Study Companion: S21 — Beams: Shear Force and Bending Moment

S20 found the forces a beam's supports must supply. Those numbers say nothing about whether the beam itself survives. A beam fails inside, at some cross-section, when the internal forces there exceed what its material and shape can carry. This lesson cuts the beam open: at every section the part on one side pushes and twists the part on the other with a shear force V and a bending moment M. Plotting them along the beam gives the shear-force and bending-moment diagrams, the starting point of every beam design, and three short equations tie them to the load.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 5/7 ("Beams—Internal Effects"), with Sample Problems 5/13–5/15. Harper, Solving Statics Problems in MATLAB, "Beams-Internal Effects" (Problem 5/136). Meriam's Sample Problem 5/15 is in US units, converted here: 1 lbf = 4.448\,\mathrm N, 1 ft = 0.3048\,\mathrm m, 1 lb·ft = 1.356\,\mathrm{N\,m}. Axes: x along the beam to the right, y up. Sign convention (Meriam's, Fig. 5/23, used throughout and drawn on the first figure): on the cut face of the left segment, positive shear V acts downward and positive moment M counter-clockwise; on the cut face of the right segment, positive V acts upward and positive M clockwise. Positive M bends the beam concave upward (sagging). A distributed load w is positive downward. Equilibrium moments on free bodies are counter-clockwise positive.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A 3\,\mathrm m plank bridges a ditch, resting on its two ends. An 80\,\mathrm{kg} person stands 1\,\mathrm m from the left end. The inset shows the sign convention for the internal forces used in this lesson.

If the plank cracks, where will it crack, and how large is the bending moment there? The plausible guess is "in the middle, where a beam is always weakest". The answer is in Nugget 03.

Nugget 01: Cut a Beam and Each Section Carries a Shear Force and a Bending Moment

Claim. At any cross-section, the material on one side acts on the other with a system of forces that reduces to a force and a couple. For loads in one plane its components are the shear force V across the section, the bending moment M, and an axial force along the beam; twisting loads add a torsional moment T.

The picture. Meriam's Figs. 5/22 and 5/24.

Left: the section's resultant split into shear, axial force and bending couple. Right: why M is a couple. In an H-section beam bent by end moments, the top flange shortens and the bottom flange stretches; their compressive and tensile forces are equal and opposite, a distance d apart.

Check with numbers. A beam section with flanges 0.2\,\mathrm m apart carries a bending moment of 10\,\mathrm{kN\,m}. Each flange then carries F = M/d = 10\times10^{3}/0.2 = 50\,\mathrm{kN}, the top one in compression and the bottom one in tension, although the section's net axial force is zero.

The idea. Meriam (5/7): besides tension or compression, a beam resists shear, bending and torsion (Fig. 5/22). The force V is the shear force, the couple M the bending moment, the couple T the torsional moment; they are the vector components of the resultant of the forces acting on a transverse section. For the H-section (Fig. 5/24) the thin web is neglected: the compressed upper flange and the stretched lower flange form a couple equal to the bending moment. A beam of another cross-section loaded the same way distributes the force over its section differently, but the resultant is the same couple. This lesson treats loads in a single plane, so V and M (and, for a curved member, the axial force).

The equation (physical law: equilibrium of the part on one side of the cut, the method of sections of S14 applied to a beam):

\text{forces on a section} \;\equiv\; V \text{ (shear)} + \text{axial force} + M \text{ (bending couple)}

Worked number. The same 10\,\mathrm{kN\,m} on a deeper section, flanges 0.4\,\mathrm m apart: 25\,\mathrm{kN} per flange. Depth halves the flange forces, which is why beams are deep rather than wide.

Real example (one you can see). Bend a foam ruler or a thick sponge: the top surface wrinkles (compression) while the bottom stretches (tension). An I-beam puts its material where those forces are, in the flanges.

Plausible wrong answer. "A beam with no axial load has no tension or compression in it." Bending puts the top in compression and the bottom in tension; they cancel as forces and add up as a couple.

Check. A shelf bracket's arm is bent downward by a load at its tip. Which side of the arm is in tension? (The top: the arm is hogging.)

Narration spine. A beam under load; a saw cuts it and the two parts float apart. Arrows appear on the cut faces: shear across, a curved couple. The section zooms into an H-shape; arrows squeeze the top flange and pull the bottom, and they pair into the couple M = Fd.

