Bending Moment vs. Shear Force in Beams: Reading the Diagrams
Before a structural engineer can size a beam, select reinforcement, or check deflection, they must understand the internal forces acting along the member. Shear force and bending moment diagrams (SFD and BMD) make those internal forces visible. These diagrams are not just academic exercises: they directly determine where a concrete beam needs the most flexural steel, where stirrups must be closely spaced, and where a steel beam section may be reduced if loads are light. Mastery of SFD and BMD is the foundation of every beam design calculation.
Free Body Diagrams and Support Reactions
The starting point is always equilibrium. A beam in static equilibrium satisfies three conditions: the sum of horizontal forces is zero, the sum of vertical forces is zero, and the sum of moments about any point is zero. For a simply supported beam these three equations yield the two unknown support reactions without ambiguity; the beam is statically determinate.
Consider a simply supported beam of span L with a uniformly distributed load w (force per unit length, in kips per foot or kN/m). Taking moments about the left support A sets the right reaction RB:
RB × L = w × L × (L/2) ⇒ RB = wL/2
By symmetry (or by summing vertical forces), RA = wL/2 as well. Both reactions equal half the total applied load, which makes intuitive sense for a symmetrically loaded span.
For a beam with a single concentrated (point) load P at distance a from the left support, reactions are RA = P(L-a)/L and RB = Pa/L. The asymmetric placement shifts load toward the nearer support. These reactions are the inputs to constructing the internal force diagrams.
Shear Force: Definition and How to Draw the Diagram
Shear force V at any cross-section is defined as the algebraic sum of all vertical forces to one side of the cut. Using the left-side convention (sum of upward forces to the left of the section minus downward forces to the left), V is positive when the left-side resultant pushes upward and the right-side resultant pushes downward.
For the uniformly loaded simply supported beam, at distance x from the left support:
V(x) = RA − wx = wL/2 − wx
At x = 0: V = +wL/2 (the left reaction). At x = L/2: V = 0. At x = L: V = −wL/2 (approaching the right reaction from the left). The shear diagram is therefore a straight line with a positive value at the left, crossing zero at midspan, and ending negative at the right. The slope of the shear diagram equals −w (the distributed load intensity): this is the fundamental relationship dV/dx = −w.
A concentrated point load causes a vertical jump in the shear diagram equal to the magnitude of the load. A downward load of 20 kips at a section causes the shear value to drop abruptly by 20 kips at that location. An upward reaction causes an upward jump. Reading the SFD, every discontinuity corresponds to a concentrated force at that point, and the magnitude of the jump equals the force magnitude.
The governing relationship between shear and moment is dM/dx = V. This means the slope of the bending moment diagram at any point equals the shear force at that point. Where V = 0, the moment diagram reaches a local maximum or minimum. Where shear is large, the moment diagram changes rapidly. This relationship is the fastest tool for sketching a qualitatively correct BMD from a known SFD.
Bending Moment: Definition and How to Draw the Diagram
The bending moment M at a cross-section is the net moment of all forces to one side of the cut about the centroid of that cross-section. Using the left-side sign convention, a moment that causes the beam to sag (concave upward, tension at the bottom fiber) is positive; a hogging moment (concave downward, tension at the top fiber) is negative.
For the uniformly loaded simply supported beam:
M(x) = RA × x − w × x × (x/2) = (wL/2)x − wx2/2
This is a parabola opening downward. At x = 0 and x = L, M = 0 (moment is zero at a simple pin or roller support). The maximum moment occurs where V = 0, at x = L/2:
Mmax = wL2/8
This result, wL squared over 8, is one of the most frequently used expressions in structural engineering. It represents the maximum sagging moment in a uniformly loaded simply supported beam and appears in the sizing of almost every floor beam in a building. For a 30-foot span carrying 3 kips per foot, Mmax = 3 × 302/8 = 337.5 kip-feet.
For a beam with a single midspan point load P, the moment diagram is triangular: M rises linearly from zero at the left support to PL/4 at midspan, then decreases linearly back to zero at the right support. There is no parabola because the load is concentrated, not distributed.
Cantilevers and Continuous Beams
A cantilever beam (fixed at one end, free at the other) is the mirror image of a simply supported beam in terms of where the critical forces occur. For a cantilever of length L with uniform load w, the maximum shear is at the fixed support (Vmax = wL, the full reaction must be provided by the wall) and the maximum bending moment is also at the fixed support:
Mmax = wL2/2
Note this is four times larger than the simply supported case for the same span and load. At the free end, both V and M are zero. The moment is entirely negative (hogging), meaning tension develops in the top fiber along the entire length. This has immediate implications for reinforced concrete: a cantilever slab needs top bars over its full length, not bottom bars.
Continuous beams span over multiple supports and are statically indeterminate: the internal forces cannot be determined from equilibrium alone but require compatibility equations or matrix structural analysis. The signature characteristic of a continuous beam is the development of negative (hogging) moment over interior supports. At an interior support, the beam is bent concave downward, putting the top fiber in tension.
For a two-span continuous beam with equal spans L and uniform load w, approximate analysis (or exact analysis by the three-moment equation) gives: maximum positive moment near midspan of approximately 0.070 wL2 to 0.096 wL2 depending on support conditions, and negative moment over the interior support of approximately 0.125 wL2 (wL2/8 for uniform loading). The interior support region thus sees moment similar to a midspan simply supported beam, but with the stress reversed.
This pattern fundamentally governs rebar placement in concrete floors. A one-way continuous slab requires bottom bars at midspan (where positive moment peaks) and top bars over supports (where negative moment peaks). The structural drawings will show these zones clearly, often with bar cutoff points determined by where the moment diagram crosses zero (the inflection point).
How Diagrams Drive Design Decisions
The bending moment diagram controls flexural design: the member must develop enough moment capacity (phi Mn in ACI strength notation, or Mn/Omega in ASD) to exceed the maximum factored moment Mu at every cross-section. For a steel beam, the section modulus S must satisfy S ≥ Mu/Fb (ASD) or the plastic section modulus Z must satisfy Z ≥ Mu/(phi Fy) (LRFD). For a concrete beam, the area of tension steel and its moment arm to the compression resultant must produce adequate moment capacity at the section of maximum M.
The shear force diagram controls transverse reinforcement. In reinforced concrete beams, closed stirrups resist the diagonal tension that shear creates. ACI 318-19 Chapter 22 requires that stirrups be provided where the factored shear Vu exceeds half the concrete shear capacity phi Vc. The required stirrup spacing is tightest (stirrups closest together) where the SFD shows the highest shear, which is almost always near the supports. At midspan, shear is low, and stirrup spacing can be relaxed or stirrups omitted entirely in some lightly loaded members.
The distinction between elastic and plastic analysis matters for indeterminate structures. Elastic analysis, which assumes linear material behavior throughout, gives moment diagrams that peak at the most highly stressed locations. Plastic analysis allows moments to redistribute once a section reaches its plastic moment capacity (forming a plastic hinge): the structure can carry additional load as other sections reach their capacity sequentially. AISC 360 and ACI 318 both allow limited plastic redistribution under specific conditions, but elastic analysis remains the standard basis for most building design and the reference framework for reading and interpreting shear and moment diagrams.