Question 5 of 9: Two-beam structure by Castigliano’s least-work theorem

Question 5: Two-beam structure by Castigliano’s least-work theorem (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. Cantilever 1–2 of length $a=3\ \text{m}$ built in at joint 1; beam 2–3 of length $L=6\ \text{m}$ built in at joint 3 and carrying $w=4\ \text{kN/m}$; the two are connected at joint 2 by a roller, which transmits a vertical force only. Both members have the same $EI$ and are inextensible.

Find. The interaction force at the roller, and the shear force and bending moment diagrams of both beams with their maximum and minimum ordinates.

Approach. The structure is one degree indeterminate; take the roller force $R$ as the redundant, write the total strain energy of the two beams in terms of $R$, and apply $\partial U/\partial R=0$ (least work), which is the statement that the two tips deflect together.

1. Degree of indeterminacy. Two built-in ends give six restraints and the roller one internal force; with three equations of equilibrium available for each of the two free bodies, $$\text{DSI}=(3+3+1)-(3+3)=\boxed{1}$$ so one redundant — the roller force $R$ — is enough.
2. Internal moments in terms of the redundant. Measuring $s$ from the free end of each member, $$M_{12}(s)=-R\,s\quad (0\le s\le 3),\qquad M_{23}(s)=R\,s-\tfrac{w s^{2}}{2}\quad (0\le s\le 6)$$ The cantilever is pushed down by the roller; beam 2–3 is held up by it.
3. Least work. With $U=\int M^{2}\,ds/(2EI)$, Castigliano’s second theorem in its least-work form requires $$\frac{\partial U}{\partial R}=\frac{1}{EI}\left[\int_0^{3}(-Rs)(-s)\,ds +\int_0^{6}\left(Rs-\tfrac{w s^{2}}{2}\right)s\,ds\right]=0$$ because the roller is an internal redundant that does no net external work.
4. Evaluate the integrals. $$\int_0^{3}R s^{2}\,ds=\frac{R\,3^{3}}{3}=9R,\qquad \int_0^{6}R s^{2}\,ds=\frac{R\,6^{3}}{3}=72R,\qquad \int_0^{6}\frac{w s^{3}}{2}\,ds=\frac{w\,6^{4}}{8}=648$$ so that $$9R+72R-648=0\;\Longrightarrow\;R=\frac{648}{81}=\boxed{8\ \text{kN}}$$ Equivalently, the tip of the cantilever and the end of the propped beam must deflect equally: $\dfrac{R a^{3}}{3EI}=\dfrac{wL^{4}}{8EI}-\dfrac{R L^{3}}{3EI}$.
5. Beam 1–2. It is a cantilever carrying the 8 kN roller reaction at its tip: $$V_{12}=8\ \text{kN}\ \text{(constant)},\qquad M_1=-8(3)=\boxed{-24\ \text{kN}\cdot\text{m}}$$ with the moment varying linearly to zero at the roller.
6. Beam 2–3. Measuring $x$ from joint 2, $$V(x)=8-4x,\qquad M(x)=8x-2x^{2}$$ The shear vanishes at $x=2\ \text{m}$, where $$M_{\max}=8(2)-2(2)^{2}=\boxed{+8\ \text{kN}\cdot\text{m}}$$ and at the built-in end $$V_3=8-4(6)=-16\ \text{kN},\qquad M_3=8(6)-2(6)^{2}=\boxed{-24\ \text{kN}\cdot\text{m}}$$
7. Check. Total load $4(6)=24\ \text{kN}$ is carried as $8\ \text{kN}$ through the cantilever and $16\ \text{kN}$ at joint 3; taking moments of the whole structure about joint 3 closes to zero. The two built-in ends happen to carry the same $24\ \text{kN}\cdot\text{m}$, a coincidence of the 3 m : 6 m proportion, not a general result.