While bending failure is usually found with long beam spans carrying uniform loads, shear is an abrupt force of actually tearing a beam in half. Think of a sheet of cardboard; bending failure would occur after you fold the cardboard in half, and fold it back again, and continue until the cardboard has yielded so much that it no longer has any strength left and is relatively easy to break (we’ve all done this as kids with sticks or anything else we found breakable). Shear would be the force of just tearing the card board in half.
The bending force takes a long time to break something and therefore is allowed a relative low safety factor. The shear force is immediate and abrupt, and similar to wire cables, or plastics has a high safety factor:
Note: It is important to know that shear force will normally not govern over bending force, unless the member in question is very short in length, with very high loads. This is due to the fact that the bending stress will normally increase exponentially with the length of a beam while shear stress will only increase if the Force acting on the beam is increased.
1) Find f_{v}
where:
Note: For a W member (aka Ibeam) the shear area, A, is the area of the web (A_{web}= t_{web} * d)
2) Find F_{v}
The allowable stress of a compact Ibeam (F_{v}) is: F_v = {F_y \over \Omega_v} = \frac {F_y}{2.5} = 0.4F_y

where:
3) Show that f_{v} < F_{v}
Following the AISC steel code^{[1]}, the beams shear capacity must exceeds it's demand, in other words:
You have an A36 W10*26 spanning 4’ with a distributed load of 2000 plf. What is the shear design ratio?
1) Find V = 2000 plf * 4’ / 2 = 4000 lbs
2) Find A = t_{w}* d =
3) f_v = {V\over A} = {1000 lbs}
4) F_{v} = 0.4 * F_{y} = 0.4 * 36 ksi = 14.4 ksi > f_{y} =
5) Shear Design Ratio = {f_v \over F_v} = { \over 14.4 ksi} =
The member is at xx.x% capacity
Will the Shear Design Ratio or the Bending Design Ratio govern for the following scenario?
1) Find V = 2000 plf * 4’ / 2 = 4000 lbs