Question 3 of 9: Castigliano's theorem of least work on a two-member frame (16 marks)

Question 3: Castigliano's theorem of least work on a two-member frame (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.

Given. A two-member rigid frame with a built-in support at joint 1 and a horizontal-plane roller at joint 3.

Find. The bending moment and the shear force carried by member 1–2 at joint 1.

Approach. The frame is one degree indeterminate; take the roller reaction as the redundant, write the bending moment in every member as a linear function of it, and set $\partial U/\partial V_{3}=0$.

1. Establish the degree of indeterminacy. The built-in support supplies three components and the roller one, and the frame has no closed loop, so $$\text{DSI}=r-3=4-3=1.$$ One redundant — take the roller reaction $V_{3}$ (upward positive).
2. Express the reactions in terms of the redundant. Horizontal equilibrium gives $H_{1}=0$ immediately (the roller cannot push sideways and there is no horizontal load). Then $$V_{1}=32-V_{3},\qquad M_{1}=32(3)-V_{3}(6)=96-6V_{3}\ \text{kN}\cdot\text{m}.$$
3. Write the bending moment fields. Measure $s$ from joint 3 along member 2–3 and $t$ from joint 1 along member 1–2; both members have direction cosines $(0.6,\,0.8)$, so a point at $s$ sits $0.6s$ horizontally from joint 3. Taking the far free body in each case, $$M_{23}(s)=0.6\,V_{3}\,s,\qquad M_{12}(t)=V_{3}\!\left(6-0.6t\right)-32\!\left(3-0.6t\right).$$ At $t=5$ both give $3V_{3}$, so the two expressions agree at the apex.
4. Differentiate with respect to the redundant. $$\frac{\partial M_{23}}{\partial V_{3}}=0.6s,\qquad \frac{\partial M_{12}}{\partial V_{3}}=6-0.6t.$$ Because the members are inextensible only the bending strain energy contributes, so least work reads $$\frac{\partial U}{\partial V_{3}}=\frac{1}{EI}\int_{0}^{5}M_{12}\,\frac{\partial M_{12}}{\partial V_{3}}\,dt+\frac{1}{EI}\int_{0}^{5}M_{23}\,\frac{\partial M_{23}}{\partial V_{3}}\,ds=0.$$ $EI$ is the same in both members, so it cancels — the answer needs no numerical stiffness.
5. Evaluate the integrals. The second integral is $0.36V_{3}\int_{0}^{5}s^{2}ds=15V_{3}$. Writing $M_{12}=a+bt$ with $a=6V_{3}-96$ and $b=19.2-0.6V_{3}$, the first integral is $22.5a+50b=105V_{3}-1200$. Adding, $$120\,V_{3}-1200=0\;\Rightarrow\;\boxed{V_{3}=10.0\ \text{kN (up)}}$$
6. Recover the remaining reactions. $$V_{1}=32-10=22.0\ \text{kN (up)},\qquad H_{1}=0,\qquad M_{1}=96-6(10)=36.0\ \text{kN}\cdot\text{m}.$$ The built-in support therefore carries a hogging moment of 36.0 kN·m and a vertical force of 22.0 kN.
7. Resolve the 22 kN into the member axes. Member 1–2 runs along $\mathbf{e}_{1}=(0.6,\,0.8)$, so its shear direction is $\mathbf{e}_{2}=(-0.8,\,0.6)$. With the joint force $(0,\,22)$ acting on the member, $$V=(0,22)\cdot(-0.8,\,0.6)=\boxed{13.2\ \text{kN}},\qquad N=(0,22)\cdot(0.6,\,0.8)=17.6\ \text{kN (compression)}.$$ The resultant closes exactly, $\sqrt{13.2^{2}+17.6^{2}}=22.0$ kN, which is the cheapest available check on the decomposition.
8. Bending moment at joint 1. The moment on the member at joint 1 is the support moment itself, $$\boxed{M_{1}=36.0\ \text{kN}\cdot\text{m (hogging)}}$$ and the moment falls linearly along member 1–2 to $3V_{3}=30.0$ kN·m at the apex, then linearly to zero at the roller.