Beam deflection by virtual work (22 marks): 16-Civ-A1 Elementary Structural Analysis, December 2019, Q8

Question 8 — Beam deflection 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 12 m two-beam structure carrying 24 kN/m over its entire length, with a pin at A $(x=0)$, a roller at C $(x=6)$, an internal hinge at D $(x=8)$ and a roller at E $(x=12)$. Point B is at $x=3$ m. $EI = 8.1\times10^{3}\ \text{kN}\cdot\text{m}^{2}$ throughout.

Find. The vertical deflection at B.

Q8 — two-beam structure: 24 kN/m over the full 12 m, pin at A, rollers at C and E, internal hinge at D. The deflection is wanted at B.

Approach. Split at the hinge to get the real moment diagram, apply a unit downward load at B to get the virtual diagram, then evaluate $\delta_B = \int Mm\,\mathrm{d}x/EI$.

With $r = 2+1+1 = 4$ and one hinge, $\mathrm{DSI} = 0$: the structure is determinate, so both moment diagrams follow from statics alone.

Real system: the suspended span D–E. This 4 m piece carries $24(4) = 96\ \text{kN}$ and is held by the hinge at D and the roller at E. Taking moments about E, $$V_D(8-12) + 96(2) = 0 \;\Rightarrow\; V_D = 48\ \text{kN}$$ so $R_E = 96 - 48 = 48\ \text{kN}$ and the main span must push 48 kN up at the hinge — equivalently the suspended span hangs 48 kN on the overhang tip.
Real system: the main span A–D. It carries $24(8) = 192\ \text{kN}$ at $x=4$ plus the 48 kN transferred at $x=8$, supported at A and C. Taking moments about A, $$6R_C = 192(4) + 48(8) = 768 + 384 = 1152 \;\Rightarrow\; R_C = 192\ \text{kN}$$ $$R_A = 192 + 48 - 192 = \boxed{48\ \text{kN}\ \uparrow}$$ so the real moment is $$M(x) = 48x - 12x^{2} + 192\langle x-6 \rangle$$ giving $M = +36\ \text{kN}\cdot\text{m}$ at B, $-144\ \text{kN}\cdot\text{m}$ at C and exactly zero at the hinge D — the check that the split was handled correctly.
Virtual system: unit load at B. Remove the real load and apply 1 kN downward at $x=3$. The suspended span is now unloaded, so it delivers nothing at the hinge and the virtual structure is simply a beam on A and C. Hence $$m(x) = 0.5x \ (0 \le x \le 3), \qquad m(x) = 0.5x - (x-3) = 3 - 0.5x \ (3 \le x \le 6)$$ and $m(x) = 0$ for $x > 6$. That the virtual diagram dies at C is what makes this problem short: only the first 6 m of the beam can contribute.
Integrate. $$\delta_B = \frac{1}{EI}\int_{0}^{6} M(x)\,m(x)\,\mathrm{d}x$$ Substituting $M = 48x - 12x^{2}$ over this stretch and the two straight branches of $m$, $$\int_{0}^{3}(48x-12x^{2})(0.5x)\,\mathrm{d}x + \int_{3}^{6}(48x-12x^{2})(3-0.5x)\,\mathrm{d}x = 81\ \text{kN}^{2}\cdot\text{m}^{3}$$
Evaluate the deflection. $$\delta_B = \frac{81}{8.1\times10^{3}} = 0.0100\ \text{m}$$ $$\delta_B = \boxed{10.0\ \text{mm downward}}$$ The positive result confirms the movement is in the direction of the unit load, so B deflects down.

Q8 real-load moment diagram. M is zero at the hinge D, as it must be.

Q8 virtual-system moment diagram for a unit downward load at B. Because m vanishes beyond C, only the 0–6 m stretch contributes to the integral.

Question 8 — results
Hinge force at D	48 kN
$R_A$ / $R_C$ / $R_E$	48 kN / 192 kN / 48 kN, all upward
$M$ at B / at C / at D	+36 / −144 / 0 kN·m
$\int_{0}^{6} Mm\,\mathrm{d}x$	81 kN$^2$·m$^3$
$\delta_B$	10.0 mm downward

Topic. Virtual work (unit-load method) for beam deflections on a structure containing an internal hinge.

Key relations. $1 \cdot \delta = \displaystyle\int \frac{Mm}{EI}\,\mathrm{d}x$, where $M$ is the bending moment under the real loading and $m$ that under a unit load applied at the point and in the direction of the required deflection; the moment is identically zero at an internal hinge in both systems.

Why this works. Equating external virtual work to internal virtual strain energy isolates one deflection component. The economy here comes from the virtual system: with no real load applied, the suspended span D–E transmits nothing through the hinge, so the virtual moment diagram exists only between A and C. Since the integrand is a product, any region where either diagram vanishes drops out entirely, reducing a 12 m integral to a 6 m one. Recognising that before integrating is the difference between a five-minute answer and a twenty-minute one.

Common pitfall. Integrating over the whole 12 m and mis-handling the region beyond C, or forgetting the hinge force when building the real diagram — omitting the 48 kN transferred at D changes $R_A$ and $R_C$ and hence every ordinate. Always confirm that the real moment diagram passes through zero at the hinge before integrating. The diagram-multiplication (Vereshchagin) shortcut is available here too, but with a parabolic $M$ against a triangular $m$ the direct integration shown is less error-prone.

Source. Hibbeler, Structural Analysis 10th ed., §9.4–9.7 (virtual work for beams and frames); Kassimali 6th ed., §7.4–7.6.

Question 8 of 8: Beam deflection by virtual work (22 marks)

Question 8 — Beam deflection by virtual work (22 marks)