Shear and moment diagrams for determinate beams

You drew shear force and bending moment diagrams in mechanics of solids. In structural analysis they become everyday tools. They show where a beam needs the most strength, where its reinforcement must change faces, where splices and connections can go, and how it will deflect. This lesson reviews the method quickly and then extends it to the loadings that actually occur in buildings: point loads from framing beams, linearly varying loads from snow drifts and soil, overhangs, and beams with internal hinges. It also introduces moment diagrams drawn by parts, which you will need for deflections in lesson 10.

Sign convention and the basic relations

We keep the convention from CIV-221. Positive shear acts downward on the right face of a cut section's left part, and positive moment bends the beam concave upward (sagging, tension at the bottom). With a distributed load \(w(x)\) taken positive downward, equilibrium of a short element gives

A downward point load causes a downward jump in the shear diagram equal to the load, and a clockwise applied couple causes an upward jump in the moment diagram equal to the couple (reading left to right). These rules let you draw both diagrams from left to right using areas, with equilibrium equations written only to find the reactions.

The shapes follow from the order of each curve. With no load between points, shear is constant and moment linear. Under a uniform load, shear is linear and moment parabolic. Under a linearly varying load, shear is parabolic and moment cubic. The moment reaches a local maximum or minimum where the shear crosses zero, and its slope changes abruptly at point loads, where the shear jumps.

Girders with framing beam reactions

Most building girders are loaded by the reactions of the beams that frame into them, so their loading is a set of point loads. Between the loads the moment diagram is a series of straight lines, and the maximum moment is at one of the load points, the one where the shear changes sign.

Linearly varying loads

Snow drifts against a higher roof, water and soil pressure on walls, and the load on a beam carrying part of a triangular tributary area all vary linearly. For a triangular load rising from zero at A to \(w_0\) at B over span L, the resultant is \(w_0 L/2\) acting at \(2L/3\) from A, so the reactions are \(w_0L/6\) at A and \(w_0L/3\) at B. At a distance x from A the load intensity is \(w_0x/L\), and the load to the left of the cut has resultant \(w_0x^2/(2L)\) acting at \(2x/3\) from A, a third of the way back from the cut. Hence

Setting \(V = 0\) gives \(x = L/\sqrt{3} = 0.577L\), and substituting gives \(M_{\max} = w_0L^2/(9\sqrt{3}) = 0.0642\,w_0L^2\). The maximum is shifted toward the heavy end, not at midspan.

Overhangs and points of inflection

An overhang creates negative moment over the support next to it, while the main span may still sag. Between the two regions the moment passes through zero at a point of inflection. Overhangs are used deliberately in canopies, cantilevered balconies and in continuous-looking beams, because the support moment reduces the sagging moment in the span.

The deflected shape from the moment diagram

The moment diagram tells you the curvature of the beam, because curvature equals \(M/EI\). Where the moment is positive the beam is concave upward; where it is negative, concave downward; and at a point of inflection the curvature changes sign. Combined with the support conditions (zero deflection at supports, zero slope at fixed ends), this lets you sketch the deflected shape before computing anything. Sketching it is a good check on your diagrams and on later deflection calculations: the canopy beam sags between A and the inflection point, curves the other way near B, and its tip at C rises or drops depending on the relative stiffness and loads, which lesson 10 will tell us how to calculate.

Moment diagrams by parts

For deflection calculations it is often easier to draw the moment diagram as a sum of simple pieces than as one complicated curve. The idea is superposition: since the structure is linear, the moment from several loads equals the sum of the moments from each load alone. Drawing each piece as a cantilever from a chosen reference point, usually a support, gives shapes whose areas and centroids are standard: rectangles from couples, triangles from point loads, and parabolic spandrels from uniform loads.

For a simply supported beam with a uniform load, for example, the moment by parts about the left support consists of a positive triangle from the right reaction, \(R_BL\) at the right end, and a negative parabolic spandrel from the load, \(-wL^2/2\). Their sum is the familiar parabola, but each piece has an area and centroid you can write down without integration. Lesson 10 uses exactly these pieces.

Inclined beams

Rafters, stair stringers and ramps are inclined. A gravity load on an inclined member can be specified per unit horizontal length (snow, which falls on the plan area) or per unit length along the slope (self-weight, which depends on the member's true length). Convert the latter to a load per horizontal length by multiplying by \(1/\cos\theta\). For a simply supported inclined beam with vertical reactions, the bending moment at any point is then the same as for a horizontal beam of the same horizontal span, while the shear and axial force are the vertical shear resolved perpendicular and parallel to the member. This result saves a lot of work on pitched roofs.

Diagrams by computer

Sofia's program handles any mix of point loads, uniform loads and linearly varying loads on a simply supported beam with optional overhangs. It finds the reactions, evaluates shear and moment along the beam, and reports the extreme values and points of inflection. Here it is applied to the drift beam with its uniform snow and dead load added.

# Simply supported beam A (x=0) to B (x=Lab) with optional overhang to Ltot; loads downward +
Lab, Ltot = 20.0, 20.0
points = []                                   # (x, P)
dist = [(0.0, 20.0, 1.2*0.40 + 1.6*0.84, 1.2*0.40 + 1.6*0.84),   # factored uniform dead + snow
        (0.0, 20.0, 0.0, 1.6*0.80)]                               # factored drift, 0 at A to peak at B

def pieces():
    for x, P in points: yield ("P", x, P)
    for a, b, w1, w2 in dist: yield ("D", a, b, w1, w2)

def left_effects(x):
    """Shear-like force and moment about x of loads left of x."""
    F = M = 0.0
    for p in pieces():
        if p[0] == "P" and p[1] < x:
            F += p[2]; M += p[2] * (x - p[1])
        elif p[0] == "D" and x > p[1]:
            a, b, w1, w2 = p[1:]; e = min(x, b); wl = w1 + (w2 - w1) * (e - a) / (b - a)
            R = (w1 + wl) / 2 * (e - a)                           # trapezoid resultant
            c = a + (e - a) * (w1 + 2 * wl) / (3 * (w1 + wl)) if w1 + wl else a
            F += R; M += R * (x - c)
    return F, M

tot = 0.0; mom = 0.0
for p in pieces():
    if p[0] == "P": tot += p[2]; mom += p[2] * p[1]
    else:
        a, b, w1, w2 = p[1:]; R = (w1 + w2) / 2 * (b - a)
        c = a + (b - a) * (w1 + 2 * w2) / (3 * (w1 + w2)); tot += R; mom += R * c
RB = mom / Lab; RA = tot - RB
def VM(x):
    F, M = left_effects(x)
    V = RA - F + (RB if x > Lab else 0.0)
    Mx = RA * x - M + (RB * (x - Lab) if x > Lab else 0.0)
    return V, Mx
xs = [i * Ltot / 6000 for i in range(6001)]
vals = [VM(x) for x in xs]
Mmax, xm = max((m, x) for (v, m), x in zip(vals, xs))
print(f"RA = {RA:.2f} kips, RB = {RB:.2f} kips")
print(f"Maximum moment {Mmax:.1f} kip*ft at x = {xm:.2f} ft (beam length {Lab} ft)")

RA = 22.51 kips, RB = 26.77 kips
Maximum moment 123.4 kip*ft at x = 10.43 ft (beam length 20.0 ft)

The peak moment falls a little past midspan, pulled toward the wall by the drift, as expected. Replacing the inputs with the canopy's overhang, or the girder's point loads, reproduces the hand results above.