Schematic shear force and bending moment diagrams: 16-Civ-B1 Advanced Structural Analysis, May 2014, Q2

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. (a) A beam encastré at A, carried on rollers at B, C and D at spacings $L$, with two point loads $P$ at $L/3$ and $2L/3$ into span BC and a load $2P$ on a free overhang $L/3$ beyond D. (b) An L-shaped frame encastré at A, with a horizontal member AB of length $L$, a vertical member BC of length $L$ carrying a roller at its foot C, and a horizontal cantilever CD of length $L$ under a UDL $w$. All members share one $EI$ and are inextensible.

Find. The shape and the governing ordinates of the shear force and bending moment diagrams for both structures. Structure (a) is twice redundant and structure (b) once redundant, so the ordinates must come from an analysis, not from statics alone.

Structure (a): propped beam on three rollers with an overhang.

Approach. Both structures are analysed by the displacement method (equivalently three-moment or moment distribution for (a)); the determinate overhang and cantilever are replaced first by their statically equivalent end actions, which fixes one diagram ordinate before any analysis is done.

Fix the determinate ends of (a) by inspection. The overhang beyond D carries $2P$ at $L/3$, so
Solve the two redundancies. Taking the reactions at C and D as redundants (or equivalently writing three-moment equations for the interior supports) and using $EI$ constant gives the support reactions
Build the shear diagram of (a). Starting from $V=-0.3077P$ at A the shear is constant to B, jumps by $R_B$ to $+1.2564P$, steps down by $P$ at each point load, jumps by $R_C$ at C and by $R_D$ at D, and closes at $-2P$ at the free end.
Build the moment diagram of (a). The moment is piecewise linear because every load is a point load. The governing ordinates are $+0.103PL$ at the built-in end A, $-0.205PL$ over B, $+0.299PL$ under the second point load, $+0.051PL$ over C and $-0.667PL$ over D.
Reduce (b) to one unknown. Cantilever CD is determinate: at C it delivers a shear $wL$ and a hogging moment $wL^{2}/2$. Because the roller at C supplies no horizontal reaction and no horizontal load acts, the shear in column BC is identically zero, so BC carries a constant bending moment.
Solve (b). With that constant-moment column, the single compatibility condition gives $M_B=M_C=0.5wL^{2}$ hogging, a shear of $0.75wL$ in member AB and a sagging moment of $0.25wL^{2}$ at the built-in end A. Vertical equilibrium then returns $R_C=1.75wL$ upwards and a hold-down of $0.75wL$ at A.

The determinate overhang ordinate of (a) is

$$M_D=-2P\left(\frac{L}{3}\right)=\boxed{-0.667\,PL}$$

and the analysis of the twice-redundant portion returns

$$R_A=-0.308P,\quad R_B=1.564P,\quad R_C=0.026P,\quad R_D=2.718P$$

whose sum is exactly the applied $4P$. Notice that the reaction at A is a hold-down: the heavy overhang load lifts the built-in end, which is why the fixed end carries a sagging moment of $+0.103PL$ rather than the hogging moment a candidate would expect.

Structure (a): shear force diagram (multiples of P).

Structure (a): bending moment diagram (sagging positive, multiples of PL).

Structure (b): L-frame with a roller foot and a loaded cantilever.

For structure (b) the cantilever fixes the diagram before any compatibility is written,

$$M_C=-\frac{wL^{2}}{2}=\boxed{-0.500\,wL^{2}}$$

and the remaining analysis distributes that moment back into the horizontal member, which finishes with $+0.25wL^{2}$ sagging at the encastré end A.

Structure (b): shear force developed along A-B-C-D.

Structure (b): bending moment developed along A-B-C-D.

Quantity	Structure (a)	Structure (b)
Reaction at the encastré end	$R_A=-0.308P$ (downward)	$R_A=-0.750wL$ (downward)
Moment at the encastré end (sagging +ve)	$+0.103PL$	$+0.250wL^{2}$
Interior reactions	$R_B=1.564P$, $R_C=0.026P$, $R_D=2.718P$	$R_C=1.563wL$
Largest hogging moment	$-0.667PL$ over D	$-0.500wL^{2}$ at C
Largest sagging moment	$+0.299PL$ at $x=5L/3$	$+0.250wL^{2}$ at A
Column / interior shear	peak $+2.00P$ on the overhang	column BC shear $=0$ (constant moment)

Topic. Reading indeterminate diagrams from their determinate anchors.

Key relations. Overhangs and cantilevers are statically determinate: $M=-Pa$ or $M=-wc^{2}/2$ at the support. Shear jumps by exactly the reaction at every support; between point loads the moment is linear and under a UDL it is parabolic with $\mathrm{d}M/\mathrm{d}x=V$. A member with no transverse load and zero end shear carries a constant moment.

Why this works. A schematic diagram is worth marks only if the shape is right, and shape is controlled by three facts that need no analysis: the determinate ordinate at a free overhang, the continuity of moment through a rigid joint, and the slope-equals-shear rule. Fixing those first turns a twice-redundant beam into a diagram with two unknown ordinates rather than eight. The counter-intuitive results here — a hold-down reaction and a sagging moment at a built-in end — both follow from the overhang load, which hogs the far span so hard that the beam lifts off its own fixed end.

Common pitfall. Assuming a built-in end must hog. Whenever a heavy overhang sits beyond the far support, check the sign of the end reaction before drawing; if it is negative the moment at that end usually reverses too.

Source. Hibbeler, Structural Analysis, 10th ed., Ch. 4 (internal loadings) and §12.1–12.3 (moment distribution for beams).

Question 2 of 9: Schematic shear force and bending moment diagrams

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