Shear Moment Diagram Calculator
Calculate shear force and bending moment diagrams for simply supported and cantilever beams. Find reactions, max shear, max moment, and critical sections.
Shear Moment Diagram Calculator
Calculator
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Formula: V(x) = RA - Sum of loads to left of x | M(x) = Integral of V(x) dx
Worked example — RA = RB = 5 kN | Max Shear: 5 kN | Max Moment: 15 kN-m at center
Formula
V(x) = RA - Sum of loads to left of x | M(x) = Integral of V(x) dx
The shear force at any section is the algebraic sum of all vertical forces to one side of that section. The bending moment is the algebraic sum of moments of all forces to one side. For a point load P at distance a: RA = Pb/L, RB = Pa/L. For UDL w: RA = RB = wL/2 (full span). Max moment occurs where shear equals zero.
Worked Examples
Example 1: Simply Supported Beam with Central Point Load
Problem:A 6 m simply supported beam carries a 10 kN point load at the center (3 m from each support). Determine the reactions, maximum shear, and maximum bending moment.
Solution:Reactions: RA = RB = P/2 = 10/2 = 5 kN (symmetric loading) Shear: V = +5 kN from left support to load, V = -5 kN from load to right support Max Shear = 5 kN (at supports) Max Moment = PL/4 = 10 x 6/4 = 15 kN-m (at center) Zero shear occurs at x = 3 m (load point)
Result:RA = RB = 5 kN | Max Shear: 5 kN | Max Moment: 15 kN-m at center
Example 2: Simply Supported Beam with Full UDL
Problem:A 6 m beam carries a uniformly distributed load of 5 kN/m over its entire length. Find the maximum shear and moment.
Solution:Total load = 5 x 6 = 30 kN Reactions: RA = RB = 30/2 = 15 kN Max Shear = wL/2 = 5 x 6/2 = 15 kN (at supports) Max Moment = wL^2/8 = 5 x 36/8 = 22.5 kN-m (at midspan) Shear is zero at midspan (x = 3 m)
Result:RA = RB = 15 kN | Max Shear: 15 kN | Max Moment: 22.5 kN-m at midspan
Frequently Asked Questions
What are shear force and bending moment diagrams used for in structural analysis?
Shear force diagrams (SFD) and bending moment diagrams (BMD) are fundamental tools in structural engineering that graphically represent the internal forces and moments acting along the length of a beam. The shear force diagram shows how the internal vertical force varies at each cross-section, which is critical for designing against shear failure and determining required web thickness in steel beams or stirrup spacing in reinforced concrete. The bending moment diagram reveals the internal moment distribution, which determines the maximum bending stress and governs the required section modulus or reinforcement for the beam. Together, these diagrams enable engineers to identify critical sections, size structural members, and verify that designs meet safety requirements.
How are support reactions calculated for a simply supported beam?
For a simply supported beam, the support reactions are determined using the equations of static equilibrium: the sum of vertical forces equals zero and the sum of moments about any point equals zero. For a point load P located at distance a from the left support on a beam of length L, taking moments about the left support gives RB = P times a divided by L, and from vertical equilibrium, RA = P minus RB. For a uniformly distributed load w over the entire span, each reaction equals wL divided by 2 due to symmetry. For partial distributed loads, the resultant force acts at the centroid of the loaded region, and moments are taken about one support to find the other. These reactions serve as the starting values for constructing shear and moment diagrams.
What is the relationship between shear force and bending moment?
Shear force and bending moment are mathematically related through calculus. The bending moment at any point is the integral (area under the curve) of the shear force diagram up to that point. Conversely, the shear force is the derivative (rate of change) of the bending moment. This means the bending moment reaches its maximum or minimum value where the shear force equals zero or changes sign. The distributed load intensity equals the negative derivative of the shear force. These relationships provide important shortcuts: the change in moment between two points equals the area under the shear diagram between those points, and the slope of the moment diagram at any point equals the shear force at that point.
How does a cantilever beam differ from a simply supported beam in analysis?
A cantilever beam is fixed at one end and free at the other, creating fundamentally different shear and moment distributions compared to a simply supported beam. The fixed support provides both a vertical reaction force and a reaction moment, whereas simple supports only provide vertical forces. In a cantilever, the maximum bending moment always occurs at the fixed support and decreases toward the free end. For a point load at the free end, the moment varies linearly from maximum at the fixed end to zero at the load. The shear force is constant along the beam. Cantilevers experience greater deflections than simply supported beams of the same span and load because there is no intermediate support to limit displacement.
Where does maximum bending stress occur in a beam and how is it calculated?
Maximum bending stress occurs at the cross-section where the bending moment is largest, specifically at the outermost fibers (top and bottom) of that cross-section. The bending stress formula is sigma equals M times y divided by I, where M is the bending moment, y is the distance from the neutral axis to the point of interest, and I is the second moment of area (moment of inertia) of the cross-section. The maximum stress occurs when both M and y are at their maximum values. For a rectangular cross-section of width b and height h, the section modulus S equals I divided by y-max equals bh-squared divided by 6, simplifying the calculation to sigma-max equals M divided by S. This stress must remain below the allowable stress determined by the factor of safety.
What sign convention is used for sagging and hogging moments?
The common beam convention calls a moment positive when the member sags, putting the top fibres in compression and the bottom in tension, and negative when it hogs, as over a cantilever's fixed end or an interior support. Shear is positive when the left-hand portion tends to slide upward relative to the right. Some texts plot the moment diagram on the tension side, which flips the drawing without changing the physics, so always check the stated convention before comparing two sets of diagrams.
What shapes should the shear and moment curves take for each load type?
Because V is the integral of the load and M is the integral of V, each integration raises the polynomial degree by one. A span with no load has constant shear and a straight-sloping moment. A uniformly distributed load gives shear that varies linearly and a parabolic moment. A triangular load gives parabolic shear and a cubic moment. A concentrated load creates a vertical jump in the shear diagram equal to the load, and an applied couple creates a vertical jump in the moment diagram equal to that couple.
What is a point of contraflexure and why does it matter?
A point of contraflexure, or inflection point, is a section where the bending moment passes through zero and the curvature reverses from sagging to hogging. On an overhanging or continuous beam it separates the region where the bottom steel carries tension from the region where the top steel does. Reinforcement detailers use these locations to decide where bars may be curtailed or lapped, and designers sometimes place construction joints or hinges near them because the section carries shear but essentially no moment.
Why can statics alone not solve a fixed-fixed or propped beam?
A simply supported or cantilever beam has exactly as many unknown reactions as there are equilibrium equations, so it is statically determinate and Shear Moment Diagram Calculator's approach works. Add restraint and it is not: a propped cantilever, fixed at one end with a roller at the other, carries one redundant reaction, and a beam fixed at both ends carries two more than vertical and moment equilibrium can resolve. Those extra conditions come from deformation compatibility, through slope-deflection, moment distribution, or a stiffness solver. A fixed-fixed beam under full uniform load carries wL^2/12 at each end and wL^2/24 at midspan, against wL^2/8 at midspan when simply supported.
How do you turn the maximum moment into a required beam size?
Take the peak moment from the diagram and divide by the allowable bending stress to get the required section modulus, S = M / sigma. A midspan moment of 22.5 kN-m with an allowable stress of 165 MPa needs S = 22.5 x 10^3 / 165 x 10^6, which is 1.36 x 10^-4 m^3, or 136 cm^3, so you pick a rolled section whose tabulated S exceeds that. Then check the peak shear against the web capacity and verify deflection separately, since a section strong enough in bending can still be too flexible.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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