Shear and Moment Diagrams: How to Draw Them

Internal shear and moment

Cut a beam at a distance x from the left end. To keep the left piece in equilibrium, the cut face must carry an internal shear force V and an internal bending moment M. Writing the equilibrium equations for that piece gives V(x) and M(x) for the segment where the cut lies.

The usual beam sign convention: positive shear acts downward on the left face of the cut (so it tends to rotate the segment clockwise), and positive moment bends the beam concave up, compressing the top fibres. A simply supported beam under downward loads has positive moment; a cantilever has negative moment.

The area method

Instead of writing equations for every segment, use the relationships between load, shear and moment. The slope of the shear diagram is the distributed load, and the slope of the moment diagram is the shear. Integrated over a segment, the change in shear is the area under the load curve and the change in moment is the area under the shear curve.

A point load makes the shear diagram jump by the load, in the direction of the load. Between point loads, with no distributed load, V is constant and M is a straight line.
A uniform distributed load makes V a sloped straight line and M a parabola.
A concentrated couple makes the moment diagram jump; it does not change V.
The maximum moment occurs where V crosses zero (or jumps through zero at a point load), and at the ends of overhangs or cantilevers.
At a free or pinned end with no applied couple, M = 0. Both diagrams must close to zero at the right end; that is your check.

Why the diagrams matter

The peak |M| sets the bending stress through σ = Mc/I, and the peak |V| sets the shear stress. Points where M changes sign are inflection points of the deflected shape, which you can compare with the beam deflection calculator.

Key formulas

Formulas used on this page, with the textbook each is cited from
Formula	Expression	Where	Cited from
Load–shear relationship	`dV/dx = −w(x), V₂ − V₁ = −∫ w dx`	V = internal shear (kN), w = distributed load intensity, positive downward (kN/m), x = position along the beam (m)	Hibbeler, Engineering Mechanics: Statics
Shear–moment relationship	`dM/dx = V, M₂ − M₁ = ∫ V dx`	M = internal bending moment (kN·m); the integral is the area under the shear diagram between x₁ and x₂ (kN·m)	Beer & Johnston, Vector Mechanics for Engineers: Statics
Jumps at concentrated loads	`ΔV = −P (downward point load), ΔM = M₀ (clockwise couple)`	P = point load (kN), M₀ = applied couple (kN·m)	Meriam & Kraige, Engineering Mechanics: Statics

Worked examples

Example: Simply supported beam with a uniform load and a point load

Given

A beam 8 m long, pinned at A (x = 0) and on a roller at B (x = 8 m).
A uniform load of 5 kN/m over the whole span and a 20 kN point load at x = 3 m.

Find

The reactions, the shear and moment diagrams, and the maximum bending moment and where it occurs.

Solution

Replace the uniform load with its resultant for the reactions: 5 × 8 = 40 kN at x = 4 m.ΣM_A: R_B(8) − 40(4) − 20(3) = 0 → R_B = 27.5 kN
Vertical balance:R_A = 40 + 20 − 27.5 = 32.5 kN
Shear from x = 0 to 3 m starts at R_A and falls at 5 kN/m:V(0) = 32.5 kN, V(3⁻) = 32.5 − 5(3) = 17.5 kN
The 20 kN load drops the shear, then it keeps falling to the right support:V(3⁺) = 17.5 − 20 = −2.5 kN, V(8) = −2.5 − 5(5) = −27.5 kN = −R_B ✓
V jumps through zero at x = 3 m, so M is largest there. Moment = area under the shear diagram from 0 to 3 m (a trapezoid):M(3) = ½(32.5 + 17.5)(3) = 75 kN·m
Check: the area from 3 m to 8 m must bring M back to zero at B.½(−2.5 − 27.5)(5) = −75 kN·m → M(8) = 75 − 75 = 0 ✓

Answer: R_A = 32.5 kN, R_B = 27.5 kN. The shear falls linearly from 32.5 kN, jumps from 17.5 to −2.5 kN at the point load, and reaches −27.5 kN at B. The moment diagram is two parabolic arcs meeting at a peak of 75 kN·m at x = 3 m.

