Moment Area Theorems

7/28/2019 Moment Area Theorems

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School of the Built Environment

Structural Analysis Notes

Moment-Area Theorems

The Moment-Area theorems provide us with further equations which we can use tosolve structural problems which are statically indeterminate. They are also used to

calculate beam deflections.

Remember that, in Plane (i.e. 2-dimensional) Statics, we have three equations of static

equilibrium:

H = 0

V = 0

M = 0

With which we can solve a maximum of three reaction forces. Many Civil

Engineering structures are built with some degree of redundancy more reactions or

more members than are needed to satisfy the requirements of stability alone. This

makes them stiffer and more resistant to damage.

Most Structural Analysis textbooks contain derivations of these theorems, so I will

not repeat them here. They are derived from the equation of circular bending with

which you are already familiar:

M = = E

I y R

And if we re-arrange the first and third terms, we get

M = 1

EI R

The deflection of a beam can be related to the curvature 1/R by examining the M/EI

diagram. This is simply the bending moment (M) diagram divided by the appropriate

value of EI at each point along the beam. In the examples which follow, we will only

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look at beams of constant EI, so the M/EI diagram will be identical in shape to the

bending moment diagram.

The two theorems can be stated as follows:

Theorem I: The change in slope of the deflected shape of a beam between points Aand B is equal to the area under the M/EI diagram between those two

points.

Theorem II: The deflection of point B from the tangent at point A is equal to the

first moment of area of the M/EI diagram between A and B about point

B.

Note that the theorems express relationships between different parts of a beam when

it is bent, not between its original and deflected shape.

Figure 1:Deflected shape and M/EI diagram.

Theorem I: angle BA = area A (units: radians)

Theorem II: deflection vBA = area A x (units: mm)

2

BA

A

deflected shape

M/EI diagram

Centroid of

shaded area

x

A = area of

shaded area


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The moment-area theorems are particularly useful when dealing with single-span

beams containing one or more built-in (fully-fixed) support. This is because at a

fully-fixed support the beam does not rotate and the tangent at this point therefore

remains stationary, providing us with a useful reference line. Look at the illustration

below, which shows how to calculate the deflection of a point on a cantilever beam.

Figure 2:Deflection of a cantilever.

To calculate the deflection of point B, the point of application of the concentrated

load:

MA = 20 8 = 160 kNm

Area under M/EI diagram (between A and B) = (160 6 )/EI = 480/EI

Hence deflection vBA = 480/EI 4 = 1920/EI

Where 4m is the distance between point B and the centroid of the M/EI diagram.

If we wished to calculate the deflection of point C:

Area under M/EI diagram (between A and C) = 480/EI

because the area under the M/EI diagram between B and C is zero, so

deflection vCA = 480/EI 6 = 2880/EI

3

8m

MA centroid

2m

20 kN

A BC

M/EI diagram

deflected shape

tangent at A

vBA v

CA

x


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where 6m is the distance from point C to the centroid of the M/EI diagram.

Equations of Compatibility

The moment-area theorems, or indeed any method of calculating deflections, becomemuch more useful when applied to the solution of indeterminate structures. Take for

instance the propped cantilever shown below:

Figure 3:Propped cantilever.

Notice that there are now four possible reactive forces three at A, plus a vertical

reaction at C, and we cannot therefore solve these unknown quantities using only the

three equations of static equilibrium. What we need is another equation. We can say

from the figure 3 that the deflection of point C is zero, so if we can write an

expression for the deflection of point C and equate it to zero, this will give us our

fourth equation. We do this by splitting up the problem into two statically

determinate parts.

4

8m

2m

20 kN

A BC

M/EI diagram


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First, remove the support at C:

Figure 4:Propped cantilever with prop released.

and then look at the effect of the support force without the load:

Figure 5:Fixing diagram.

Now, clearly the beam moves neither up nor down at point C. It is fixed here, hence

the name of the bending moment diagram in figure 5. But both of the beams in

figures 4 and 5 are statically determinate, so we can calculate the deflections vload and

vreaction. All we are then going to say is that the two added together equal zero this isour equation of compatibility (which simply states, in this case, that the amount by

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centroid

2m

20 kN

A BC

M/EI diagram

deflected shape

tangent at A

vload

x = 6m

vreaction

centroid

M/EI diagram

RCA

C

deflected shape

8m

x = 16/3 m

-160

8RC


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which the load wants to deflect point C downwards is equal and opposite to the

amount by which the reactive force wants to push it up).

Using the first moment-area theorem, we can write:

vload = (-160 6)/EI 6 = -5760/EI

vreaction = (8RC 8)/EI 16/3 = 1024RC/3EI

We then simply add these two together and equate to zero:

1024RC/3EI - 5760/EI = 0

Hence R C = 16.9 kN

Notice that the numerical value of EI is unimportant the value of RC depends solely

upon the magnitude and position of the load.

Those of you who are more adventurous may wish to use the moment-area theorems

to derive the values of end moments for a fully-fixed beam:

Figure 6:Fully-fixed beam.

Answer: MA = Pab2/L2 MB = Pa

2b?L2

Hint: use both of the moment-area theorems - you can see that the deflection of B

from the tangent at A is zero and the rotation of the tangent at B from the tangent at A

is also zero. This will give you two simultaneous equations. Several textbooks givethe solution look under the section describing Moment Distribution.

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ba

P

L

A B

MAMB

Moment Area Theorems

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