Horizontal deflection of a frame by virtual work: 07-Str-A1, Undated paper, Q8

Question 8: Horizontal deflection of a frame by virtual work (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. A T-shaped plane frame. A vertical column runs from a pin at node 1 (0, 0) up through node 2 (0, 4) to a free tip at node 3 (0, 8). A horizontal beam runs from node 2 to a roller at node 4 (8, 4), the roller providing a vertical reaction only. A uniform horizontal load of 6 kN/m acts on the column over its whole 8 m height, pushing towards the beam. All members are inextensible with $EI = 1.8 \times 10^{5}$ kN·m².

Find. The horizontal deflection of joint 2.

Question 8 — the frame, its loading and the computed reactions.

Approach. Unit-load virtual work. Solve the determinate frame for the real moment field, apply a horizontal unit load at joint 2 for the virtual field, then evaluate $\delta = \int Mm/EI\,\mathrm{d}s$ member by member; the members are inextensible, so no axial term appears.

Question 8 — the real and virtual moment diagrams. The virtual moment vanishes identically above joint 2, so the 4 m of column between joints 2 and 3 contributes nothing to the integral.

Real system. The frame has $r = 2 + 1 = 3$ reaction components against three equations, so it is determinate. The applied load totals $6(8) = 48$ kN acting at mid-height, $y = 4$ m. Horizontal equilibrium gives the pin 48 kN in the opposite sense, and moments about the pin give the roller, $$\sum M_1 = 0:\qquad 8D_y = 48(4) \quad\Rightarrow\quad D_y = 24.0\ \text{kN up},$$ $$A_x = 48.0\ \text{kN (opposing the load)},\qquad A_y = 24.0\ \text{kN down}.$$ The vertical pair is a couple that balances the roller, and the vertical reaction at the pin is downward — a detail worth stating because it is easy to sign wrongly.
Real moment field. Below joint 2, taking the free body from the pin up to height $y$, $$M(y) = 48y - 3y^{2} \qquad (0 \le y \le 4),$$ which rises from zero at the pin to $+144$ kN·m immediately below joint 2. The beam delivers $D_y \times 8 = 192$ kN·m into that joint, so immediately above it the column moment is $144 - 192 = -48$ kN·m, and above joint 2 $$M(y) = 48y - 3y^{2} - 192 \qquad (4 \le y \le 8),$$ which closes on exactly zero at the free tip — the check on the whole real system. Along the beam the moment is simply $M(x) = 24(8-x)$, i.e. $+192$ kN·m at joint 2 falling linearly to zero at the roller.
Virtual system. Remove the real load and apply a horizontal unit load at joint 2. The same three equations give $d_y = 4/8 = 0.5$ at the roller, $a_y = 0.5$ downward at the pin and a unit horizontal reaction there. The virtual moment below joint 2 is $m(y) = y$, reaching $4.00$ m at the joint, and along the beam $m(x) = 0.5(8-x)$, also $4.00$ m at joint 2.
Note that the upper column drops out. Above joint 2 the virtual free body contains the pin reaction, the unit load and the roller reaction, and their moments cancel identically: $$m(y) = -y + (y - 4) + 4 = 0 \qquad (4 \le y \le 8).$$ The 4 m of column between joints 2 and 3 therefore contributes nothing to the integral no matter what real moment it carries. Recognising that before integrating removes a third of the arithmetic, and it makes physical sense: a horizontal force at joint 2 is resisted entirely by the pin below it and the couple formed with the roller, so the cantilever tip above is never called upon.
Integrate the two surviving members. Over the loaded column, $$\int_0^{4}\!\!\left(48y - 3y^{2}\right)y\,\mathrm{d}y = \int_0^{4}\!\!\left(48y^{2} - 3y^{3}\right)\mathrm{d}y = 1024 - 192 = 832\ \text{kN}\cdot\text{m}^{3},$$ and over the beam, where both diagrams are linear and share the same zero, $$\int_0^{8} 24(8-x)\cdot 0.5(8-x)\,\mathrm{d}x = 12\!\int_0^{8}(8-x)^{2}\mathrm{d}x = 12\left(\frac{512}{3}\right) = 2048\ \text{kN}\cdot\text{m}^{3}.$$
Assemble the deflection. Summing and dividing by the flexural rigidity, $$\delta_{2h} = \frac{832 + 2048}{EI} = \frac{2880}{1.8 \times 10^{5}} = 0.0160\ \text{m}$$ $$\boxed{\delta_{2h} = 16.0\ \text{mm, in the direction of the applied load}}$$ The positive sign means the joint moves the same way the unit load was applied, i.e. horizontally away from the wind. Note where the flexibility lives: the beam supplies 2048 of the 2880 units, about 71 per cent, even though it carries no load directly. Stiffening the loaded column alone would barely move the answer.

Question 8 — results
Quantity	Value
Total applied horizontal load	48.0 kN at mid-height
Reaction at the roller, node 4	24.0 kN up
Reaction at the pin, node 1	48.0 kN horizontal, 24.0 kN down
Real moment just below / above joint 2	+144 / −48 kN·m
Real moment in the beam at joint 2	+192 kN·m
∫Mm ds, column 1–2	832 kN·m³
∫Mm ds, beam 2–4	2048 kN·m³
∫Mm ds, column 2–3	0 (virtual moment vanishes)
Horizontal deflection of joint 2	16.0 mm

Concept & Theory

Topic. The unit-load method for the sway deflection of a determinate plane frame.

Key relations. $1 \cdot \delta = \displaystyle\int \frac{Mm}{EI}\,\mathrm{d}s$ for inextensible flexural members; for a section at height $y$ under a uniform horizontal intensity $w$, $M = R_x y - wy^{2}/2$; for two linear diagrams sharing a zero, $\int Mm\,\mathrm{d}s = \tfrac{1}{3}M_{\max}m_{\max}L$.

Why this works. Virtual work equates the external work of a fictitious unit load moving through the real displacement with the internal work of the virtual internal forces moving through the real deformations. Choosing the unit load to be a single force at the point and in the direction of interest makes the external term numerically equal to the wanted deflection, so the whole calculation reduces to an integral of the product of two moment diagrams. Because the members are stated to be inextensible, the axial term $\sum NnL/AE$ is dropped by assumption rather than by neglect, which is what the question's wording is there to authorise. The vanishing of the virtual moment above joint 2 is not a coincidence but a statement about load path: whatever carries the virtual load does not pass through the upper cantilever.

Common pitfall. Integrating the upper column because its real moment is large (it runs to 48 kN·m), without checking that the virtual moment there is identically zero — wasted work that also invites a sign error. The complementary mistake is sign inconsistency between the two diagrams: $M$ and $m$ must be measured with the same convention on each member, or the beam's contribution — the larger of the two here — will be subtracted instead of added, giving 6.8 mm rather than 16.0 mm.

Source. Hibbeler, Structural Analysis, Ch. 9 (virtual work for frames, and the tabulated products of common diagram shapes); Kassimali, Structural Analysis, Ch. 7.6.

Question 8 of 8: Horizontal deflection of a frame by virtual work

Question 8: Horizontal deflection of a frame by virtual work (22 marks)