Member forces in two trusses: 07-Str-A1, May 2014, Q4

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 determinate plane trusses. Taking the left-hand bottom joint as the origin in each case:

Question 4 — joint coordinates and loading
	Joint coordinates (m)	Supports	Loads
(a)	$L_{1}(0,0)$, $L_{2}(6,0)$, $L_{3}(12,0)$, $U_{1}(0,5)$, $U_{2}(6,2.5)$	pin at $L_{1}$, roller at $L_{3}$	72 kN horizontal at $U_{1}$; 60 kN downward at $L_{2}$
(b)	$L_{1}(0,0)$, $L_{2}(6.25,0)$, $L_{3}(9.75,0)$, $L_{4}(16,0)$, $U_{1}(4,3)$, $U_{2}(8,6)$, $U_{3}(12,3)$	pin at $L_{1}$, roller at $L_{4}$	30 kN at $L_{1}$, 60 kN at $U_{1}$, 30 kN at $U_{2}$, each on a 4 vertical : 3 horizontal slope

Find. Six member forces, each with its sense stated as tension or compression.

Approach. Get the reactions from global equilibrium, then resolve at the joint that has only two unknown members and work along the truss; in (b) recognise first that the three applied loads are normal to the windward rafter, which collapses most of the arithmetic.

4(a)

Question 4(a) — the truss, its reactions (blue) and the three member forces asked for. U₁, U₂ and L₃ are collinear, which is what makes the joint solution so short.

Note the collinearity. $U_{1}(0,5)$, $U_{2}(6,2.5)$ and $L_{3}(12,0)$ lie on one straight line of slope $-5/12$, so $U_{1}-U_{2}$ and $U_{2}-L_{3}$ are two pieces of a single straight chord. Its unit vector is $$\mathbf{e} = \frac{(6,\,-2.5)}{6.5} = (0.9231,\, -0.3846)$$ because $\sqrt{6^{2} + 2.5^{2}} = 6.5$ exactly — a 5–12–13 triangle at half scale.
Reactions. Moments about $L_{1}$, with the 72 kN acting horizontally 5 m above it and the 60 kN acting 6 m to the right: $$5(72) + 6(60) = 12\,V_{L_{3}} \;\Rightarrow\; V_{L_{3}} = \frac{360 + 360}{12} = \boxed{60.0\ \text{kN} \uparrow}$$ Vertical equilibrium then gives $V_{L_{1}} = 60 - 60 = 0$, and horizontal equilibrium gives $H_{L_{1}} = 72$ kN acting to the left. A zero vertical reaction is not an error — the 72 kN couple about $L_{1}$ is exactly balanced by the 60 kN load.
Joint $U_{1}$ gives the top chord. Only two members meet there, the vertical $U_{1}-L_{1}$ and the chord $U_{1}-U_{2}$, and the 72 kN acts along $+x$: $$\sum F_{x} = 0:\quad 0.9231\,F_{U_{1}U_{2}} + 72 = 0 \;\Rightarrow\; F_{U_{1}U_{2}} = \boxed{78.0\ \text{kN (C)}}$$ and $\sum F_{y} = 0$ then gives $F_{U_{1}L_{1}} = 0.3846(78.0) = 30.0$ kN tension.
Joint $L_{1}$ gives the diagonal and the bottom chord. Three members meet there and both reaction components are now known: $$\sum F_{y} = 0:\quad 30.0 + 0.3846\,F_{L_{1}U_{2}} = 0 \;\Rightarrow\; F_{L_{1}U_{2}} = \boxed{78.0\ \text{kN (C)}}$$ $$\sum F_{x} = 0:\quad 0.9231(-78.0) + F_{L_{1}L_{2}} - 72 = 0 \;\Rightarrow\; F_{L_{1}L_{2}} = 144\ \text{kN (T)}$$
Joint $L_{2}$ carries the bottom chord through. The vertical $L_{2}-U_{2}$ takes the whole 60 kN in tension, and because no horizontal load acts at $L_{2}$ the chord force passes straight through: $$F_{L_{2}L_{3}} = F_{L_{1}L_{2}} = \boxed{144\ \text{kN (T)}}$$
Check at $L_{3}$. Working back from the roller, $0.3846\,F_{U_{2}L_{3}} + 60 = 0$ gives 156 kN compression in the end chord, and $\sum F_{x}$ at $L_{3}$ returns $F_{L_{2}L_{3}} = 0.9231(156) = 144$ kN tension — the same value reached from the other end, which closes the solution.

4(b)

Question 4(b) — roof truss with three panel loads acting normal to the windward rafter (4 vertical : 3 horizontal). Reactions in blue; the three requested member forces are marked.

