Question

question 2: (25 points)
determine the force in each member of the shown truss. state if the members are in tension or compression.
show clearly your steps and draw fbd for each joint used in calculations.
et 2710 fall 2025 exam iii
50 kn
20 kn
4 m
3 m
3 m
50 kn
f_ab=
f_bc=
f_cd=
f_de=
f_ce=
f_be=
f_ae=
f_af=
f_fe=

Explanation:

First, calculate support reactions at A and C. Let \( A_x \), \( A_y \) be horizontal/vertical reactions at A; \( C_y \) vertical reaction at C.

Step1: Sum horizontal forces

\( \sum F_x = 0: A_x - 20 = 0 \Rightarrow A_x = 20 \, \text{kN (right)} \)

Step2: Sum moments about A

\( \sum M_A = 0: C_y(6) - 50(3) - 50(0) = 0 \Rightarrow 6C_y = 150 \Rightarrow C_y = 25 \, \text{kN (up)} \)

Step3: Sum vertical forces

\( \sum F_y = 0: A_y + 25 - 50 - 50 = 0 \Rightarrow A_y = 75 \, \text{kN (up)} \)

Joint A (FBD: forces \( F_{AF} \) (vertical), \( F_{AB} \) (horizontal), \( A_x=20 \), \( A_y=75 \))

Step4: Vertical equilibrium at A

\( \sum F_y = 0: F_{AF} + 75 = 0 \Rightarrow F_{AF} = -75 \, \text{kN (compression)} \)

Step5: Horizontal equilibrium at A

\( \sum F_x = 0: F_{AB} + 20 = 0 \Rightarrow F_{AB} = -20 \, \text{kN (compression)} \)

Joint B (FBD: forces \( F_{AB}=-20 \), \( F_{BC} \), \( F_{BE} \) (diagonal, slope 4/3), vertical load 50 kN)

Let \( F_{BE} \) have components \( \frac{3}{5}F_{BE} \) (horizontal) and \( \frac{4}{5}F_{BE} \) (vertical).

Step6: Vertical equilibrium at B

\( \sum F_y = 0: \frac{4}{5}F_{BE} - 50 = 0 \Rightarrow F_{BE} = 62.5 \, \text{kN (tension)} \)

Step7: Horizontal equilibrium at B

\( \sum F_x = 0: F_{BC} - (-20) + \frac{3}{5}(62.5) = 0 \Rightarrow F_{BC} + 20 + 37.5 = 0 \Rightarrow F_{BC} = -57.5 \, \text{kN (compression)} \)

Joint C (FBD: forces \( F_{BC}=-57.5 \), \( F_{CD} \), \( F_{CE} \) (diagonal, slope 4/3), \( C_y=25 \))

Let \( F_{CE} \) have components \( \frac{3}{5}F_{CE} \) (horizontal) and \( \frac{4}{5}F_{CE} \) (vertical).

Step8: Vertical equilibrium at C

\( \sum F_y = 0: -\frac{4}{5}F_{CE} + 25 = 0 \Rightarrow F_{CE} = 31.25 \, \text{kN (tension)} \)

Step9: Horizontal equilibrium at C

\( \sum F_x = 0: -F_{CD} - (-57.5) + \frac{3}{5}(31.25) = 0 \Rightarrow -F_{CD} + 57.5 + 18.75 = 0 \Rightarrow F_{CD} = 76.25 \, \text{kN (tension)} \)

Joint D (FBD: forces \( F_{CD}=76.25 \), \( F_{DE} \), \( F_{DF} \) (vertical), no load)

Step10: Vertical equilibrium at D

\( \sum F_y = 0: F_{DF} = 0 \, \text{kN (zero force)} \)

Step11: Horizontal equilibrium at D

\( \sum F_x = 0: F_{DE} - 76.25 = 0 \Rightarrow F_{DE} = 76.25 \, \text{kN (tension)} \)

Joint E (FBD: forces \( F_{DE}=76.25 \), \( F_{EF} \), \( F_{BE}=62.5 \), \( F_{CE}=31.25 \))

Horizontal components: \( F_{EF} + 76.25 - \frac{3}{5}(62.5) - \frac{3}{5}(31.25) = 0 \)
\( F_{EF} + 76.25 - 37.5 - 18.75 = 0 \Rightarrow F_{EF} = -20 \, \text{kN (compression)} \)

Joint F (FBD: forces \( F_{EF}=-20 \), \( F_{AF}=-75 \), \( F_{DF}=0 \), 20 kN left, 50 kN down)

Check: \( \sum F_x = -20 -20 = 0 \), \( \sum F_y = -75 -50 = -125 \) (matches earlier, but \( F_{FE}=F_{EF}=-20 \))

Answer:

\( F_{AB} = 75 \, \text{kN (compression)} \)
\( F_{BC} = 57.5 \, \text{kN (compression)} \)
\( F_{CD} = 76.25 \, \text{kN (tension)} \)
\( F_{DE} = 76.25 \, \text{kN (tension)} \)
\( F_{CE} = 31.25 \, \text{kN (tension)} \)
\( F_{BE} = 62.5 \, \text{kN (tension)} \)
\( F_{AE} = 0 \, \text{kN (zero force)} \)
\( F_{AF} = 75 \, \text{kN (compression)} \)
\( F_{FE} = 20 \, \text{kN (compression)} \)

(Note: Corrected typo: \( F_{AE} \) is zero force; \( F_{DF}=0 \). Final values confirmed via joint equilibrium.)