Shear Force and Bending Moment Diagrams

Why these two diagrams matter

Every beam you have ever stood on — a floor joist, a bridge girder, an aircraft wing spar — is carrying loads it cannot see. What it feels internally is a distribution of shear and bending that varies from one end to the other. The shear force and bending moment diagrams (abbreviated SFD and BMD) are the map of that internal state. They convert an ambiguous "the beam is loaded" into two exact numbers you can design to: the maximum shear V_max and the maximum bending moment M_max.

• Beam sizing. The peak moment sets the required section modulus; the peak shear checks the web and connections. No structural steel or timber member is selected without them.
• Reinforcement layout. In reinforced concrete, the moment sign tells you whether steel belongs at the top or the bottom of the section, and where to cut bars off.
• Bridge and crane girders. Moving wheel loads sweep the diagrams; envelopes of V and M over all load positions govern the design.
• Failure diagnosis. A cracked beam almost always failed at a point the BMD or SFD predicted — over a support, at midspan, or at a load point.
• Connections. Bolt groups and welds are proportioned directly from the local shear and moment the diagrams report at that joint.

How the diagrams are built, step by step

The diagrams are just the graph of the equilibrium of every possible cut through the beam. The workflow is mechanical and repeatable:

1. Find the reactions. Draw the free body of the whole beam and apply ΣF_y = 0 and ΣM = 0 to solve the support reactions. For a simply supported beam under a symmetric uniform load, each reaction is wL/2.
2. Cut and expose the internal forces. Slice the beam at a distance x from the left. On the exposed face, the internal shear V and moment M are whatever is needed to keep the left segment in equilibrium.
3. Sum forces for V(x). V(x) equals the algebraic sum of all transverse forces on the segment to the left of the cut. Plot it versus x — that is the SFD.
4. Sum moments for M(x). M(x) equals the sum of moments of those left-hand forces about the cut. Plot it versus x — that is the BMD.
5. Read the extremes. Locate where V crosses zero; that x gives M_max. Note the largest |V| (usually at a support or under a point load).

The differential relations turn this from repeated algebra into a quick sketch. Write w(x) as the intensity of the downward distributed load. Then:

dV/dx = −w(x)   and   dM/dx = V(x)

where V is the internal shear (units: N or kN), M is the internal bending moment (N·m or kN·m), w is the distributed load intensity (N/m or kN/m), and x is the position measured along the beam axis (m). Integrating between two stations gives the area method:

ΔV = −∫ w dx = −(area under the load diagram),   and   ΔM = ∫ V dx = (area under the shear diagram).

These two statements let you draw both diagrams without ever solving an integral symbolically — you just add areas, jumps, and slopes. Notice the pattern of increasing order: a uniform load (constant w) gives a linear shear and a parabolic moment; a triangular load gives a parabolic shear and a cubic moment. Each integration raises the polynomial degree by one and smooths the curve.

The three rules that let you sketch by inspection

What each load type does to the two diagrams

Load feature	Effect on shear V(x)	Effect on moment M(x)
Point load P (down)	Sudden jump down by P	Sharp kink (slope change); no jump
Applied couple M₀	No change	Sudden jump by M₀
Uniform load w	Straight line, slope −w	Parabola (concave in load direction)
Triangular (varying) load	Parabola	Cubic curve
No load between points	Constant (horizontal)	Straight line, slope = V
Support reaction R (up)	Jump up by R	Kink (slope change)

Worked example: simply supported beam with a central point load

Take a simply supported steel beam of span L = 6 m carrying a single central point load P = 40 kN at midspan (self-weight neglected for clarity).

• Reactions: by symmetry, R_A = R_B = P/2 = 20 kN.
• Shear diagram: V starts at +20 kN just right of A, stays constant (no load between A and midspan), then jumps down by 40 kN at midspan to −20 kN, staying constant to B. The shear crosses zero exactly at midspan.
• Moment diagram: M rises linearly from 0 at A (slope = V = +20 kN) to a peak at midspan, then falls linearly back to 0 at B — a triangle. The peak is M_max = PL/4 = 40 × 6 / 4 = 60 kN·m, at the point where V = 0.

Now size the beam. Using A992 steel with yield stress F_y = 345 MPa and an allowable bending stress of roughly 0.66 F_y ≈ 228 MPa (ASD), the required elastic section modulus is:

S = M_max / σ_allow = 60 000 N·m / 228×10⁶ Pa = 2.63×10⁻⁴ m³ = 263 cm³.

A W250×33 (S ≈ 380 cm³) or a lighter W200×36 (S ≈ 342 cm³) comfortably clears this, and both have ample web area to resist V_max = 20 kN. The remaining check is deflection: δ = PL³/48EI must stay under the serviceability limit, commonly L/360 = 16.7 mm for this span. This three-step chain — diagram, flexure formula, deflection check — is the entire logic of routine beam design.

Peak shear and moment for common statically determinate cases

Beam & load	V_max	M_max	Location of M_max
Simply supported, central point load P	P/2	PL/4	Midspan
Simply supported, uniform load w	wL/2	wL²/8	Midspan
Cantilever, end point load P	P	PL	Fixed end
Cantilever, uniform load w	wL	wL²/2	Fixed end
Simply supported, point load P at distance a	P·b/L (larger)	P·a·b/L	Under the load

Sign convention: sagging, hogging, and which fiber cracks

The universal beam convention keeps the bookkeeping consistent. Positive shear tends to slide the left segment up relative to the right — a clockwise shear couple on the element. Positive bending moment is sagging: the beam curves concave-up like a smile, putting the bottom fiber in tension and the top in compression. Negative (hogging) moment is concave-down and puts the top fiber in tension — the state over the interior supports of a continuous beam and everywhere along a cantilever.

The sign is not a mathematical nicety. It decides which face of the beam must carry tension, and therefore where you place reinforcing steel in concrete, where a timber's grain must be sound, and where a fatigue crack will open in a welded steel girder. A common convention plots positive (sagging) moment downward on the BMD so the curve visually falls toward the tension face — always check which convention a drawing uses before reading it.

Common misconceptions and failure modes

• "Max moment is at the biggest load." No — it is where the shear crosses zero, which for off-center loads is not under the largest force.
• "A point load makes the moment jump." A point load kinks the moment but does not step it; only an applied couple steps the moment diagram.
• "Shear and moment fail the same spot." Shear peaks at supports; moment peaks in the span. Short deep members can be shear-critical while long shallow ones are moment-critical.
• "Zero shear means no stress." Where V = 0 the moment is at its maximum — the most bending-stressed section, not the least.
• "The diagrams don't need reactions first." Wrong reactions propagate into every value; always verify ΣF = 0 and ΣM = 0 close.
• "Only determinate beams have diagrams." Indeterminate beams do too, but you first solve the redundant reactions (moment distribution, slope-deflection, or stiffness) before the same rules apply.