1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• w = 220 N/m
• OA = 0.5 m
• AC = 0.2 m
• AB = 0.4 m
• L = 1.9 m

Unknowns:

• N_c
• V_c
• M_c

4. Approach

Use equilibrium equations. First, solve for reaction forces, then make a cut at C and solve for the internal forces.

5. Analysis

$$w=\frac{F}{L}\\F=wL\\F_R=220N/m\cdot 1.9m\\F_R=418N\\\sum F_X=0=B_X$$

Find reaction forces:

$$\sum M_A=0=B_y(0.4m)-F_R(0.45m)\\(0.4m)B_y=418N(0.45m)\\B_y=\frac{188.1 N\cdot m}{0.4m}\\B_y=470.25N$$

$$\sum F_y=0=-F_R+A_y+B_y\\A_y=F_R-B_y\\A_y=418N-470.25N\\A_y=-52.25N$$

The answer we got for A_y is negative, which means that the arrow should be drawn in the other direction. We will change it for our next sketch.

Make a cut at C:

Now solve for the internal forces:

$$\sum F_x=0\:\:;\:\:N_c=0\\\sum F_y=0=-A_y-V_c-(w\cdot L_A)\\V_c=-52.25N-(220N/m\cdot 0.7m)\\V_c=-206.25 N\\\sum M_c=A_y(0.2m)+M_c+(F_{Rc}\cdot 0.35m)\\M_c=52.25 N (0.2m)+(220N/m\cdot 0.7m\cdot 0.35m)\\M_c=64.35N\cdot m$$

Final FBD, showing the arrows in the correct directions:

6. Review

It makes sense that A_y and B_y are in different directions, because the resultant force Fr of the solar panel on the beam is not between A and B. It also makes sense that the moment at C is in the counterclockwise direction rather than the clockwise direction, when you think about the direction of the forces applied to the beam.

1. Problem

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F1 = 500 N
• F2 = 300 N

Unknowns:

• A_y
• A_x
• C_y
• V_B
• M_B
• N_B

4. Approach

Shear/moment equations, EOM equations

5. Analysis

Solve for reaction forces:

(Ax, Cy)

\begin{aligned}
\sum F_{x}=0=A_{x}=0 \\
\sum M_{A}=0 &=-F_{1} \cdot 2 m-F_{2} \cdot 5.5 m+C_{y} \cdot 8 m \\
C_{y}=&\frac{F_{1} \cdot 2 m+F_{2} \cdot 5.5 m}{8m} \\
C_{y} &=\frac{500 N \cdot 2 m+300 N \cdot 5.5 m}{8 m} \\
C_{y} &=331.25 \mathrm{~N}
\end{aligned}

(Ay)

\begin{aligned}
\sum F_{y}=0 &=A_{y}+C_{y}-F_{1}-F_{2} \\
A_{y} &=F_{1}+F_{2}-C_{y} \\
A_{y} &=500 \mathrm{~N}+300 \mathrm{~N} – 331.25 \mathrm{~N} \\
A_{y} &=468.75 \mathrm{~N}
\end{aligned}

Cut 1: at B

\begin{aligned}
\sum F_{X}=0=A_{X} &+N_{B}=0 \\
& N_{B}=0 \\
\sum F_{y}=0 &=A_{y}-V_{B}-F_{1} \\
V_{B} &=A_{y}-F_{1} \\
V_{B} &=468.75 N – 500 N \\
V_{B}=-31.25 N
\end{aligned}

\begin{aligned}
\sum M_{B}=& 0=-A_{y}(4 m)+F_{1}(2 m)+M_{B} \\
& M_{B}=A_{y}(4 m)-F_{1}(2 m) \\
& M_{B}=468.75 N(4 m)-500 N(2 m) \\
M_{B} &=875 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Cut 2: At the point where F₁ is applied

\begin{aligned}
\sum M_{1}=0 &=-A_{y}(2 m)+M_{1}=0 \\
M_{1} &=A_{y}(2 m) \\
M_{1} &=468.75 N(2 m) \\
M_{1} &=937.5 \mathrm{~N} \cdot m
\end{aligned}

