07-Str-A1 · December 2015

Question 4 of 8: Member forces in two trusses

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2015, 07-Str-A1 — Elementary Structural Analysis. Three hours, closed book, approved Sharp or Casio calculator only. Six questions constitute a complete paper: answer ALL of Questions 1 to 4, then ONLY TWO of Questions 5, 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are worked below, because the four alternatives exercise four different methods — moment distribution, three-hinged-frame statics, influence lines and virtual work — and the complete set is the more useful study resource.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3–5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8–9 (deflections and virtual work), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (virtual work), Ch. 8–9 (influence lines), Ch. 16–17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text this exam code is named after.

Check: the roller in Question 2(c). The roller under the inclined member of Question 2(c) is drawn on a bed ruled parallel to the member, so its reaction is taken normal to the member axis. A vertical-roller reading is also arithmetically self-consistent. The distinction changes only the reaction components and the axial force: the shear and bending-moment diagrams the question asks for are identical under either reading, and both sets of reactions are reported.

Question 4: Member forces in two trusses (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.

Given. Two plane roof trusses, each spanning 16 m in four 4 m panels, pinned at the left support and on a roller at the right, each carrying two 36 kN downward panel loads.

Given data for Question 4
Truss	Joint coordinates (x, y), m	Loads
(a)	L₁(0,0), L₂(4,0), L₃(8,0), L₄(12,0), L₅(16,0); U₁(4,3), U₂(8,5), U₃(12,3)	36 kN down at L₃ and at L₄
(b)	L₁(0,0), L₂(4,0), L₃(12,0), L₄(16,0); M₁(4,3), M₂(8,3), M₃(12,3); U₁(8,6)	36 kN down at U₁ and at M₃

Find. The force in each listed member, stated as tension or compression.

4(a) — the listed members by joints and sections

Figure 6 — truss 4(a). The three members asked for are drawn in red. Note that the rafter L₁–U₁ lies on a 3 : 4 slope, so its unit vector is (0.8, 0.6).

Approach. Take the reactions first, resolve at the support joint for the rafter, then pass one vertical section through the panel that contains the diagonal and take moments about a convenient pole to isolate each remaining member in turn.

Reactions. Moments about L₁ with the two 36 kN loads at 8 m and 12 m give $$ 16 V_5 = 36(8) + 36(12) = 720 \quad\Longrightarrow\quad V_5 = 45\ \text{kN}, \qquad V_1 = 72 - 45 = 27\ \text{kN}. $$ The pin carries no horizontal force, since no horizontal load is applied.
Joint L₁ for the rafter. Only two members meet the support joint: the bottom chord, which is horizontal, and the rafter L₁–U₁ on a 3 : 4 slope. Vertical equilibrium at that joint reads $$ 0.6\,F_{L_1U_1} + 27 = 0 \quad\Longrightarrow\quad \boxed{F_{L_1U_1} = 45\ \text{kN compression}} $$ and horizontal equilibrium then gives $F_{L_1L_2} = +36$ kN, a tension. Because the vertical U₁–L₂ carries no load, the bottom-chord force is unchanged across L₂, so $F_{L_2L_3} = 36$ kN tension as well.
A section for the diagonal U₁–L₃. Cut vertically at $x = 6$ m: the cut severs the top chord U₁–U₂, the diagonal U₁–L₃ and the bottom chord L₂–L₃, and no load lies on the left-hand part. The top chord runs from (4, 3) towards (8, 5), so produced backwards it crosses the bottom chord at $(-2, 0)$; taking moments about that point removes both chords at once and leaves only the diagonal, whose unit vector is $(0.8,\,-0.6)$: $$ \sum M_{(-2,0)} = 0 : \qquad 27(2) - 6\,F_{U_1L_3} = 0 \quad\Longrightarrow\quad \boxed{F_{U_1L_3} = 9\ \text{kN tension}} $$ The lever arm of 6 m is the perpendicular distance from $(-2,0)$ to the line of the diagonal.
A section for the bottom chord L₃–L₄. Cut vertically at $x = 10$ m, severing L₃–L₄, U₃–L₃ and U₂–U₃. The last two both pass through U₃(12, 3), so taking moments about U₃ for the right-hand part — which carries the 36 kN at L₄, acting through the same point, and the 45 kN reaction 4 m away — gives $$ 45(4) - 3\,F_{L_3L_4} = 0 \quad\Longrightarrow\quad \boxed{F_{L_3L_4} = 60\ \text{kN tension}} $$ the lever arm being the 3 m height of U₃ above the bottom chord.
Check by an independent route. Resolving at joint U₁ with $F_{L_1U_1} = -45$ and $F_{U_1L_3} = +9$ returns $F_{U_1U_2} = -48.3$ kN, and joint L₃ then closes with $F_{U_2L_3} = 43.2$ kN tension and $F_{U_3L_3} = 21.0$ kN compression. Global vertical equilibrium of the whole truss is satisfied to the last digit, which is the cheap confirmation that no sign has been dropped.

4(b) — a truss the method of sections cannot open

[Figure not reproduced: Figure 7 — truss 4(b). Both rafters are straight lines through three joints: L 1 , M 1 , U 1 are collinear on a 3 : 4 slope, and so are U 1 , M 3 , L 4 . The thin vertical line through M 2 in the examination figure is a dimension witness line, not a member. See the official exam paper.]

