Question 3 of 9: Least work applied to two cantilevers joined by a tension member

Question 3: Least work applied to two cantilevers joined by a tension member (18 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 bending moment and the shear force in beam (3)–(4) at its built-in left end, joint (3).

Approach. The structure is once redundant. Take the axial force $S$ in the pin-ended tie as the redundant, write the total strain energy of the two beams plus the tie in terms of $S$, and impose the least-work condition $\partial U/\partial S = 0$; the tie is a two-force member, so its own extension must be included — that term is what stops the answer being the rigid-prop value.

1. Identify the redundant and the internal forces. Both beams are built in at the wall and free at their tips; the only connection between them is the pin-ended vertical member, which can carry axial force alone. Calling that force $S$ (tension positive), the tie pulls the lower tip up and the upper tip down. Measuring $x$ from each free end, $$M_{\text{lower}}(x) = Sx - \tfrac{1}{2}wx^{2},\qquad M_{\text{upper}}(x) = -Sx$$ so that $\partial M_{\text{lower}}/\partial S = x$ and $\partial M_{\text{upper}}/\partial S = -x$.
2. Write the least-work condition. With the beams inextensible, the strain energy is bending in the two beams plus axial in the tie, and Castigliano's second theorem applied to the redundant gives $$\frac{\partial U}{\partial S} = \frac{1}{EI}\int_0^L M\, \frac{\partial M}{\partial S}\,\mathrm{d}x \;+\; \frac{Sh}{AE} = 0 .$$ Substituting the two moment expressions, $$\frac{1}{EI}\int_0^{L}\!\left(Sx - \tfrac{1}{2}wx^{2}\right)x\,\mathrm{d}x +\frac{1}{EI}\int_0^{L}\!\left(-Sx\right)\left(-x\right)\mathrm{d}x +\frac{Sh}{AE} = 0 .$$
3. Evaluate the integrals. Each bending integral contributes $SL^{3}/3$, and the load term contributes $-wL^{4}/8$, so $$S\left(\frac{2L^{3}}{3EI} + \frac{h}{AE}\right) = \frac{wL^{4}}{8EI}.$$ This is a flexibility statement in disguise: the bracket is the relative vertical flexibility of the two tips plus the stretch of the tie, and the right side is the free deflection of the loaded tip.
4. Insert the numbers. With $L = 6\ \text{m}$, $h = 5\ \text{m}$, $w = 8\ \text{kN/m}$: $$\frac{2L^{3}}{3EI} = \frac{2(216)}{3(1.8\times10^{5})} = 8.000\times10^{-4}\ \text{m/kN},\qquad \frac{h}{AE} = \frac{5}{5.0\times10^{4}} = 1.000\times10^{-4}\ \text{m/kN}$$ $$\frac{wL^{4}}{8EI} = \frac{8(1296)}{8(1.8\times10^{5})} = 7.200\times10^{-3}\ \text{m}$$ so that $$S = \frac{7.200\times10^{-3}}{9.000\times10^{-4}} = \boxed{8.00\ \text{kN}\ \ (\text{tension})}$$ The tie stretch accounts for one ninth of the total flexibility; ignoring it (that is, treating the tie as a rigid prop) would give $S = 7.200\times10^{-3}/8.000\times10^{-4} = 9.00\ \text{kN}$, an error of 12.5 per cent in the wrong direction.
5. Return to the loaded beam and take the free body. Beam (3)–(4) is now a determinate cantilever carrying the UDL downwards over 6 m and the tie force upwards at its tip. Vertical equilibrium of the whole member gives the shear at the wall, $$V_3 = wL - S = 8(6) - 8.00 = \boxed{40.0\ \text{kN}}$$ and moments about joint (3) give $$M_3 = -\frac{wL^{2}}{2} + SL = -\frac{8(36)}{2} + 8.00(6) = \boxed{-96.0\ \text{kN}\cdot\text{m}\ \ (\text{hogging})}$$ The negative sign says the top fibre of the lower beam is in tension at the wall, as a cantilever must be. For completeness the upper beam carries a pure tip load of 8.00 kN downwards, so its wall moment is $-8.00(6) = -48.0\ \text{kN}\cdot\text{m}$ and its wall shear is 8.00 kN.
6. Check the answer two ways. First, the load path: of the $48.0\ \text{kN}$ applied to the lower beam, $40.0\ \text{kN}$ goes directly into wall (3) and $8.00\ \text{kN}$ is hung off the upper beam through the tie — the two wall shears sum to the applied load. Second, compatibility: the tip of the lower beam moves down by $wL^{4}/8EI - SL^{3}/3EI = 7.200\times10^{-3} - 3.200\times10^{-3} = 4.000\times10^{-3}\ \text{m}$, the tip of the upper beam moves down by $SL^{3}/3EI = 3.200\times10^{-3}\ \text{m}$, and the difference, $0.800\times10^{-3}\ \text{m}$, is exactly the extension $Sh/AE$ of the tie.