Bending, Shear and Combined Stresses

Combined axial and bending stress

Combined axial and bending stress arises when a member carries an axial force together with a bending moment, usually because the load is applied off the centroidal axis or because a transverse load acts on the same section. By superposition, the normal stress at any fibre is the sum of a uniform direct stress and a linearly varying bending stress: \sigma = \frac{P}{A} + \frac{My}{I}, where y is measured from the centroidal axis and tension is taken as positive. An eccentric load P at offset e from the centroid is statically equivalent to a central load plus a moment M = Pe, so the extreme stress is \sigma_{max} = \frac{P}{A}\left(1 + \frac{ec}{r^2}\right), where r^2 = I/A. For a solid circular section of diameter d, r^2 = d^2/16, and the outer-fibre stress becomes \frac{P}{A}\left(1 + \frac{8e}{d}\right). An offset equal to the diameter therefore multiplies the direct stress by nine.

For a rectangular section, the two edge stresses are \frac{P}{A}\left(1 \pm \frac{6e}{h}\right), where h is the depth in the direction of the eccentricity. Since 6e/h \le 1 exactly when e \le h/6, no tension develops for a load inside the middle third, and the far face remains in compression. For example, a column of depth 60 mm with a compressive load at 5 mm offset has extreme stresses of 1.5\,P/A and 0.5\,P/A, both compressive.

A linear term in the stress distribution does not change the axial force on a section that is symmetric about its centroid. For \sigma = \alpha + \beta y, the resultant force is \int \sigma\,dA = \alpha A, because \int y\,dA = 0. The moment about the centroidal axis is \int \sigma y\,dA = \beta I, because \int y^2\,dA = I. The constant part therefore gives the axial force and the linear part gives the bending moment.

When the axial force Q passes through the centroid while a bending moment M also acts, the axial stress is uniform and the bending stress is linear. The neutral axis is the line where the total stress is zero. Its offset from the centroid is y_n = -\frac{QI}{MA}, so the shift depends only on the ratio of axial force to moment and on the section ratio I/A. For a circular section, I/A = d^2/16, which is a convenient shortcut in calculation.

Shear stress in rectangular section

The transverse shear stress in a beam is given by the shear formula \tau = \frac{VQ}{Ib}, where V is the shear force, Q = \int y\,dA is the first moment of the area above (or below) the level where the stress is sought, I is the second moment of area about the neutral axis, and b is the width of the section at that level. For a rectangle of width b and depth h, Q = \frac{b}{2}\left(\frac{h^2}{4} - y^2\right), which gives a parabolic distribution \tau = \frac{3V}{2A}\left(1 - \frac{4y^2}{h^2}\right) with A = bh. The shear stress is therefore zero at the top and bottom fibres, which are free surfaces, and maximum at the neutral axis with value \tau_{max} = \frac{3V}{2A}.

Because the average shear stress is V/A, the ratio of maximum to average shear stress for a rectangular section is \frac{3}{2} = 1.50. This is a standard result worth memorizing for quick checks. For comparison, a solid circular section has a ratio of 4/3, so the two values should not be confused.

The shear stress at a point depends on the local shear force, so it vanishes wherever V = 0. In a symmetric simply supported beam under symmetric loading, the shear force at midspan is zero, so the shear stress there is zero. Likewise, the bending normal stress is zero on the neutral axis. A point on the neutral axis at a section where the shear force is zero therefore carries neither normal nor shear stress.