# Question: Problem 7.53 Part A Draw the shear diagram for the beam. Click on “add discontinuity” to add discontinuity lines.–Free Chegg Question Answer

Transcribed text From Image:Problem 7.53 Part A Draw the shear diagram for the beam. Click on "add discontinuity" to add discontinuity lines. Then click on "add segment" button to add functions between the lines. add vertical line off add segment 0 lb/ft 200 lb ft 20 ft 0 ft Vlb 400 300 200 100 x,ft 30 10 100 Submit My Answers Give Ue
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Answer:

## Step-by-step

### Step 1 of 5

Part A

Draw the free body diagram.

Here, {B_y}By​ is the vertical component of reaction force at support B.

Consider the vertical equilibrium condition.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\{A_y} + {B_y} – \left( {50} \right)\left( {20} \right) = 0\\\\{A_y} = 1000 – {B_y}\\\end{array}∑Fy​=0Ay​+By​−(50)(20)=0Ay​=1000−By​​

Here, {A_y}Ay​ is the vertical reaction at A.

Substitute 510\;{\rm{lb}}510lb for By.

\begin{array}{l}\\{A_y} = 1000 – 510\\\\\,\,\,\,\,\,\, = 490{\rm{ lb}}\\\end{array}Ay​=1000−510=490lb​

Consider the horizontal equilibrium condition.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\{A_x} = 0\\\end{array}∑Fx​=0Ax​=0​

Here, {A_x}Ax​ is the horizontal component of reaction force at support A.Explanation | Common mistakes | Hint for next step

The vertical component of reaction force at support B is calculated by applying moment equilibrium condition about point A. Then the horizontal and vertical components of reaction force at point A are calculated by applying force equilibrium conditions along x and y axis respectively.

### Step 2 of 5

Consider a section between AB and at a distance x from end A \left( {0 \le x < 20\,{\rm{ft}}} \right)(0≤x<20ft) .

Consider equilibrium equation in vertical direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\{A_y} – V – \left( {50} \right)\left( x \right) = 0\\\\V = {A_y} – \left( {50} \right)\left( x \right)\\\end{array}∑Fy​=0Ay​−V−(50)(x)=0V=Ay​−(50)(x)​

Substitute {\rm{490}}\;{\rm{lb}}490lb for Ay.

V = 490 – 50xV=490−50x

\begin{array}{l}\\\sum {{M_x}} = 0\\\\M + \left( {50} \right)\left( x \right)\left( {\frac{x}{2}} \right) – \left( {{A_y}} \right)\left( x \right) = 0\\\\M = \left( {{A_y}} \right)\left( x \right) – 25{x^2}\\\end{array}∑Mx​=0M+(50)(x)(2x​)−(Ay​)(x)=0M=(Ay​)(x)−25x2​

Substitute {\rm{490}}\;{\rm{lb}}490lb for Ay.

M = 490x – 25{x^2}M=490x−25x2

Calculate shear force and bending moments at A and B.

At

At

\begin{array}{c}\\{M_B} = 490\left( {20} \right) – 25{\left( {20} \right)^2}\\\\ = – 200{\rm{ lb}} \cdot {\rm{ft}}\\\end{array}MB​=490(20)−25(20)2=−200lb⋅ft​

Calculate the location of the maximum bending moment.

At maximum bending moment shear force reaches to zero

\begin{array}{l}\\V = 0\\\\490 – 50x = 0\\\\x = \frac{{490}}{{50}}\\\\\,\,\,\, = 9.8{\rm{ ft}}\\\end{array}V=0490−50x=0x=50490​=9.8ft​

Calculate the magnitude of the bending moment.

M = 490x – 25{x^2}M=490x−25x2

\begin{array}{l}\\{M_{\max }} = 490\left( {9.8} \right) – 25{\left( {9.8} \right)^2}\\\\\,\,\,\,\,\,\,\,\,\,\,\, = 2401{\rm{ lb}} \cdot {\rm{ft}}\\\end{array}Mmax​=490(9.8)−25(9.8)2=2401lb⋅ft​Explanation | Common mistakes | Hint for next step

The values of shear force and bending moment between section AB are calculated by using force and moment equilibrium conditions.

### Step 3 of 5

Consider a section between BC and at a distance x from end A \left( {20\,{\rm{ft}} \le x < 30\,{\rm{ft}}} \right)(20ft≤x<30ft) .

Consider equilibrium equation in vertical direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\ – V + {A_y} + {B_y} – \left( {50} \right)\left( {20} \right) = 0\\\\V = 490 + 510 – 1000\\\\\,\,\,\,\, = 0\\\end{array}∑Fy​=0−V+Ay​+By​−(50)(20)=0V=490+510−1000=0​

\begin{array}{l}\\\sum {{M_x}} = 0\\\\M + \left( {50} \right)\left( {20} \right)\left( {x – 10} \right) – \left( {{A_y}} \right)\left( x \right) – \left( {{B_y}} \right)\left( {x – 20} \right) = 0\\\\M = 490x + 510\left( {x – 20} \right) – 1000\left( {x – 10} \right)\\\\\,\,\,\,\,\, = 490x + 510x – 10200 – 1000x + 10000\\\\\,\,\,\,\,\,\, = – 200{\rm{ lb}} \cdot {\rm{ft}}\\\end{array}∑Mx​=0M+(50)(20)(x−10)−(Ay​)(x)−(By​)(x−20)=0M=490x+510(x−20)−1000(x−10)=490x+510x−10200−1000x+10000=−200lb⋅ft​

At , the shear force and bending moment are,

{M_B} = – 200{\rm{ lb}} \cdot {\rm{ft}}MB​=−200lb⋅ft

At ,

{M_C} = – 200{\rm{ lb}} \cdot {\rm{ft}}MC​=−200lb⋅ftExplanation | Common mistakes | Hint for next step

The values of shear force and bending moment between section BC are calculated by using force and moment equilibrium conditions.

### Step 4 of 5

Draw the shear force diagram as follows:

[Part A]

Part APart A

The shear force diagram is as follows:

Explanation | Hint for next step

The shear force values are calculated at different points of the beam. Then the shear force diagram is drawn by plotting the shear force values on the y-axis and beam length on the x-axis.

### Step 5 of 5

Draw the bending moment diagram of the beam as follows:

[Part B]

Part BPart B

The bending moment diagram is as follows:

Explanation

The bending moment values are calculated at different points of the beam. Then the bending moment is drawn by plotting the bending moment values on the y-axis and beam length on the x-axis.