Chapter 4, Problem 4.5.27P

The simple beam ACE shown in the figure is subjected to a triangular load of maximum intensity q₀= 200 lb/ft at a = 8 ft and a concentrated moment M = 400 Ib-ft at A.

1. Draw the shear-force and bending-moment diagrams for this beam,
2. Find the value of distanced that results in the maximum moment occurring at L/2. Draw the shear-force and bending-moment diagrams for this case.
3. Find the value of distance a for which M_maxis the largest possible value.

  

To determine

To draw: Shear force and bending moment diagrams for simply supported beam.

Answer to Problem 4.5.27P

The $SFD$ and $BMD$ are shown in explanation part.

Explanation of Solution

Given Information:

Max load $q_0=200\text{lb/ft}$

Distance $a=8\text{ ft}$

Moment $M=400\text{ lb-ft at A}$

Length $L=10\text{ ft}$

Concept Used:

Shear forces and bending moments at various points shall be calculated.

Calculation:

Draw free body diagram

From equilibrium

  $R_A+R_B=\frac{1}{2}(200\times 8+200\times 2)=1000\text{ lb }\mathrm{............}(1)$

Also,

$\sum M_B=0$

$\begin{array}{l}{R}_A\times 10+400-[\frac{1}{2}\times 200\times 8\times (\frac{8}{3}+10-8)]-\frac{1}{2}\times 200\times 2\times \frac{2\times 2}{3}=0\\ R_A\times 10+400-3733.3333-266.6667=0\\ R_A=360\text{ lb}\end{array}$

From equation $\left(1\right)$

  $\begin{array}{l}{R}_A+R_B=1000\\ 360+R_B=1000\\ R_B=640\end{array}$

Shear Force calculation

  $\begin{array}{l}S{F}_B=-R_B=-640\\ S{F}_{C{B}_x}=0.5x\times 100x-R_B=50x^2-640\\ S{F}_C={[}_{50}=50\times 2^2-640=-440\\ S{F}_A=R_A=360\end{array}$

SFD

Bending Moment calculation

  $\begin{array}{l}B{M}_B=0\\ \text{BM between C and B when x from right}\\ B{M}_{C{B}_x}=R_B\times x-\frac{1}{2}\times x\times 100x\times \frac{x}{3}=640x-\frac{50}{3}{x}^3\\ B{M}_C={[}_{640}=640\times 2-\frac{50}{3}\times 2^3=1146.67\\ \text{BM between A and C when x from left}\\ B{M}_{A{C}_x}=R_A\times x+400-\frac{1}{2}\times x\times 25x\times (\frac{x}{3})=360x+400-4.167x^3\\ B{M}_A=400\end{array}$

BMD

Conclusion:

The $SFD$ and $BMD$ are shown in explanation part.

To determine

To find: The value of $a$ for maximum bending moment at $L/2$

Answer to Problem 4.5.27P

The distance is $a=5\text{ ft}$

Explanation of Solution

Given Information:

Max load $q_0=200\text{lb/ft}$

Distance $=a$

Moment $M=400\text{ lb-ft at A}$

Length $L=10\text{ ft}$

Concept Used:

Shear forces and bending moments at various points shall be calculated.

Calculation:

The free body diagram is as follows:

From equilibrium

  $R_A+R_B=\frac{1}{2}[200\times a+200\times (10-a)]=1000\text{ lb }\mathrm{............}(1)$

Also,

$\sum M_B=0$

$\begin{array}{l}{R}_A\times 10+400-[\frac{1}{2}\times 200\times a\times (10-0.67a)]-\frac{1}{2}\times 200\times (10-a)\times \frac{(10-a)\times 2}{3}=0\\ R_A\times 10+400-1000a+67a^2-6666.67+1333.33a-66.67a^2=0\\ R_A=-33.333a+626.667\text{ lb}\end{array}$

From equation $\left(1\right)$

  $\begin{array}{l}{R}_A+R_B=1000\\ -33.33a+626.67+R_B=1000\\ R_B=33.33a+373.33\text{ lb}\end{array}$

Bending moment

$\begin{array}{l}B{M}_{L/2}=R_A\times 0.5L+400-\frac{1}{2}\times 0.5L\times \frac{200}{a}\times 0.5L\times (\frac{0.5L}{3})\\ B{M}_{L/2}=(-33.33a+626.67)\times 5+400-\frac{1}{2}\times 5\times \frac{200}{a}\times 5\times (\frac{5}{3})\\ B{M}_{L/2}=-166.65a+3133.35-4166.67/a\\ \\ \text{For max moment}\\ \frac{d}{da}(B{M}_{L/2})=0\\ \frac{d}{da}(-166.65a+3133.35-4166.67/a)=0\\ -166.65+4166.67/a^2=0\\ -166.65a^2+4166.67=0\end{array}$

On solving above equation, we get:

$\begin{array}{l}a=\pm 5\\ a=5\text{ ft}\end{array}$

Shear Force calculation:

  $\begin{array}{l}S{F}_B=-R_B=-(33.33a+373.33)=540\text{ lb}\\ S{F}_{C{B}_x}=0.5x\times \frac{200}{5}x-R_B=20x^2-540\\ S{F}_C={[}_{20}=20\times 5^2-540=-40\text{ lb}\\ S{F}_A=R_A=-33.333a+626.66=460\end{array}$

SFD

Bending Moment calculation:

  $\begin{array}{l}B{M}_B=0\\ \text{BM between C and B when x from right}\\ B{M}_{C{B}_x}=R_B\times x-\frac{1}{2}\times x\times 40x\times \frac{x}{3}=540x-\frac{20}{3}{x}^3\\ B{M}_C={[}_{540}=540\times 5-\frac{20}{3}\times 5^3=-1866.67\\ B{M}_A=-400\end{array}$

BMD

Conclusion:

The distance is $a=5\text{ ft}$ .

To determine

To find: The value of $a$ for maximum bending moment.

Answer to Problem 4.5.27P

The distance is $a=4.7005\text{ ft}$ .

Explanation of Solution

Given Information:

Max load $q_0=200\text{lb/ft}$

Distance $=a$

Moment $M=400\text{ lb-ft at A}$

Length $L=10\text{ ft}$

Concept Used:

Bending moment shall be calculated.

Calculation:

The free body diagram:

From part $b$ we get,

$R_A=-33.333a+626.667\text{ lb}$

  $R_B=33.33a+373.33\text{ lb}$

Bending moment

  $\begin{array}{l}\text{BM between C and B when x from right:}\\ B{M}_{C{B}_x}=R_B\times x-\frac{1}{2}\times x\times \frac{200}{10-a}x\times \frac{x}{3}=(33.33a+373.33)x-\frac{33.33}{10-a}{x}^3\\ \text{But }x=10-a\\ B{M}_{C{B}_x}=[33.33(10-x)+373.33]x-33.33x^2\\ B{M}_{C{B}_x}=[333.3x-33.33x^2+373.33x-33.33x^2]=-66.66x^2+706.63x\\ \\ \text{for max bending moment :}\\ \frac{d}{dx}(B{M}_{C{B}_x})=0\\ \frac{d}{dx}(-66.66x^2+706.63x)=0\\ 133.33x-706.63=0\\ x=5.2995\\ \\ a=10-x\\ a=10-5.2995=4.7005\end{array}$

And maximum bending moment is $M_{\max }=1872.67$

Conclusion:

The distance is $a=4.7005\text{ ft}$ .