Truss deflection by virtual work (14 marks): 16-Civ-A1 Elementary Structural Analysis, December 2019, Q3

Question 3 — Truss deflection by virtual work (14 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. A truss cantilevered from a wall, pinned to it at both $U_1$ and $L_1$. Taking $L_1$ as the origin, the joint coordinates in metres are:

$L_1$	$(0,\ 0)$	pinned to the wall
$U_1$	$(0,\ 6.5)$	pinned to the wall
$L_2$	$(7.8,\ 0)$	mid bottom-chord joint
$U_2$	$(7.8,\ 3.25)$	40 kN applied downward
$U_3$	$(15.6,\ 0)$	free tip

The six members are $U_1U_2$, $U_2U_3$, $L_1L_2$, $L_2U_3$, $U_2L_2$ and $L_1U_2$, with $AE = 84.5\times10^{3}\ \text{kN}$ throughout. The top chord $U_1$–$U_2$–$U_3$ is straight, so $U_1U_2$ and $U_2U_3$ each have length $\sqrt{7.8^{2}+3.25^{2}} = 8.45\ \text{m}$.

Find. (a) the vertical deflection of $U_3$ under the 40 kN load at $U_2$; (b) the vertical deflection of $U_2$ when that same load is moved to $U_3$.

Q3 — truss pinned to the wall at U1 and L1. The 40 kN load acts at U2; part (a) asks for the vertical deflection at U3.

Approach. Use the unit-load (virtual work) method, $\delta = \sum NnL/AE$, solving the truss twice — once for the real 40 kN load and once for a unit load at the joint and in the direction of the required deflection — and then answer part (b) by Maxwell’s reciprocal theorem rather than by re-analysing.

Determinacy first: $m=6$ members, $n=5$ joints and $r=4$ (two pins), so $m+r-2n = 6+4-10 = 0$ and the truss is statically determinate, as the virtual-work method requires.

Real forces $N$ under the 40 kN load at $U_2$. Only two members meet at $U_2$ other than the unloaded pair towards $U_3$ and $L_2$, and joint equilibrium at $U_2$ resolves the load entirely into the two members running back to the wall. With direction cosines $7.8/8.45 = 0.923$ and $3.25/8.45 = 0.385$, $$N_{U_1U_2} = \frac{40 \times 8.45}{2(3.25)} = +52\ \text{kN (T)}, \qquad N_{L_1U_2} = -52\ \text{kN (C)}$$ and every other member carries zero. Joints $U_3$ and $L_2$ are unloaded and each connects only to zero-force members, which confirms the result.
Virtual forces $n$ for a unit downward load at $U_3$. Repeating the joint analysis with 1 kN at the tip gives $$n_{U_1U_2} = +2.6, \quad n_{U_2U_3} = +2.6, \quad n_{L_1L_2} = n_{L_2U_3} = -2.4, \quad n_{U_2L_2} = n_{L_1U_2} = 0$$ A quick global check: the wall couple is resisted over the 6.5 m depth, and $1 \times 15.6 / 6.5 = 2.4$, consistent with the chord forces above.
Form the sum $\sum NnL/AE$. A product $Nn$ is non-zero only where both systems load the member. Here that happens for $U_1U_2$ alone: $U_2U_3$ has $n=2.6$ but $N=0$, while $L_1U_2$ has $N=-52$ but $n=0$. So a single term survives:

Virtual-work tabulation (only non-zero contributions shown)
Member	$N$ (kN)	$n$ (kN/kN)	$L$ (m)	$NnL/AE$ (mm)
$U_1U_2$	+52	+2.6	8.45	13.52
$U_2U_3$	0	+2.6	8.45	0
$L_1U_2$	−52	0	8.45	0
$L_1L_2$, $L_2U_3$, $U_2L_2$	0	—	—	0
Sum				13.52

Evaluate the deflection. $$\delta_{U_3} = \sum \frac{NnL}{AE} = \frac{(52)(2.6)(8.45)}{84.5\times10^{3}} = \frac{1142.44}{84\,500} = 0.01352\ \text{m}$$ $$\delta_{U_3} = \boxed{13.52\ \text{mm downward}}$$ The positive sign means the deflection is in the same direction as the assumed unit load, so $U_3$ moves down.
Part (b) by reciprocity. Maxwell’s reciprocal theorem states that for a linearly elastic structure the deflection at $j$ caused by a load at $k$ equals the deflection at $k$ caused by the same load applied at $j$: $$\delta_{jk} = \delta_{kj}$$ Part (a) is exactly $\delta_{U_3 U_2}$ — the deflection at $U_3$ due to 40 kN at $U_2$. Part (b) asks for $\delta_{U_2 U_3}$, the deflection at $U_2$ due to 40 kN at $U_3$. These are the same number: $$\delta_{U_2} = \boxed{13.52\ \text{mm downward}}$$ No second analysis is needed. (Carrying one out anyway reproduces 13.52 mm to the last digit.)

Question 3 — results
Determinacy check	$m+r-2n = 6+4-10 = 0$ (determinate)
$N_{U_1U_2}$ / $N_{L_1U_2}$ (real)	+52 kN (T) / −52 kN (C)
$n_{U_1U_2}$ (unit load at $U_3$)	+2.6
(a) $\delta_{U_3}$	13.52 mm downward
(b) $\delta_{U_2}$	13.52 mm downward (by Maxwell’s reciprocal theorem)

Topic. Virtual work (unit-load method) for truss deflections, and Maxwell’s reciprocal theorem.

Key relations. $1 \cdot \delta = \sum \dfrac{NnL}{AE}$, where $N$ are the member forces under the real loading and $n$ those under a unit load placed at the point and in the direction of the wanted deflection; Maxwell’s theorem $\delta_{jk} = \delta_{kj}$ for any linearly elastic structure.

Why this works. Equating the external virtual work done by the unit load moving through the real deflection to the internal virtual work done by the virtual member forces moving through the real member extensions $NL/AE$ isolates the single deflection component you want. Part (b) is the real point of the question: reciprocity follows from the symmetry of the flexibility matrix, which in turn follows from the symmetry of the elastic strain-energy expression, so it holds for any linear elastic structure regardless of geometry. Recognising it converts a second full analysis into a single sentence.

Common pitfall. Grinding through part (b) from scratch and losing ten minutes — the examiner has deliberately paired the two joints so reciprocity applies. The other frequent error is a sign slip in $n$: the unit load must act at the joint and in the direction of the deflection sought, and a negative final sum means the joint moves opposite to that assumed direction. Note too how many members here carry zero force in one system or the other, so the tabulation collapses to one term — always check for zero-force members before tabulating.

Source. Hibbeler, Structural Analysis 10th ed., §8.1–8.4 (virtual work for trusses) and §9.6; McCormac, Structural Analysis 4th ed., Ch. 13 for Maxwell’s law and Betti’s generalisation.

Question 3 of 8: Truss deflection by virtual work (14 marks)

Question 3 — Truss deflection by virtual work (14 marks)