Question 1 of 9: Schematic shear force and bending moment diagrams (12 marks)

Question 1: 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. Three plane structures drawn on page 2, all of uniform $EI$ and all inextensible.

Find. The shear force and bending moment diagrams of each structure, with the governing ordinates labelled.

Part (a) — two-span beam with an overhang

Approach. The beam is two degrees statically indeterminate, so treat it as a two-span continuous beam AB–BE with a determinate overhang EF and solve for the two joint rotations by slope-deflection.

1. Reduce the overhang to a known end moment. The 3 m tip carries $w=P/3$, so the bending moment delivered to E is $$M_{\text{sag}}(E)=-\frac{wL_{oh}^{2}}{2}=-\frac{(P/3)(3)^{2}}{2}=-1.5P\ \text{(hogging)}$$ and a shear of $wL_{oh}=P$ enters the beam at E.
2. Fixed-end moments of span BE. Two point loads act at $a=2$ m and $a=4$ m from B on a 6 m span, so with $\mathrm{FEM}_{i}=+Pab^{2}/L^{2}$, $$\mathrm{FEM}_{BE}=\frac{P(2)(4)^{2}}{36}+\frac{P(4)(2)^{2}}{36}=\frac{32P+16P}{36}=+1.3333P,\qquad \mathrm{FEM}_{EB}=-1.3333P$$ Span AB carries no load, so its fixed-end moments are zero.
3. Write the slope-deflection equations. No support settles, so every $\psi=0$ and $\theta_A=0$: $$M_{AB}=\tfrac{EI}{2}\theta_B,\quad M_{BA}=EI\theta_B,\quad M_{BE}=\tfrac{EI}{3}\bigl(2\theta_B+\theta_E\bigr)+1.3333P$$ $$M_{EB}=\tfrac{EI}{3}\bigl(2\theta_E+\theta_B\bigr)-1.3333P,\qquad M_{EF}=+1.5P$$
4. Enforce joint equilibrium at B and E. Summing the end moments meeting at each joint, $$1.6667\,EI\theta_B+0.3333\,EI\theta_E=-1.3333P,\qquad 0.3333\,EI\theta_B+0.6667\,EI\theta_E=-0.1667P$$ $$\boxed{EI\theta_B=-0.8333P,\qquad EI\theta_E=+0.1667P}$$
5. Back-substitute for the bending moments. The sagging moments at the five stations follow directly: $$M_A=+0.4167P,\quad M_B=-0.8333P,\quad M_C=+0.9444P,\quad M_D=+0.7222P,\quad M_E=-1.5P$$ The value at A is exactly $-\tfrac12 M_B$: span AB is unloaded, so the moment applied at B carries over to the fixed end with the usual factor of one half.
6. Differentiate for the shears. Each bay is straight, so the shear is the moment gradient: $$V_{AB}=\frac{M_B-M_A}{4}=-0.3125P,\quad V_{BC}=+0.8889P,\quad V_{CD}=-0.1111P,\quad V_{DE}=-1.1111P$$ and the overhang runs from $+1.0P$ at E to zero at the free tip.
7. Recover the reactions and check global equilibrium. Jumps in the shear diagram give $$\boxed{R_A=-0.3125P\ (\text{downward}),\quad R_B=+1.2014P,\quad R_E=+2.1111P}$$ $$\textstyle\sum V:\;-0.3125+1.2014+2.1111=3.000P=P+P+wL_{oh}\quad\checkmark$$

The fixed end pulls down. That is not an error: span AB is unloaded and is bent only by the hogging moment handed to it at B, so it acts as a lever that must be held down at A.

Part (b) — beam hinged at both ends

Approach. Identify what the hinges release before analysing anything: a member hinged at both ends transmits no moment, which collapses this frame to a simple beam.

1. Read the releases. Both ends of the beam carry the drawn hinge circle, so $M=0$ at each end. The beam is therefore a simply supported span of 6 m under $w$, regardless of what the columns do.
2. Analyse the beam. With $L=6$ m, $$V_{\text{end}}=\frac{wL}{2}=3w,\qquad \boxed{M_{\max}=\frac{wL^{2}}{8}=4.5w\ \text{at midspan}}$$ The shear falls linearly from $+3w$ to $-3w$ and the moment is the usual parabola.
3. Show that the columns are unbent. Each beam end delivers a purely vertical force $3w$ to a vertical member, along the member axis. A force acting along a straight member's own axis produces axial force only, so no horizontal thrust is called for anywhere; the beam's axial force is zero and both columns carry $$\boxed{V=0,\qquad M=0,\qquad N=3w\ \text{compression}}$$
4. Confirm stability. The left column is built in top and bottom and the right column is built in at its base, so nothing is a mechanism even though the beam is doubly hinged. The right column's head is held laterally by the inextensible beam.

The whole of part (b) is therefore a simple-beam diagram drawn between two hinges, with both columns blank. Recognising that from the hinge symbols, rather than launching a moment distribution, is what the twelve marks reward.

Part (c) — bracket frame, three degrees indeterminate

Approach. The two horizontals are geometrically and mechanically identical, so the vertical carries no axial force; that single deduction reduces three unknowns to two and closes the problem by hand.

1. Kinematics. Every member is inextensible: the horizontals fix $u_B=u_C=0$ and the vertical fixes $v_B=v_C=v$. The unknowns are therefore $\theta_B$, $\theta_C$ and the common vertical translation $v$, with $\psi_{AB}=\psi_{DC}=v/4$ and $\psi_{BC}=0$.
2. Show that the two horizontals behave identically. They have the same length, the same $EI$, the same fixity and the same load, and their far ends are joined by one member; equal end moments in the vertical ($M_{BC}=M_{CB}$) require $\theta_B=\theta_C=\theta$. Vertical equilibrium of the top member then gives $V_A+N=4w$ and of the bottom member $V_D-N=4w$; since $V_A=V_D$, the column axial force is $$\boxed{N=0}$$ so each horizontal carries its own $4w$ straight into the wall.
3. First equation — the wall shear. For a member with end sagging moments, $$V_A=\frac{M_{AB}+M_{BA}}{L}+\frac{wL}{2}=4w \;\Longrightarrow\; M_{AB}+M_{BA}=8w$$ Substituting the slope-deflection expressions ($\mathrm{FEM}=\pm wL^{2}/12=\pm 4w/3$), $$1.5\,EI\theta-3\,EI\psi=8w$$
4. Second equation — joint B. With $M_{BC}=\tfrac{EI}{2}(2\theta+\theta)=1.5\,EI\theta$, the balance $M_{BA}+M_{BC}=0$ gives $$2.5\,EI\theta-1.5\,EI\psi=\frac{4w}{3}$$
5. Solve. The pair yields $$\boxed{EI\theta=-\frac{32w}{21}=-1.5238w,\qquad EI\psi=-\frac{24w}{7}=-3.4286w}$$ so both joints sink by $v=4\psi=-96w/(7EI)$.
6. Assemble the diagrams. Back-substitution gives, identically for the top and the bottom member, $$\boxed{M_{\text{wall}}=-\frac{40w}{7}=-5.714w\ \text{(hogging)},\qquad M_{B}=M_{C}=+\frac{16w}{7}=+2.286w}$$ with the shear running from $4w$ at the wall to zero at the far end, and the vertical carrying a constant shear of $8w/7=1.143w$ with equal and opposite end moments $\pm 16w/7$.
7. Check the joints. At B, $+16w/7$ from the beam plus $-16w/7$ from the column sums to zero; the same holds at C. Globally $\sum V=2(4w)=8w$ and $\sum H=0$ because the two wall thrusts $8w/7$ oppose one another.