Question 4 of 8: Truss member forces by the method of sections/joints

Question 4: Truss member forces by the method of sections/joints (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Calculate the forces in U₁–U₂, L₂–U₂ and L₂–L₃

Given. A 24 m Warren truss, height 4 m, panels of 6 m (top chord U₁–U₄ at x=3,9,15,21; bottom chord L₁–L₅ at x=0,6,12,18,24); pin at L₁, roller at L₅; 24 kN horizontal at U₁; 32/40/48 kN downward at L₂/L₃/L₄.

Find. Forces in U₁U₂, L₂U₂, L₂L₃ (tension/compression).

Approach. Find the reactions from overall equilibrium, then cut a section through the panel containing L2/U2 (severing U1U2, L2U2 and L2L3) and take moments/force-sums of the left free body.

1. Reactions. $$\Sigma F_x=0:\ R_{x,L1}=-24\text{ kN}$$ $$\Sigma M_{L1}=0:\ 24R_{y,L5}=24(4)+32(6)+40(12)+48(18)=1632\Rightarrow R_{y,L5}=\boxed{68\text{ kN}}$$ $$R_{y,L1}=32+40+48-68=\boxed{52\text{ kN}}$$
2. Section cut through the L2/U2 panel — left free body {L1, U1, L2}. $$\Sigma F_y=0:\ 52-32+0.8\,F_{L2U2}=0\Rightarrow F_{L2U2}=\boxed{-25\text{ kN (25 kN compression)}}$$
3. Moments about U₂ (9,4) (eliminates U₁U₂ and L₂U₂, both passing through it): $$4F_{L2L3}=468\Rightarrow F_{L2L3}=\boxed{+117\text{ kN (tension)}}$$
4. Then $$\Sigma F_x=0:\ -24+F_{U1U2}+0.6(-25)+117=0\Rightarrow F_{U1U2}=\boxed{-102\text{ kN (102 kN compression)}}$$

(b) Calculate the forces in L₁–U₂, L₁–L₂ and U₁–L₃

Given. A 2-panel (3 m each), 4 m tall truss pinned to a rigid wall at both U₁ and L₁; verticals at each panel point; diagonals L₁U₂ and U₁L₃ (the counter-diagonal spanning both panels) and L₂U₃; 20 kN down at U₂, 36 kN horizontal at U₃, 60 kN down at each of L₂ and L₃.

Find. Forces in L₁U₂, L₁L₂, U₁L₃ (tension/compression).

Approach. Two pins into one rigid wall give 4 reaction components — one more than the 3 a single rigid body needs — but that redundancy lives entirely in the vertical U₁L₁ member and the reaction split; it never enters the equilibrium equations of the four away-from-the-wall joints (U₂, U₃, L₂, L₃), so those 8 equations solve the other 8 members — including all three asked for — uniquely by the method of joints.

1. Joint U₂ (members U₁U₂, U₂U₃, L₁U₂; load 20 kN down). Direction L₁→U₂ is (3,4)/5. $$\Sigma F_y=0:\ -0.8F_{L1U2}-20=0\Rightarrow F_{L1U2}=\boxed{-25\text{ kN (25 kN compression)}}$$
2. Joint L₃ (members L₂L₃, L₃U₃ vertical, U₁L₃; load 60 kN down). Direction L₃→U₁ is (−6,4)/√52. Working the four free joints together (L₂’s and U₃’s equilibrium first fixes L₂U₃=75 kN (T) and U₃L₃=0) gives, at L₃: $$0.5547\,F_{U1L3}=120\Rightarrow F_{U1L3}=60\sqrt{13}=\boxed{+216.3\text{ kN (tension)}}$$
3. Joint L₂ (members L₁L₂, L₂L₃, L₂U₃; load 60 kN down) with L₂L₃=−180 kN (from L₃’s $$\Sigma F_x$$): $$F_{L1L2}=F_{L2L3}+0.6(75)=-180+45=\boxed{-135\text{ kN (135 kN compression)}}$$