Cut 3: At the point where F₂ is applied

\begin{aligned}
\sum M_{2}=0 =-A_{y}(5.5 \mathrm{~m})+F_{1}(3.5 \mathrm{~m})+M_{2} \\
M_{2} =A_{y}(5.5 \mathrm{~m})-F_{1}(3.5 \mathrm{~m}) \\
M_{2} =468.75 \mathrm{~N}(5.5 \mathrm{~m})-500 \mathrm{~N}(3.5 \mathrm{~m}) \\
M_{2} =828.125 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Answer: N_B = 0, V_B = -31.25 N, M_B = 875 Nm

6. Review

The reaction forces make sense as they offset the applied forces. The shear/moment diagrams returned to zero, so they are correct too. The moment found at B is in the moment diagram, it is smaller than the maximum.

1. Problem 

2. Draw 

3. Knowns and Unknowns 

Knowns:

• F_g = 30N
• r _AB =  r _BC =  0.04 m

Unknowns:

• Reaction forces = R_Ax, R_Ay, R_Cy
• For the segment of beam AB nd BC,
• Shear = V, V₂
• Bending Moment = M, M₂
• Normal force = N, N₂

4. Approach 

Find reaction forces from equilibrium equations. Use them to find the shear and bending moment of segment AB and BC. Draw the V-M diagram from the answers.

5. Analysis 

Finding reaction forces:

$$\sum F_x=0=R_{Ax}\\R_{Ax}=0N$$

Solving for RC_y:

$$\sum M_A=0= -(r_{AB}\cdot F_g)+(r_{AC}\cdot R_{Cy})\\r_{AB}\cdot F_{g}=r_{AC}\cdot R_{Cy} $$

$$R_{Cy}=\frac{r_{AB}\cdot F_g}{r_{AC}}\\R_{By}=\frac{0.04m\cdot 30 N}{0.08 m} $$

$$R_{Cy}=15 N$$

Solving for RA_y:

$$\sum F_y=0=R_{Cy}+R_{Ay} -F_g\\R_{Ay}=F_g-R_{Cy}\\R_{Ay}= 15 N$$

Finding shear and Moment from A to B :

$$\sum F_x=0\\N=0N$$

Solving for V

$$\sum F_y=0=R_{Ay}+-V\\ V= 15 N$$

Solving for M :

$$\sum M_A=0= -( V\cdot 0.04) +M\\M = 15\cdot 0.04\\M=0.6N $$

Finding shear and Moment from B to C:

$$\sum F_x=0\\N=0N$$

Solving for V

$$\sum F_y=0=R_{Cy}+V_2\\ V_2= -15 N$$

Solving for M :

$$\sum M_C=0= -( V_2\cdot(0.04) -M_2\\M_2 = 15\cdot 0.04\\M=0.6N $$

V-M diagram:

For V-x graph, from A to B there constant shear that is 15 N then from B to C the shear has same magnitude but different direction.  To plot the M-x graph, substitute x in the equation for M and M₂ to find the peak bending moment. Also, because M is the integral of shear force, horizontal plots in shear correspond to a linear plot in the M-x graph.

6. Review 

The answers make sense. The moment, shear, and reaction forces are equal in both segments of the beam. This is as expected because of the symmetry of the beam.

1. Problem

Roughly sketch the shear and moment diagram for the following structure. Measure from point A.

2. Sketch

N/A, Sketch Provided.

3. Knows and Unknowns:

Knowns:

• Distributed Load: w
• Dimensions and shown

Unknown:

• Shear/Moment Diagram

4. Approach:

Break the beam into sections as forces change.

Determine the plotted shape of the shear and moment of each section.

Draw the V/M diagrams using these shape diagrams.

5. Analysis:

A distributed load creates a linear section across it, with the magnitude of the shear force increasing across the load. The maximum shear will be at the end of the distributed load. With no other forces being applied except for reactions, section B-C should be constant. Since the beam is static, the reactions must bring the shear back to zero.

The moment can be drawn using the knowledge that dM = V. Section A-B in the shear diagram is linear; integrating a linear function gives a quadratic function, thus, section A-B in the moment diagram must be curved. Section B-C is constant in shear. Integrating a constant function gives a linear function; therefore, the moment across this section must be linear, and since the shear is negative, the slope of the moment must be negative. Finally, because the beam is static, the moment must return to zero due to the reaction forces.

