Question 2 of 9: Schematic shear force and bending moment diagrams (12 marks)

Question 2: Schematic shear force and bending moment diagrams (12 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. Two unloaded-dimension “schematic” structures. Part (a) carries the bay lengths listed below; part (b) prints no dimensions at all, so the drawn proportions are adopted.

Find. The shear force and bending moment diagrams of both structures, with the ordinates that the shapes turn on.

Approach. Use the hinge to peel off the statically determinate front end by hand, then carry the remaining four-times indeterminate frame with the displacement method; for (b) apply the slope-deflection equations to the prescribed support movement.

1. (a) Count the indeterminacy. The frame has no closed loop, so $$\text{DSI}=r-3-c=(1+3+2+2)-3-1=4,$$ with $r$ the roller, the built-in end and the two pinned column bases, and $c=1$ the moment condition at the hinge. Four redundants means the back half must be solved by a stiffness (or flexibility) analysis — but the front half need not be.
2. (a) The hinge isolates a determinate front end. Take the free body from the tip to the hinge and set the moment there to zero. The overhang resultant $3w$ acts $6.5\text{ m}$ from the hinge and the roller is $5\text{ m}$ from it: $$R_{3}(5)-3w(6.5)=0\;\Rightarrow\;\boxed{R_{3}=3.90\,w}$$ so this reaction, and everything to the left of the hinge, is independent of the column height.
3. (a) Front-end ordinates. The shear grows linearly to $V(3^{-})=-3.00\,w$ and jumps to $V=+0.90\,w$ at the roller, holding that value all the way to the hinge; the moment reaches $M(3)=-w(3)^{2}/2=-4.50\,w$ at the roller and climbs linearly to zero at the hinge, $$M(8)=-4.50\,w+0.90\,w(5)=0\;\checkmark$$ The hinge therefore hands a shear of $0.90\,w$ (and no moment) to the frame behind it.
4. (a) Solve the four-times indeterminate remainder. With $h=5$ m the stiffness solution gives the beam moments $+1.80\,w$ just left of the first column and $-2.65\,w$ just right of it — a genuine step of $4.45\,w$, which is exactly the moment carried into the column head. The same thing happens at the second column: $-5.15\,w$ on the left, $-2.71\,w$ on the right, a step of $2.44\,w$. The largest sagging moment in the loaded 8 m bay is $$M_{\max}=+4.15\,w \text{ at } x=13.68\text{ m},$$ and the built-in end finishes at $+1.36\,w$.
5. (a) Assemble the diagrams. The shear runs $0\to-3.00\,w$ over the overhang, jumps to $+0.90\,w$ at the roller and stays there through the hinge to the first column, jumps to $+3.69\,w$ where that column delivers its reaction, falls linearly under the UDL to $-4.31\,w$ at the second column, and jumps to a constant $+0.68\,w$ over the last bay. Every jump equals the reaction or column force applied there, and the vertical reactions sum to $11\,w$, the total load — the arithmetic check worth doing before the diagram is drawn.
6. (b) Kinematics of the settling portal. The columns are vertical and inextensible, so $v_{B}=v_{A}=0$ and $v_{C}=v_{D}=-\Delta$; the beam is horizontal and inextensible, so $u_{B}=u_{C}=\delta$, one unknown sway. Writing $\rho=\delta/L$ and $\tau=\Delta/L$, the chord rotations in the clockwise-positive form are $\psi_{AB}=\psi_{DC}=\rho$ and $\psi_{BC}=\tau$.
7. (b) Slope-deflection and joint equilibrium. With $k=2EI/L$ and $\theta_{A}=\theta_{D}=0$ (both bases are fixed and neither rotates), joint equilibrium at $B$ and $C$ reads $$4\theta_{B}+\theta_{C}=3(\rho+\tau),\qquad 4\theta_{C}+\theta_{B}=3(\rho+\tau),$$ whose difference gives $\theta_{B}=\theta_{C}=\theta$ and therefore $\theta=\tfrac{3}{5}(\rho+\tau)$.
8. (b) The storey-shear equation closes it. There is no horizontal load, so the two column shears must sum to zero, which requires $(M_{AB}+M_{BA})+(M_{DC}+M_{CD})=0$, i.e. $\theta=2\rho$. Combining with the joint result, $$1.4\rho=0.6\tau\;\Rightarrow\;\rho=\tfrac{3}{7}\tau,\qquad \theta=\tfrac{6}{7}\tau.$$
9. (b) Back-substitute. Every one of the six end moments comes out with the same magnitude, $$\boxed{|M|=\frac{6EI\Delta}{7L^{2}}\approx 0.857\,\frac{EI\Delta}{L^{2}}}$$ and because $M_{AB}=-M_{BA}$ in each column the column shear is exactly zero — there is no horizontal reaction at either base. The beam carries the whole effect as a constant shear $$V_{BC}=\frac{12EI\Delta}{7L^{3}},$$ which is also the axial force in each column and the (equal and opposite) vertical reactions.
10. (b) Diagram shapes. The bending moment diagram is a straight line on every member: each column runs from $+6EI\Delta/7L^{2}$ at one end to $-6EI\Delta/7L^{2}$ at the other with the contraflexure point at mid-height; the beam is a straight line of constant slope between the two joints, with no point of contraflexure. The shear diagram is zero on both columns and a constant $12EI\Delta/7L^{3}$ on the beam.