Triangular Truss — Member Forces and Joint Deflection: 07-Str-A1, May 2017, Q6

The examination requires one only of Questions 6, 7 and 8. All three are worked below.

Question 6: Triangular Truss — Member Forces and Joint Deflection (9 + 13 = 22 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.

Item	Value
Joints (m)	$L_1(0,0)$, $L_2(6,0)$, $L_3(12,0)$, $M_1(3,4)$, $M_2(9,4)$, $U_1(6,8)$
Members (9)	$L_1L_2$, $L_2L_3$, $L_1M_1$, $M_1U_1$, $U_1M_2$, $M_2L_3$, $M_1M_2$, $M_1L_2$, $M_2L_2$
Supports	Pin at $L_1$, roller at $L_3$
Loads	90 kN horizontal (to the right) at each of $U_1$, $M_1$ and $M_2$
Axial rigidity	$AE = 1.5\times10^{5}$ kN, the same for every member

Every inclined member spans 3 m by 4 m, so its length is exactly 5 m; the horizontal $M_1M_2$ and both bottom-chord members are 6 m.

Find. (a) the forces in $L_2M_1$, $L_2M_2$ and $M_1M_2$; (b) the horizontal deflection of $U_1$ by virtual work.

Question 6: the triangular truss, its supports and the three horizontal 90 kN loads. The dashed vertical is the drawing's centre line, not a member — see the note below.

Check — the vertical line on the drawing. The line from $U_1$ down to $L_2$ is read here as a centre line, not a member: it is drawn in two segments broken at $M_1M_2$, and the determinacy count settles it. With nine members, $m + r = 9 + 3 = 12 = 2j = 2(6)$ and the truss is determinate, which is what a 9-mark statics question and a $\sum NnL/AE$ deflection both require. Counting the vertical as a tenth member gives $m + r = 13 > 2j$: the panel $M_1$–$U_1$–$M_2$–$L_2$ would then carry both its diagonals and the truss would be indeterminate to the first degree, so neither part (a) nor part (b) could be answered as asked.

6(a) Member forces

Approach. Joint $U_1$ carries only two members, so start there; then take the reactions from global equilibrium, walk in from both supports, and close at $L_2$.