6. Review

Using principles of shear force (V) and moments and their integral/derivative relationship, the shear and moment diagrams make sense for their general shape.

1. Problem

Find the reaction forces and create the shear and moment diagrams for the beam.

2. Draw

$$ 4N/m \cdot \frac{5.5m}{2} = 11N$$

$$ \bar{x} = 1/3 \cdot 5.5m = 1.83m$$

$$ \bar{x} \ from  A = 4m + 4.5m + 1.83m = 10.3m$$

3. Knowns and Unknowns

Knowns:

• Applied Forces:
• F_B = -20N
• F_C = -400N
• F_D = 500N
• F_Load = -4N/m
• Dimensions as shown

Unknowns:

• Reaction Forces: R_AX, R_AY, M_A
• Shear/Moment Diagram

4. Approach

1. Use equilibrium equation to find the reaction forces
2. Use beam analysis/cutting between forces to find equations for the shear/moment diagram.

5. Analysis

First, use the sum of forces and the sum of moments to determine R_Ay and M_A.

$$ \sum F_x = R_{Ax} = 0$$

$$ \sum F_y = -R_{Ay} – 20N – 400N – 11N +500N = 0$$

$$ R_{Ay} = 69N$$

$$ \sum M_A = – M_A – (20N \cdot 4m) – (400N \cdot 8.5m) – (11N \cdot 10.3m) + (500 N \cdot 14m) = 0 $$

$$ M_A = 3406.7Nm$$

Next, cut the beam between A and B and analyze the internal forces.

$$ \sum F_x = N + 0 = 0 \\ N = 0$$

$$ \sum F_y = -69N -V = 0 \\ V= -69N$$

$$ \sum M_{cut} =  M + (69N \cdot x) – (3406.7Nm) = 0 \\ M = -69x + 3406.7$$

Repeat the analysis of internal forces from A to C.

$$ \sum F_x = N + 0 = 0 \\ N = 0$$

$$ \sum F_y = -69N – 20N -V = 0 \\ V = -89N$$

$$ \sum M_{cut} = M + (69N \cdot x) + (20N \cdot{(x-4m)}) – 3406.7Nm = 0 \\ M+89x-3326.7Nm = 0 \\ M = -89x + 3326.7$$

Finally,  analyze the beam in the area from C to D. Ensure that the value of X used is now taken from point D.

$$ \sum F_x = -N = 0 \\ N = 0$$

Since the section does not reach the end of the distributed load, the slope of the load must be calculated:

$$\frac{4}{5.5} = \frac{m}{14-x} \\ m = \frac{56-4x}{5.5}$$

This can be used to find the equivalent point load in terms of x.

$$ F_{eq} = \frac{\frac{56-4x}{5.5} \cdot (14-x)}{2} \\ F_{eq} = \frac{4x^2}{11} – \frac{112x}{11} + \frac{784}{11}$$

This point load can be used when summing forces to find v.

$$\sum F_y = 0 = V – F_{eq} + 500 \\ V = F_{eq} – 500 \\ V = \frac{4x^2}{11} – \frac{112x}{11} + \frac{784}{11} – 500 \\ V = \frac{4x^2}{11} – \frac{112x}{11} – 428.7$$

To save time, rather than use the sum of moments to determine the equation for moments in this section, the moment can be found by the integral of V.

$$ M = \int \frac{4x^2}{11} – \frac{112x}{11} – 428.7 dx  \\ M = \frac{4x^3}{33} – \frac{112x^2}{22} – 428.7x +C$$

We know that for this beam to be static, the moment at x=14m must be 0. This can be used to determine C.

$$ M(14) = 0 =  \frac{4(14)^3}{33} – \frac{112(14)^2}{22} – 428.7(14) +C \\ C = 6667 \\ Therefore: M_{cd}(x) = \frac{4x^3}{33} – \frac{112x^2}{22} – 428.7x + 6667$$

With the values for internal forces found between each change in force, the internal forces can be plotted.

6. Review

Using principles of shear force (V) and moments and their integral/derivative relationship, the shear and moment diagrams make sense for their general shape. The numerical values calculated align with the variables provided.

