Question 4 of 9: Least Work Analysis of a Tied Trapezoidal Frame

Question 4: Least Work Analysis of a Tied Trapezoidal Frame (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. A symmetric tied trapezoidal frame, supported by a pin at joint 2 and a roller at joint 3, with the two free lower ends joined by a pin-ended tie.

Find. The force in the tie and hence the maximum bending moment in the horizontal beam 2–3.

Approach. Take the tie force $T$ as the single redundant, express every member's bending moment in terms of $T$ and the applied load, then impose the least-work condition $\partial U/\partial T=0$ and back substitute.

1. Count the redundants and use the symmetry. Externally the frame is determinate: the pin at joint 2 and the roller at joint 3 give three reaction components. Internally the pin-ended tie closes the chain 1–2–3–4 and supplies exactly one redundancy, so the frame is $\boxed{1}$ degree statically indeterminate with $T$ as the natural redundant. Because both the structure and the load are symmetric about midspan, the two legs carry identical moments and only one of them needs integrating.
2. Reactions are independent of the redundant. The tie pulls the two free ends towards each other with equal and opposite horizontal forces, so it contributes nothing to the global equations. Moments about joint 2 give $V_3=w L_b/2=12.0$ kN, and by symmetry $V_2=12.0$ kN with $H_2=0$.
3. Bending moments in terms of $T$. Each leg is a cantilever hanging from its support joint, loaded only by the tie force at its free end, so at a height $y$ above the tie the moment is $M=Ty$; along the leg, measured by $s$ from the free end, $M_{\text{leg}}=0.8\,Ts$ with $0\le s\le 3.0$, reaching $2.4\,T$ at the support. For the beam, with $\xi$ measured from joint 2, $$M_{\text{beam}}(\xi)=2.4\,T-12\xi+3\xi^{2} \quad\text{(sagging}\;=\;-M_{\text{beam}}\text{)}.$$
4. Impose least work. The total complementary energy is stationary with respect to the redundant, so $$\frac{\partial U}{\partial T}= \frac{2}{3EI}\int_0^{3}(0.8Ts)(0.8s)\,\mathrm{d}s +\frac{1}{EI}\int_0^{4}\left(2.4T-12\xi+3\xi^{2}\right)(2.4)\,\mathrm{d}\xi +\frac{T L_t}{EA}=0.$$ Evaluating the integrals, $$\frac{3.84\,T}{EI}+\frac{23.04\,T-76.8}{EI}+\frac{7.6\,T}{EA}=0 \;\Longrightarrow\; \frac{26.88\,T-76.8}{9280}+\frac{7.6\,T}{1900}=0.$$
5. Solve for the tie force. Collecting terms, $0.0028966\,T+0.0040000\,T=0.0082759$, hence $$\boxed{T=1.20\ \text{kN (tension)}}.$$ Note the two stiffness terms are of the same order: the tie's own extensibility supplies more than half the flexibility of the redundant path, so it cannot be treated as rigid.
6. Maximum moment in the horizontal beam. Substituting $T=1.20$ kN into the beam expression and converting to sagging moments, $$M_{\text{sag}}(\xi)=-2.88+12\xi-3\xi^{2},$$ which gives $-2.88\ \text{kN}\cdot\text{m}$ (hogging) at each end and, at midspan where the shear vanishes, $$\boxed{M_{\max}=+9.12\ \text{kN}\cdot\text{m}\ \text{sagging at midspan of }2\text{--}3.}$$ The check is joint 2: the leg delivers $2.4T=2.88\ \text{kN}\cdot \text{m}$, exactly the beam's end moment, and the pin support carries no moment of its own.