Short Answer:
Shear stress distribution in a circular beam describes how internal tangential stress varies across a circular cross-section when the beam carries a transverse shear force . For a solid circular section of radius the shear stress varies parabolically with depth, is zero at the outer surface and reaches its maximum at the neutral axis (mid-depth). The general expression is
where is the distance from the centroid (neutral axis).
The maximum shear stress occurs at (neutral axis) and equals , which is times the average shear . This result follows from the general shear formula applied to the circular geometry.
Detailed Explanation :
Shear Stress Distribution in Circular Beam
Overview
When a beam section carries a transverse shear force , internal shear stresses develop to resist the tendency of one part of the section to slide relative to another. The local shear stress at a horizontal layer depends on the shape of the cross-section. For a solid circular section (radius , area ), the shear stress distribution is smooth and symmetric about the neutral axis; it reaches its maximum at the centroidal plane and falls to zero at the outer surface.
Starting formula
The standard relation from elementary beam theory for shear stress at a layer a distance from the neutral axis is
where
- = internal shear force at the section,
- = first moment of the area of the portion of the cross-section on one side of the layer about the neutral axis,
- = second moment of area (moment of inertia) of the whole cross-section about the neutral axis,
- = width (chord length) of the cross-section at that layer (length of the slice parallel to the neutral axis),
- = vertical coordinate measured from the neutral axis (centroidal axis).
Geometry and expressions for a circle
For a solid circle of radius with centroid at :
- Local chord width at level :
- Moment of inertia about centroidal axis:
- First moment : choose the area above the horizontal layer (from to ). Its area is
and the first moment about the neutral axis is
Evaluate the integral by substitution , . This gives
(As a check: at , , which equals the first moment of the top half-area .)
Substitute into shear formula
Now substitute , and into :
Factor to get the common normalized form:
Key features of the distribution
- Parabolic variation in : is a quadratic function of (maximum at , zero at ).
- Maximum shear at neutral axis:
- Shear at surface zero: .
- Relation to average shear: average shear . Thus (i.e. 1.333… times the average).
Practical interpretation and use
- For design checks in beams with circular cross-sections (shafts, round bars, piles), compare with the allowable shear strength.
- The fact that shear concentrates near the centroid means the central region (not the outer fibers) controls shear failure for solid circular members.
- For hollow circular tubes, the same procedure applies but with modified , and ; typically the web area (annulus) distribution changes the numeric factor.
- The derivation assumes elastic behavior, small deformations, and plane sections remaining plane. For deep beams or materials with significant shear deformation, more advanced theories (e.g., Timoshenko beam theory) are needed for deflection—but the stress-distribution approach above remains the standard check for shear capacity.
Example (quick numeric check)
For a circular bar of radius mm carrying kN:
Conclusion
The shear stress distribution in a solid circular section is parabolic across the depth and given by . The maximum shear occurs at the neutral axis, and the stress falls to zero at the outer surface. This closed-form result is commonly used when checking shear capacity of round members.