Bridge. The two faces of a cut carry equal and opposite V and M. A convention must say which is "positive".

Nugget 02: Draw V and M in Their Positive Senses and Let the Signs Speak

Claim. Meriam's convention defines positive V and M by what they do to the beam, so the same sign means the same thing on both faces of a cut, even though the arrows on the two faces point opposite ways. Draw them positive on every free-body diagram; a negative answer means the opposite sense.

The picture. Meriam's Fig. 5/23, and the shape positive M gives a beam.

On the left segment's face, +V points down and +M turns counter-clockwise; on the right segment's face, +V points up and +M turns clockwise: action and reaction. A positive M at both ends of a beam bends it concave upward, top in compression.

Check with numbers. A 4\,\mathrm m simple beam with 8\,\mathrm{kN} at mid-span; reactions 4\,\mathrm{kN} up each. Cut at x = 1\,\mathrm m. Left segment: \Sigma F_y: 4 - V = 0, V = +4\,\mathrm{kN}; \Sigma M about the cut: M - 4(1) = 0, M = +4\,\mathrm{kN\,m}. Right segment (from x = 1 to 4): \Sigma F_y: V - 8 + 4 = 0, V = +4\,\mathrm{kN}; \Sigma M about the cut: -M - 8(1) + 4(3) = 0, M = +4\,\mathrm{kN\,m}. Both segments give the same signed values, as they must.

The idea. Meriam (5/7): for forces in a single plane, the conventions of Fig. 5/23 are the ones generally used. By action and reaction, V and M act in opposite directions on the two sections. It is frequently impossible to tell without calculation whether they are positive or negative at a given section, so they are drawn in their positive directions and the algebraic signs of the results give their true directions. The calculations at every section must be consistent with this convention.

The equation (definition: Meriam's sign convention):

\text{left segment: } +V \downarrow,\ +M \circlearrowleft; \qquad \text{right segment: } +V \uparrow,\ +M \circlearrowright; \qquad +M \Rightarrow \text{sagging}

Worked number. A cantilever fixed at its left end with 2\,\mathrm{kN} at its tip, 1.5\,\mathrm m out. Cut near the wall and keep the right segment: V = +2\,\mathrm{kN}, and \Sigma M_{\text{cut}}: -M - 2(1.5) = 0, M = -3\,\mathrm{kN\,m}. Negative: the cantilever hogs, concave down.

Real example. Structural engineers' software and design codes use this convention (sagging positive) so that a bending-moment diagram reads the same in every report: positive means the bottom fibres are in tension, where concrete beams get their reinforcing bars.

Plausible wrong answer. "V is positive when it points up." On the left segment's face, positive V points down; the convention is about the pair, not one arrow.

Check. On the right segment's cut face, which way do positive V and M act? (Up, and clockwise.)

Narration spine. A beam cut in two. On each face a down or up arrow and a curved arrow appear, labelled +V and +M, mirror images of each other. The two pieces rejoin, and positive moments at the ends bend the beam into a smile, its top fibres glowing in compression.

Bridge. With a convention fixed, V and M can be computed at every section.

Nugget 03: Cut, Isolate, Solve: Shear and Moment Along the Beam

Claim. Find all the reactions first. Then cut at a general position x in each interval between loads, keep the simpler side, draw V and M positive, and solve its equilibrium: V(x) and M(x) for that interval. Plotted, they are the shear-force and bending-moment diagrams. Never cut exactly at a concentrated load or couple.

The picture. Meriam's Sample Problem 5/13: a 10\,\mathrm m simple beam with 4\,\mathrm{kN} at 6\,\mathrm m.

Check with numbers. Reactions: R_1 = 4(4)/10 = 1.6\,\mathrm{kN}, R_2 = 4(6)/10 = 2.4\,\mathrm{kN}.

0 < x < 6\,\mathrm m, left segment: \Sigma F_y: 1.6 - V = 0, V = 1.6\,\mathrm{kN}; \Sigma M about R_1: M - 1.6x = 0, M = 1.6x.
6 < x < 10\,\mathrm m, right segment: \Sigma F_y: V + 2.4 = 0, V = -2.4\,\mathrm{kN}; \Sigma M about R_2: -2.4(10 - x) + M = 0, M = 2.4(10 - x).

M rises to 9.6\,\mathrm{kN\,m} under the load, where V changes sign, then falls to zero.

