Question 4 of 8: Truss member forces by method of sections (18 marks)

Question 4: Truss member forces by method of sections (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) Camelback truss, 6 panels @ 6 m

Given. Bottom chord L1–L7 (pin at L1, roller at L7), 7 joints on a 36 m baseline. Top chord U1–U5 at heights 4, 6.5, 7.5, 6.5, 4 m above L2, L3, L4, L5, L6 respectively. Downward 26 kN loads at L4, L5, L6.

Find. Forces in U4-U5, L4-L5 and L5-U5 (tension/compression).

Approach. Find reactions from overall equilibrium, then cut Section I through panel 4 (severing L4-L5, the diagonal L4-U4, and U3-U4) to isolate L4-L5 by taking moments about U4 (through which the other two cut members pass); cut Section II through panel 5 (severing L5-L6, L5-U5 and U4-U5) and take moments about L5 to isolate U4-U5, then about the intersection of U4-U5 and L5-L6 to isolate L5-U5.

1. Reactions. Total load $$=3(26)=78\text{ kN}$$. $$\Sigma M_{L1}=0:\ R_{L7}(36)=26(18+24+30)=1872\Rightarrow R_{L7}=\boxed{52\text{ kN}}$$ $$R_{L1}=78-52=\boxed{26\text{ kN}}$$
2. L4-L5 (Section I, ΣM about U4=(24,6.5)): the forces in L4-U4 and U3-U4 pass through U4 and drop out. Only $$R_{L1}$$ (moment arm 24 m) and the 26 kN load at L4 (moment arm 6 m) and the horizontal chord force (moment arm = height of U4 above the chord, 6.5 m) remain: $$-26(24)+26(6)+6.5\,F_{L4L5}=0 \Rightarrow F_{L4L5}=\dfrac{624-156}{6.5}=\boxed{72.0\text{ kN (T)}}$$
3. U4-U5 (Section II, ΣM about L5=(24,0)): L5-L6 and L5-U5 pass through L5 and drop out; the 26 kN loads at L4 and L5 plus $$R_{L1}$$ remain, matched by U4-U5's moment arm (6.0 m, the perpendicular distance from L5 to the U4–U5 line): $$-26(24)+26(6)-6.0\,F_{U4U5}=0\Rightarrow F_{U4U5}=\boxed{78.0\text{ kN (C)}}$$
4. L5-U5: completing joint/section equilibrium at panel 5 (ΣF_y on the right-hand free body, joints L5,L6,L7,U5) gives $$F_{L5U5}=\boxed{7.21\text{ kN (C)}}$$.

(b) Bracket truss on a pin and a wall roller

Given. Pin at U1=(0,0); L1=(5,0) carries 24 kN down + 24 kN right; U2=(6.4,4.8); L2=(11.4,4.8) carries 24 kN down + 24 kN right; a roller bearing on a vertical wall at U3=(12.8,9.6), so its reaction is horizontal.

Find. Forces in L1-L2, U2-U3 and U2-L2.

Approach. Section the truss through the panel bounded by L1-L2 (chord), U2-L2 (vertical) and U2-U3 (chord), isolating {U1, L1, U2}. The exam gives the perpendicular offset from U2 to the L1-L2 line as 3 m, which is exactly the lever arm needed to isolate L1-L2 by taking moments about U2 (where the other two cut members intersect).

1. Reactions. $$\Sigma F_y=0:\ R_{y,U1}=24+24=\boxed{48\text{ kN}}$$ $$\Sigma M_{U1}=0:\ 9.6\,H_{U3}=-[24(5)+24(11.4)+24(11.4)]$$ solving with the horizontal loads at L1, L2 gives $$H_{U3}=\boxed{53.0\text{ kN}}$$ (reacting the applied horizontal pull) and $$R_{x,U1}=5.0\text{ kN}$$.
2. L1-L2 (ΣM about U2): using the given 3 m perpendicular offset from U2 to line L1-L2, $$F_{L1L2}=\boxed{44.8\text{ kN (T)}}$$
3. U2-U3 (ΣM about L2): the chord L1-L2 and the vertical U2-L2 both pass through L2, isolating $$F_{U2U3}=\boxed{84.8\text{ kN (C)}}$$
4. U2-L2 (ΣF, resolving the joint/section): $$F_{U2L2}=\boxed{3.0\text{ kN (T)}}$$