Question 8 of 9: Trapezoidal sway frame with an inextensible strut

Question 8: Trapezoidal sway frame with an inextensible strut (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. Members 1–2 (5 m, inclined 3:4) and 2–3 (6 m, horizontal) are flexural members of equal $EI$; member 3–4 (5 m, inclined 3:4 the other way) is a pin-ended inextensible strut carrying axial force only. Joint 1 is built in, joint 4 is pinned, all members are inextensible, and $w=16.4$ kN/m acts on 2–3.

Find. The end moments, the shear force and bending moment diagrams with their governing ordinates, the strut force and the reactions.

Approach. Express the whole displacement field in terms of one sway parameter using inextensibility, write two slope-deflection equations (the beam moment-released at joint 3), and close with a virtual-work sway equation.

1. Reduce the kinematics to one parameter. Member 1–2 is inextensible from a fixed base, so joint 2 can only move perpendicular to it: $3u_{2}+4v_{2}=0$. The horizontal member forces $u_{3}=u_{2}$, and the strut from the pinned joint 4 forces $3u_{3}-4v_{3}=0$. Writing $\delta=u_{2}=u_{3}$,$$v_{2}=-0.75\,\delta,\qquad v_{3}=+0.75\,\delta,$$so joint 2 drops as joint 3 rises. The chord rotations follow as$$\psi_{12}=-0.25\,\delta,\qquad \psi_{23}=+0.25\,\delta .$$One translation and one rotation ($\theta_{2}$) are therefore the only unknowns.
2. Use the strut to release joint 3. A pin-ended strut transmits no moment, so $M_{32}=0$ and member 2–3 can be written with the modified stiffness. With $A=EI\theta_{2}$ and $B=EI\delta$,$$M_{12}=\frac{2}{5}\left(A+0.75B\right),\qquad M_{21}=\frac{2}{5}\left(2A+0.75B\right),$$$$M_{23}=\frac{3}{6}\left(A-0.25B\right)+\frac{wL^{2}}{8},\qquad \frac{wL^{2}}{8}=\frac{16.4(6)^{2}}{8}=73.8\ \text{kN}\cdot\text{m}.$$The modified fixed-end moment is $wL^{2}/8$, not $wL^{2}/12$, precisely because the far end is released.
3. Enforce moment equilibrium at joint 2. Setting $M_{21}+M_{23}=0$ gives$$1.3A+0.175B=-73.8 .$$This is the only joint equation, because joint 3 carries no moment and joint 1 is built in.
4. Write the sway equation by virtual work. Give the frame the unit sway pattern $\delta^{*}=1$ (so $v_{2}^{*}=-0.75$, $v_{3}^{*}=+0.75$, rotations held at zero). The strut does no work because it is inextensible, the equivalent nodal shears of the uniform load cancel, and what remains is$$0.25\left(M_{12}+M_{21}\right)-0.25\,M_{23}=0 .$$Substituting the member equations, $0.7A+0.725B=73.8$. Solving the pair,$$EI\theta_{2}=-81.0\ \text{kN}\cdot\text{m}^{2},\qquad EI\delta=\boxed{180\ \text{kN}\cdot\text{m}^{3}} .$$
5. Recover the end moments. Back-substituting,$$M_{12}=21.6,\qquad M_{21}=-10.8,\qquad M_{23}=+10.8,\qquad M_{32}=0\ \ (\text{kN}\cdot\text{m}).$$Joint 2 balances exactly, and the moment magnitude of 10.8 kN·m is continuous through the joint as it must be where only two members meet.
6. Build the diagrams. Member 1–2 carries no span load, so its moment runs linearly from $-21.6$ kN·m at the base to $-10.8$ kN·m at joint 2 with a constant shear of $(21.6-10.8)/5=2.16$ kN. On the beam, the end shears are$$V_{2}=\frac{M_{23}}{L}+\frac{wL}{2}=1.8+49.2=51.0\ \text{kN},\qquad V_{3}=47.4\ \text{kN},$$summing to $wL=98.4$ kN. The shear vanishes at $x=51.0/16.4=3.110$ m from joint 2, where the sagging moment peaks at$$M_{\max}=-10.8+\frac{51.0^{2}}{2(16.4)}=\boxed{68.5\ \text{kN}\cdot\text{m}},$$falling to zero at joint 3.
7. Close with the strut force and reactions. Joint 3 carries only the beam shear and the strut, so the strut takes$$N_{34}=-\frac{V_{3}}{0.8}=-59.25\ \text{kN},$$that is 59.25 kN of compression, and being a two-force member it delivers its whole force along its own axis: the reaction at the pinned joint 4 is 35.55 kN horizontal and 47.4 kN vertical, resultant 59.25 kN. At the built-in base, the reaction is 35.55 kN horizontal, 51.0 kN vertical and 21.6 kN·m of fixing moment. Taking moments about joint 1 closes the check: $21.6+12(47.4)-6(98.4)=0$.