Consider a cantilever beam with a **symmetric** cross section
subjected to a lateral (or transverse) force P at the tip.
The shear and normal stresses induced by this force are required to be below the
corresponding elastic limits of the material.

Next, we consider a transverse section located at some distance x from the free end of the beam as shown above. The bending moment at this section is related to the normal stress according to the equation

In order to go any further, we need to know how s
_{x} varies along the section. Recall from previous chapter that if
the section is in the state of pure bending, then the normal strain vaires
linearly. Furthermore, if the section is in elastic condition and the
material is linearly elastic, then according to Hooke's law, normal stress
would vary linearly as well. With that in mind, we need to determine
whether the normal strain varies linearly in the presence of transverse
shear force. Without getting too deep into the theory of elasticity, it
suffices to say that if the beam is long compared to its cross section, then
the normal strain does vary linearly. On the other hand, if the beam is
short, then normal strain variation is in fact nonlinear when transverse
shear force is present.

So we either have to limit our analysis to long beams or assume that the normal strain variation is linear regardless of the length of the beam. With the requirement of elastic condition and the use of linearly elastic materials, the general elastic bending formula described in the previous chapter can be used. In the presence of one moment component and zero product of inertia, the normal stress equation reduces to

The distance y is measured from the centroidal z axis (which in
this case coincides with the neutral axis), I_{z} is the moment of
inertia of the section about the z axis.

Relating the bending moment to the applied shear force P gives

The maximum normal stress, at section x, will occur at the farthest distance from the neutral axis. In this case the stress distribution indicates that distance to be to the top surface of the section as shown below.

To examine the shear force and associated shear stress created as a result of
transverse force P, let's consider a beam segment highlighted in the figure
below. While the left edge of the beam segment is free from any force in the x
direction, the right edge is not
due to the presence of s_{x}. The question
is how can equilibrium be maintained in this case? This points to the
existance of a horizontal shear force H along the bottom surface
(**longitudinal surface**) of the highlighted segment.
The important point to notice first is that
a transverse shear force results in the creation of a
*longitudinal shear
force.*

Now the equation for equilibrium of forces in the x direction is written as

where A^{*} represents the area of the beam section
from y_{1} to c. Making appropriate substitutions and simplifying the
resulting equation gives

The integral term represents a geometric quantity referred to as the *moment
of area,*and is denoted by *Q*_{z}. The subscript z
indicates that the moment of area is obtained about the z axis. The
equation for H can be written as

This equation indicates that the longitudinal shear force is an explicit
function of x and an implicit function of y through the moment of area.
At the tip section x is zero, hence H is zero for any value y. However, for
any positive value of x, the variation in H depends on the location of
y_{1}. It can be shown that for y_{1} = 0, H reaches its
maximum value whereas at y_{1}
corresponding to the top and bottom surfaces, H goes to zero. Therefore, the
longitudinal shear force H varies linearly with respect to x and quadratically
with respect to y.

Now, divide both sides of the equation by x

Note that P times x represents the moment M_{z} at
position x, and that in the limit as x approaches zero, Px/x simply represents
the change in the bending
moment which is simply equal to the transverse shear force at that location.
Therefore, we can rewrite this equation as

The ratio of H over x represents the quantity known as the **shear flow**
denoted by *q.* In this case, shear flow is acting along the longitudinal surface
located at distance y_{1}. For the problem described here, the shear
flow can be written as

At a given position x, q varies according
to the variation of Q. The equation for Q reveals that at the NA corresponding
to y_{1} = 0, Q reaches its maximum value, and at y_{1}
corresponding to the top and bottom surfaces, it will be zero.

To obtain the flexural shear stress along the longitudinal surface
corresponding to y = y_{1}, we simply divide q by the width of
the longitudinal surface. If this dimension is denoted by t, then

The subscripts on t indicates that the
longitudinal surface has a normal vector in the y direction and that the
shear stress component is in the direction of x axis. Since shear stress
in the longitudinal plane must be equal to that in the transverse plane, then
the two subscripts can change places without any change in the right hand
side of the equation. The equation above is known as the *average flexural
shear stress formula.*

Although the value of shear flow is always maximum at the neutral axis location, we cannot conclude that shear stress will also be maximum there. The reason is that t may vary from one point to another, and that it is possible for a point other than the neutral axis to have the maximum shear stress. In order to calculate the maximum shear stress, the following relationship should always be followed.

__Restrictions:__

Before we can apply the average flexural shear stress formula, we must consider the restrictions that apply to this equation:

1. The section has to be homogeneous (made of a single material).

2. The section has to have an axis of symmetry, therefore, product of
inertia is zero.

3. The shear force V passes through the shear center of the section,
and is parallel to one of the two principal centroidal axes.

4. The shear stress at every point is below the corresponding elastic limit
of the material.

What if the longitudinal surface we considered in above derivation was not parallel to z axis? What would change in that case?

**Answer:**

To answer these questions, consider the figure shown below.

In the case of horizontal longitudinal surface, we will be calculating the
yx or xy component of shear stress. To calculate Q_{z} we use the area
above the longitudinal surface. t in this case is equal to the distance from
the left edge to the right edge of the top section. In the case of the vertical
longitudinal surface, we will be calculating a different component of shear
stress, i.e., t_{zx} which is the same as
t_{xz} on the transverse surface.
In this case Q_{z} can be found by considering
the area to the right side of the longitudinal surface, and t would be equal to
the flange thickness.
This difference is demonstrated in the following discussion:

Shear Stress Components
t_{yx} and t_{zx}

If V is along z direction, then the moment of inertia is that about the
perpendicular axis or y axis in this case. For V_{z} loading, the
flexural shear stress formula can be written as

Therefore, at a given point on the cross section we can calculate two components of shear stress depending on the orientation of the longitudinal surface. Generally speaking, one component is always much larger than the other thus, that would be the one of most interest in the analysis.

Table below gives equations for
maximum shear stress for specific cross sections.
The distance **e** is defined as the distance from
the neutral axis to the point of maximum shear stress.
The shear force V is normal to the NA and it acts along the vertical axis of
symmetry, hence through the shear center.
The maximum shear stress is given in terms of the ratio V/A.

**Shear Stress Variation:**

If we obtain the equation that describes the shear stress as a function of position on the cross section, it would be easy to find how shear stress varies from point to point as shown in the example figure below.

Knowing that the ratio V/I is constant for a given section, we need to only concentrate on the variation described by Q/t. For example, if we follow the derivation steps, we find the average flexural shear stress variation for a beam with rectangular cross section, as shown above, will be parabolic according to the equation

**Shear Flow Variation:**

The shear flow distribution at a given cross section is determined by writing the equation VQ/I, which is basically the shear stress multiplied by 't'. The shear flow distribution calculation can be seen in the example problems given at the bottom of this page.

Here are some examples of shear flow diagrams for sections with at least one axis of symmetry. Each of these cases show the shear flow when the shearing load passes though the shear center.

In each of these shear flow diagrams, the maximum shear flow occurs at the neutral axis, which passes through the centroid of the section. This will also be the location of the maximum shear stress as the wall thickness is constant. Also notice how the shear load is perpendicular to the NA.

EXAMPLE PROBLEMS

- Example 1 Maximum shear and flexural stresses in a beam with T cross-section
- Example 2 Average shear stress in a beam with an uneven I cross section

To Section III.4

To Index Page of Transverse Shear Loading of Open Sections