Approach. Every vertical cut between L₂ and L₃ severs four members — the bottom chord, a diagonal, the sub-chord and a rafter — so three equations cannot settle it and the method of sections is unusable. Walk the joints instead, starting from the two support joints where only two members meet.

Reactions. Moments about L₁ with the loads at 8 m (at U₁) and 12 m (at M₃) give $$ 16 V_4 = 36(8) + 36(12) = 720 \quad\Longrightarrow\quad V_4 = 45\ \text{kN}, \qquad V_1 = 27\ \text{kN}. $$
Joint L₄. The roller joint carries the bottom chord and the right-hand rafter, whose unit vector towards M₃ is $(-0.8,\,0.6)$. Vertical equilibrium gives $0.6\,F_{L_4M_3} + 45 = 0$, so $F_{L_4M_3} = 75$ kN compression, and horizontal equilibrium gives $F_{L_3L_4} = 60$ kN tension.
Joint U₁, the apex. Only the two rafters meet there, and they are symmetric about the vertical, so the 36 kN load splits equally: $$ -1.2\,F = 36 \quad\Longrightarrow\quad F_{U_1M_1} = F_{U_1M_3} = 30\ \text{kN compression}. $$
Joint M₃. Four members meet here — the two collinear rafter lengths, the sub-chord M₂–M₃ and the vertical M₃–L₃ — but two of the four are now known, so the joint closes. Horizontal equilibrium reads $$ -75(0.8) + (-30)(-0.8) - F_{M_2M_3} = 0 \quad\Longrightarrow\quad F_{M_2M_3} = -36\ \text{kN}, $$ 36 kN compression, and vertical equilibrium, which must also carry the 36 kN panel load, gives $F_{M_3L_3} = 9$ kN compression.
Joint L₃. With the bottom chord to the right known at 60 kN tension and the vertical at 9 kN compression, vertical equilibrium of the diagonal M₂–L₃, whose unit vector towards M₂ is $(-0.8,\,0.6)$, gives $$ -9 + 0.6\,F_{L_3M_2} = 0 \quad\Longrightarrow\quad F_{L_3M_2} = 15\ \text{kN tension}, $$ and horizontal equilibrium then delivers the first of the three answers, $$ 60 - F_{L_2L_3} - 15(0.8) = 0 \quad\Longrightarrow\quad \boxed{F_{L_2L_3} = 48\ \text{kN tension}} $$
Joint M₂, which closes the set. The joint carries no load and joins two horizontal sub-chord lengths and two diagonals. Vertical equilibrium of the two diagonals, which have equal and opposite vertical components, gives $$ 15(-0.6) + F_{M_2L_2}(-0.6) = 0 \quad\Longrightarrow\quad \boxed{F_{L_2M_2} = 15\ \text{kN compression}} $$ and horizontal equilibrium closes on the sub-chord: $$ -36 + 15(0.8) - F_{M_1M_2} + (-15)(-0.8) = 0 \quad\Longrightarrow\quad \boxed{F_{M_1M_2} = 12\ \text{kN compression}} $$
Why the section fails, in one line worth writing down. Any vertical line drawn between L₂ and L₃ crosses L₂–L₃, L₂–M₂, M₁–M₂ and the rafter M₁–U₁: four unknowns against three equations. Recording that observation and switching to joints shows the examiner the reasoning; writing three equations for four unknowns produces confident nonsense.

Question 4 — forces in the listed members
Truss	Member	Force	Sense
(a)	L₁–U₁	45 kN	Compression
	U₁–L₃	9 kN	Tension
	L₃–L₄	60 kN	Tension
(b)	M₁–M₂	12 kN	Compression
	L₂–M₂	15 kN	Compression
	L₂–L₃	48 kN	Tension

Topic. Plane truss analysis by the method of joints and the method of sections, and the choice between them.

Key relations. At a joint, $\sum F_x = 0$ and $\sum F_y = 0$, each member force acting along its own axis with unit vector $(\cos\theta,\ \sin\theta)$. On a section, $\sum M_O = 0$ about a pole chosen so that all but one of the cut members pass through it — often the intersection of two of them, which need not lie inside the truss. Tension is taken positive throughout, so a negative answer is compression.

Why this works. A section is only usable when the cut severs at most three members with non-concurrent lines of action, because a free body in the plane offers exactly three equations. Truss (a) satisfies that everywhere and falls out in two moment equations. Truss (b) does not, because the sub-strut system puts a fourth member across every panel cut between L₂ and L₃; the fix is to start at a support joint with only two members, then follow a route that never presents more than two unknowns at once — here L₄, apex, M₃, L₃, M₂. Recognising which of the two methods the geometry permits is most of the marks on this question.

Common pitfall. Reading the thin vertical line through M₂ in truss (b) as a member. Adding it would make $m + r = 17 > 2j = 16$, the truss would be indeterminate, and the question would be unanswerable by statics — which is the tell that the line is a dimension witness. The second pitfall is missing the collinearity of L₁, M₁ and U₁: treating M₁ as a kink invents a vertical component that is not there.

Source. Hibbeler, Structural Analysis, 10th ed., §3.4 (joints) and §3.5–3.6 (sections, zero-force members); Kassimali, Structural Analysis, 6th ed., §4.5–4.7.

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