SF, BM, slope and deflection


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This section is to show you that there is a mathematical relationship between, Load, Shear Force, Bending Moment, Slope and Deflection.  It is expected that you have some basic knowledge of Integral and differential calculus.

Consider a beam carrying a distributed load which is not necessarily of uniform intensity. 

Let its value at distance x from some datum be  ω kN per unit length (i.e., kN/m). 

 Let AB be a small section of the beam of length δx

The load on this section will = ω δx 

Let      F = shearing force and M = bending moment acting at the end A of the element. 

(F + δF) and (M + δM) are the corresponding values at the end B.  In other words, the shear force and bending moment have increased by δF and δM respectively. 

For section AB to be in equilibrium, the upward force must equal the downward force. 

Thus:  F = (F + δF) + ω δx 

Removing the brackets we get:      F = F + δF + ω δx

                                    Leaving:         δF= - ω δx 


and in the limiting case, as δx becomes smaller and approaches zero, 

                     . Eqn 1 

i.e., the slope or gradient of the shear force diagram at any point is equal to the load intensity at this point.

Furthermore, the algebraic sum of all the moments about any point must be zero for equilibrium. 

The load ω δx is assumed to act at the mid-point.  i.e. at its centre of gravity or centroid. 

Taking moments about A 

            M (M + δM) + (ω δx)(δx/2) + (F + δF) δx  = 0 

And so:         

            (M + δM) = M + (ω δx)(δx/2) + (F + δF) δx                    remove the brackets, 

so        M + δM = M + ω δx2/2 + F δx + δF δx                            M cancels to leave           

             δM = ω δx2/2 + F δx + δF δx 

Since δx and δF are both very small, then δx2 and δF δx are even smaller and thus negligible compared with the other terms, leaving: 

           δM = F δx 

From which                         and in the limiting case, as δx approaches zero


                                              .  Eqn 2. 

i.e. the slope or gradient of the bending moment curve at any point is equal to the shear force at that point. 

Note:  Some text books may show equations 1 and 2 as                       and

Provided there is consistency with the signs, the same answers will be obtained.

The following general conclusions can be drawn from eqns 1 and 2. 

If the load per unit length ω can be expressed as a simple function of x, expressions giving the shearing force and bending moment at any point can be obtained by integration.

These expressions will involve constants of integration which can be evaluated by using the known conditions at particular points along the beam. 

If the load is uniformly distributed then ω is a constant.

                     By integration of eqn 1              F = - ωx + A           

A is the constant of integration

And from eqn 2                                                but    F = - ωx + A

Substituting for F, we get        

Integrating this equation, we now get            M = - ωx2/2 + Ax + B

Where A and B are constants of integration.

So, where there is a UDL on the beam, the shearing force varies linearly and the bending moment variation is parabolic because of the x2 term.

For a part of a beam which carries no load (ω = 0 (zero))

So:                   F =              and      M = Ax + B                        

Thus over any part of the beam which has no load, the shearing forces are constant and the bending moment varies linearly.

See diagram below.

Over a part of the span where the shearing force is zero, (F=0) the bending moment is constant.  (See below)

At the point where the shearing force passes through zero.


Which is a condition for M to be a mathematical maximum or minimum.



as  changes from positive to negative or negative to positive,  will =0 giving a maximum Bending Moment on the beam.

Now, equation 2:       may be written as dM = F dx

Let Mx1 be the bending moment at some point x1 on the beam and Mx2 be the bending moment at a point x2, then:

                   or Mx2 Mx1  =        ..  Eqn 3

i.e. The change in bending moment between two points x1 and x2 on the beam is equal to the area under the shear force curve bounded by x1 and x2.

If at x1 = 0Mx1 = 0 then equation 3 reduces to Mx =

In other words, the bending moment at any section x is equal to the area under the shear curve between 0 and x.

When considering the area under the shear curve to the left of x, areas above the reference, or zero axis, are positive and areas below are negative.  When considering areas to the right of x, areas above are negative and areas below are positive. 

Look at the example below.  It is a simply supported beam with a mid-span concentrated load of 20 kN.


