Question 7 of 8: Vertical Deflection of a Frame by Virtual Work

Question 7: Vertical 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.

Find. The vertical deflection of node 4, with its direction.

Approach. Confirm the frame is determinate, solve the real system, then apply a unit vertical load at node 4 and evaluate $\delta_4 = \int Mm\,\mathrm{d}s/EI$ over every member. Almost all of that work disappears once it is noticed which members carry no moment at all.

1. Check determinacy, and notice the hinge. With $m = 4$ members, $j = 5$ joints, $r = 4$ reaction components and $c = 1$ release at the node-2 hinge, $$i = 3m + r - 3j - c = 12 + 4 - 15 - 1 = 0$$ so the frame is statically determinate. The hinge at node 2 is what makes it so, and it also makes column 1–2 a two-force member: it is pinned at node 1, hinged at node 2 and carries no load between, so it transmits a purely vertical force and no moment.
2. Solve the real system. Since column 1–2 can deliver only a vertical force at node 2, there is no horizontal reaction anywhere, and the beam 2–3–4 is simply supported at nodes 2 and 3 with an unloaded overhang. Taking moments about node 3, $$V_1(4) = 6(4)(2) = 48 \;\Longrightarrow\; \boxed{V_1 = V_5 = 12.0\ \text{kN}\ \uparrow}$$
3. Write the real bending moments. On the span, with $x$ from node 2, $$M(x) = 12x - 3x^2, \qquad 0 \le x \le 4$$ which peaks at $M(2) = 12.0\ \text{kN}\cdot\text{m}$ and returns to zero at node 3. The overhang carries nothing, so $M = 0$ there; the moment is therefore continuous and zero through node 3, which leaves column 3–5 with no end moment either, and column 1–2 has none because it is a two-force member. The entire real moment diagram lives on the single 4 m span.
4. Apply the unit load and write the virtual moments. A unit downward load at node 4 acts 1 m beyond node 3, so moments about node 3 give $$v_1 = -\frac{1(1)}{4} = -0.25 \quad(\text{i.e. }0.25\ \text{kN downward at node 1}),\qquad v_5 = +1.25\ \text{kN}$$ Along the span, $m(x) = -0.25x$, reaching $m = -1.0\ \text{kN}\cdot\text{m}$ at node 3, which is exactly the hogging moment the 1 m overhang delivers there. On the overhang the virtual moment runs from $0$ at node 4 to $-1.0$ at node 3, and both columns again carry no moment.
5. Evaluate the single surviving integral. Only span 2–3 has both $M \ne 0$ and $m \ne 0$: $$\int_0^4 M m\,\mathrm{d}x = -0.25\int_0^4\left(12x^2 - 3x^3\right) \mathrm{d}x = -0.25\left[4x^3 - \tfrac{3}{4}x^4\right]_0^4 = -16.0$$ in $\text{kN}^2\cdot\text{m}^3$.
6. Divide by $EI$ and state the direction. $$\delta_4 = \frac{-16.0}{2\,000} = -0.0080\ \text{m}$$ $$\boxed{\delta_4 = 8.0\ \text{mm}\ \text{UPWARD}}$$
7. Confirm it geometrically. The overhang is rigid and unloaded, so node 4 simply follows the rotation of node 3. For a simply supported span under a uniform load that end rotation is $$\theta_3 = \frac{wL^3}{24EI} = \frac{6(4)^3}{24(2\,000)} = 0.0080\ \text{rad}$$ and the tip rises by $\theta_3 \times 1\ \text{m} = 8.0\ \text{mm}$, reproducing the virtual-work result exactly.