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: ퟏ ퟑ (4) MC = M + 2x*x*1/2 * 1/3*x – 48x = 0 and solving for M we have 푴 = ퟒퟖ풙 − 풙 … measured in ft-lbs. ퟑ
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:
푽 = ퟒퟖ − 풙ퟐ
Matches exactly the formula we found for the moment:
ퟏ 푴 = ퟒퟖ풙 − 풙ퟑ ퟑ
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.