1. Problem

A person who weighs 140lbs is standing on the edge of a 10ft long diving board. The board is secured by a pin at 0ft and supported by a roller at 4ft.

a. Calculate the reaction forces on the board.

b. Calculate the internal forces at the midpoint between A and B.

c. Calculate the internal forces at the midpoint between B and C.

d. Draw the shear and moment diagram.

source: https://www.flickr.com/photos/lac-bac/40229806264

2. Draw 

3. Knowns and Unknowns 

Knowns:

• F_g = 140 lbs
• X_A= 0 ft
• X_B= 4 ft
• X_C= 10 ft

Unknowns:

• R_Ax
• R_Bx
• R_Ay
• Internal forces N, V, M between B and C, & A and B

4. Approach 

Calculate the reaction forces using equilibrium equations. Cut the diving board between A & B, and use equilibrium equations to find internal forces. Do the same to find internal forces between B and C. Use the values to draw the shear and moment diagram.

5. Analysis

a.

[latex]\sum{F_{x}} = 0=R_{Ax}[/latex]

[latex]\sum{F_{y}}=0=R_{Ay}+R_{By}-140 lb \\R_{Ay}+R_{By}=140 lb[/latex]

[latex]\sum{M_{A}}=\left(X_{B} \cdot R_{By}\right)-\left(X_{c} \cdot F_{g}\right)=0[/latex]

[latex]R_{By}= \frac{X_{c}\cdot F_{g}}{X_{B}}\\R_{By}= \frac{10 ft\cdot 140 lb}{4 ft}[/latex]

[latex]R_{By}=350 lb[/latex]

now, [latex]R_{Ay}=140 lb - R_{By}\\ \kern 1pc=140 lb - 350 lb \\ \kern 1pc =-210 lb[/latex]

The negative sign for [latex]R_{Ay}[/latex] means that the direction of [latex]R_{Ay}[/latex]  is not +y but in the -y direction.

Thus, the reation forces are [latex]R_{Ay}=-210 lb, R_{By} = 350 lb, R_{Ax}=0[/latex]

b. Finding internal forces between A and B 

To find the internal forces between A and B, cut in the middle of A and B, giving [latex]x = \frac{X_{B}}{2} =2 ft[/latex]

[latex]\sum{F_{x}} = 0=R_{Ax} + N\\ N = 0[/latex]

[latex]\sum{F_{y}}=0=-V - R_{Ay}\\ V = -R_{Ay}\\V=-210 lb[/latex]

[latex]\sum{M_{A}}=0\\ \kern 2pc =M-xV\\ \kern 2pc = M- \frac{4 ft}{2}\left(-210 lb\right)\\  M =-420 lb ft[/latex]

Internal forces between A and B are [latex]N = 0, V= -210 lb, M=-420 lb ft[/latex]

c.  Finding internal forces between B and C

Cut at the midpoint between B and C to find the internal forces.

Finding x

To apply the equilibrium equation, you need to find x in this scenario.

[latex]x= X_{B}+\frac{X_{c}-X_{B}}{2}\\ = 4 ft +\frac{10 ft -4 ft}{2}\\ x =7 ft[/latex]

[latex]\sum{F_{x}} = 0=R_{Ax} + N\\ N = 0[/latex]

[latex]\sum{F_{y}}=0=R_{By}-R_{Ay}-V\\ \kern 1pc V=R_{By} -R_{Ay} \\ \kern 1pc V = 350 lb -210 lb\\ V= 140 lb[/latex]

now,

[latex]\sum{M_{A}}=0[/latex]

[latex]M =- \left( X_{B} \cdot R_{By} \right) + xV\\ \kern 1 pc = -\left(4 ft \cdot 350 lb\right) +\left( 7 ft \cdot140 lb\right) \\ \kern 1 pc = 980 lb ft- 1400 lb ft[/latex]

[latex]M =-420 lb ft[/latex]

Internal forces between C and B are [latex]N = 0, V= 140 lb, M=-420 lb ft[/latex]

d. Shear and Moment diagram 

6. Review 

There is no force in the x direction; therefore, it makes sense that the reaction force, R_Ax, is zero. Because the moment at A is generated by the person’s weight, R_By should be considerably large. As R_By is larger than the person’s weight, R_Ay should act downwards and would be of the same magnitude as the difference between R_By and the person’s weight to make the system stable.

