A visual description of Shear Stress
Fig. 1: Shear Break in a Concrete sample

Shear stress will generally not govern (although as seen in the case of the I35 Mississippi River bridge when shear is the cause of failure it is usually an immediate failure). It governs on short spans with high loads, concrete beams, wood, and thin tubes. Since shear failure is a sudden brittle failure, it has a larger safety factor than bending.

Shear stress (fv) is related to shear force by the using the members area. Shear stress is simply the shear force at a given location in the beam, divided by it's area. Put into equation form:

f_v = \tau = \frac {V}{A}

This will be explained in greater detail below.

Understanding Shear Stress
  1. How to calculate the shear stress
  2. How to calculate the allowable shear capacity
  3. Example #1
  4. Example #2

How to Calculate Shear Stress in a member:

Once the shear force is obtained, the shear stress acting on a beam (fv) is relatively easy to find:

f_v = \tau = \frac {V}{A}

where:

fv = the shear stress acting on the member (ksi)
V = the shear force acting on the member (examples of how to calculate shear force can be found here.)
A = the area of the shear area (usually taken as the cross sectional area of the beam)

Note: For a flanged beams (I-beams, etc.) the shear area, A, is the area of the web (Aweb = tweb * d)

Calculating the Allowable Shear Capacity in a Member

Calculating the shear capacity for a member is all based on the type of material you're working with. For example concrete, steel, wood, and every other material will calculate it out differently. For the sake of the examples below we will use steel.

The allowable stress of a standard steel beam (Fv) is (in ASD):

F_v = {F_y \over \Omega_v} = \frac {F_y}{2.5} = 0.4F_y

A beam will not fail in shear if, fv ≤ Fv

Examples

Example #1:

You have an A36 W6*20 spanning 4 with a distributed load of 2000 plf. What is the shear design ratio?

Use the following steps:

1) Find V = wl/2 = 4000 plf * 4 / 2 = 8000 lbs
2) Find Aweb = t_w_ * d = 0.258 in * 6.2 in = 1.55 in2
3) Find fv = \frac{V}{A_{web}} = \frac{8000 lbs}{1.55 in^2} = 5161 psi
4) Find Fv = 0.4 * Fy = 0.4 * 36 ksi = 14.4 ksi > fy = 5.2 ksi
5) Shear Design Ratio = \frac{f_v}{F_v} = \frac{5.2 ksi}{14.4 ksi} = 0.36

Example #2

Will the Shear Design Ratio or the Bending Design Ratio govern for the following scenario?


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