Question 5 of 9: Slope-deflection analysis of a propped gable frame

Question 5: Slope-deflection analysis of a propped gable 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 symmetric-looking gable frame with equal rafters, propped at its apex by a fixed-base column, with short free stubs beyond each springing and a UDL over the right-hand half.

Find. All member end moments, hence the shear force and bending moment diagrams with their maximum and minimum ordinates.

Approach. Prove first that inextensibility forbids sway, so the analysis carries only joint rotations; replace the two free stubs by their statically equivalent joint actions; then write slope-deflection equations for the rafters and the column and solve the two independent joint-moment equations.

1. Kill the sway before writing anything. The column ③–⑥ is vertical and inextensible with a fixed base, so $v_3=0$; the pin at ④ gives $u_4=v_4=0$; the inextensible rafter ③–④ then forces $\mathbf{d}_3\cdot\mathbf{e}_{34}=0$, i.e. $u_3=0$. Joint ③ is therefore fully restrained in translation and, working back along rafter ②–③, so is ②. Every chord rotation vanishes.
2. Deal with the two free stubs. Stub ①–② carries nothing, so member ①–② is a zero-force member and $M_{21}=0$, which forces $M_{23}=0$ and lets the rafter be handled with the modified stiffness $3EI/L$. Stub ④–⑤ is a 2 m cantilever under the UDL and applies a known hogging moment $20(2)(1)=40$ kN·m and a shear of 40 kN to joint ④.
3. Get the fixed-end moments of the loaded rafter. A UDL specified per metre of horizontal projection on an inclined member has fixed-end moments $w\,L_{\text{proj}}^{2}/12$ — the projection, not the sloping length — because the transverse component is $w\cos^{2}\alpha$ and $L\cos\alpha=L_{\text{proj}}$.
4. Write the slope-deflection equations. With $k=EI/6.5$ and all $\psi=0$: $M_{32}=3k\theta_3$ (modified stiffness), $M_{36}=4k\theta_3$, $M_{63}=2k\theta_3$, $M_{34}=2k(2\theta_3+\theta_4)+60$ and $M_{43}=2k(2\theta_4+\theta_3)-60$.
5. Impose joint equilibrium. At ③, $M_{32}+M_{34}+M_{36}=0$ gives $11k\theta_3+2k\theta_4=-60$; at ④, the cantilever fixes $M_{43}=-40$ and gives $2k\theta_3+4k\theta_4=20$.
6. Solve and back-substitute. The two-by-two system returns $k\theta_3=-7.0$ and $k\theta_4=+8.5$ kN·m, whence all five end moments follow directly and the joint check $-21+49-28=0$ closes exactly.

The fixed-end moment of the loaded rafter is

and solving the two joint equations gives the end moments

Converted to ordinary sagging moments, rafter ②–③ runs from zero at the roller to $-21.0$ kN·m at the apex, while rafter ③–④ starts at $-49.0$ kN·m on the other side of the same joint. The 28.0 kN·m step between them is the column moment, and that step is joint equilibrium rather than an error. Between the two ends the rafter carries the free parabola of the projected UDL, peaking at $+45.6$ kN·m at $x=11.08$ m.

The reactions follow from member equilibrium and confirm the analysis: $R_2=-3.50$ kN (a hold-down at the roller), $V_4=101.2$ kN, $H_4=-6.46$ kN, and at the column base $V_6=62.3$ kN, $H_6=+6.46$ kN with $M_6=-14.0$ kN·m. Vertical equilibrium closes on the applied $20(8)=160$ kN and the two horizontal reactions cancel, as they must under purely vertical loading.