Question 4 of 8: Deflection of a cable-propped beam

Question 4: Deflection of a cable-propped beam (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.

(a) Vertical deflection at C

Given. A horizontal beam pinned at A, with points B, C and D at 3, 6 and 9 m from A. At D the beam is held by a vertical cable 2 m long running up to a fixed anchor. A single 18 kN downward load acts at B. Flexural rigidity $EI = 21\,000$ kN·m² for the beam; axial rigidity $AE = 2000$ kN for the cable.

Find. The vertical deflection of point C, including the contribution of the cable's own stretch.

Approach. Unit-load virtual work. Build the real force system, build the virtual system for a unit vertical load at C, then evaluate $\delta = \int Mm/EI\,\mathrm{d}x + \sum NnL/AE$, the second term capturing the cable.

1. Real system. The structure is determinate: the pin supplies two components, the cable one, against three equations. Taking moments about A, the cable force follows at once, $$\sum M_A = 0:\qquad 18(3) = T(9) \quad\Rightarrow\quad T = 6\ \text{kN (tension)},$$ $$\sum F_y = 0:\qquad A_y = 18 - 6 = 12\ \text{kN (up)} .$$ The real bending moment is therefore $M = 12x$ for $0 \le x \le 3$ and $M = 54 - 6x$ for $3 \le x \le 9$, peaking at $+36$ kN·m under the load and returning to zero at D.
2. Virtual system. Remove the real load and apply a downward unit load at C, 6 m from A. The same two equations give the virtual cable force and pin reaction, $$t = \frac{6}{9} = 0.667,\qquad a_y = \frac{3}{9} = 0.333 ,$$ so the virtual moment is $m = x/3$ for $0 \le x \le 6$ and $m = 6 - 2x/3$ for $6 \le x \le 9$, peaking at $m = 2.00$ m under the unit load.
3. Integrate the flexural term. Both $M$ and $m$ are piecewise linear, so each 3 m segment can be evaluated with the trapezoidal product rule $$\int_0^{L} M m\,\mathrm{d}x = \frac{L}{6}\bigl(2M_1m_1 + 2M_2m_2 + M_1m_2 + M_2m_1\bigr).$$ Segment A–B ($M: 0 \to 36$, $m: 0 \to 1$) gives 36; segment B–C ($M: 36 \to 18$, $m: 1 \to 2$) gives 117; segment C–D ($M: 18 \to 0$, $m: 2 \to 0$) gives 36. Summing, $$\int M m\,\mathrm{d}x = 36 + 117 + 36 = 189\ \text{kN}\cdot\text{m}^{3}$$ $$\delta_{\text{beam}} = \frac{189}{21\,000} = 0.00900\ \text{m} = 9.00\ \text{mm}$$
4. Add the cable term. The cable is a single axial member of length 2 m carrying $N = 6$ kN in the real system and $n = 0.667$ in the virtual system, so $$\delta_{\text{cable}} = \frac{N n L}{AE} = \frac{6(0.667)(2)}{2000} = 0.00400\ \text{m} = 4.00\ \text{mm}.$$ It is worth confirming this kinematically rather than trusting the formula: the cable actually stretches $TL/AE = 6(2)/2000 = 6.00$ mm, which drops point D by 6.00 mm; with the beam rigid and hinged at A that drop moves C, at two-thirds of the span, down by $\tfrac{2}{3}(6.00) = 4.00$ mm. The two routes agree exactly, which is the check that the virtual cable force was assigned the right sign.
5. Combine. Virtual work adds the two mechanisms directly: $$\delta_C = \delta_{\text{beam}} + \delta_{\text{cable}} = 9.00 + 4.00 = \boxed{13.0\ \text{mm downward}}$$ Roughly 31 per cent of the movement at C comes from the cable, not from bending; on a structure of this proportion the prop is emphatically not a rigid support, and omitting its axial term would under-predict the deflection by that margin.

(b) Load moved to C: deflection at B

Given. The same structure, with the 18 kN load now applied at C instead of B.

Find. The vertical deflection at B, without further calculation, and the theorem that justifies it.

The answer is 13.0 mm downward — the same number as part (a), and the theorem is Maxwell's law of reciprocal deflections (the special case of Betti's law for two single loads). Maxwell's law states that for a linear elastic structure the deflection at point B caused by a load applied at point C equals the deflection at point C caused by the same load applied at point B, that is $\delta_{BC} = \delta_{CB}$. It follows from Betti's reciprocal-work theorem: the work done by force system 1 moving through the displacements caused by system 2 equals the work done by system 2 moving through the displacements caused by system 1, which is true whenever the material is linear elastic, the displacements are small and the supports do not settle.

The result holds here even though the structure mixes bending and axial action, because the flexibility coefficient that reciprocity concerns is the total one: $f_{BC} = \int M_B m_C / EI\,\mathrm{d}x + \sum N_B n_C L / AE$, and both integrands are symmetric under exchange of the subscripts. It is also worth stating what reciprocity does not promise: the two load cases have different internal force distributions, so the cable tension, the peak bending moment and the shape of the deflected beam are all different. Only the one paired deflection number is shared.