Roymech engineering encyclopedia

Beam Shear Force and Bending Moment Analysis


Shear force and bending moment diagrams are fundamental tools for analysing beams subjected to transverse loading. They show how the internal shear force and bending moment vary along the beam and allow critical sections to be identified for subsequent stress and deflection calculations.

Need help choosing a beam calculation? See the Beam Design and Analysis Guide for guidance on beam types, loading, bending, shear, deflection and selecting the appropriate RoyMech calculation method.

This page covers calculation of support reactions, the sign convention used for shear force and bending moment, construction of shear force and bending moment diagrams, and the mathematical relationships between distributed load, shear force and bending moment.

Contents

Nomenclature

Symbol Description Typical SI Unit
F Concentrated force applied to beam N
L Beam length or span m
M Bending moment N m
R Support reaction N
S Internal shear force N
W Total applied load N
w Distributed load per unit length N/m
x Distance along beam m

The symbols E, I, Z, stress and strain are required when the resulting bending moments are subsequently used for stress or deflection calculations, but they are not required to construct the basic shear force and bending moment diagrams.

Calculation of External Forces

Before the internal shear force and bending moment can be determined, the external forces acting on the beam must normally be established. A free-body diagram is produced in which the beam supports are replaced by their corresponding reaction forces and, where applicable, reaction moments.

Free-body diagram of a simply supported beam showing applied load and support reactions

For a beam in plane static equilibrium, the applied forces and moments must satisfy the equilibrium equations:

∑ Fx = 0
∑ Fy = 0
∑ M = 0

For the simply supported beam shown above, consider a concentrated load W positioned a distance a from the left-hand support. Taking moments about the left-hand reaction R1:

R2 L − W a = 0

Therefore:

R2 = W a L

Vertical force equilibrium then gives:

R1 + R2 − W = 0

and hence:

R1 = W − R2
Support conditions matter. A simple support generally provides force reactions, whereas a fixed or built-in support can provide both force reactions and a reaction moment. The appropriate reactions must be included in the free-body diagram before the beam is analysed.

Shear Force and Bending Moment Diagrams

A shear force diagram (SFD) shows the variation of internal shear force along the length of a beam.

A bending moment diagram (BMD) shows the corresponding variation of internal bending moment along the beam.

These diagrams are normally constructed beneath the beam free-body diagram using the same horizontal scale. This makes the relationship between the applied loading, shear force and bending moment easier to identify.

The diagrams are particularly useful for locating maximum bending moments and changes in shear force. These values can then be used in the appropriate beam stress and deflection calculations.

Sign Convention

Different engineering texts and software packages may adopt different sign conventions for shear force and bending moment. The important requirement is that the selected convention is used consistently throughout the analysis.

The sign convention used on this page is illustrated below.

Sign convention used for positive and negative beam shear force and bending moment

When comparing equations or diagrams from different references, check the sign convention before comparing numerical signs. Two analyses may describe the same physical beam behaviour while using opposite algebraic signs.

Typical Shear Force and Bending Moment Diagrams

A convenient way of determining the internal forces is to imagine a section moving along the beam from left to right. At any position x, the forces and moments acting on one side of the section are used with the equilibrium equations to determine the internal shear force and bending moment at that position.

Concentrated Loads

A concentrated transverse force produces a sudden change in the shear force diagram. Between concentrated forces, where there is no distributed load, the shear force remains constant.

The bending moment changes continuously through a concentrated force unless an applied concentrated moment is also present.

The following sketches show simply supported beams subjected to concentrated loading.

Simply supported beam with concentrated load showing corresponding shear force and bending moment diagrams

Cantilever Beams

For a cantilever, the fixed support provides both a reaction force and a reaction moment. The shear force and bending moment distributions depend on the applied loading arrangement.

The following sketches illustrate typical cantilever loading cases and their associated diagrams.

Cantilever beams with typical applied loads and corresponding shear force and bending moment diagrams

Distributed Loads

For a distributed load, the equivalent resultant force over a selected length is equal to the area under the load-intensity diagram. The resultant acts through the centroid of that loaded area.

For a uniform distributed load w acting over a length l:

W = w l

and the resultant acts at the centre of the uniformly loaded length.

Use force units in beam calculations. A value quoted as a mass, for example 50 kg, is not a force. The corresponding weight near the Earth's surface is approximately 490 N using g = 9.81 m/s2. Distributed loading should similarly be expressed in force per unit length, such as N/m or kN/m.

Load, Shear Force and Bending Moment Relationships

The relationships between distributed load, shear force and bending moment can be obtained by considering equilibrium of a differential beam element of length dx.

Differential beam element showing distributed load, shear forces and bending moments

Let the shear force and bending moment at the left side of the element be S and M. At the right side they become S + dS and M + dM.

Distributed Load and Shear Force

Vertical force equilibrium gives:

S − w dx − ( S + dS ) = 0

Therefore:

dS dx = − w

Thus the slope of the shear force diagram is equal to the negative of the distributed load intensity, using the sign convention adopted here.

Shear Force and Bending Moment

Taking moments about one side of the differential element gives:

M + S dx − ( M + dM ) − w ( dx ) 2 2 = 0

In the limit as dx tends to zero, the second-order term becomes negligible and:

dM dx = S

Therefore the slope of the bending moment diagram is equal to the shear force.

Integral Relationships

Between two positions x1 and x2, the change in shear force is:

S2 − S1 = − ∫ x1 x2 w dx

Similarly, the change in bending moment is:

M2 − M1 = ∫ x1 x2 S dx

The area under the shear force diagram therefore gives the change in bending moment between two positions.

Useful Consequences

The differential relationships provide several useful checks when constructing beam diagrams:

  • Where w = 0, the shear force is constant and the bending moment varies linearly.
  • Under a uniform distributed load, the shear force varies linearly and the bending moment varies quadratically.
  • Where the shear force is zero within a continuous region, the bending moment has a stationary value and may therefore be a local maximum or minimum.
  • A concentrated transverse force produces a step change in the shear force diagram.
  • An externally applied concentrated moment produces a step change in the bending moment diagram.

Related Beam References



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