Question 8 of 9: Symmetric frame with a Gerber top member (24 marks)

Question 8: Symmetric frame with a Gerber top member (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 symmetric frame of 22 m overall projection. Pins at joints 1 and 7 carry 3.75 m outer columns of stiffness $EI$; inclined members of stiffness $1.3EI$ rise 2.5 m over a 6 m run to joints 3 and 5 at 6.25 m; inner columns of stiffness $EI$ drop 5 m from joints 3 and 5 to fixed bases at joints 4 and 8, which sit 1.25 m above the outer supports. The 10 m top member (stiffness $1.3EI$) carries internal hinges 2 m in from each end. A UDL $w=30$ kN/m acts over the whole 22 m projection.

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

Approach. Strip the Gerber top member into a determinate suspended span plus two cantilevers, then use symmetry to show that the remaining half-frame cannot sway, leaving two rotations.

1. Take the suspended span first. The 6 m length between the two hinges has zero moment at both ends, so it is a simple span: $$R_{\text{hinge}}=\frac{30(6)}{2}=90\ \text{kN},\qquad M_{\text{mid}}=\frac{30(6)^{2}}{8}=\boxed{135\ \text{kN}\cdot\text{m}}$$
2. Turn each 2 m stub into a known joint action. The stub between joint 3 and the first hinge is a determinate cantilever carrying its own UDL plus the 90 kN handed over by the suspended span: $$M_{3,\text{beam}}=\frac{30(2)^{2}}{2}+90(2)=60+180=\boxed{240\ \text{kN}\cdot\text{m}},\qquad V_{3}=30(2)+90=150\ \text{kN}$$
3. Show the half-frame cannot sway. Every horizontal displacement in the chain 2–3–5–6 is the same because the top member and both inclined members are inextensible with vertical columns at their ends; symmetry requires $u_{5}=-u_{3}$, so both are zero. The inner column fixes $v_{3}=0$ and the outer column fixes $v_{2}=0$. Only $\theta_{2}$ and $\theta_{3}$ survive.
4. Write the fixed-end moment of the inclined member. The UDL is quoted per metre of horizontal projection, so the moment diagram along the member is the diagram of a horizontal beam of span 6 m; the flexibility integrals all pick up the same factor $L/L_{p}$, which cancels: $$\mathrm{FEM}_{23}=\frac{wL_{p}^{2}}{12}=\frac{30(6)^{2}}{12}=90\ \text{kN}\cdot\text{m}$$ The member stiffness, however, still uses the true length 6.5 m.
5. Write the two joint equations. With $k_{12}=3EI/3.75=0.8EI$ (pinned at joint 1), $k_{23}=2(1.3EI)/6.5=0.4EI$ and $k_{34}=2EI/5=0.4EI$, $$\text{joint 2:}\quad 1.6\,EI\theta_{2}+0.4\,EI\theta_{3}=-90$$ $$\text{joint 3:}\quad 0.4\,EI\theta_{2}+1.6\,EI\theta_{3}=90-240=-150$$
6. Solve. $$EI\theta_{2}=-35,\qquad EI\theta_{3}=-85$$
7. Recover the end moments. $$M_{21}=0.8(-35)=\boxed{-28.0},\qquad M_{23}=0.4(-155)+90=\boxed{+28.0}$$ $$M_{32}=0.4(-205)-90=\boxed{-172.0},\qquad M_{34}=0.8(-85)=\boxed{-68.0},\qquad M_{43}=0.4(-85)=\boxed{-34.0}$$ all in kN·m, with $M_{12}=0$ at the pin. Joint 3 balances: $-172-68+240=0$.
8. Find the peak sagging moment in the inclined member. Working on the equivalent 6 m horizontal span with end sagging moments $-28$ and $-172$, the end shear is $R=(540-172+28)/6=66$ kN, so the shear vanishes $2.2$ m along the projection: $$M_{\max}=-28+66(2.2)-15(2.2)^{2}=\boxed{+44.6\ \text{kN}\cdot\text{m}}$$

Two features of the answer are easy to mistake for errors. First, the bending moment steps at joint 3: the top member arrives with 240 kN·m while the inclined member leaves with 172 kN·m, and the 68 kN·m difference is exactly the moment driven into the inner column. Second, the load is shared very unevenly. The two inner columns take 260.9 kN each while the outer pins take only 69.1 kN each, because the stiff inner columns sit directly under the heavy end of the roof. The horizontal reactions are small and self-cancelling: 7.47 kN at each pin and 20.40 kN at each fixed base.