Example: Beam with an overhang and negative moment

Given

A beam supported by a pin at A (x = 0) and a roller at B (x = 4 m), overhanging to a free end C at x = 6 m.
A uniform load of 6 kN/m between A and B, and a 10 kN point load at C.

Find

The reactions, the maximum positive and negative moments, and the inflection point.

Solution

Moments about A (uniform load resultant 24 kN at x = 2 m):R_B(4) − 24(2) − 10(6) = 0 → R_B = 27 kN
Vertical balance:R_A = 24 + 10 − 27 = 7 kN
Between A and B, V(x) = 7 − 6x. It crosses zero where the positive moment peaks:7 − 6x = 0 → x = 1.167 m
Maximum positive moment, using M(x) = 7x − 3x² on AB:M(1.167) = 7(1.167) − 3(1.167)² = 4.08 kN·m
Moment at B, from the left segment and checked from the overhang:M(4) = 7(4) − 3(4)² = −20 kN·m; from the right: −10 × 2 = −20 kN·m ✓
Shear jumps at B and stays constant on the overhang:V(4⁻) = 7 − 24 = −17 kN, V(4⁺) = −17 + 27 = 10 kN, V(6⁺) = 10 − 10 = 0 ✓
The inflection point, where M changes sign on AB:7x − 3x² = 0 → x = 7/3 = 2.333 m

Answer: R_A = 7 kN, R_B = 27 kN. Maximum positive moment 4.08 kN·m at x = 1.167 m; maximum negative moment −20 kN·m over support B. M changes sign at x = 2.333 m. The overhang moment governs the design, which is easy to miss if you only look for where V = 0.

Common mistakes

Drawing the diagrams before finding the reactions. Every diagram starts at the left reaction. Solve the reactions first and check ΣFy = 0.
Using the resultant of a distributed load inside the diagram. The resultant is only for the reactions. In the diagram, a uniform load makes V slope steadily; it is not a point load at the centre.
Only checking where V = 0 for the maximum moment. Also check jumps through zero at point loads and the moment over supports next to an overhang. The largest |M| can be negative.
Mixing sign conventions between textbooks. State the convention on your sketch (positive moment = concave up) and use it for every segment.
Diagrams that do not close. V and M must return to zero at a free or simply supported end. If they do not, an area or a reaction is wrong.

In industry

Frequently asked questions

Which way is positive shear?

In the common beam convention, shear is positive when it acts downward on the left face of a cut (upward on the right face). A positive reaction at the left end therefore starts the shear diagram above the axis.

Why is the moment diagram a parabola under a uniform load?

Under a uniform load the shear changes linearly with x, and the moment is the integral of the shear, so it is quadratic.

Where is the maximum bending moment?

Where the shear diagram crosses or jumps through zero, at supports next to an overhang, or at a fixed end. Compare the magnitudes at all of them.

Do I need equations for V(x) and M(x)?

Only when you need values at arbitrary points or the exact location of a peak. The area method gives the shape and the key values faster.

Keep learning

All Statics lessons · All engineering subjects

Sources

R. C. Hibbeler, Engineering Mechanics: Statics, Pearson
F. P. Beer, E. R. Johnston, D. F. Mazurek, Vector Mechanics for Engineers: Statics, McGraw Hill
J. L. Meriam, L. G. Kraige, Engineering Mechanics: Statics, Wiley

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Shear and Moment Diagrams

Internal shear and moment

The area method

Why the diagrams matter

Key formulas

Worked examples

Example: Simply supported beam with a uniform load and a point load

Given

Find

Solution

Example: Beam with an overhang and negative moment

Given

Find

Solution

Common mistakes

In industry

Frequently asked questions

Keep learning

See also

Sources

Generate yours automatically from your existing docs