For section b, the shear force between A and B is equal and opposite to the reaction force R_Ay. Similarly, with the concept of equilibrium, shear is equal and opposite to R_Ax, and the moment in this section is the negative of the moment generated at A.

It also makes sense that the shear between B and C is equal to the difference between R_By and R_Ay, and as expected, acting downward, thus it equals the weight of the person.

1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F=200N
• L= 30cm => 0.3m

Unknowns:

• Internal normal force, shear force and bending moment at point B
• Shear + moment diagram

4. Approach

Solve for reaction forces at A, then take arbitrary sections between A and B and between B and C to derive V and M equations. Plot results.

5. Analysis

Solve for reaction forces:

\begin{aligned}
\sum M_{A} &= 0 = -200\,\mathrm{N} \cdot 0.3\,\mathrm{m} – M_A \\
M_A &= -60\,\mathrm{Nm}
\end{aligned}

\begin{aligned}
\sum F_{y} &= 0 = A_y – 200\,\mathrm{N} \\
A_y &= 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum F_{x} &= 0 = N \\
N &= 0
\end{aligned}

No force in the x-direction

Cut 1: With reaction forces known, cut a section between A and B

0m<x<1.5m

\begin{aligned}
\sum F_{y} &= 0 = A_y – V \\
V &= A_y = 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum M_x &= 0 = -M_A – A_y \cdot x +M(x) \\
M(x) &= 200x-60\,\mathrm{Nm}
\end{aligned}

You can also cut the section between B and C

0.15m<x<3m

\begin{aligned}
\sum F_{y} &= 0 = A_y – V \\
V &= A_y = 200\,\mathrm{N}
\end{aligned}

\begin{aligned}
\sum M_x &= 0 = M_x – M_A- A_y \cdot x \\
M(x) &= 200x – 60\,\mathrm{Nm}
\end{aligned}

\begin{aligned}
\text{At } x = 0, \quad &M = 0 \\
\text{At } x = 0.15\,\mathrm{m}, \quad &M = -30\,\mathrm{Nm} \\
\text{At } x = 0.3\,\mathrm{m}, \quad &M = 0
\end{aligned}

Answer: N = 0, V = 200 N, M_x = 200x-60 Nm

6. Review

Negative sign for M_A indicates that it should be in the other direction, as shown in the final diagram above. The shear force remains constant at 200N from A to C, then drops to zero at the end where the load is applied. The shear force is constant, hence it makes sense that the moment is linear.

1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• Mass of frame (𝑀𝑓) = 3kg
• Length of frame (𝐿𝑓) = 0.8m
• Mass of each lightbulb (𝑀𝑏) = 0.5kg

Unknowns:

• Tension in each rod (𝑇₁ 𝑎𝑛𝑑 𝑇₂)
• Internal forces at point D (𝑁_D, 𝑉_D, 𝑎𝑛𝑑 𝑀_D)

4. Approach

Assuming the vertical rods are positioned centrally between the bulbs, and the internal forces at points C and D are the same (use any point to get the internal forces). First, find the tension and then use it in calculating internal forces.

5. Analysis

Weight of frame (𝑊𝑓) = 𝑀𝑓𝑔 = 3kg × 9.81m/s² = 29.43N

Weight of one bulb (𝑊𝑏) = 𝑀𝑏𝑔 = 0.5kg × 9.81m/s² = 4.905N

The bulbs are evenly spaced, so the distance between 2 bulbs (from centre to centre):

[latex]\frac{0.8\,\mathrm{m}}{4} = 0.2\,\mathrm{m}[/latex]

To get the tension in each rod:

[latex]+\uparrow \sum F_y = 0 \quad ; \quad T_1 + T_2 - 5W_b - W_f = 0[/latex]

Since the rods are connected symmetrically, they are equal (𝑇₁ = 𝑇₂ ), therefore:

To get the internal forces at point D, taking the right side of the diagram:

Since the rods are positioned centrally between the bulbs, the distance from point D to the bulb is

[latex]\frac{0.2\,\mathrm{m}}{2} = 0.1\,\mathrm{m}[/latex]

No force in the x-direction

6. Review

To check, taking the left side of the diagram