Lecture 39

Lecture 39:

Shearing Stresses

For small members in shear, it is customary to assume shear is uniformly distributed over the entire area. However, this assumption is not permissible for beam cross sections. The vertical shearing stress at any point in a beam may be determined from the horizontal shearing stress at that cross-section.

The infinitesimally small particle "A", whose dimensions are a, b, and 1, lies in a plane subject to shear.
fs represents the vertical shearing stress and fv represents the horizontal shearing stress.
fsb1 is the total shearing force on one vertical surface (force is equal to stress fs times the area b1); since the particle is in equilibrium an equal and opposite force must be acting on the other parallel face.
Similarly, fva1 represents the total shearing force on each of the horizontal planes.
The particle is in equilibrium, therefore the sum of the moments about any axis must equal zero. Take moments about axis "O":

This shows that at any point, the horizontal shearing stress is equal to the vertical shearing stress.

The general shear formula will find the shearing stress at any point in a beam. It is:

fv = V A' y / I b

V = the absolute value of vertical shear from the shear diagram at the point where shear is being investigated.
A' = the area above or below the horizontal plane being investigated.
y = the distance from the NA to the centroid of A'.
I = the moment of inertia of the entire section
b = the width of beam at the point where shear is being investigated.

Before deriving the formula, we will look at an application. If we wanted to find the maximum stress in the glue-line for this 2 x 10 composite beam:

First draw the V diagram.
Next, determine the geometry and assiciated properties of the beam and the appropriate value for y.
Then plug into the equation to determine the stress at the glue-line.
In this case, the stress would be 3000# (4")(2")(3") / (1/12)(2")(10^3)(2") or 216 psi.

Note: This is not a bending stress, it is a shearing stress.

SHEAR FORMULA
Development of the general shear formula:

Figure "A" shows a loaded beam and its V & M diagrams.
Figure "B" represents a section of the beam with an infinitesimally small block of length Delta L removed.
Figure "C" shows the small block that has been removed from the beam. The bending moment is greater at section 2 than at section 1; so the stresses are greater at section 2 than at section 1. Therefore due to the difference in stresses the force F2 is greater than force F1.
The small block is in equilibrium in the beam, therefore it must be in equilibrium as a free body; the sum of horizontal forces must be equal to zero. The horizontal shearing stress fv acting over the area bDL is the third force, which when added to F1 balances F2. Therefore, F1 + fvbDL = F2. (as shown in figure "C"), or fvbDL = F2 - F1.

To develop the general shear formula express the forces F1 and F2 in terms of the bending moments:
From the flexure formula: f =
or fy =
Stress on area "a" at "y" distance from NA.: fy =
Force on area "a" (F2) equals stress times area: F2=
Total force F2 =
The first static moment of an area (from centroids):
Therefore, the total force F2 =
and F1 =
Substituting in the formula developed in step 4: fvbDeltaL = F2 - F1
From figure "A" we note that the change in moment (M) between two sections (1 & 2) is equal to the area of the shear diagram (V L) between the same two sections.
Combining steps 12 & 13 gives the general shear formula:

fvbDeltaL =

fv=

fv=
Review of the most important aspects of the general shear formula:

fv = V A' y / I b

V is always the absolute value of the total shear (from the V diagram) on the beam at the section being investigated (this is usually the point where the shear is greatest).
A' is always the area on the face of the structural section on either side of the plane at which shear is being investigated. For most beam applications this is the area above (or below) a horizontal plane through the neutral axis of a beam.
y is always the distance from the neutral axis to the centroid of the area A'.
I is always the moment of inertia of the entire cross sectional area of the beam with respect to the neutral axis.
b is always the total width of the beam at the plane where the shear is being investigated (this is usually the width at the neutral axis).
The general shear formula is often written with the static moment A'y represented as Q in which the general shear formula is:

fv = V Q / I b

Plotting Shear Forces

Plotting Shearing Stresses

The second example illustrates that the maximum horizontal shearing stress in a rectangular beam occurs at the neutral axis. We can easily develop a formula for this very special case which is most useful for wood beams.

shear formula for a rectangular cross section

This formula applies only to the horizontal shearing stress in a rectangular beam and then only at the neutral axis.

IT APPLIES TO NOTHING ELSE!

HORIZONTAL SHEAR FORMULA FOR A STEEL SECTION
The approximate web-shear formula recognizes that most of the shear stress in a steel section is taken by the web. It gives values within 10-15% of the actual true value from the general shear formula. (Fv = 14.5 ksi, A-36 steel)
fv = V / tw d where

tw is the thickness of the web (read from the section tables)
d is the depth of the beam.

STRESS COMPARISON

The following is a comparison of bending and shearing stresses in a rectangular beam. It is an important table to memorize!