Resolve the loads once. Each load is drawn on a 4 vertical : 3 horizontal slope triangle, so its unit vector is $(0.6,\, -0.8)$. The left rafter runs from $L_{1}(0,0)$ to $U_{2}(8,6)$, whose unit vector is $(0.8,\, 0.6)$; the dot product of the two is zero, so all three loads act normal to the windward rafter. In components the loads are $(18,\,-24)$ kN at $L_{1}$, $(36,\,-48)$ kN at $U_{1}$ and $(18,\,-24)$ kN at $U_{2}$, totalling $(72,\,-96)$ kN.
Reactions. Moments about $L_{1}$, using $M = xF_{y} - yF_{x}$ for each load, $$M_{U_{1}} = 4(-48) - 3(36) = -300, \qquad M_{U_{2}} = 8(-24) - 6(18) = -300$$ $$16\,V_{L_{4}} = 600 \;\Rightarrow\; V_{L_{4}} = \boxed{37.5\ \text{kN} \uparrow}$$ so $V_{L_{1}} = 96 - 37.5 = 58.5$ kN upward and $H_{L_{1}} = 72$ kN to the left.
Joint $L_{1}$. The net external force there is the applied load plus the reaction, $(18 - 72,\; -24 + 58.5) = (-54,\; 34.5)$ kN. The rafter $L_{1}-U_{1}$ has unit vector $(0.8,\,0.6)$ and the bottom chord runs along $+x$, so $$\sum F_{y} = 0:\quad 0.6\,F_{L_{1}U_{1}} + 34.5 = 0 \;\Rightarrow\; F_{L_{1}U_{1}} = \boxed{57.5\ \text{kN (C)}}$$ $$\sum F_{x} = 0:\quad 0.8(-57.5) + F_{L_{1}L_{2}} - 54 = 0 \;\Rightarrow\; F_{L_{1}L_{2}} = \boxed{100\ \text{kN (T)}}$$
Joint $U_{1}$ — where the normal loading pays off. $U_{1}$ is the midpoint of the rafter, and the web member $U_{1}-L_{2}$ runs from $(4,3)$ to $(6.25,0)$, a direction of $(0.6,\,-0.8)$: the same direction as the load. Resolving along and normal to the rafter, the along-rafter equation gives $F_{U_{1}U_{2}} = F_{L_{1}U_{1}} = -57.5$ kN, and the normal equation gives $$F_{U_{1}L_{2}} = -60\ \text{kN}\ \text{i.e.}\ 60\ \text{kN (C)}$$ — the 60 kN panel load is carried entirely by that one web member.
Joint $L_{2}$ gives the requested web force. With $F_{L_{2}U_{1}} = -60$ kN known and $L_{2}-U_{2}$ running from $(6.25,0)$ to $(8,6)$, a length of exactly 6.25 m and unit vector $(0.28,\,0.96)$: $$\sum F_{y} = 0:\quad 0.8(-60) + 0.96\,F_{L_{2}U_{2}} = 0 \;\Rightarrow\; F_{L_{2}U_{2}} = \boxed{50.0\ \text{kN (T)}}$$
Check the bottom chord. Horizontal equilibrium at $L_{2}$ gives $F_{L_{2}L_{3}} = 100 - 0.6(60) - 0.28(50) = 50.0$ kN tension. The chord force halves across the panel because the inclined web members pick up the difference, which is the expected behaviour under a one-sided roof load.

Question 4 — requested member forces
Truss	Member	Force	Sense
(a)	$U_{1}-U_{2}$	78.0 kN	Compression
	$L_{1}-U_{2}$	78.0 kN	Compression
	$L_{2}-L_{3}$	144 kN	Tension
(b)	$L_{1}-U_{1}$	57.5 kN	Compression
	$L_{1}-L_{2}$	100 kN	Tension
	$L_{2}-U_{2}$	50.0 kN	Tension

Topic. Plane truss analysis by the method of joints, with inclined and normal panel loading.

Key relations. At every pin, $\sum F_{x} = 0$ and $\sum F_{y} = 0$; a member force resolves as $F\mathbf{e}$ with $\mathbf{e}$ the unit vector from the joint toward the far end; tension is positive, meaning the force pulls away from the joint.

Why this works. Because every member is a two-force member, each joint supplies two independent scalar equations, and a determinate truss can always be entered at a joint with no more than two unknowns and unwound from there. Two geometric observations shorten this paper enormously. In (a) three joints are collinear, so the top chord is one straight member and its direction cosines are used twice. In (b) the loads are normal to the rafter, so the along-rafter component of each load is zero, the rafter force does not change across a loaded joint, and the whole panel load goes into the single web member that happens to lie along the normal.

Common pitfall. Reading the 72 kN in (a) as acting to the left. It points into joint $U_{1}$ from the left, so it acts to the right; the reaction at $L_{1}$ is what points left. Reversing it changes every sign in the answer. In (b) the standard error is treating the loads as vertical, which destroys the normal-load simplification and gives roughly 15 per cent errors in the chord forces.

Source. Hibbeler, Structural Analysis 10th ed., §3.4–3.6; Kassimali, Structural Analysis 6th ed., §4.5–4.7; Norris, Wilbur and Utku, Elementary Structural Analysis 4th ed., Ch. 4.

Question 4 of 8: Member forces in two trusses

Question 4: Member forces in two trusses (18 marks)

4(a)

4(b)