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:

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 R_{1} and R_{2 }

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

R_{1} + R_{2} - 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).

**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 C_{1} and C_{2
}. Refer again to the diagram.

When x = 0 deflection y = 0. Substituting x = 0 and y = 0 in the expression above
for y gives **C**_{2}** = 0**.

When x = L deflection y =0. Substituting these values in the
expression above for y and with C_{2} = 0 gives:

Substituting for C_{1} and C_{2} 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 C_{1} 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.

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.

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.

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