Questions: Method of Sections 1. The given truss members have equal lengths. Assume that the force at B, E, and D has the magnitude of FB=10 kN, FE=20 kN, and FD=30 kN, respectively. (a) Draw a free-body diagram for the truss. (b) Use the diagram to calculate the reaction forces at A and C. (c) Use the method of sections to compute the forces in the members of the truss AB, BE, and ED. (d) Identify which force in the members of the truss is a compression or a tension.

Method of Sections
1. The given truss members have equal lengths. Assume that the force at B, E, and D has the magnitude of FB=10 kN, FE=20 kN, and FD=30 kN, respectively.
(a) Draw a free-body diagram for the truss.
(b) Use the diagram to calculate the reaction forces at A and C.
(c) Use the method of sections to compute the forces in the members of the truss AB, BE, and ED.
(d) Identify which force in the members of the truss is a compression or a tension.

Transcript text: Method of Sections 1. The given truss members have equal lengths. Assume that the force at $\mathrm{B}, \mathrm{E}$, and D has the magnitude of $F_{B}=10 \mathrm{kN}, F_{E}=20 \mathrm{kN}$, and $F_{D}=30 \mathrm{kN}$, respectively. (a) Draw a free-body diagram for the truss. (b) Use the diagram to calculate the reaction forces at A and C . (c) Use the method of sections to compute the forces in the members of the truss $\mathrm{AB}, \mathrm{BE}$, and ED. (d) Identify which force in the members of the truss is a compression or a tension.

Solution

Solution Steps

Step 1: Draw a free-body diagram for the truss

Draw the truss structure with all the given forces $ F_B = 10 \text{ kN} $, $ F_E = 20 \text{ kN} $, and $ F_D = 30 \text{ kN} $ applied at points B, E, and D respectively.
Indicate the reaction forces at supports A and C. Assume $ A_y $ and $ C_y $ are the vertical reaction forces at A and C, respectively, and $ C_x $ is the horizontal reaction force at C.

Step 2: Calculate the reaction forces at A and C

Apply the equilibrium equations to the entire truss:
- Sum of vertical forces: $ \sum F_y = 0 $ \[ A_y + C_y - F_B - F_E - F_D = 0 \] \[ A_y + C_y - 10 - 20 - 30 = 0 \] \[ A_y + C_y = 60 \text{ kN} \]
- Sum of horizontal forces: $ \sum F_x = 0 $ \[ C_x = 0 \]
- Sum of moments about point A: $ \sum M_A = 0 $ \[ -F_B \cdot L + F_E \cdot 2L + F_D \cdot 3L - C_y \cdot 4L = 0 \] \[ -10L + 20 \cdot 2L + 30 \cdot 3L - C_y \cdot 4L = 0 \] \[ -10L + 40L + 90L - 4L \cdot C_y = 0 \] \[ 120L = 4L \cdot C_y \] \[ C_y = 30 \text{ kN} \]
- Substitute $ C_y $ back into the vertical force equation: \[ A_y + 30 = 60 \] \[ A_y = 30 \text{ kN} \]

Step 3: Use the method of sections to compute the forces in the members of the truss AB, BE, and ED

Cut the truss through members AB, BE, and ED and consider the left section.
Apply the equilibrium equations to the left section:
- Sum of vertical forces: $ \sum F_y = 0 $ \[ A_y - F_{AB} \sin \theta - F_{BE} \sin \theta = 0 \]
- Sum of horizontal forces: $ \sum F_x = 0 $ \[ F_{AB} \cos \theta - F_{BE} \cos \theta = 0 \]
- Sum of moments about point E: $ \sum M_E = 0 $ \[ A_y \cdot 2L - F_{AB} \cdot L \sin \theta = 0 \] \[ 30 \cdot 2L - F_{AB} \cdot L \sin \theta = 0 \] \[ 60L = F_{AB} \cdot L \sin \theta \] \[ F_{AB} = 60 \text{ kN} \]