Joint $U_1$. The two members leave $U_1$ symmetrically, in directions $(-3,-4)/5$ and $(3,-4)/5$, and a 90 kN horizontal load is applied. Vertical equilibrium gives $F_{U_1M_1} = -F_{U_1M_2}$, and substituting into the horizontal equation, $$\tfrac{6}{5}F_{U_1M_2} = -90.0 \;\Rightarrow\; F_{U_1M_2} = -75.0, \qquad F_{U_1M_1} = +75.0$$ so $M_1U_1$ carries 75.0 kN tension and $U_1M_2$ 75.0 kN compression — the windward leg is stretched, the leeward leg squashed.
Reactions. Horizontally, $H_{L_1} = -270.0$ kN, i.e. 270.0 kN to the left. Taking moments about $L_1$, each horizontal load contributes $-y_i(90.0)$: $$-8(90.0) - 4(90.0) - 4(90.0) + 12\,V_{L_3} = 0 \;\Rightarrow\; V_{L_3} = \frac{1440}{12} = 120.0\ \text{kN}\ \uparrow$$ and hence $V_{L_1} = -120.0$ kN, i.e. 120.0 kN downward. A horizontal load set on a symmetric truss produces an uplift/hold-down couple, not a net vertical force.
Joint $L_1$. Two members meet at the pin, the bottom chord $(1,0)$ and the leg $(3,4)/5$. Vertically, $$\tfrac{4}{5}F_{L_1M_1} - 120.0 = 0 \;\Rightarrow\; F_{L_1M_1} = 150.0\ \text{kN (T)}$$ and horizontally, $$F_{L_1L_2} + \tfrac{3}{5}(150.0) - 270.0 = 0 \;\Rightarrow\; F_{L_1L_2} = 180.0\ \text{kN (T)}$$
Joint $L_3$. Similarly, with the roller pushing up 120.0 kN, $$\tfrac{4}{5}F_{L_3M_2} + 120.0 = 0 \;\Rightarrow\; F_{L_3M_2} = -150.0 \ \text{kN, i.e. } 150.0\ \text{kN (C)}$$ $$-F_{L_3L_2} - \tfrac{3}{5}(-150.0) = 0 \;\Rightarrow\; F_{L_2L_3} = 90.0\ \text{kN (T)}$$
Joint $M_1$ — vertical equilibrium gives $L_2M_1$. Four members meet at $M_1$: $M_1L_1$ and $M_1U_1$ (collinear, direction $\pm(3,4)/5$), the horizontal $M_1M_2$, and the diagonal $M_1L_2$ in direction $(3,-4)/5$. The 90 kN load is horizontal, so it drops out of the vertical equation: $$-\tfrac{4}{5}F_{M_1L_1} + \tfrac{4}{5}F_{M_1U_1} - \tfrac{4}{5}F_{M_1L_2} = 0 \;\Rightarrow\; F_{M_1L_2} = F_{M_1U_1} - F_{M_1L_1} = 75.0 - 150.0$$ $$F_{L_2M_1} = -75.0 \;\Rightarrow\; \boxed{75.0\ \text{kN\ (C)}}$$
Joint $L_2$ gives $L_2M_2$. At $L_2$ the two bottom-chord members are horizontal and the two diagonals rise symmetrically, so vertical equilibrium requires $$\tfrac{4}{5}F_{L_2M_1} + \tfrac{4}{5}F_{L_2M_2} = 0 \;\Rightarrow\; F_{L_2M_2} = -F_{L_2M_1} = \boxed{75.0\ \text{kN\ (T)}}$$
Joint $M_1$ — horizontal equilibrium gives $M_1M_2$. Substituting the four known forces, $$-\tfrac{3}{5}(150.0) + \tfrac{3}{5}(75.0) + F_{M_1M_2} + \tfrac{3}{5}(-75.0) + 90.0 = 0$$ $$-90.0 + 45.0 + F_{M_1M_2} - 45.0 + 90.0 = 0 \;\Rightarrow\; F_{M_1M_2} = \boxed{0\ \ (\text{zero-force member})}$$
Independent check at $M_2$ and $L_2$. Horizontal equilibrium at $M_2$ reads $0.6(-150.0) - 0.6(-75.0) - 0 - 0.6(75.0) + 90.0 = 0$, and at $L_2$, $-180.0 + 90.0 + 0.6(75.0) + 0.6(75.0) = 0$. Both close exactly, confirming the whole force set.

Member	Force	Sense
$L_2-M_1$	75.0 kN	Compression
$L_2-M_2$	75.0 kN	Tension
$M_1-M_2$	0	Zero-force member
(check) $L_1-M_1$ / $M_2-L_3$	150.0 kN / 150.0 kN	T / C
(check) $M_1-U_1$ / $U_1-M_2$	75.0 kN / 75.0 kN	T / C
(check) $L_1-L_2$ / $L_2-L_3$	180.0 kN / 90.0 kN	both T
Reactions	$L_1$: 270.0 kN ← and 120.0 kN ↓; $L_3$: 120.0 kN ↑

6(b) Horizontal deflection of $U_1$ by virtual work

Approach. Apply a 1 kN horizontal dummy load at $U_1$ in the direction of the required deflection, obtain the virtual forces $n$ by the same joint walk, and sum $NnL/AE$ over the nine members.