The puzzle. Plank 3\,\mathrm m, weight P = 80(9.81) = 784.8\,\mathrm N at 1\,\mathrm m: R_A = 784.8(2)/3 = 523.2\,\mathrm N, R_B = 261.6\,\mathrm N. For x < 1\,\mathrm m, M = 523.2x; for x > 1\,\mathrm m, M = 261.6(3 - x). The moment peaks under the person: M = 523.2\,\mathrm{N\,m}, where V jumps from +523.2 to -261.6\,\mathrm N. At mid-span it is only 261.6(1.5) = 392.4\,\mathrm{N\,m}. A uniform plank cracks under the person, not in the middle.

The idea. Meriam (5/7): the variation of V and M along the beam is what design needs, and the largest magnitude of M usually governs the choice of beam, so its value and position must be found. Steps: (1) all external reactions from the whole beam; (2) isolate the portion to the left or right of an arbitrary section and apply equilibrium, which gives V and M at the cut; the side with fewer forces is usually simpler. Avoid a section at a concentrated load or couple, which is a point of discontinuity (Meriam's hint to SP 5/13); and keep each calculation consistent with Fig. 5/23. In SP 5/13 the maximum moment occurs where the shear changes direction, and, moving right from x = 0, M is the accumulated area under the shear diagram (1.6 \times 6 = 9.6).

The equation (physical law: equilibrium of the segment on one side of x):

\Sigma F_y = 0 \Rightarrow V(x), \qquad \Sigma M_{\text{cut}} = 0 \Rightarrow M(x) \quad \text{in each interval between loads}

Worked number. A point load P at distance a from one support of a simple span L (b = L - a): M_{\max} = Pab/L under the load. Puzzle: 784.8(1)(2)/3 = 523.2\,\mathrm{N\,m}. Standing at the middle: PL/4 = 588.6\,\mathrm{N\,m}, the worst place to stand.

Real example (one you can see). Step onto a springy plank and walk across: it bends most sharply under your feet, and a cracked plank usually shows the crack where someone stood.

Plausible wrong answer. "A beam breaks in the middle." The moment is largest under the load; for an off-centre load that is not the middle.

Check. In SP 5/13, what is M at x = 8\,\mathrm m? (2.4(10 - 8) = 4.8\,\mathrm{kN\,m}.)

Narration spine. The plank, the person, the reactions. A cutting plane slides from the left, and beneath the beam the shear and moment diagrams draw themselves behind it. The moment peaks under the feet; a hairline crack flickers there while the mid-span marker shows the smaller 392\,\mathrm{N\,m}.

Bridge. Cutting interval by interval works but is slow. The load, shear and moment are linked by derivatives.

Nugget 04: The Load Is the Slope of the Shear, and the Shear the Slope of the Moment

Claim. For an element dx of a loaded beam, equilibrium gives dV/dx = -w and dM/dx = V. So V changes by minus the area under the load, and M changes by the area under the shear diagram. A concentrated load makes V jump; a concentrated couple makes M jump.

The picture. Meriam's Fig. 5/25: an element of length dx, load w\,dx, with V and M positive on both faces.

Check with numbers. A simple beam 4\,\mathrm m long under w = 3\,\mathrm{kN/m}: R_A = R_B = 6\,\mathrm{kN}. V = 6 - 3x (slope -3 = -w), zero at x = 2\,\mathrm m; M is the area under V: M = 6x - 1.5x^2, and dM/dx = 6 - 3x = V. At mid-span M = 6\,\mathrm{kN\,m} = wL^2/8, the area of the shear triangle \frac12(6)(2).

The idea. Meriam (5/7). Over dx, w may be taken as constant. Vertical forces: V - w\,dx - (V + dV) = 0, so

w = -\frac{dV}{dx} \qquad (5/10)

The slope of the shear diagram is the negative of the load. Integrating, V = V_0 - \int_{x_0}^{x} w\,dx: V_0 minus the area under the loading curve. Moments about the element's left side: M + w\,dx\,\frac{dx}{2} + (V + dV)\,dx - (M + dM) = 0. The M's cancel; w(dx)^2/2 and dV\,dx are of higher order and are dropped:

V = \frac{dM}{dx} \qquad (5/11)

