Vertical deflection of the apex joint of a symmetrical roof truss: 07-Str-A1, May 2013, Q3

Question 3: Vertical deflection of the apex joint of a symmetrical roof truss (16 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 symmetrical roof truss of 15.6 m span, apex 3.25 m above the bottom chord, with two 26 kN loads applied normal to the rafters at U₁ and U₃. Joints U₁ and U₃ lie at the feet of the perpendiculars dropped from the centre joint L₂ onto the rafters, which is what the two right angles marked on the figure state.

Quantity	Symbol	Value
Half span, L₁ to L₂	a	7.80 m
Rise at the apex, U₂ above L₂	h	3.25 m
Rafter length, L₁ to U₂	L_r	8.45 m (7.80² + 3.25² = 8.45² exactly)
Distance L₁ to U₁ along the rafter	—	7.80² / 8.45 = 7.20 m
Web member U₁ – L₂ (perpendicular to the rafter)	—	7.80 × 3.25 / 8.45 = 3.00 m
Applied load at U₁ and at U₃	P	26 kN, normal to the rafter
Axial rigidity, all members	EA	6.5 × 10⁵ kN

Find. The vertical deflection of the apex joint U₂, in millimetres, stating its direction.

The real load system. Each 26 kN load acts at right angles to its rafter, so it resolves into 10 kN horizontally and 24 kN vertically.

Approach. Solve the truss for the real loads, solve it again for a single downward unit load at U₂, and combine the two sets of bar forces through the unit-load (virtual work) equation $\Delta = \sum N n L / EA$; symmetry halves the work in both systems.

Resolve the inclined loads and find the reactions. Each rafter rises 3.25 m in a length of 8.45 m, so its direction cosines are $7.8/8.45$ and $3.25/8.45$. A 26 kN force normal to the left rafter therefore has components $$\begin{aligned} P_x &= 26\left(\frac{3.25}{8.45}\right) = 10.0\ \text{kN}, \\ P_y &= -26\left(\frac{7.8}{8.45}\right) = -24.0\ \text{kN} \end{aligned}$$ and the load at U₃ is its mirror image, $(-10.0,\ -24.0)$ kN. The two horizontal components cancel and the two vertical components total 48 kN, so by symmetry $$\begin{aligned} R_{L1} &= R_{L3} = \boxed{24.0\ \text{kN}\ (\uparrow)}, \\ H_{L1} &= 0 \end{aligned}$$
Set out the member lengths, which the geometry makes exact. Because U₁ is the foot of the perpendicular from L₂, the triangle L₁U₁L₂ is right angled at U₁, so $L_1U_1 = a^2/L_r = 7.8^2/8.45 = 7.20$ m and $U_1L_2 = ah/L_r = 7.8(3.25)/8.45 = 3.00$ m; the remainder of the rafter is $U_1U_2 = 8.45 - 7.20 = 1.25$ m. As a check, $1.25^2 + 3.00^2 = 10.5625 = 3.25^2$, which is the length of the central vertical U₂L₂.
Real bar forces N. At L₁ only the rafter and the bottom chord meet, so resolving vertically and then horizontally, $$N_{L1U1}\left(\frac{3.25}{8.45}\right) + 24.0 = 0 \;\Rightarrow\; N_{L1U1} = -62.4\ \text{kN}$$ $$N_{L1L2} = -N_{L1U1}\left(\frac{7.8}{8.45}\right) = +57.6\ \text{kN}$$ At U₁ the two rafter segments are collinear and the applied load is perpendicular to them, so the whole of the 26 kN is taken by the web member, $N_{U1L2} = -26.0$ kN, and the rafter force passes through unchanged. At the apex the two rafters, each carrying 62.4 kN of compression, push the joint upward with a combined vertical force of $2(62.4)(3.25/8.45) = 48.0$ kN, which the central vertical must hold down: $N_{U2L2} = +48.0$ kN. The truss is symmetric, so the right half mirrors the left.
Virtual bar forces n. Remove the real loads and apply 1 kN downward at U₂. The reactions become 0.5 kN each, and the same two joint resolutions give $$\begin{aligned} n_{L1U1} &= -\frac{0.5(8.45)}{3.25} = -1.30, \\ n_{L1L2} &= 1.30\left(\frac{7.8}{8.45}\right) = +1.20 \end{aligned}$$ The unit load sits at the apex, where the only members are the two rafters and the central vertical; with the load taken entirely by the rafters, joint U₁ then has two collinear rafter segments and one web member and nothing else, so $$n_{U1L2} = n_{U3L2} = n_{U2L2} = 0$$ Three of the nine members are therefore idle in the virtual system and cannot contribute to the deflection however large their real forces are — this is what "take advantage of symmetry" is pointing at.
Assemble the unit-load summation. The virtual work equation for a pin-jointed truss is $$1 \cdot \Delta = \sum \frac{N n L}{EA}$$ and with $EA$ the same for every member it factors straight out of the sum.
Evaluate and convert. Summing the last column of the table below gives $\sum N n L = 2449.2$ kN²·m, so $$\Delta_{U2} = \frac{2449.2}{6.5 \times 10^{5}} = 0.003768\ \text{m} = \boxed{3.77\ \text{mm downward}}$$ The result is positive, so the apex moves in the direction of the assumed unit load, that is vertically downward. Expressed as a fraction of the span it is 15.6/0.003768, or about span/4140, which is far inside any serviceability limit and confirms that the answer is of a sensible order.