Virtual force system. The dummy load is $1$ kN to the right at $U_1$ only — the loads at $M_1$ and $M_2$ are not scaled, because virtual work needs a unit load at the one point of interest. Repeating steps 1–4 with that single load gives $$n_{M_1U_1} = +\tfrac{5}{6}, \quad n_{U_1M_2} = -\tfrac{5}{6}, \quad n_{L_1M_1} = +\tfrac{5}{6}, \quad n_{M_2L_3} = -\tfrac{5}{6},$$ $$n_{L_1L_2} = n_{L_2L_3} = +0.5, \qquad n_{M_1M_2} = n_{M_1L_2} = n_{M_2L_2} = 0$$ The three members that are unloaded in the virtual system drop out of the sum altogether — including $M_1L_2$, which carries 75 kN in the real system.
Assemble the sum member by member.

Member	$L$ (m)	$N$ (kN)	$n$ (kN/kN)	$NnL$ (kN²m)
$L_1L_2$	6	+180.0	+0.500	540.0
$L_2L_3$	6	+90.0	+0.500	270.0
$L_1M_1$	5	+150.0	+0.8333	625.0
$M_1U_1$	5	+75.0	+0.8333	312.5
$U_1M_2$	5	−75.0	−0.8333	312.5
$M_2L_3$	5	−150.0	−0.8333	625.0
$M_1M_2$	6	0	0	0
$M_1L_2$	5	−75.0	0	0
$M_2L_2$	5	+75.0	0	0
Total $\sum NnL$				2685.0

Divide by the axial rigidity. Because $AE$ is the same for every member it comes outside the sum: $$\Delta_{U_1} = \frac{1}{AE}\sum N n L = \frac{2685.0}{1.5\times10^{5}} = 0.017900\ \text{m}$$ $$\boxed{\Delta_{U_1} = 17.90\ \text{mm to the RIGHT}}$$ The result is positive, so the joint moves in the direction of the dummy load, i.e. with the applied loads — which is the expected sense.
Where the flexibility comes from. The two bottom-chord members supply 810 of the 2685 units and the four inclined chord members the remaining 1875; the three interior web members contribute nothing at all, because the virtual system leaves them unstressed. That is worth stating in an answer: stiffening the diagonals would not reduce this deflection.

Quantity	Value
$\sum N n L$	2685.0 kN²m
$AE$	$1.5\times10^{5}$ kN
Horizontal deflection of $U_1$	17.90 mm to the right

Topic. Determinate truss under horizontal load; zero-force members; joint deflection by the unit-load method.

Key relations. Joint equilibrium $\sum F_x = \sum F_y = 0$ with member forces along member axes; $\Delta = \displaystyle\sum \frac{N n L}{AE}$, where $N$ is the real member force, $n$ the force under a unit load applied at the joint in the direction of the wanted deflection, and the sum runs over every member.

Why this works. The unit-load theorem is the virtual-work statement for a pin-jointed assembly: the external work $1 \times \Delta$ equals the internal work $\sum n \,(NL/AE)$ of the virtual member forces moving through the real member extensions. Because the virtual system need only be any system in equilibrium with the unit load, the same determinate truss is the natural choice, and members carrying no virtual force drop out however heavily they are loaded in reality. That also explains why $M_1M_2$ being a zero-force member in part (a) is not an accident of this load case — joint $M_1$'s vertical equation involves only the three inclined members, and the collinearity of $L_1M_1U_1$ is what forces the horizontal member's share to vanish.

Common pitfall. Applying the virtual load at all three loaded joints, or scaling the real load case by $1/90$, instead of putting a single unit load at the joint whose deflection is wanted. The result would be the wrong quantity entirely (a weighted combination of three deflections). The second pitfall is treating the drawing's centre line as a member, which makes the truss indeterminate and the question unanswerable.

Source. Hibbeler, Structural Analysis 10th ed., §3.4 (joints), §3.3 (zero-force members), §8.7 and §9.2 (truss deflection by virtual work); Kassimali 6th ed., §7.3.

Question 6 of 8: Triangular Truss — Member Forces and Joint Deflection

Question 6: Triangular Truss — Member Forces and Joint Deflection (9 + 13 = 22 marks)

6(a) Member forces

6(b) Horizontal deflection of $U_1$ by virtual work