The shear is the slope of the moment curve, and M = M_0 + \int_{x_0}^{x} V\,dx, the area under the shear diagram; with no applied couple at x_0 = 0, the moment at a section is the area under V up to it. Eq. 5/10 fails at a concentrated load (where V jumps by the load) and Eq. 5/11 at a concentrated couple (where M jumps by the couple), but both hold on either side. Since each integration raises the degree by one, V is one degree higher in x than w and M two degrees higher: a linear load (w = kx) gives a quadratic V and a cubic M. Combined,

\frac{d^2M}{dx^2} = -w \qquad (5/12)

so for a continuous w, M follows by two integrations with the limits evaluated properly each time. For a discontinuous w, special singularity functions can write V and M across the discontinuities; Meriam does not use them. Bending in more than one plane is analysed plane by plane and combined vectorially.

The equation (physical law: equilibrium of the element; with Meriam's convention and w positive downward):

\frac{dV}{dx} = -w, \qquad \frac{dM}{dx} = V, \qquad \frac{d^2M}{dx^2} = -w

Worked number. SP 5/13 by areas alone: M(6) = 0 + 1.6(6) = 9.6\,\mathrm{kN\,m}; M(10) = 9.6 + (-2.4)(4) = 0, as the free end of a simple span requires.

Real example. A bridge engineer sketches a moment diagram from a shear diagram in seconds by summing areas, then checks the zero at each free end: a quick error check before any computer run.

Plausible wrong answer. "Where the load is largest, the moment is largest." The moment is largest where its slope, the shear, is zero.

Check. A beam carries a uniform load. What shapes are its shear and moment diagrams? (A straight sloping line; a parabola.)

Narration spine. A thin slice glows; arrows V and V + dV, M and M + dM attach to its faces, and w\,dx presses on top. Two equations write themselves. Then three stacked plots: load, shear, moment; the shaded area under each turns into the height of the next.

Bridge. The design number is the largest moment, and Eq. 5/11 says where to look.

Nugget 05: The Largest Moment Sits Where the Shear Changes Sign

Claim. Where V passes continuously through zero, dM/dx = 0 and M is a maximum or minimum; where V jumps across zero at a concentrated load, M peaks there too. Overhangs make negative moments over the supports, which may be the larger magnitude.

The picture. Meriam's Sample Problem 5/15, in SI: a beam on supports 3.048\,\mathrm m apart with a 0.610\,\mathrm m overhang.

Check with numbers. Loads: a triangle rising to 1459\,\mathrm{N/m} (100 lb/ft) over the first 1.219\,\mathrm m (4 ft), 889.6\,\mathrm N at 0.813\,\mathrm m; 1459\,\mathrm{N/m} from 1.219 to 2.438\,\mathrm m, 1779\,\mathrm N at 1.829\,\mathrm m; 1334\,\mathrm N (300 lb) at the free end, 3.658\,\mathrm m. Reactions: R_1 = 1097\,\mathrm N (247 lb), R_2 = 2906\,\mathrm N (653 lb). Interval by interval (Meriam's expressions in lb and ft, here in SI):

0 < x < 1.219\,\mathrm m: V = 1097 - 598.5x^2 (the triangle's area up to x is \frac12(1459x/1.219)x).
1.219 < x < 2.438\,\mathrm m: V = 1097 - 889.6 - 1459(x - 1.219), zero at x = 1.361\,\mathrm m (4.47 ft).
2.438 < x < 3.048\,\mathrm m: V = -1572\,\mathrm N (-353 lb). Beyond R_2: V = +1334\,\mathrm N.

M_{\max} = 991\,\mathrm{N\,m} (731 lb·ft) at x = 1.361\,\mathrm m; over the right support M = -1334(0.610) = -813.5\,\mathrm{N\,m} (-600 lb·ft). Meriam prints 732 lb·ft, from its coefficients rounded to 247 and 447.

The idea. Meriam (5/7): when V passes through zero and is a continuous function with dV/dx \ne 0, M is a maximum or minimum, since dM/dx = 0 there; critical values of M also occur where V crosses zero discontinuously, under concentrated loads. SP 5/15 finds the reactions from the resultants of the distributed loads, then works interval by interval from free-body diagrams, using the right-hand portion for the third interval and inspection for the overhang; it checks that M for x < 4 ft is the area under the shear curve, \int_0^x(247 - 12.5x^2)\,dx = 247x - 4.17x^3.

