Shear Force and Bending Moment Diagram

Cut an imaginary slice through a loaded beam and ask what internal force and twisting effect are needed to hold one piece still. Slide that cut along the beam and those two values trace out two curves: the shear force diagram and the bending moment diagram. Read where the shear jumps and where the moment peaks, and you already know which section of the beam is working hardest.

The formulas and what the symbols mean

For a beam carrying transverse loads, imagine cutting at position $x$ and isolating one side:

• The shear force $V(x)$ is the internal force across the cut.
• The bending moment $M(x)$ is the internal turning effect across the cut.

For a simply supported beam of span $L$ with a central downward point load $P$, the support reactions follow from symmetry:

The shear and moment are then the piecewise functions

with peak moment

under the usual sagging-positive convention.

Why the diagrams take these shapes

The shapes are not arbitrary: a handful of rules link loads to slopes. A point load or support reaction creates a sudden jump in shear; an applied concentrated moment creates a sudden jump in the bending moment. In a region with no distributed load the shear stays constant, and wherever shear is constant the bending moment changes linearly. Under a constant distributed load the shear changes linearly and the moment curves. One more consequence is worth memorizing: the largest bending moment often occurs where the shear changes sign, reliable in the usual cases where $M(x)$ stays continuous and there is no applied concentrated moment there. With one sign convention, positive moment means sagging, the beam curving like a shallow smile; another convention flips the diagram vertically, so always check which one your course or software uses.

Worked example: simply supported beam with a center load

Take the beam above. By symmetry $R_A = R_B = P/2$. Just to the right of the left support the shear is $+P/2$. At the midpoint the downward load makes it drop by $P$ to $-P/2$, and the right reaction returns it to zero, which is exactly the piecewise $V(x)$ given earlier. Because the shear is constant on each half, the moment rises linearly from zero at the left support to $PL/4$ at the center, then falls linearly back to zero at the right support, giving a triangular bending moment diagram. The peak sits where the shear flips sign, right under the load.

Practice it yourself

Keep the same simply supported beam but move the point load off-center, say to $x = L/3$. Work it in order: find the reactions first ($R_A = 2P/3$, $R_B = P/3$ for that position), draw the shear from left to right, then build the moment from the shear. As a self-check, the moment is still zero at both simple supports, but the peak now sits under the load rather than at midspan, and its value is the product of the smaller reaction and its distance to the load.

Calculation traps to avoid

• Confusing the load diagram with the shear diagram. A downward load does not make the shear graph slope down the same way; shear and moment are responses, not copies of the loading.
• Forgetting the support reactions. Wrong reactions corrupt every later value in both diagrams.
• Making the moment jump at a point load. A point load jumps the shear; only a concentrated applied moment jumps the bending moment.
• Ignoring the sign convention. Two correct solutions can look vertically flipped. Compare magnitudes, jump sizes, and zero locations only after confirming the sign rule.

A second worked variation: an overhang

The center-load case is the cleanest, but it helps to see the rules survive a different geometry. Imagine the same beam extended past the right support by a short cantilever overhang carrying a downward tip load. The reactions are no longer equal: the support nearest the overhang carries more, and the far support can even pull downward. Walking the cut from left to right, the shear still holds constant between concentrated loads, jumps by the size of each load or reaction it passes, and the moment still bends only where the shear is non-zero. The moment becomes negative (hogging) over the support next to the overhang, which is the structural reason cantilevers crack on top rather than underneath. None of the rules changed; only the loading did, which is exactly the point of learning the rules instead of memorizing one diagram.

Where shear and bending moment diagrams are used

These diagrams appear in beam design, structural analysis, and mechanics courses. They locate critical sections, connect loading to internal stresses, and check whether a support-and-load arrangement makes physical sense. The bending moment in particular feeds straight into the bending-stress check, since the largest moment usually marks the section where the material is most likely to yield or fail. Even if you never design a structure, the diagrams are a clean way to see how local internal forces emerge from global equilibrium, and how a single misplaced load can shift the critical section to a completely different point along the span.