Question 6 of 9: Slope-deflection with a uniformly distributed load and a support that moves outward (24 marks)

Question 6: Slope-deflection with a uniformly distributed load and a support that moves outward (24 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 frame built in at joint 1 and pinned at joint 3, with the pin displaced horizontally outward.

Find. Shear force and bending moment diagrams with the maximum and minimum ordinate on each member.

Approach. Inextensibility of the two members turns the 16 mm support movement into a prescribed displacement of joint 2, leaving a single unknown rotation.

1. Fix the translations from inextensibility. Member 1–2 is horizontal and anchored at the built-in end, so $u_{2}=u_{1}=0$. Member 2–3 runs along $\mathbf{e}_{1}=(0.6,\,-0.8)$, and its length cannot change, so $$\bigl(\mathbf{D}_{3}-\mathbf{D}_{2}\bigr)\cdot\mathbf{e}_{1}=0\;\Rightarrow\;(0.016)(0.6)+(0-v_{2})(-0.8)=0,$$ giving $$\boxed{v_{2}=-0.012\ \text{m}}$$ The knee drops 12 mm. Both translations are now known, so the frame — which is two degrees statically indeterminate — is only one degree kinematically indeterminate.
2. Chord rotations. For the horizontal member the transverse displacement is the 12 mm drop over 8 m, $$\psi_{12}=+\frac{0.012}{8}=+1.500\times10^{-3}\ \text{rad (clockwise)}.$$ For the inclined member the transverse component of $\mathbf{D}_{3}-\mathbf{D}_{2}=(0.016,\,0.012)$ along $\mathbf{e}_{2}=(0.8,\,0.6)$ is $0.0200$ m over 5 m, so $$\psi_{23}=-4.000\times10^{-3}\ \text{rad (clockwise positive)}.$$
3. Fixed-end moments of the loaded member. The 12 kN/m load on the 8 m span gives, in the clockwise-positive form, $$M^{F}_{12}=-\frac{wL^{2}}{12}=-\frac{12(8)^{2}}{12}=-64.0\ \text{kN}\cdot\text{m},\qquad M^{F}_{21}=+64.0\ \text{kN}\cdot\text{m}.$$ Member 2–3 carries no span load.
4. Slope-deflection equations. With $2(1.2EI)/8=0.3EI$ on the beam and the modified $3EI/5=0.6EI$ on the member that is pinned at its far end, $$M_{12}=0.3EI\bigl(\theta_{2}-3\psi_{12}\bigr)-64,\qquad M_{21}=0.3EI\bigl(2\theta_{2}-3\psi_{12}\bigr)+64,$$ $$M_{23}=0.6EI\bigl(\theta_{2}-\psi_{23}\bigr),\qquad M_{32}=0 .$$
5. Joint 2 equilibrium gives the rotation. Setting $M_{21}+M_{23}=0$, $$1.2EI\,\theta_{2}+0.00105\,EI+64=0\;\Rightarrow\;\theta_{2}=-\frac{74.5}{1.2(10\,000)}=\boxed{-6.2083\times10^{-3}\ \text{rad}}$$ (clockwise positive, so the knee rotates counter-clockwise).
6. End moments. Substituting back, $$M_{12}=-96.13\ \text{kN}\cdot\text{m},\qquad M_{21}=+13.25\ \text{kN}\cdot\text{m},\qquad M_{23}=-13.25\ \text{kN}\cdot\text{m},\qquad M_{32}=0.$$ Joint 2 balances exactly, and as sagging ordinates the built-in end carries $$\boxed{M_{1}=-96.13\ \text{kN}\cdot\text{m (hogging)}}$$ while the knee carries $-13.25$ kN·m on both members.
7. Shears on the beam. With $M_{\text{sag}}$ known at both ends of the 8 m span, $$V_{1}=\frac{wL}{2}+\frac{M_{\text{sag}}(2)-M_{\text{sag}}(1)}{L}=48+\frac{-13.25+96.13}{8}=58.36\ \text{kN},$$ and $V_{2}=58.36-12(8)=-37.64$ kN. The shear passes through zero at $$x=\frac{58.36}{12}=4.863\ \text{m},$$ where the sagging moment peaks at $$\boxed{M_{\max}=+45.78\ \text{kN}\cdot\text{m}}$$ and the diagram crosses the axis at $x=2.10$ m and $x=7.62$ m.
8. The inclined member and the reactions. Member 2–3 carries no span load, so its moment falls linearly from 13.25 kN·m at the knee to zero at the pin, and its shear is constant, $$V_{23}=\frac{13.25}{5}=2.65\ \text{kN},$$ with an axial compression of 49.04 kN. Resolving that member force at joint 3 gives the reactions $$H_{1}=31.54\ \text{kN},\quad V_{1}=58.36\ \text{kN},\quad M_{1}=96.13\ \text{kN}\cdot\text{m};\qquad H_{3}=-31.54,\quad V_{3}=37.64\ \text{kN}.$$ The vertical reactions add to 96.0 kN, which is $12\times8$ — the load, exactly.