagram as in Figure 2.

Location of the maximum bending moment:

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

Refer to Figure 2;

Use the similar triangle concept.

$\begin{array}{l}\frac{0.25w}{x}=\frac{0.25w}{0.5-x}\\ 0.5-x=x\\ 2x=0.5\\ x=0.25\text{m}\end{array}$

The maximum bending moment occurs at a distance of 0.45 m from left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_B-M_A=-\frac{1}{2}\times 0.20w\times 0.2\\ M_B=-0.02w+M_A\\ =-0.02w+0\\ =-0.02w\end{array}$

$\begin{array}{c}{M}_{\max }-M_B=\frac{1}{2}\times 0.25w\times 0.25\\ M_{\max }=0.03125w+M_B\\ =0.03125w-0.02w\\ =0.01125w\end{array}$

$\begin{array}{c}{M}_C-M_{\max }=-\frac{1}{2}\times 0.25w\times 0.25\\ M_C=-0.03125w+M_{\max }\\ =-0.03125w+0.01125\\ =-0.02w\end{array}$

$M_D=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	–0.02w
Max BM	0.01125w
C	–0.02w
D	0

Plot the bending moment diagram as in Figure 3.

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

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;

$\begin{array}{l}{A}_1=\left(20\times 60\right){\text{mm}}^2;y_1=50\text{mm}\\ A_2=\left(60\times 20\right){\text{mm}}^2;y_2=10\text{mm}\end{array}$

Substitute $\left(20\times 60\right){\text{mm}}^2$ for $A_1$, 50 mm for $y_1$, $\left(60\times 20\right){\text{mm}}^2$ for $A_2$, and 10 mm for $y_2$.

$\begin{array}{c}\overline{y}=\frac{\left(20\times 60\times 50\right)+\left(60\times 20\times 10\right)}{\left(20\times 60\right)+\left(60\times 20\right)}\\ =30\text{mm}\end{array}$

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

$I=\frac{d_1{b}_1^3}{12}+A_1{\left(y_1-\overline{y}\right)}^2+\frac{d_2{b}_2^3}{12}+A_2{\left(y_2-\overline{y}\right)}^2$

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 60 mm for $d_1$, 20 mm for $b_1$, $\left(20\times 60\right){\text{mm}}^2$ for $A_1$, 50 mm for $y_1$, 30 mm for $\overline{y}$, 20 mm for $d_2$, 60 mm for $b_2$, $\left(60\times 20\right){\text{mm}}^2$ for $A_2$, and 10 mm for $y_2$.

$\begin{array}{c}I=\frac{60\times {20}^3}{12}+20\times 60{\left(50-30\right)}^2+\frac{20\times {60}^3}{12}+60\times 20{\left(10-30\right)}^2\\ =40,000+480,000+360,000+480,000\\ =1.36\times {10}^6{\text{mm}}^4\times {\left(\frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^4\\ =1.36\times {10}^{-6}{\text{m}}^4\end{array}$

Refer to Figure 4;

$\begin{array}{l}{y}_{bottom}=30\text{mm}\\ {\text{y}}_{top}=50\text{mm}\end{array}$

Tension at Points B and C:

Refer to Figure 3;

Determine the distributed load (w) using the relation.

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

Substitute –0.02w for $M_B\text{and}{M}_C$, 80 MPa for ${\left({\sigma }_{all}\right)}_T$, $1.36\times {10}^{-6}{\text{m}}^4$ for I, and 50 mm for $y_{top}$.

$\begin{array}{l}-0.02w=-\frac{80\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 1.36\times {10}^{-6}}{50\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ w=108.8\text{kN/m}\end{array}$

Compression at Points B and C:

Refer to Figure 3;

Determine the distributed load (w) using the relation.

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

Substitute –0.02w for $M_B\text{and}{M}_C$, –130 MPa for ${\left({\sigma }_{all}\right)}_C$, $1.36\times {10}^{-6}{\text{m}}^4$ for I, and –30 mm for $y_{bottom}$.

$\begin{array}{l}-0.02w=-\frac{-130\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 1.36\times {10}^{-6}}{-30\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ w=294.7\text{kN/m}\end{array}$

Tension at Points E:

Refer to Figure 3;

Determine the distributed load (w) using the relation.

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

Substitute 0.01125w for $M_E$, 80 MPa for ${\left({\sigma }_{all}\right)}_T$, $1.36\times {10}^{-6}{\text{m}}^4$ for I, and –30 mm for $y_{bottom}$.

$\begin{array}{l}0.01125w=-\frac{80\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 1.36\times {10}^{-6}}{-30\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ w=322.4\text{kN/m}\end{array}$

Compression at Points E:

Refer to Figure 3;

Determine the distributed load (w) using the relation.

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

Substitute 0.01125w for $M_E$, –130 MPa for ${\left({\sigma }_{all}\right)}_C$, $1.36\times {10}^{-6}{\text{m}}^4$ for I, and –50 mm for $y_{top}$.

$\begin{array}{l}0.01125w=-\frac{-130\text{MPa}\times \frac{{10}^3{\text{kN/m}}^2}{1\text{MPa}}\times 1.36\times {10}^{-6}}{50\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}}\\ w=314.3\text{kN/m}\end{array}$

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

Therefore, the largest permissible distributed load (w) is $\underset{\_}{108.8\text{kN/m}}$.