By inspection, shear force is maximum at the support at each end, of 10 kN.

If the bending moment at mid-span is required, calculate the area of the shear force diagram from mid-span to the left hand support.  This is 10 x 3.0 = 30 kNm.  If the bending moment at 5.0 m from left hand support is required, then the area of the shear force diagram from 5.0 m to the left hand support is (10 x 3.0) (10 x 2.0) = 10 kNm

You should try some examples for yourself.



Beams subjected to externally applied transverse loads deflect and change shape as shown above.

The magnitude of any change in shape will depend upon:

  1.             the length of the beam
  2.             the external bending moment
  3.             the shape and size of the beam cross-section
  4.             the modulus of elasticity of the beam material

In considering deflection, the same assumptions are taken as in the derivation of the bending formula.           

The diagrams above show an enlarged portion of the loaded beam.

The arc AB of length δs is on the neutral axis of the beam and subtends angle δi at the centre O of curvature.

The length of arc AB = δs = R δi                (where δi is in radians, NOT degrees)

Thus                                      (where R is the radius of curvature)

The slope or gradient of arc AB =                   

(where δy is a small change in the vertical deflection)

Since the angle is small, tan i approximates to i radians. 

Therefore the slope of AB = = i

In the limit, as δy approaches zero, i        ,           ds = dx           and

Thus,           which becomes =   which becomes = d2y/dx2

So  = d2y/dx2

From the simple bending equation,                   

Thus, substituting for                  = d2y/dx2

                         M = bending moment

                        E = Modulus of elasticity

                        I  = Second moment of area

                       = curvature of beam

Thus               M = EI d2y/dx2

This is the elastic equation for bending.

The product EI in the elastic equation for bending is known as the flexural stiffness or flexural rigidity of the beam with units Nm2 and is a measure of the resistance of the beam to a change in shape.

Provided the bending moment M can be expressed as a function of x, the equation can be integrated to give the slope i.

 Thus:              slope =

And a second integration gives the deflection

So,                   deflection =  

Where A and B are constants of integration.

The bending moment diagram is always plotted on the tension side of the beam, although mathematically negative, the bending moment is positive if sagging.

Therefore:                  EI d2y/dx2 = -M

Combining the differential equation of flexure (deflection) with the general relationships between bending moment M, shear force F and intensity of load w, leads to the following results for uniform beams and cantilevers.

bulletDeflection = y = EIy
bulletSlope = i = dy/dx = EI dy/dx
bulletBending moment = M = -EI d2y/dx2
bulletShearing force = F = dm/dx = - EI d3y/dx3
bulletLoad per unit length w = -dF/dx = EI d4y/dx4

An example will show you how all this works.

See below.


                        dF/dx = - w = EI d4y/dx4

                 Integrating to get  SF = -wx + A

                        So        EI d3y/dx3 = -wx  + A

            From the shear force diagram it is clear that  SF = 0 when x = L/2

                                    Thus 0 = -                which makes A =

Integrating again gives M = -EI d2y/dx2 = -

Thus M =                        Boundary conditions gives M = 0 when x = 0

So B = 0

And M =                       

Maximum bending moment occurs at mid-span when x = L/2

Substitute x for L/2 gives BM max =

Simplify the equation to give 

Thus BM max = -     (negative because the diagram is drawn on tension side of beam)

Going back to the general equation for M =

Integrating once more to get the slope equation gives:

Slope = EI dy/dx =

Maximum deflection occurs at mid-span and so the slope at this position will be zero.

i.e. Slope is equal to zero when x = L/2                          Substitute x for L/2 gives

0 =                             Thus C =

Finally integrating to get deflection

EIy =      Boundary conditions give zero deflection when x = 0

Thus D = 0

Therefore the general deflection equation for a simply supported beam with a UDL

is         EIy =

Maximum deflection occurs at mid-span when x = L/2

Substitute x for L/2 gives:

            EIymax =       =         =

Thus ymax =          

but total load on the beam = W = wL

and so the equation for maximum deflection on a simply supported beam with a total UDL = W is :

        ymax =



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Last Edited :  10 March 2015 10:42:11