How Shear and Moment Work?

Beams are in every home and in one sense are

quite simple. The weight placed on them causes

them to bend, and you just need to choose a

large enough beam to withstand it.

As shown in the diagram below, when bending,

the top portion is resisting a desire to shorten

while the bottom portion is resisting a desire to

lengthen.

So how do we make sense of what is happening inside the beam deeply enough to quantify

it and choose the right beam?

After all you cannot wait

until the house is built to find

out the beam is too small.

Solution to How Shear and

Moment Work:

There are 2 basic forces at play to

consider, bending and shear.

Bending is happening due to the

beams desire to crack in the middle under load, both halves rotating inward. Engineers call this desire to

bend, moment. The integrity of the wood fibers resist the crushing force on the top half of the beam and the

separating force on the bottom half of the beam.

Shear is simply the vertical weight on a

beam at any given point.

Let’s consider an interesting example where

a 12’ beam is supporting a continuous load

that increases from 0 lbs/ft on the left to 24

lbs/ft on the right such as a rake wall would

produce. Wall load is typically considered to

be 8lbs/ft2 as it is composed of wood and

drywall.

Engineers would show this continuous load

with a multitude of forces acting downward

on the beam.

In order for the post to support the beam at

A we need to know how much weight the

beam would put on it.

Moment = force x distance … but a continuous load has infinitely many forces?

Thanks to calculus, which makes short work of infinity, and the mathematics of the center of mass, we can consider the

whole weight on the beam to act at a distance 2/3 of the way across the beam from the left (since this is a triangular

load). We don’t want the beam to rotate so the sum of the moments should be 0.

The total weight on the beam is:

(1) 24 lbs/ft * 12 ft * ½ = 144 lbs … (as a result of the area of triangle formula) and it acts at a point 8’ from A.

Moment around A:

(1) MA = FB*12 - 144*8 = 0 … and solving for FB we get 96 lbs.

Moment around B:

(2) MB = FA*12 - 144*4 = 0 … and solving for FA we get 48 lbs.

Now on the beautiful mathematics of shear and moment. For a beam to resist shear and bending away from the

supports we have to make a cut some distance in and consider the internal forces on the beam.

Shear:

If we measure in a distance (x) and consider the vertical forces, again they must sum to 0 if the beam is going to remain

in equilibrium (aka not move). Notice the forces of the posts upward add up to the 144 lbs of weight on the beam.

Notice the 12’ beam has a force of 24 lbs/ft at its

end. This is twice its length, so if our cut is at x then

the weight at the cut would be 2x lbs/ft.

Notice also that the fibers of the beam will have to

be capable of resisting the shear force at the cut

labeled V in the diagram.

As before the total weight on the beam is found

with the area of the triangle formula, and we can now find a formula for V based on the location of the cut x.

(3) 48 – 2x*x*1/2 - V = 0 … and solving for V we have V = 48 – x2 measured in lbs … a lovely parabola.

Notice that this would indicate there is

no shear strength required of the beam

when x ≈ 6.93 ft.

Notice also that the maximum shear

strength required of the beam is 96 lbs

(which not coincidentally matches the

force we found the post exerts at B. The

versa-lam design table from Boise

Cascade at the end of this document,

lists the allowable shear strength of

different beam designs.

We now have one force on a beam figured out and a rough idea of how to find a beam capable of resisting it.

Now let’s move on to the second … moment.

Moment:

Now we reconsider our section of the beam and

consider the internal moment the beam will have to

resist based on its fiber strength to keep the beam

from actually cracking at the distance x.

As before, the sum of the moments in our section

has to be 0 to maintain equilibrium. The M in the

figure is the moment around the cut that the beam

will have to be capable of producing to counteract

the rotation the weight and the post want to cause.

Moment around the cut:

(4) MC = M + 2x*x*1/2 * 1/3*x – 48x = 0 and solving for M we have 𝑴 = 𝟒𝟖𝒙 −𝟏

𝟑𝒙𝟑 … measured in ft-lbs.

Notice this indicates the maximum moment of 222 ft-lbs occurs when x = 6.93 ft … which coincides with the location of

minimum shear.

Now for the really beautiful part. The units on the

shear diagram would indicate that the shaded area

would have a unit in ft-lbs. And sure enough the

integral of the shear formula:

𝑽 = 𝟒𝟖 − 𝒙𝟐

Matches exactly the formula we found for the

moment:

𝑴 = 𝟒𝟖𝒙 −𝟏

𝟑𝒙𝟑

The versa-lam design table from Boise Cascade at the

end of this document, also lists the allowable moment

for different beam designs.

So there you have it. Calculate the maximum shear

and moment that a beam will have to bear based on

the forces that are on it and look in the table to find a

beam capable of resisting them.