Question 7 of 9: Slope-Deflection Analysis with a Jacked Support and a Span Load

Question 7: Slope-Deflection Analysis with a Jacked Support and a Span Load (22 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 with a pinned foot that has been jacked sideways after erection, plus a gravity load on the horizontal member.

Find. The member-end moments, and the shear force and bending moment diagrams of both members with their maximum and minimum ordinates.

Approach. Use inextensibility to convert the jacking displacement into a vertical movement of joint 2 and hence into chord rotations, add the fixed-end moments of the uniformly distributed load, and solve the two conditions $M_{12}=0$ (pin) and $M_{21}+M_{23}=0$ (joint 2). Superposition is legitimate throughout because the response is linear.

1. Propagate the jack through the kinematics. Member 2–3 is horizontal, inextensible, and anchored at the built-in joint 3, so $u_2=u_3=0$: joint 2 cannot move sideways at all. Member 1–2 is then inextensible along $(0.6,\,0.8)$, and with $u_1=+0.024$ m, $$(u_2-u_1)(0.6)+(v_2-0)(0.8)=0 \;\Longrightarrow\;\boxed{v_2=+18.0\ \text{mm (upward)}.}$$ Pushing the foot towards the frame lifts the knee — a pure geometry result, independent of $EI$.
2. Chord rotations. With $\mathbf{e}_2$ the member axis turned $+90^\circ$, $$\psi_{12}=\frac{(-0.024)(-0.8)+(0.018)(0.6)}{6.0} =\frac{0.030}{6.0}=+5.00\times10^{-3}\ \text{rad},\qquad \psi_{23}=\frac{0-0.018}{12.0}=-1.50\times10^{-3}\ \text{rad}.$$
3. Fixed-end moments of the span load. For the horizontal member under $w=12$ kN/m over 12 m, $$\mathrm{FEM}_{23}=+\frac{wL^{2}}{12}=+144\ \text{kN}\cdot\text{m},\qquad \mathrm{FEM}_{32}=-144\ \text{kN}\cdot\text{m}.$$
4. Slope-deflection equations and the two conditions. With $k_{12}=2EI/6=6666.7$ and $k_{23}=2EI/12=3333.3\ \text{kN}\cdot\text{m}$, the pin condition $M_{12}=0$ gives $\theta_1=(0.015-\theta_2)/2$, whence $M_{21}=10\,000\,\theta_2-50$. Joint 2 equilibrium then reads $$\left(10\,000\,\theta_2-50\right)+ \left(6666.7\,\theta_2+159\right)=0 \;\Longrightarrow\; \theta_2=-6.540\times10^{-3}\ \text{rad},$$ and back substitution gives $\theta_1=+1.0770\times10^{-2}$ rad.
5. Member-end moments. $$M_{12}=0,\qquad \boxed{M_{21}=-115.4\ \text{kN}\cdot\text{m}},\qquad M_{23}=+115.4\ \text{kN}\cdot\text{m},\qquad \boxed{M_{32}=-150.8\ \text{kN}\cdot\text{m}.}$$ Both the load and the jack push the same way here, which is why the built-in end carries more than the $wL^{2}/12=144$ it would see if joint 2 were fully fixed.
6. Shear force diagram. Member 1–2 carries no span load, so its shear is the constant $$V_{12}=\frac{|0-115.4|}{6.0}=19.23\ \text{kN}.$$ For member 2–3 the sagging end moments are $-115.4$ at joint 2 and $-150.8$ at joint 3, so $$V_2=\frac{wL}{2}+\frac{M_{\text{sag},3}-M_{\text{sag},2}}{L} =72.0-2.95=\boxed{+69.05\ \text{kN}},\qquad V_3=69.05-144=\boxed{-74.95\ \text{kN}}.$$
7. Bending moment diagram. The shear crosses zero at $x=69.05/12=5.754$ m from joint 2, where the sagging moment peaks at $$M_{\max}=-115.4+69.05(5.754)-6(5.754)^{2} =\boxed{+83.3\ \text{kN}\cdot\text{m}}.$$ Points of contraflexure occur at $x=1.97$ m and $x=9.54$ m along member 2–3. As a global check, the vertical reactions $69.05+74.95=144.0$ kN balance the total load $12\times12=144$ kN exactly, and taking moments about joint 3 with the reaction $(75.83,\,69.05)$ kN at the pin closes to zero against the $150.8\ \text{kN}\cdot\text{m}$ fixing moment.