The equation (mathematical identity: the extremum condition, from Eq. 5/11):

\frac{dM}{dx} = V = 0 \;\Rightarrow\; M = M_{\max} \text{ or } M_{\min}

Worked number. The magnitude that governs the design of SP 5/15's beam is 991\,\mathrm{N\,m} (sagging, mid-span region); the hogging 813.5\,\mathrm{N\,m} over the support is close behind (991/813.5 = 1.22); a heavier overhang load raises the hogging moment and lowers the span moment, so a fifth more would make the support the critical section.

Real example (one you can see). A balcony slab overhanging its supporting wall hogs over the wall: its cracks, when they appear, are on the top surface above the wall, which is why its reinforcement runs along the top there.

Plausible wrong answer. "Only the positive peak matters." The negative moment over a support can be as large or larger, and it puts tension on the opposite face.

Check. Where on SP 5/15's beam is the shear zero besides 1.361\,\mathrm m? (Nowhere continuously; it jumps through zero at R_2, a second critical point, the hogging peak.)

Narration spine. The overhanging beam with its three loads. The shear diagram draws, crossing zero at 1.361\,\mathrm m; a vertical line drops to the moment diagram and lands on its summit at 991\,\mathrm{N\,m}. A second line drops at the support where V jumps, to the trough at -813.5\,\mathrm{N\,m}.

Bridge. When the load is a smooth function, Eq. 5/12 does it all by integration.

Nugget 06: A Smooth Load Gives V and M by Integration

Claim. For a continuous load w(x), integrate dV = -w\,dx and then dM = V\,dx, starting from known end values. A cantilever's root values come from the whole beam's equilibrium; the free end has V = M = 0, a check.

The picture. Meriam's Sample Problem 5/14: a cantilever of length l under w = w_0\sin(\pi x/l), and its dimensionless V and M.

Check with numbers. Whole beam, V_0 and M_0 drawn positive at the wall (x = 0): \Sigma F_y: V_0 = \int_0^l w_0\sin(\pi x/l)\,dx = 2w_0l/\pi. \Sigma M about the wall: -M_0 - \int_0^l x\,w\,dx = 0, M_0 = -w_0l^2/\pi (negative: hogging). Then

V - V_0 = -\int_0^x w_0\sin\frac{\pi x}{l}\,dx \;\Rightarrow\; \frac{V}{w_0l} = \frac1\pi\Big(1 + \cos\frac{\pi x}{l}\Big),

M - M_0 = \int_0^x V\,dx \;\Rightarrow\; \frac{M}{w_0l^2} = \frac1\pi\Big(\frac xl - 1 + \frac1\pi\sin\frac{\pi x}{l}\Big).

At the wall: 0.637 and -0.318; at x = l both are zero, as at any free end.

The idea. Meriam's hints to SP 5/14: by symmetry the resultant 2w_0l/\pi acts at mid-span, so M_0 = -(2w_0l/\pi)(l/2) = -w_0l^2/\pi, the minus sign meaning the moment at the wall is opposite to the sense drawn; and the free-body diagram reminds us to account for the limits of integration for V as well as for x. Eq. 5/12 needs w continuous, which the sine is.

The equation (mathematical identity: Eqs. 5/10–5/11 integrated):

V(x) = V_0 - \int_0^x w\,dx, \qquad M(x) = M_0 + \int_0^x V\,dx

Worked number. At mid-length, x = l/2: V = w_0l/\pi = 0.318w_0l and M = (w_0l^2/\pi)(-\frac12 + \frac1\pi) = -0.0578w_0l^2. For w_0 = 2\,\mathrm{kN/m}, l = 3\,\mathrm m: V = 1.91\,\mathrm{kN}, M = -1.04\,\mathrm{kN\,m}; at the wall M_0 = -5.73\,\mathrm{kN\,m}.

Real example. A wing or a turbine blade is a cantilever under a smoothly varying aerodynamic load; its root bending moment, found by exactly these integrals, sizes the spar and the root fittings.

Plausible wrong answer. "M_0 is positive because it was drawn positive." The sign of the result decides: -w_0l^2/\pi, a hogging moment.

Check. What are V and M at the free tip of any cantilever with no tip load? (Both zero.)

Narration spine. The sine-shaped load sits on the cantilever. Two integral signs sweep from the wall; the shear curve falls from 0.637 to zero, the moment curve climbs from -0.318 to zero, both landing exactly at the tip.

