Internal forces: shear and bending moment diagrams

Lessons 4 and 5 found the forces acting on a body from outside: support reactions and, in trusses, the axial forces carried by each two-force member. But a beam does not fail at its supports; it fails somewhere along its length, where the internal forces are too large for its cross-section. To choose the size of a beam, a shaft or a frame member, an engineer must know how the internal forces vary along it and where they are greatest. This lesson shows how to find the internal axial force, shear force and bending moment at any section by the method of sections, how to draw shear and bending moment diagrams for a whole beam, and how the simple relationships between load, shear and moment make those diagrams quick to sketch and check. These diagrams are the bridge between statics and the design of structural members in mechanics of materials.

Internal forces at a section

Imagine cutting a beam at some point. For each piece to stay in equilibrium, the material on one side of the cut must exert forces on the material on the other side. In a two-dimensional member, these internal forces reduce to three components at the cut:

• The normal (axial) force N, along the member's axis, tension or compression.
• The shear force V, perpendicular to the axis, which tends to slide one side of the cut past the other.
• The bending moment M, a couple that tends to bend the member.

By Newton's third law, the forces on the two faces of the cut are equal and opposite. To find them, cut the member, draw a free-body diagram of either piece including the unknown N, V and M at the cut, and apply the equilibrium equations. Choose the piece with fewer loads on it.

A sign convention is needed so that results from different sections can be compared. The usual convention in North American textbooks is: tension positive for N; shear positive when it pushes the left piece up (and the right piece down); and moment positive when it bends the member concave upward, called sagging, with compression on top and tension on the bottom. Negative moment, hogging, bends the member concave downward, as over a support or at a cantilever's root.

Shear and bending moment diagrams

Repeating the method of sections at a general position x gives V(x) and M(x) as functions, which are plotted below a sketch of the beam as the shear diagram and the moment diagram. Between loads, the functions are simple: constant or linear shear, linear or parabolic moment. Each change of loading starts a new segment, so a beam with point loads and partial distributed loads needs a separate expression for each region.

Relationships between load, shear and moment

Considering equilibrium of a tiny slice of beam of length dx, with distributed load w(x) acting downward, gives two relationships that make diagrams much faster to draw:

In words: the slope of the shear diagram equals minus the load intensity, and the slope of the moment diagram equals the shear. Integrating, the change in shear between two points equals minus the area of the load diagram between them, and the change in moment equals the area of the shear diagram between them. Point loads and couples, which act over zero length, cause jumps:

• A downward point force P makes the shear jump down by P.
• A concentrated couple M₀ makes the moment jump, up for a clockwise couple with the convention above.

Several practical rules follow. Where there is no distributed load, shear is constant and moment varies linearly. Under a uniform load, shear varies linearly and moment parabolically. The maximum moment occurs where the shear passes through zero (or at a point load, support or free end). At a free end or a pinned end without a couple, the moment is zero.

# Shear and bending moment along the overhanging beam of Lesson 4
# 8 m beam: pin at A (x = 0), roller at B (x = 6), 4 kN/m on A to B, 10 kN down at x = 8
w, P, xB, L = 4.0, 10.0, 6.0, 8.0
By = (w * xB * xB / 2 + P * L) / xB          # moments about A
Ay = w * xB + P - By
print(f"Reactions: Ay = {Ay:.2f} kN, By = {By:.2f} kN")

def V(x):                                     # shear at x (right of the support at B); positive when the left part is pushed up
    v = Ay - w * min(x, xB)
    if x >= xB:
        v += By
    return v

def M(x):                                     # bending moment at x (sagging positive)
    a = min(x, xB)
    m = Ay * x - w * a * (x - a / 2)
    if x > xB:
        m += By * (x - xB)
    return m

print("   x (m)   V (kN)   M (kN m)")
for x in [0, 1, 2, 3, 4, 5, 6, 7, 8]:
    print(f"{x:7.1f} {V(x):8.2f} {M(x):10.2f}")
x0 = Ay / w                                   # where the shear crosses zero
print(f"Shear is zero at x = {x0:.3f} m: maximum sagging moment {M(x0):.2f} kN m")
print(f"Moment over support B: {M(xB):.2f} kN m (hogging)")
xi = 2 * Ay / w                               # where M returns to zero between A and B
print(f"Point of inflection at x = {xi:.2f} m")

Reactions: Ay = 8.67 kN, By = 25.33 kN
   x (m)   V (kN)   M (kN m)
    0.0     8.67       0.00
    1.0     4.67       6.67
    2.0     0.67       9.33
    3.0    -3.33       8.00
    4.0    -7.33       2.67
    5.0   -11.33      -6.67
    6.0    10.00     -20.00
    7.0    10.00     -10.00
    8.0    10.00       0.00
Shear is zero at x = 2.167 m: maximum sagging moment 9.39 kN m
Moment over support B: -20.00 kN m (hogging)
Point of inflection at x = 4.33 m

The output traces the diagrams for the beam of Lesson 4. Shear starts at +8.67 kN at A and falls at 4 kN/m, crossing zero at 2.17 m, where the moment peaks at +9.39 kN·m (sagging). The shear keeps falling to −15.33 kN just left of B, then jumps up by the 25.33 kN reaction to +10 kN, which stays constant along the unloaded overhang until the 10 kN load at the end brings it back to zero. The moment, after peaking, falls through zero at 4.33 m, a point of inflection where the beam changes from sagging to hogging, reaches −20 kN·m over support B, and rises linearly to zero at the free end. The largest moment in magnitude is the hogging moment over the support, not the sagging moment in the span.

Standard cases

A few cases occur so often that engineers know their results by heart, and design handbooks tabulate many more:

Comparing them is instructive. A uniform load on a simply supported span produces half the maximum moment of the same total load concentrated at midspan (wL²/8 against PL/4 with P = wL). A cantilever with a uniform load has four times the maximum moment of the same beam simply supported. Superposition lets you add the diagrams for separate loads on the same beam, because statics is linear.

Why the diagrams matter

Shear and moment diagrams tell a designer where a member is most heavily stressed and how much strength is needed along it. Bending stress in a beam is proportional to the bending moment divided by a property of the cross-section that depends on its second moment of area (Lesson 11); shear stress is proportional to the shear force. So the maximum moment usually sets a beam's depth and flange size, while the maximum shear, often near the supports, sets the web thickness and the connections. Reinforced concrete designers place tension steel on the side of the member where the moment diagram says tension occurs: at the bottom in sagging regions and at the top over supports, where hogging moment puts the top in tension. A beam with steel only in the bottom would crack at the top over an interior support. Points of inflection mark where the bars can change sides.

Members of frames carry axial force, shear and moment together, and the same method applies to each member, using the forces at its joints found by the methods of Lesson 5. In a portal frame or a crane jib, the moment diagram shows where the frame is stressed most, typically at the knees where beams meet columns.