Chapter 9, Problem 9.1P

Determine the reactions at the suooorts, then draw the shear and moment diagrams. Assume the support at A is fixed and B is a roller, EI is constant.

To determine

The reactions at the supports and to draw the shear and moment diagrams.

Answer to Problem 9.1P

The vertical reaction at support A is $2400\text{lb}$.

The horizontal reaction at support A is $0\text{lb}$.

The reaction moment at support A is $4800\text{lb-ft}$.

The vertical reaction at support B is $600\text{lb}$.

The shear diagram is shown below.

The moment diagram is shown below.

Explanation of Solution

Concept Used:

Write the expression for the net force balance in the vertical direction of the beam.

${\displaystyle \sum F_Y=0}$ .. (I)

Here, ${\displaystyle \sum F_Y}$ is the sum of all the vertical forces.

Write the expression for the net force balance in the horizontal direction in the beam.

${\displaystyle \sum F_X=0}$ .. (II)

Here, ${\displaystyle \sum F_X}$ is the sum of all the horizontal forces.

Write the expression for the net moment about end $A$ of the beam.

${\displaystyle \sum M_A=0}$ .... (III)

Here, ${\displaystyle \sum M_A}$ is the sum of all the moments taken about the end $A$.

Calculations:

The free body diagram for the beam is shown below.

       Figure (1)

Here, the vertical reaction at point $A$ is $A_y$, the vertical reaction at point $B$ is $B_y$ and the horizontal reaction at point $A$ is $A_x$ and the reaction moment at A is $M_A$.

Calculate the support reactions using Equation (II).

$\begin{array}{l}{A}_y+B_y-\left(500\times \frac{12}{2}\right)\text{lb}=0\\ A_y+B_y=\left(500\times \frac{12}{2}\right)\text{lb}\\ A_y+B_y=3000\text{lb}\end{array}$. (IV)

The uniformly varying load is replaced by a concentrated force of magnitude $3000\text{lb}$ as shown below.

       Figure (2)

Consider the moment at point A using Equation (III).

$\begin{array}{l}{M}_A+\left(B_y\times 12\right)-\left(3000\times \frac{12}{3}\right)=0\\ M_A+12B_y-12000=0\end{array}$ (V)

The displacement of the beam for the given load is shown below.

       Figure (3)

Calculate the displacement of the beam for the given load.

$v_B^{\text{'}}=\frac{w{L}^4}{30EI}$ (VI)

Here, the displacement of the beam for the given load is $v_B^{\text{'}}$.

The displacement of the beam for the reaction at point B. is shown below.

Calculate the displacement of the beam for the given load.

$v_B^{"}=-\frac{B_y{L}^3}{3EI}$ (VII)

Here, the displacement of the beam for the reaction at point B is $v_B^{"}$.

Add Equation (VI) and Equation (VII) for the displacement values according to the compatibility condition.

$\begin{array}{l}{v}_B^{\text{'}}+v_B^{"}=0\\ \frac{w{L}^4}{30EI}+\left(-\frac{B_y{L}^3}{3EI}\right)=0\\ \frac{B_y{L}^3}{3EI}=\frac{w{L}^4}{30EI}\\ B_y=\frac{w{L}^4}{30EI}\times \frac{3EI}{L^3}\end{array}$

$B_y=\frac{w{L}^{}}{10}$ (VIII)

Substitute $500\text{lb}/\text{ft}$ for $w$ and $12\text{ft}$ for L in Equation (VIII).

$\begin{array}{l}{B}_y=\frac{500\text{lb}/\text{ft}\times 12\text{ft}}{10}\\ B_y=600\text{lb}\end{array}$ (IX)

Calculate the vertical reaction at A using Equation (IV).

Substitute $600\text{lb}$ for $B_y$ in Equation (IV).

$\begin{array}{l}{A}_y+600\text{lb}=\text{3000}\text{lb}\\ A_y=2400\text{lb}\end{array}$

Calculate the bending moment at A using Equation (V).

Substitute $600\text{lb}$ for $B_y$ in Equation (V).

$\begin{array}{l}{M}_A+\left(12\times 600\right)\text{lb-ft}-12000\text{(lb-ft)}=0\\ M_A=12000\text{lb-ft}-\left(12\times 600\right)\text{lb-ft}\\ M_A=12000\text{lb-ft}-72\text{00}\text{lb-ft}\\ M_A=4800\text{lb-ft}\end{array}$

Consider the beam as shown below.

       Figure (5)

Write the equation to determine the shear in the beam.

$\begin{array}{l}V+B_y=0.5\times x\times 500\left(\frac{x}{12}\right)\\ V+600=\frac{250x^2}{12}\\ V=\frac{250x^2}{12}-600\end{array}$ (X)

Calculate shear force at a distance $x=0\text{ft}$.

Substitute $0\text{ft}$ for $x$ in Equation (V).

$\begin{array}{c}V=\left(\frac{250\times 0^2}{12}-600\right)\text{lb}\\ =-600\text{lb}\end{array}$

Calculate shear force at a distance $x=3\text{ft}$.

