The T section shown below is used to explain the difference between the two components of shear stress, i.e., t

__SOLUTION:__

With the shear force along the y axis, the average flexural shear stress formula for the two components of interest is written as

Although the general form of the equation is the same for both components,
there is a difference in the way Q_{z} and t are found in each case. With
V_{y} given, we proceed with the calculation of centroidal position.

The centroid is located at 3.25 in. from the bottom edge of the section. The moment of inertia about the z axis is found as

To calculate the t_{yx} component of
shear stress, the longitudinal surface has to have a normal in the y
direction. The intersection of this longitudinal surface and the transverse
surface is shown by a solid blue line in the figure below.

Depending on the position of the longitudinal surface along the y axis, the
corresponding value for Q_{z} is found as

The equation for Q_{z} is quadratic. Q_{zmax} =
5.28 in^{3}, and it occurs at the
NA (y_{2} = 0). The variation of Q_{z} is shown below.

With the variation of Q_{z} known, t_{yx
} is found as

Keep in mind
that t is the **width of the longitudinal surface**. Along -3.25 < y < 0.75,
t = 1 in. and along 0.75 < y < 1.75, t = 4 in. As is shown in the figure above,
there is a discontinuity in the shear stress at y = 0.75 in. due to the discontinuity in t.
The maximum value of t_{yx} is found to be
290.64 psi and it occurs at the NA. The values of t_{yx} over the longitudinal surface are equal in magnitude to the
corresponding t_{xy} values on the transverse
surface.

It is crucial to pay attention to t_{yx}
variation at the junction. The value of 68.81 psi found in the analysis refers
to the shear stress in the horizontal flanges just above the junction point.
As there can be no shear stress on a free surface, the shear stress
t_{yx} and its counterpart
t_{xy} are zero at points B and
C but not at point A. As indicated by its plot,
t_{yx} is more significant in the vertical
flange as opposed to horizontal flanges.

To calculate the t_{zx} component of
shear stress, the longitudinal surface has to have a normal in the z
direction. The intersection of this longitudinal surface and the transverse
surface is shown by a solid blue line in the figure below.

Depending on the position of the longitudinal surface along the z axis, the
corresponding value for Q_{z} is found as

The equation for Q_{z} is linear in this case. At the section
intersecting the corner of the horizontal and vertical flanges i.e.,
z_{z} = +/-0.5, Q_{z} =
1.875 in^{3}.
The variation of Q_{z} is shown below.

The blue dashed line gives the Q_{z} variation at the junction of
the horizontal and vertical flanges. This value is used to find the shear
flow at that location.

With the variation of Q_{z} known, t_{zx
} is found as

Keep in mind
that t is the width of the longitudinal surface. Along -2 < z < -0.5,
t = 1 in. As is shown in the figure above,
t_{zx} reaches its maximum value of
103.21 psi in the
horizontal flanges at the junction with the vertical flange. In the vertical
flange this component of shear stress is zero as shown below.

Notice that there is a counterpart shear stress (t_{xz}) on the transverse surface.
Also in this case t_{zx} is more
significant in the horizontal flanges as opposed to the vertical flange.

**key observation:**

At any point on the boundary the resultant shear stress is parallel to the boundary. This fact helps us to sketch the figures shown below as a general depiction of shear flow and shear stress variations in this example.

To Section III.3

To Index Page of Transverse Shear Loading of Open Sections