5.2 Axial, Bending, and Torsional Stresses

Axial, Bending, and Torsional Stresses

Every mechanical component you design will experience some combination of axial, bending, and torsional loading. These three stress types describe how internal forces are distributed through a material when external loads are applied. Getting comfortable with calculating each one is fundamental to sizing components, selecting materials, and predicting failure.

This topic covers the formulas and reasoning behind each stress type, along with key geometric properties like moment of inertia and polar moment of inertia that show up repeatedly in design calculations.

Axial and Normal Stresses

Stress Fundamentals

Axial stress is the simplest type: a force applied along the longitudinal axis of a member, pulling it in tension or pushing it in compression. Because the force acts perpendicular (normal) to the cross-section, axial stress is a type of normal stress.

The calculation is straightforward:

$\sigma = \frac{F}{A}$

where $F$ is the applied force and $A$ is the cross-sectional area perpendicular to that force. This formula assumes the load is applied uniformly across the section and acts through the centroid, so there's no bending involved.

• Tensile stress stretches the member (positive by convention)
• Compressive stress shortens it (negative by convention)
• Units are pascals (Pa) in SI or pounds per square inch (psi) in US customary

One thing to watch: this formula gives you the average stress over the cross-section. Near holes, notches, or sudden geometry changes, local stresses can be significantly higher due to stress concentrations.

Stress-Strain Relationship

When you apply stress to a material, it deforms. Strain quantifies that deformation as a dimensionless ratio:

$\epsilon = \frac{\Delta L}{L}$

where $\Delta L$ is the change in length and $L$ is the original length.

Within the elastic region of a material, stress and strain are linearly related by Hooke's Law:

$\sigma = E\epsilon$

The constant $E$ is the modulus of elasticity (Young's modulus), which measures a material's stiffness. Steel, for example, has $E \approx 200 \text{ GPa}$, while aluminum is around $E \approx 70 \text{ GPa}$. A higher $E$ means less deformation for the same applied stress.

Hooke's Law only holds up to the proportional limit. Beyond that, the material begins to yield and deform permanently, and the linear relationship breaks down.

Bending Stresses

Bending Stress Fundamentals

Bending stress develops when a moment is applied to a member, causing it to curve. Unlike axial stress, bending stress isn't uniform across the cross-section. It varies linearly from the neutral axis (where stress is zero) outward to the extreme fibers (where stress is maximum).

The flexure formula captures this:

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

• $M$ = internal bending moment at the section of interest
• $y$ = distance from the neutral axis to the point where you're calculating stress
• $I$ = moment of inertia of the cross-section about the neutral axis

On one side of the neutral axis, the material is in tension; on the other side, it's in compression. For a simply supported beam loaded from above, the top fibers compress and the bottom fibers stretch.

The maximum bending stress occurs at the point farthest from the neutral axis (the extreme fiber), where $y$ is largest.

Moment of Inertia and Section Modulus

The moment of inertia $I$ is a geometric property that tells you how the cross-sectional area is distributed relative to the neutral axis. Material farther from the neutral axis contributes more to bending resistance, which is why I-beams are so efficient: they concentrate material in the flanges, far from the center.

Formally:

$I = \int y^2 \, dA$

For common shapes, you'll use tabulated values rather than integrating each time. For a solid rectangle of width $b$ and height $h$: $I = \frac{bh^3}{12}$. For a solid circle of diameter $d$: $I = \frac{\pi d^4}{64}$.

The section modulus $S$ simplifies the maximum bending stress calculation by combining $I$ and the extreme fiber distance $c$:

$S = \frac{I}{c}$

This lets you write the maximum bending stress as:

$\sigma_{max} = \frac{M}{S}$

The section modulus is especially useful when comparing different cross-section shapes for beam design, since a larger $S$ means lower peak stress for the same moment.

Beam Deflection

Beyond stress, you also need to check that a beam doesn't deflect too much under load. Excessive deflection can cause functional problems (misalignment, vibration) even if the stress is well within limits.

Deflection depends on the load, the beam's length and cross-section, and the material stiffness. For a simply supported beam with a single concentrated load $P$ at midspan:

$\delta_{max} = \frac{PL^3}{48EI}$

• $P$ = applied load
• $L$ = beam span
• $E$ = modulus of elasticity
• $I$ = moment of inertia

Notice that deflection scales with the cube of the length. Doubling the span increases deflection eightfold, which is why long beams need much stiffer sections. Also note that both $E$ and $I$ appear in the denominator: you can reduce deflection by choosing a stiffer material (higher $E$) or a cross-section with a larger $I$.

Different loading and support conditions produce different deflection formulas, so always confirm which case applies before plugging in numbers.

Torsional and Shear Stresses

Torsional Stress Fundamentals

Torsional stress arises when a torque (twisting moment) is applied about the longitudinal axis of a member. Think of a drive shaft transmitting power from an engine to the wheels. The shear stress distribution in a solid circular cross-section varies linearly from zero at the center to a maximum at the outer surface:

$\tau = \frac{Tr}{J}$

• $T$ = applied torque
• $r$ = radial distance from the center of the cross-section
• $J$ = polar moment of inertia

The polar moment of inertia is the torsional equivalent of $I$ for bending. It describes how the cross-sectional area is distributed relative to the center:

$J = \int r^2 \, dA$

This linear stress distribution (zero at center, max at surface) applies specifically to circular cross-sections. Non-circular sections have more complex torsional behavior and require different analysis methods.

Shear Stress Fundamentals

Shear stress acts parallel to a surface rather than perpendicular to it. In beams, transverse shear stress develops from the internal shear force $V$ that acts on each cross-section.

The shear stress distribution across a beam's cross-section is parabolic, not linear. Maximum shear stress occurs at the neutral axis, and shear stress drops to zero at the top and bottom surfaces. This is the opposite pattern from bending stress, which is zero at the neutral axis and maximum at the extreme fibers.

A quick estimate uses the average shear stress:

$\tau_{avg} = \frac{V}{A}$

For a rectangular cross-section, the actual maximum shear stress at the neutral axis is 1.5 times the average:

$\tau_{max} = \frac{3V}{2A}$

In most beam design situations, bending stress governs over shear stress. But shear becomes critical in short, heavily loaded beams and near supports or concentrated loads.

Torsion of Circular Shafts

Circular shafts are the most common geometry for transmitting torque because their symmetric cross-section produces a straightforward, predictable stress distribution.

For a solid circular shaft of diameter $d$:

$J = \frac{\pi d^4}{32}$

The maximum shear stress occurs at the outer surface ($r = d/2$):

$\tau_{max} = \frac{T(d/2)}{J} = \frac{16T}{\pi d^3}$

For a hollow circular shaft with outer diameter $d_o$ and inner diameter $d_i$:

$J = \frac{\pi(d_o^4 - d_i^4)}{32}$

Hollow shafts are more material-efficient because the core of a solid shaft carries very little stress anyway. Removing that low-stress material saves weight with minimal loss in torsional strength.

The angle of twist tells you how much the shaft rotates under a given torque:

$\phi = \frac{TL}{GJ}$

• $L$ = shaft length
• $G$ = shear modulus (modulus of rigidity) of the material

The shear modulus $G$ is related to Young's modulus by $G = \frac{E}{2(1+\nu)}$, where $\nu$ is Poisson's ratio. For steel, $G \approx 80 \text{ GPa}$.

Just like beam deflection, the angle of twist is a serviceability check. Even if the stress is acceptable, too much twist in a shaft can cause vibration, misalignment, or timing errors in the system it drives.