Substitute $3\text{ft}$ for $x$ in Equation (V).

$\begin{array}{c}V=\left(\frac{250\times 3^2}{12}-600\right)\text{lb}\\ =\left(\frac{2250}{12}-600\right)\text{lb}\\ =\left(187.5-600\right)\text{lb}\\ =-412.5\text{lb}\end{array}$

Calculate shear force at a distance $x=6\text{ft}$.

Substitute $6\text{ft}$ for $x$ in Equation (V).

$\begin{array}{c}V=\left(\frac{250\times 6^2}{12}-600\right)\text{lb}\\ =\left(\frac{9000}{12}-600\right)\text{lb}\\ =\left(750-600\right)\text{lb}\\ =150\text{lb}\end{array}$

Calculate shear force at a distance $x=9\text{ft}$.

Substitute $9\text{ft}$ for $x$ in Equation (V).

$\begin{array}{c}V=\left(\frac{250\times 9^2}{12}-600\right)\text{lb}\\ =\left(\frac{20250}{12}-600\right)\text{lb}\\ =\left(1687.5-600\right)\text{lb}\\ =1087.5\text{lb}\end{array}$

Calculate shear force at a distance $x=12\text{ft}$.

Substitute $12\text{ft}$ for $x$ in Equation (V).

$\begin{array}{c}V=\left(\frac{250\times {12}^2}{12}-600\right)\text{lb}\\ =\left(\frac{36000}{12}-600\right)\text{lb}\\ =\left(3000-600\right)\text{lb}\\ =2400\text{lb}\end{array}$

Substitute $0$ for $V$ in Equation (X) to get location of maximum moment.

$\begin{array}{l}0=\frac{250x^2}{12}-600\\ \frac{250x^2}{12}=600\\ x^2=\frac{600\times 12}{250}\\ x^2=28.8{\text{ft}}^{\text{2}}\end{array}$

$x=5.366\text{ft}$

Write the expression for the bending moment.

$\begin{array}{l}\left(600\times x\right)-\left(0.5\times x\times \frac{500x}{12}\right)\left(\frac{x}{3}\right)+M=0\\ M=\left(0.5\times x\times \frac{500x}{12}\right)\left(\frac{x}{3}\right)-\left(600\times x\right)\\ M=\left(\frac{250\times x^3}{36}-\left(600\times x\right)\right)\text{lb-ft}\end{array}$. (XI)

Calculate moment at a distance $x=0\text{ft}$.

Substitute $0$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times 0^3}{36}-\left(600\times 0\right)\right)\text{lb-ft}\\ =0\text{lb-ft}\end{array}$

Calculate moment at a distance $x=3\text{ft}$.

Substitute $3\text{ft}$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times 3^3}{36}-\left(600\times 3\right)\right)\text{lb-ft}\\ =\left(\frac{6750}{36}-1800\right)\text{lb-ft}\\ =\left(187.5-1800\right)\text{lb-ft}\\ =-1612.5\text{lb-ft}\end{array}$

Calculate moment at a distance $x=6\text{ft}$.

Substitute $6\text{ft}$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times 6^3}{36}-\left(600\times 6\right)\right)\text{lb-ft}\\ =\left(\frac{54000}{36}-3600\right)\text{lb-ft}\\ =\left(1500-3600\right)\text{lb-ft}\\ =-2100\text{lb-ft}\end{array}$

Calculate moment at a distance $x=9\text{ft}$.

Substitute $9\text{ft}$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times 9^3}{36}-\left(600\times 9\right)\right)\text{lb-ft}\\ =\left(\frac{182250}{36}-5400\right)\text{lb-ft}\\ =\left(5062.5-5400\right)\text{lb-ft}\\ =-337.5\text{lb-ft}\end{array}$

Calculate moment at a distance $x=12\text{ft}$.

Substitute $12\text{ft}$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times {12}^3}{36}-\left(600\times 12\right)\right)\text{lb-ft}\\ =\left(\frac{432000}{36}-7200\right)\text{lb-ft}\\ =\left(12000-7200\right)\text{lb-ft}\\ =4800\text{lb-ft}\end{array}$

Calculate moment at a distance $x=5.366\text{ft}$.

Substitute $3\text{ft}$ for $x$ in Equation (XI).

$\begin{array}{c}M=\left(\frac{250\times {5.366}^3}{36}-\left(600\times 5.366\right)\right)\text{lb-ft}\\ =\left(\frac{38627.091}{36}-3219.6\right)\text{lb-ft}\\ =\left(1072.97-3219.6\right)\text{lb-ft}\\ =-2146.63\text{lb-ft}\end{array}$

Conclusion:

The vertical reaction at support A is $2400\text{lb}$.

The horizontal reaction at support A is $0\text{lb}$.

The reaction moment at support A is $4800\text{lb-ft}$.

The vertical reaction at support B is $600\text{lb}$.

The shear diagram is shown below

       Figure (6)

The moment diagram is shown below

       Figure (7)