Short Answer:

The maximum shear stress in a rectangular section occurs at the neutral axis (mid-depth) and has a parabolic distribution across the depth. For a rectangle of width and depth carrying a shear force , the maximum shear stress is

This value is 1.5 times the average shear and falls to zero at the top and bottom fibers.

Knowing is important for checking shear capacity and sizing members. The result is obtained from the general shear formula by evaluating the first moment , the inertia , and the local width for a rectangular cross-section.

Detailed Explanation :

Maximum Shear Stress in Rectangular Section

What we want to find
We want the largest value of the internal shear stress that acts on a rectangular beam cross-section when a transverse shear force is applied at the section. This maximum value controls shear design and determines whether the section will resist sliding failure.

Basic starting point — general shear formula
The shear stress at a horizontal layer in any cross-section is given by the standard relation:

where

= internal shear force at the section,
= first moment of area of the portion of the section on one side of the horizontal layer about the neutral axis,
= second moment of area (moment of inertia) of the whole cross-section about the neutral axis,
= width of the cross-section at the level of the layer,
= distance from the neutral axis to the layer being considered.

This formula assumes elastic behaviour, small deformations and that plane sections remain plane.

Apply to a rectangle
Consider a rectangle of width (into the page) and total depth . Place the neutral axis at (mid-depth). For a layer at coordinate (measured from neutral axis), take the area above that layer as the area whose first moment we compute.

Area of portion above layer: .
Centroid of that portion from neutral axis: .
So the first moment:

Moment of inertia of the whole rectangle about neutral axis:

Width at the layer is constant: .

Substitute into the shear formula:

Replace by and simplify to the standard form:

Distribution and maximum value
This expression is a parabola in . It gives:

At neutral axis :

At extreme fibers :

So the shear stress is maximum at the centroidal plane and decreases to zero at the top and bottom surfaces. The average shear stress over the whole cross-section is ; thus the maximum is 1.5 times the average:

Interpretation and design use

The parabolic shape means shear is concentrated near the neutral axis. For rectangular members, the most critical location for shear checks is at mid-depth.
In thin-webbed sections (I, T, box), the web carries most shear; flange regions carry less. For those shapes a different and local width must be used; often designers check web shear with using the web thickness as .
The formula assumes elastic behavior and neglects shear deformation effects important in short, deep beams; if shear deformations are significant, Timoshenko beam theory should be used for deflection, but the stress distribution approach above still guides shear capacity checks.

Units and practical calculation

in N, and in m gives in N/m² (Pa).
Example: rectangular beam m, m, shear kN:

Limitations and practical notes

The derived is valid for uniform, prismatic, elastic rectangular sections under transverse shear.
For composite or non-prismatic sections, compute , , and local appropriately.
Shear yielding and web buckling are separate failure modes; shear check may require comparing with allowable shear strength or checking shear buckling for thin webs.

Conclusion

For a rectangular cross-section carrying shear force , the shear stress varies parabolically with depth and reaches its maximum at the neutral axis. The closed form result,

is widely used in beam design to check shear capacity. It is 1.5 times the average shear , zero at the extreme fibers, and should be compared with allowable shear strength or used to size section dimensions.