Bridge. A beam need not be straight. Harper's curved member adds a third internal force.

Nugget 07: A Curved Member Carries an Axial Force as Well

Claim. In a curved member the direction "along the member" changes with position, so the internal force at a section has a component along the member (axial, here compressive C) as well as across it (V), and both, with M, vary with the section's angle. Harper finds them as functions of the angle and locates the largest moment.

The picture. Harper's Problem 5/136: a uniform quarter-circular member of mass m and radius r, hinged at A, held at B by a roller against a wall, loaded only by its weight.

Check with numbers. Whole member, moments about A: the weight acts through the arc's centroid, 2r/\pi from the vertical radius OA (S18), and B acts horizontally at height r above A: Br - mg(2r/\pi) = 0, B = 2mg/\pi. Segment BS, angle \theta: weight W = (2\theta/\pi)mg, centroid at \bar r = 2r\sin(\theta/2)/\theta from O on the bisector. Moments about S: M + W(\bar r\cos(\theta/2) - r\cos\theta) - Br\sin\theta = 0, which simplifies to

\frac{M}{mgr} = \frac2\pi\theta\cos\theta, \qquad \frac{C}{mg} = \frac2\pi(\theta\cos\theta + \sin\theta), \qquad \frac{V}{mg} = \frac2\pi(\cos\theta - \theta\sin\theta),

the last two from \Sigma F_x: C\sin\theta + V\cos\theta - B = 0 and \Sigma F_y: C\cos\theta - V\sin\theta - W = 0. dM/d\theta = 0 gives \cos\theta = \theta\sin\theta, \theta = 0.860\,\mathrm{rad} (49.3^\circ), and M_{\max} = 0.357mgr. At that angle V = 0: Nugget 05's rule, dM/ds = V, holds along the arc too.

The idea. Harper (5.3) sets up the two free-body diagrams, solves symbolically, plots V/mg, C/mg and M/mgr against \theta, and finds the maximum by solving dM/d\theta = 0. The symbolic solver cannot find the root of \cos\theta - \theta\sin\theta, so it uses fzero from a starting guess near 45^\circ (read off the plot), getting 0.8603 rad and 0.3572. It then compares MATLAB's max on the sampled values, 0.3572 at 0.860, and notes that max finds only the largest computed value, not a true mathematical maximum.

The equation (physical law: equilibrium of the segment BS, three equations for V, C, M):

\frac{M}{mgr} = \frac2\pi\theta\cos\theta, \qquad \frac{dM}{d\theta} = 0 \Rightarrow \theta\tan\theta = 1, \ \theta = 49.3^\circ

Worked number. A 20\,\mathrm{kg} curved steel handrail of radius 1.5\,\mathrm m mounted this way: M_{\max} = 0.357(20)(9.81)(1.5) = 105\,\mathrm{N\,m}, about 49^\circ round from the wall support; B = 2(196.2)/\pi = 124.9\,\mathrm N.

Real example. Arches, crane hooks and curved brackets are designed from exactly these three internal forces; an arch is shaped so that M stays small and the material works mostly in compression.

Plausible wrong answer. "The largest moment is at the hinge or at the wall." Both ends have M = 0 (\theta = 0 at B, \cos\theta = 0 at A); the peak is in between at 49.3^\circ.

Check. What are V and C at the hinge A (\theta = 90^\circ)? (V/mg = -1, C/mg = 2/\pi = 0.637: the hinge's vertical mg across the member and its horizontal 2mg/\pi along it, since the member is horizontal at A.)

Narration spine. The quarter-circle hangs between its hinge and the wall. A section point S slides round from B to A; three gauges, V, C and M, swing with it and draw their curves, the moment peaking at 49.3^\circ just as the shear gauge passes zero.

Bridge. Beams carry loads by bending. A cable cannot bend at all; it carries load by changing its shape.

Synthesis

The plank cracks under the person, where the moment is 523\,\mathrm{N\,m}, not in the middle (392\,\mathrm{N\,m}). The shear diagram said so before any moment was computed: it changes sign under the load, and the moment's slope is the shear.

The whole lesson is one free-body diagram applied repeatedly. Cut the beam, draw V and M positive by Meriam's convention, and solve the piece on one side; do it at a general x and the diagrams appear. Shrink the piece to an element dx and its equilibrium becomes dV/dx = -w and dM/dx = V: load, shear and moment are each the slope of the next, so the diagrams can be drawn by areas, and the largest moment sits where the shear changes sign. Inside a beam, the shear is the slope of the bending moment and the load the slope of the shear; the moment peaks where the shear crosses zero.

