12.3 Combined loading in beams

Combined Bending and Shear Loading

Simultaneous Application of Bending Moments and Shear Forces

In real structures, beams rarely experience pure bending or pure shear alone. Combined loading refers to the simultaneous presence of bending moments and shear forces at a given cross-section. This means both normal stresses (from bending) and shear stresses (from transverse shear) act on the same cross-section at the same time.

Combined loading shows up in all common beam configurations: cantilevers, simply supported beams, and continuous beams. Consider a simply supported beam carrying a distributed load along its length with a concentrated load at midspan. At most cross-sections along that beam, both a nonzero bending moment and a nonzero shear force exist simultaneously.

Stress Distributions in Beams under Combined Loading

The two stress components follow distinctly different distribution patterns across the cross-section:

Normal stress varies linearly with distance from the neutral axis. Bending causes compression on one side and tension on the other. The maximum normal stress occurs at the extreme fibers (top and bottom), and it equals zero at the neutral axis.

Shear stress varies parabolically across the depth of the cross-section (for common shapes like rectangles and wide-flange sections). Shear stress is zero at the top and bottom free surfaces and reaches its maximum at the neutral axis.

This creates an important tradeoff across the cross-section:

• At the extreme fibers: maximum normal stress, zero shear stress
• At the neutral axis: zero normal stress, maximum shear stress
• At points in between: some combination of both normal and shear stress

The resultant stress state at any point is the combination of the normal and shear components acting there. Because the two distributions peak at different locations, the most critical point for failure depends on the failure criterion being used and the relative magnitudes of bending moment and shear force.

Principal Stresses in Beams

Concept of Principal Stresses

At any point in a beam under combined loading, you can find special orientations where the stress state is purely normal, with no shear component. The normal stresses on these planes are called principal stresses, and the planes themselves are called principal planes.

Two key properties of principal planes:

• Shear stress is zero on every principal plane
• The two principal planes at any point are always perpendicular to each other

The principal stresses represent the absolute maximum tensile and maximum compressive normal stresses at that point. This makes them essential for checking failure criteria, since most materials fail based on maximum normal stress or a combination of principal stresses.

Calculation of Principal Stresses

For a 2D (plane stress) state with normal stresses $\sigma_x$ and $\sigma_y$ and shear stress $\tau_{xy}$, the principal stresses are:

$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

where:

• $\sigma_x$ = normal stress in the x-direction (typically the bending stress $\sigma = My/I$)
• $\sigma_y$ = normal stress in the y-direction (often zero for a typical beam element)
• $\tau_{xy}$ = shear stress on the element (from transverse shear, $\tau = VQ/It$)

The orientation of the principal planes relative to the original axes is found from:

$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}$

This equation gives two values of $\theta_p$ separated by 90°, corresponding to the two principal planes.

For a typical beam element where $\sigma_y = 0$, the formulas simplify to:

$\sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2}$

Steps to find principal stresses at a point in a beam:

1. Calculate the bending stress: $\sigma_x = My/I$
2. Calculate the shear stress: $\tau_{xy} = VQ/It$
3. Set $\sigma_y = 0$ (unless another loading produces a vertical normal stress)
4. Substitute into the principal stress formula
5. Find $\theta_p$ to determine the orientation of the principal planes

Maximum Stresses in Beams

Maximum Normal Stress

The maximum normal stress from bending alone occurs at the extreme fibers and is calculated with the flexure formula:

$\sigma = \frac{My}{I}$

• $M$ = bending moment at the section
• $y$ = distance from the neutral axis to the point of interest (use $y = c$ for the extreme fiber)
• $I$ = second moment of area (moment of inertia) of the cross-section about the neutral axis

At the extreme fibers, shear stress is zero, so the bending stress is the principal stress there. This location often governs design for beams where bending dominates.

Maximum Shear Stress

The maximum transverse shear stress occurs at the neutral axis and is calculated with the shear stress formula:

$\tau = \frac{VQ}{It}$

• $V$ = internal shear force at the section
• $Q$ = first moment of area of the portion of the cross-section above (or below) the point of interest, taken about the neutral axis
• $I$ = second moment of area of the entire cross-section
• $t$ = width of the cross-section at the point where shear stress is being calculated

At the neutral axis, normal bending stress is zero, so the stress state is pure shear. The principal stresses there turn out to be equal in magnitude to $\tau$ but oriented at 45° to the beam axis (one tensile, one compressive). This is why you sometimes see diagonal cracking near the neutral axis in concrete beams: the concrete fails in tension along those 45° principal planes.

Normal vs. Shear Stress Interaction

Simultaneous Action of Normal and Shear Stresses

At most points in a beam cross-section (anywhere between the extreme fibers and the neutral axis), both normal and shear stresses act simultaneously. This combined stress state means the principal stresses are not aligned with the beam's longitudinal and transverse axes. Instead, the principal planes are rotated by the angle $\theta_p$, which depends on the ratio of shear stress to normal stress at that point.

The presence of shear stress always increases the magnitude of the maximum principal stress beyond what the normal stress alone would produce. Even a modest shear stress combined with a normal stress shifts the critical plane to an oblique orientation.

Mohr's Circle Representation

Mohr's circle provides a graphical way to visualize the complete stress state at a point and find principal stresses, maximum shear stress, and their orientations.

How to construct Mohr's circle for a beam element:

1. Plot point A at $(\sigma_x, \tau_{xy})$ representing the stress state on the x-face. (Convention: plot shear that would cause clockwise rotation as positive upward.)
2. Plot point B at $(\sigma_y, -\tau_{xy})$ representing the stress state on the y-face.
3. Connect A and B. The midpoint of this line is the center of the circle, located at $\left(\frac{\sigma_x + \sigma_y}{2},\ 0\right)$.
4. Draw the circle through points A and B.

Reading results from the circle:

• Principal stresses $\sigma_1$ and $\sigma_2$: where the circle crosses the horizontal axis (shear = 0)
• Maximum in-plane shear stress: the radius of the circle, equal to $\sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$
• Principal angle $\theta_p$: half the angle measured on the circle from point A to the $\sigma_1$ intersection, with angles on the circle being twice the physical rotation angle

Mohr's circle is especially useful for building intuition. You can see directly how adding shear stress (increasing the radius) increases the difference between $\sigma_1$ and $\sigma_2$, and how the principal planes rotate as the ratio of shear to normal stress changes. For beam design, this helps you identify which points in the cross-section are most vulnerable and what type of stress (tension, compression, or shear) governs failure at each location.