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.