Central idea: internal V and M at a section from equilibrium of one side.
Key relationship: w = -dV/dx, V = dM/dx, d^2M/dx^2 = -w (Eqs. 5/10–5/12).
Most important assumption: loads in one plane, a determinate beam, and a consistent sign convention (Meriam's Fig. 5/23).
Open question: a flexible cable cannot carry M or V at all. What shape must it take to carry its load? (S22.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.

A computer draws the diagrams the way the relations w = -dV/dx and V = dM/dx say, using a fine grid of x. V is the left reaction minus every load to the left, a running integral of w. M is the running area under V. The largest |M| and the zero of V are then read off the arrays.

Python (numpy, scipy). S21_shear_and_moment.py has one function, diagrams, for beams on two supports or built in, built on cumulative_trapezoid. It runs Meriam's beams and Harper's quarter-circle member, where the grid maximum is checked against a bounded search.

Part 1: Meriam's beams (N, m or kN, m)
  plank, 784.8 N at 1 m of 3 m: V from 523.2 to -261.6; largest |M| = 523.2 at x = 1.000
    at mid-span M = 392.4
  SP 5/13, 4 kN at 6 m of 10 m: V from 1.6 to -2.4; largest |M| = 9.6 at x = 6.000
  SP 5/15: V from 1097.2 to 1334.5; largest |M| = 990.9 at x = 1.361
    V = 0 at x = 1.361 m, M there 990.9 N m; over the support M = -813.5 N m
  SP 5/14, w0 = 2 kN/m, l = 3 m: V(0) = 3.820, M(0) = -5.730; mid V = 1.910, M = -1.041
Part 2: Harper's quarter-circle member, M / mgr = (2 / pi) theta cos theta
  grid max 0.3572 at 0.860 rad
  bounded search 0.3572 at 0.8603 rad = 49.3 deg
  20 kg handrail, r = 1.5 m: M_max = 105 N m, B = 124.9 N

Sample Problem 5/15 is entered in its own units, converted exactly (100\,\mathrm{lb/ft}, 300\,\mathrm{lb}, feet), so it lands on the lesson's 991 and 813.5\,\mathrm{N\cdot m}.

MATLAB (Octave-compatible). S21_shear_and_moment.m does the same with trapz, cumtrapz, max and fminbnd.

SP 5/15: R_A = 1097 N, R_B = 2906 N, M max 990.9 N m at 1.361 m
  over the support M = -813.5 N m
SP 5/14: V(0) = 3.820, M(0) = -5.730, mid V = 1.910, M = -1.041
grid max 0.3572 at 0.860 rad; fminbnd 0.3572 at 0.8603 rad

Traps

Calling a bending moment "no axial force". The flanges carry equal and opposite forces. (From Nugget 01.)
Drawing V and M with guessed senses. Draw them positive and read the sign. (From Nugget 02.)
Using one sense on both faces. They are opposite on the two faces. (From Nugget 02.)
Cutting at a concentrated load. V is discontinuous there; cut on either side. (From Nugget 03.)
Expecting the maximum moment at mid-span. It is where V changes sign. (From Nugget 03.)
Forgetting the minus sign in dV/dx = -w. A downward load makes the shear fall. (From Nugget 04.)
Ignoring jumps. A concentrated load jumps V; a concentrated couple jumps M. (From Nugget 04.)
Missing the hogging moment over a support. Check both signs' peaks. (From Nugget 05.)
Integrating without the starting values. V_0 and M_0 come from the whole beam. (From Nugget 06.)
Treating a curved member like a straight one. It carries an axial force that varies along it. (From Nugget 07.)

