Deflection of beams in bending

In previous tutorials we considered normal and shear bending stresses in beams under load which allows us to determine whether the beam has sufficient strength to carry the expected load.

Another consideration for design is that the deflection under load is acceptable.  In this tutorial we look at methods of determining deflections of beams in bending.

In a previous tutorial  we noted the relationship between bending moment M and radius of curvature ρ of the neutral axis x-x in the plane of bending as shown in this (greatly exaggerated) diagram:

radius of curvature in bending

  E is the modulus of elasticity and I is the second moment of area about the z axis.


In bending dy/dx is small and thus the denominator of the above expression reduces to 1 and we obtain the relationship:

The diagram below shows the slope (dy/dx) and deflection (y) of the neutral axis xx  in bending. 

Thus if we have an expression for bending moment M as a function of x we can obtain expressions for the slope and deflection by successive integrations.

We will now derive expressions for slope (dxydx) and deflection (y) for a simply supported beam with uniform distributed load shown below.

The first step is to derive a general expression in x for bending moment M.

Firstly find reaction forces R1 and R2  

Consider Σ forces = 0  (sign convention according to the diagram)

 R1 + R2 - wL = 0 ------------ (i)

Consider  Σ moments at x = 0 at position x = 0   (+ve moments are clockwise)

To find shearing force V and bending moment M consider a section of beam at position x shown below (we do not in fact require to find V but will do so the sake of revising previous tutorials).

forces and moments on beam section

To find V consider Σ forces = 0


What is the sign of V?

If x < L/2    V is -ve on y axis and +ve by convention for shear force.

If x > L/2    V is +ve on y axis and -ve by convention for shear force.

We must use the convention for shear force .

To find M consider Σ moments at x


Because the load results in positive bending M must be +ve by the convention for bending moments.

Note that since x ≤ L the sign of M is -ve (anti-clockwise) by the convention for general moments but is +ve by the convention for bending moments.


The starting point for successive integrations is:



We must now find the values of constants C1 and C.  Refer again to the diagram.

When x = 0 deflection y = 0.  Substituting x = 0 and y = 0 in the expression above for y gives C2 = 0.

When x = L deflection y =0. Substituting these values in the expression above for y and with C2 = 0 gives:

Substituting for C1 and C2 in the expression for y above gives:

Using this expression we can calculate deflection y for any value of x from x =0 to x = L. 

We know intuitively that the maximum deflection occurs where x = L/2. This can be confirmed by substituting  x = L/2 and C1 in the expression above for dy/dx which shows that dy/dx = 0 for these values.

Substituting x = L/2 into the expression above for y gives:

At this point it is instructive to look at our tutorial: bending moments and shearing forces in beams - singularity functions.  There we perform progressive integrations to find shearing force and bending moment equations for a loaded beam using the rules for singularity functions.

We can apply additional integration steps using singularity functions to find firstly the slope θ (dy/dx) of the beam and then the deflection y.  The complete sequence of integrations beginning with the loading function q(x) is:


In effect, singularity functions execute the same integrations as the indefinite integrals used in the example above but  handle complex configurations of loading more effectively as well as resolving all the important parameters associated with bending in sequence.

Superposition of loads

A useful principle concerning deflection of beams is superposition of loads which states that the resultant deflection produced by multiple loads is the sum of deflections produced by each load acting alone, provided the force and resultant deflection are linearly related.

superposition of loads 

In the above example deflections for point loads F1 and F2 and for distributed load w can be obtained individually for points along the beam as required and then summed.  In this context it is extremely useful to use published tables of solutions to beam equations for simple loading configurations.  

There is also a method for deriving deflection of beams using the concept of strain energy.  However, as this requires more advanced knowledge of calculus I mention it for information only.

I welcome feedback at: 

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Tutorials - stregnth of materials

Stress and strain - Basic concepts
Beams - Terminology and sign conventions
Beams - bending moments and shearing forces - simply supported beam with point loads
Beams - bending moments and shearing forces - cantilever with uniform distributed load
Beams - bending moments and shearing forces - simply supported beam with variable distributed load
Bending moments and shearing forces in beams - Singularity functions
Normal bending stresses in beams
Shear bending stresses in beams
Deflection of beams in bending
Forces in structural trusses
Torsional stresses
Buckling of columns

Tutorials - mechanics of machines

Crank mechanism kinematics - classic analysis
Crank mechanism kinematics - velocity and acceleration diagrams
Crank mechanism kinematics - vector equations
Crank mechanism - free body diagrams
Crank mechanism - inertia forces and crankshaft torque
Crank mechanism - crank effort and balancing
Crank mechanism - offsets and eccentrics

Tutorials - robotics

Kinematics of robot manipulators - introduction
Kinematics of robot manipulators - forward kinematics 1 (homogeneous transformation matrices)
Kinematics of robot manipulators - forward kinematics 2 (Denavit Hartenberg method)
Kinematics of robot manipulators - inverse kinematics

Tutorials - mechanical vibrations

Mechanical vibrations - introduction and overview
Mechanical vibrations - maths tutorial
Free vibrations
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Alistair's tutorials 2019