The virtual system: a single downward unit load at U₂. The three web members carry no virtual force, so they drop out of the summation.

Member	L (m)	N (kN)	n (kN/kN)	N n L (kN²·m)
L1 – U1	7.200	-62.40	-1.300	584.06
U1 – U2	1.250	-62.40	-1.300	101.40
U2 – U3	1.250	-62.40	-1.300	101.40
U3 – L3	7.200	-62.40	-1.300	584.06
L1 – L2	7.800	57.60	1.200	539.14
L2 – L3	7.800	57.60	1.200	539.14
U1 – L2	3.000	-26.00	0.000	0.00
U2 – L2	3.250	48.00	0.000	0.00
U3 – L2	3.000	-26.00	0.000	0.00
Sum				2449.20

Quantity	Value
Reactions at L₁ and L₃	24.0 kN upward at each
Σ N n L	2449.2 kN²·m
EA (all members)	6.5 × 10⁵ kN
Vertical deflection of U₂	3.77 mm downward

Topic. Deflection of a statically determinate pin-jointed truss by the unit-load (virtual work) method.

Key relations. $1 \cdot \Delta = \sum \dfrac{N n L}{AE}$, where $N$ are the bar forces under the real loads and $n$ the bar forces under a unit load applied at the point and in the direction of the required displacement; a load applied normal to a member of slope $h : a$ resolves into $P h / L_r$ and $P a / L_r$; the foot of the perpendicular from L₂ to the rafter divides it into $a^2/L_r$ and $h^2/L_r$.

Why this works. The unit-load equation is a statement of the principle of virtual forces: the external work done by a virtual unit load moving through the real displacement equals the internal work done by the virtual bar forces acting through the real bar extensions $NL/AE$. Because the virtual system need only be in equilibrium, any convenient statically admissible force field will do, and for a determinate truss the obvious choice is the truss itself carrying the unit load. The method is exact for axial deformation and requires no compatibility reasoning at all. Symmetry is worth exploiting twice over here: the real system is symmetric, so only half the bar forces need be computed, and the virtual system leaves every web member unstressed, so five of the nine products vanish or repeat.

Common pitfall. Two traps. The first is to resolve the 26 kN loads as though they were vertical: they are applied at right angles to the rafters, and treating them as gravity loads changes both the bar forces and the answer. The second is a sign slip — a compression member must be entered with a negative $N$ and a negative $n$, whose product is positive, so a truss in which the loaded and the virtual systems put the same members in compression gives a positive, that is a downward, deflection. Cancelling those signs by "taking magnitudes" is the single most common way this calculation goes wrong.

Source. Hibbeler, Structural Analysis, 10th ed., Ch. 8 (virtual work for trusses); Kassimali, Structural Analysis, 6th ed., Section 7.3.

Question 3 of 8: Vertical deflection of the apex joint of a symmetrical roof truss

Question 3: Vertical deflection of the apex joint of a symmetrical roof truss (16 marks)