Self-check

SP 5/13's beam with the 4\,\mathrm{kN} load moved to mid-span (5\,\mathrm m). Find V, M and M_{\max}.
A simple beam of span L carries a uniform load w. Find V(x), M(x) and M_{\max}; evaluate for w = 3\,\mathrm{kN/m}, L = 4\,\mathrm m.
A 3\,\mathrm m cantilever, fixed at its left end, carries 2\,\mathrm{kN/m}. Find V and M at the wall, with signs.
In the puzzle, where should the person stand to make the plank's largest moment as large as possible, and how large is it?
A 4\,\mathrm m simple beam carries only a counter-clockwise couple of 6\,\mathrm{kN\,m} at mid-span. Find the reactions, and M just left and just right of the couple.
A simple beam of span 3\,\mathrm m carries a load rising from 0 at the left to 6\,\mathrm{kN/m} at the right. Find where M is largest and its value. What degree is M in x?
SP 5/14: evaluate V/(w_0l) and M/(w_0l^2) at x = l/2.
In SP 5/15 (SI), what is M at the right support, and which face of the beam is in tension there?
Harper's member: find V/mg and C/mg at \theta = 0 (at B) and at \theta = 90^\circ (at A).
Using areas only, find M at x = 6\,\mathrm m and at x = 10\,\mathrm m in SP 5/13.
The overhanging beam of S20's self-check 10 (supports at 0 and 3\,\mathrm m, 0.5\,\mathrm{kN/m} on the span and 2\,\mathrm{kN/m} on the overhang to 4\,\mathrm m): find M over the right support.
On the right-hand segment of a cut beam, V is computed as -3\,\mathrm{kN}. Which way does the shear force on that face actually point?

Fully worked solutions

1. R_1 = R_2 = 2\,\mathrm{kN}. V = +2\,\mathrm{kN} for x < 5, -2\,\mathrm{kN} for x > 5; M = 2x, then 2(10 - x); M_{\max} = 10\,\mathrm{kN\,m} at mid-span (PL/4).

2. R = wL/2 each. Left segment: V = wL/2 - wx, M = \frac{wL}{2}x - \frac{wx^2}{2} = \frac{wx(L - x)}{2}. V = 0 at x = L/2: M_{\max} = wL^2/8. For 3\,\mathrm{kN/m} and 4\,\mathrm m: 6\,\mathrm{kN\,m}.

3. Keep the right segment from the cut at the wall: \Sigma F_y: V - 2(3) = 0, V = +6\,\mathrm{kN}; \Sigma M_{\text{cut}}: -M - 6(1.5) = 0, M = -9\,\mathrm{kN\,m} (hogging).

4. M_{\max} = Pa(L - a)/L is largest at a = L/2: 784.8(3)/4 = 588.6\,\mathrm{N\,m}, at the middle.

5. \Sigma M_A: 4R_B + 6 = 0, R_B = -1.5\,\mathrm{kN} (down); R_A = +1.5\,\mathrm{kN} (up). Left of the couple: M = 1.5x, 3\,\mathrm{kN\,m} at x = 2^-. Right of it, left segment including the couple: M - 1.5x + 6 = 0, M = 1.5x - 6, -3\,\mathrm{kN\,m} at x = 2^+. M jumps by 6\,\mathrm{kN\,m} at the couple, and V = 1.5\,\mathrm{kN} throughout.

6. R_A = 3\,\mathrm{kN}, R_B = 6\,\mathrm{kN} (S20). V = 3 - x^2 (kN, x in m): zero at x = \sqrt3 = 1.732\,\mathrm m. M = 3x - x^3/3: M_{\max} = 3(1.732) - 1.732^3/3 = 3.464\,\mathrm{kN\,m} (w_0L^2/(9\sqrt3)). The load is degree 1, so M is degree 3.

7. V/(w_0l) = (1 + \cos\frac\pi2)/\pi = 1/\pi = 0.318; M/(w_0l^2) = (\frac12 - 1 + \frac1\pi)/\pi = -0.0578.

8. From the overhang: M = -1334(0.610) = -813.5\,\mathrm{N\,m}, hogging: the top face is in tension.

9. \theta = 0: V/mg = 2/\pi = 0.637, C = 0 (the member is vertical at B and B is horizontal, across it). \theta = 90^\circ: V/mg = (2/\pi)(-\pi/2) = -1, C/mg = 2/\pi = 0.637.

10. M(6) = 1.6(6) = 9.6\,\mathrm{kN\,m}; M(10) = 9.6 - 2.4(4) = 0.

11. Right segment from just left of the support: the overhang load 2\,\mathrm{kN} at 0.5\,\mathrm m: -M - 2(0.5) = 0, M = -1\,\mathrm{kN\,m}.

12. Positive V on the right segment's face points up, so -3\,\mathrm{kN} means 3\,\mathrm{kN} downward.