Horizontal Deflection of a Portal Frame by Virtual Work: 07-Str-A1, December 2017, Q7

Question 7: Horizontal Deflection of a Portal 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 portal frame with pinned bases at ①$(0,\,0)$ and ④$(20,\,0)$, columns 10 m high to ②$(0,\,10)$ and ③$(20,\,10)$, and a 20 m beam between them. The connection at ② is a hinge (the small circle in the figure); the knee at ③ is a rigid corner. A single vertical load of 20 kN acts at the midpoint of the beam. $EI = 2.0 \times 10^5$ kN.m$^2$ throughout, and only flexural strain is to be counted.

Find. The horizontal displacement of joint ③, with its direction.

Check: the thin horizontal line drawn between the two base pins is the witness line of the 10 m height dimension, not a tie member — it stops short of the left pin circle and continues past the right one to the dimension tick. Taking it as a member would make the frame indeterminate and the question unanswerable in the marks available. With the hinge at ② counted, $r = 4$ against three equations plus one condition, so the frame is exactly determinate.

Approach. Solve the real structure by statics, exploiting the fact that the left column is a two-force member; apply a unit horizontal load at ③ and solve that system the same way; then evaluate $\delta = \int M m / EI\,ds$ over the members that carry moment in both systems.

Question 7 — the frame, and the real and virtual bending moment diagrams on the beam.

Recognise the two-force column. Member ①–② is pinned at ① and hinged at ② and carries no load between, so it transmits force along its own axis, which is vertical. The reaction at ① is therefore purely vertical: $H_1 = 0$, and global horizontal equilibrium then gives $H_4 = 0$ as well. This one observation is most of the question.
Vertical reactions. Moments about ④ give $$V_1 = \frac{20(10)}{20} = 10\ \text{kN} \quad\text{and}\quad V_4 = 10\ \text{kN}$$ both upward.
Real bending moment diagram. With no horizontal reaction anywhere, the right column ③–④ carries no shear, and being pinned at its base it carries no moment either, so $M = 0$ along it and hence $M = 0$ at the knee ③. The left column is a strut and also carries no moment. The beam is therefore a simply supported span with a central point load: $$M_{\text{beam}}(s) = 10s \;(0 \le s \le 10), \qquad M_{\max} = \frac{PL}{4} = \boxed{+100\ \text{kN.m at midspan}}$$
Virtual system: unit horizontal load at ③. Apply 1 kN acting to the right at ③. Member ①–② is still a two-force member, so $h_1 = 0$ and the whole unit load is taken by the right pin, $h_4 = 1$ kN to the left. Moments about ④ give $v_1 = -0.5$ (downward) and $v_4 = +0.5$. The left column again carries no moment, while along the beam $$m(s) = -0.5s, \qquad m(20) = -10\ \text{m}$$ and the right column carries a linear moment from zero at the pin to that same value at the knee.
Evaluate the virtual-work integral. The columns contribute nothing, because the real moment is zero in the left column and zero in the right column too, so only the beam is left: $$\int_0^{20} Mm\,ds = \int_0^{10}(10s)(-0.5s)\,ds + \int_{10}^{20} 10(20-s)(-0.5s)\,ds = -1666.7 - 3333.3 = -5000$$ in units of kN$^2$.m$^3$. Dividing by the rigidity, $$\delta_3 = \frac{-5000}{2.0\times 10^5} = -0.025\ \text{m} = \boxed{25\ \text{mm to the left}}$$ the negative sign meaning the movement opposes the assumed unit load, which pointed to the right.
Kinematic check. Because the right column carries no moment it stays straight and simply rotates about its pin, so the knee moves horizontally by $10\theta_3$, where $\theta_3$ is the end rotation of the simply supported beam: $$\theta_3 = \frac{PL^2}{16EI} = \frac{20(20)^2}{16(2.0\times 10^5)} = 2.5 \times 10^{-3}\ \text{rad}, \qquad 10\theta_3 = 25\ \text{mm}$$ in exact agreement, and the sense of the beam-end rotation confirms that the movement is to the left.

Quantity	Value
Reaction at ①	$H_1 = 0$, $V_1 = 10$ kN up
Reaction at ④	$H_4 = 0$, $V_4 = 10$ kN up
Real moment, columns	zero throughout
Real moment, beam	$+100$ kN.m at midspan, zero at both ends
Virtual moment, beam	$0$ at ② to $-10$ m at ③
$\int Mm\,ds$	$-5000$ kN$^2$.m$^3$
Horizontal deflection at ③	25 mm to the left

Topic. Deflection of a determinate plane frame by the unit-load method, counting flexural strain only.

Key relations. $1 \cdot \delta = \displaystyle\int \frac{Mm}{EI}\,ds$ summed over all members. A member pinned or hinged at both ends and unloaded between them is a two-force member and carries axial force alone; a member with a moment-free end and no transverse load carries no bending at all if its shear is zero.

Why this works. The integral is a sum over the whole structure, but a term survives only where both diagrams are non-zero. Identifying the left column as a two-force member kills the horizontal reactions in both the real and the virtual system, which in turn kills the real moment in the right column, and what began as a three-member integral collapses to a single simply supported span. That is the general lesson: spend the first minute of a virtual-work question looking for two-force members and moment-free reaches, because each one removes a member from the arithmetic. The kinematic check in the last step is worth the two lines it takes, since it reaches the same number by an entirely different route and settles the direction beyond argument.

Common pitfall. Reading the hinge at ② as a rigid corner. That would make the frame indeterminate to the first degree, so the horizontal reactions would no longer vanish, the columns would bend, and the question could not be answered by virtual work alone. The second trap is sign: the integral came out negative, and reporting 25 mm without saying “to the left” throws away the part of the answer the direction of the unit load was chosen to determine.

Source. Hibbeler, Structural Analysis, 10th ed., §9-6 (virtual work applied to frames); Kassimali, 6th ed., §7.6; Leet, Uang and Gilbert, 5th ed., §9.5.

Question 7 of 8: Horizontal Deflection of a Portal Frame by Virtual Work

Question 7: Horizontal Deflection of a Portal Frame by Virtual Work (22 marks)