Castigliano’s theorem — horizontal deflection at joint 3 (16 marks): 07-Str-A4, December 2017, Q3

Question 3: Castigliano’s theorem — horizontal deflection at joint 3 (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 symmetric two-member frame: a pin at joint 1, a roller at joint 3 on the same level, and a rigid apex at joint 2 carrying a 60 kN vertical load.

Question 3 — data
Quantity	Value
Horizontal run of each member	4 m
Rise of the apex above the supports	3 m
Length of each member	5 m
Applied load at joint 2	60 kN downward
Flexural rigidity of both members	$EI=2.0\times10^{4}$ kN·m$^{2}$

Find. The horizontal deflection of joint 3.

Q3: symmetric two-member frame, pin at joint 1 and roller at joint 3, 60 kN applied vertically at the apex. Each member is 5 m long (4 m run, 3 m rise).

Approach. Add a dummy horizontal force $Q$ at the roller, write the bending moment in each member as a function of $Q$, and evaluate $\delta=\partial U/\partial Q$ at $Q=0$.

Find the reactions. With the dummy force $Q$ applied to the right at joint 3, horizontal equilibrium requires the pin to supply $Q$ to the left. Both horizontal forces act at support level, so neither exerts a moment about the other support and the vertical reactions are untouched: $$V_{1}=V_{3}=\frac{60}{2}=30\ \text{kN}$$
Write the bending moment. Measure $s$ along a member from its support, so a point on the left member sits at $(0.8s,\,0.6s)$. The vertical reaction contributes $30\times0.8s$ and the collinear horizontal pair contributes $-Q\times0.6s$, because a horizontal force at support level has a lever arm equal to the height of the section: $$M(s)=-\left(24+0.6Q\right)s,\qquad 0\le s\le 5\ \text{m}$$ Symmetry makes the right member identical, which is the simplification the question invites.
Differentiate with respect to the dummy force. $$\frac{\partial M}{\partial Q}=-0.6\,s$$ on both members, so the integrand is the product of two linear functions of $s$ and the integral is elementary.
Integrate over both members and set $Q=0$. $$\delta_{3H}=\frac{2}{EI}\int_{0}^{5}(24s)(0.6s)\,ds=\frac{2}{EI}\left[14.4\,\frac{s^{3}}{3}\right]_{0}^{5}=\frac{1200}{EI}$$
Substitute the stiffness. With $EI=2.0\times10^{4}$ kN·m$^{2}$, $$\delta_{3H}=\frac{1200}{2.0\times10^{4}}=\boxed{0.060\ \text{m}=60\ \text{mm outward}}$$

The result is positive in the direction of the dummy force, so joint 3 moves away from the frame: the two members flatten under the load and the base spreads by 60 mm. Because the members are declared inextensible, no axial term appears in the strain energy; had they been elastic, the collinear pair would also have stretched the line 1–3 and added a small extra spread. As a check on the bending field, the moment at the apex is $M(5)=-120$ kN·m, that is 120 kN·m with the outside face in tension, and a direct-stiffness solution of the same frame with axially rigid members returns 60.0 mm and 120 kN·m exactly.

Question 3 — results
Quantity	Value
Vertical reaction at each support	30 kN
Horizontal reaction with no dummy load	0
Bending moment at the apex	120 kN·m
$\int M\,(\partial M/\partial Q)\,ds$ over both members	1200 kN·m$^{3}$
Horizontal deflection of joint 3	60 mm outward

Topic. Castigliano’s second theorem with a dummy load on a determinate frame.

Key relations. $\delta_{i}=\partial U/\partial P_{i}$ with $U=\int M^{2}ds/(2EI)$, hence $\delta_{i}=\int (M/EI)\,(\partial M/\partial P_{i})\,ds$; for an inextensible member the axial term $\int N(\partial N/\partial P)\,ds/EA$ vanishes.

Why this works. A horizontal dummy force applied at a roller sitting level with the pin produces a self-equilibrating collinear pair, so its entire effect on the bending moment is $-Qy$ with $y$ the height of the section. That makes $\partial M/\partial Q$ a pure function of geometry and reduces the whole question to one integral, which is precisely the economy the phrase “take advantage of symmetry” is pointing at.

Common pitfall. Assuming the dummy force alters the vertical reactions. It does not here, because it acts through the line joining the supports; assume otherwise and the moment field picks up spurious terms that survive differentiation.

Source. Kassimali, Structural Analysis, 6th ed., §7.6–7.7 (Castigliano’s second theorem); Hibbeler, 10th ed., §9.5–9.6.

Question 3 of 9: Castigliano’s theorem — horizontal deflection at joint 3 (16 marks)

Question 3: Castigliano’s theorem — horizontal deflection at joint 3 (16 marks)