Shear
General Discussion | Analysis Example
- Shear generally controls the design of short (l/d £ 10), heavily loaded beams.
- Shown in the figure above is an element taken at the n.a. of the beam where shear stresses are maximized.
- Recall, shear stresses develop because:
- The fibers, which are grossly modeled as discrete planks below, try to move by sliding past one and another.
- The internal resistance to this sliding results in shear stresses.
- Therefore at extreme fibers, there are no other fibers to slide past Þ no shear stresses.
- Likewise, at mid-depth, half of the beam is trying to slide past the other half and shear stresses are maximized.
- Recall, shear stresses develop because:
- In addition to horizontal shear stresses, vertical shear stresses of the same magnitude simultaneously occur.
- From mechanics of materials:
- fv = VQ / Ib
- V = shear load as a f(span, loads, supports).
- b = beam width at the point you are investigating.
- Q = 1st moment of the area sliding past the point investigating.
- I = moment of inertia.
- fv = VQ / Ib
- For rectangular cross section with fvmax:
- It is permissible, by the '94 UBC 2306.3 and the 1997 NDS, to use a reduced value for V in fv calculations if:
- In the usual loading case, the beams are fully supported for bearing on one side and the loads are applied to the other edge.
- This reduced value is the shear at a distance "d" from face of support, which is called V'.
- For example, consider the shear diagram for a simply supported, uniformly loaded beam.
- A reduced V can be used in fv because loads are transmitted to support by diagonal compression, not shear, within this "d" zone (i.e. diagonal compression, the critical mode).
- For example, consider the shear diagram for a simply supported, uniformly loaded beam.
- For design:
- fv £ Fv'
- Where Fv' = Fv CD CM Ct CH
- With Fv = tabulated horizontal shear stress:
- The shear strength of wood parallel to grain, (horizontal), is generally much weaker than the shear strength perpendicular to the grain, (vertical).
- CH = shear strength factor
- ³ 1.0 for solid-sawn.
- Not widely used in design, because it requires knowledge of split lengths and anticipated behavoir of splits.
- May, however, be useful in evaluating an existing in-service condition.
- = 1.0 for glulams (recognizing that extensive splitting doesn't occur in a manufactured product).
- ³ 1.0 for solid-sawn.
Analysis Example: Determine if the following beam is adequate in shear.
