Stress at a point

Cut any loaded member with an imaginary plane and the two halves must still hold each other up — internal forces spread across that cut, and their intensity per unit area is stress. It is the single idea that turns "how big a force" into "how big a force for this member's size," which is what actually decides whether something breaks. This node opens the mechanics-of-materials half of ME10001's "Properties of Areas and Axial Loading" unit, following directly on cross-section geometry (moments of inertia). You will define average normal stress $\sigma = P/A$ and average shear stress $\tau = V/A$ on a stated plane, compute both for axially and transversely loaded members, and connect stress to material strength through ultimate stress, allowable stress, and factor of safety — the vocabulary every later strength calculation in the course builds on.

The core insight

Stress is internal force per unit area on a cut through a loaded body — normal stress $\sigma = P/A$ acts perpendicular to the cut, shear stress $\tau = V/A$ acts parallel to it, and a member is safe when its stress stays below an allowable stress set by dividing the material's ultimate stress by a chosen factor of safety.

The core idea

When an external force acts on a body, internal forces develop across any imaginary section cut through the material — this is how a free-body diagram of "half the member" stays in equilibrium at all. Stress is the intensity of that internal force: force per unit area at a point on the cut. For a straight, prismatic member carrying an axial force $P$ spread uniformly over a cross-sectional area $A$ (a good approximation away from points of load application, per Saint-Venant's principle), the average normal stress is

\sigma = \frac{P}{A}

where: $\sigma$ is normal stress (Pa), $P$ is the internal axial force transmitted across the cut (N), and $A$ is the cut's cross-sectional area (m²). Normal stress acts perpendicular to the cut surface; by convention $\sigma > 0$ is tension (the material pulled apart) and $\sigma < 0$ is compression. If the internal force instead acts parallel to the cut — a bolt loaded transversely, or the bonded faces of a lap joint — it produces average shear stress:

\tau = \frac{V}{A}

where: $\tau$ is shear stress (Pa), $V$ is the shear force transmitted across the plane (N), and $A$ is the area of that plane. Both $\sigma$ and $\tau$ carry SI units of pascals (Pa = N/m²); engineering members usually report stress in MPa (1 MPa = 10⁶ Pa = 1 N/mm²). These are average values over the section — true point-wise stress can vary across the area, especially near holes or load-application points. "Stress at a point" is, strictly, the limit of force over area as the sampled area shrinks toward that point; for uniform axial or shear loading, the average value is used directly as the stress everywhere on the section.

Figure

A bar cut by an imaginary plane showing the internal normal force spread as uniform stress arrows over the cross section and a separate bolt cut by a shear plane showing the internal shear force spread as parallel stress arrows

Once a stress is known it is compared against the material's strength. The ultimate stress $\sigma_{ult}$ is the stress at which a material fails, measured in a tension test. Engineers never design up to $\sigma_{ult}$ ; instead they specify an allowable (working) stress $\sigma_{allow}$ , related through the factor of safety:

FS = \frac{\sigma_{ult}}{\sigma_{allow}}

where: $FS$ is dimensionless and always greater than one, $\sigma_{ult}$ is the material's failure stress, and $\sigma_{allow}$ is the maximum stress the design permits. Its value is chosen based on uncertainty in loading, material variability, and the consequences of failure. A member is safe if the actual computed stress does not exceed $\sigma_{allow}$ .

Derivation ASizing a member from a target factor of safety

The goal is to turn a required factor of safety and a material's known ultimate stress into the minimum cross-sectional area — and, for a circular rod, the minimum diameter — that safely carries a specified axial load, using nothing beyond the two definitions $\sigma = P/A$ and $FS = \sigma_{ult}/\sigma_{allow}$ rearranged and chained together.

Step 1. Fix the allowable stress from the target factor of safety. Rearranging $FS = \sigma_{ult}/\sigma_{allow}$ gives $\sigma_{allow} = \sigma_{ult}/FS$ . This is the ceiling the design must respect: the material can sustain $\sigma_{ult}$ before failing, but the design only permits stress up to $\sigma_{ult}$ divided by the safety margin.

Step 2. Convert the stress ceiling into an area requirement. Safety demands $\sigma = P/A \le \sigma_{allow}$ , so the smallest acceptable area is the one that makes this an equality: $A_{min} = \dfrac{P}{\sigma_{allow}} = \dfrac{P \cdot FS}{\sigma_{ult}}$ . Any area at or above $A_{min}$ keeps the actual stress at or below the allowable stress.

Step 3. Specialize to a circular cross-section. Since $A = \pi d^2/4$ , solving for the diameter that produces exactly $A_{min}$ gives $d_{min} = \sqrt{\dfrac{4A_{min}}{\pi}}$ . This is the smallest rod diameter that meets the target factor of safety for the given load and material.

Step 4. Apply it to the rod material from the worked example below: $\sigma_{ult} = 400$ MPa, $P = 40$ kN, and suppose a target $FS = 2.5$ is now specified instead of simply checking a fixed 16 mm rod. Then $\sigma_{allow} = 400/2.5 = 160$ MPa, $A_{min} = (40\times10^3)/(160\times10^6) = 2.5\times10^{-4}\text{ m}^2$ , and $d_{min} = \sqrt{4(2.5\times10^{-4})/\pi} = 0.01784$ m $= 17.8$ mm. A 16 mm rod (area $2.011\times10^{-4}\text{ m}^2$ , computed below) is therefore too small for $FS = 2.5$ ; it only delivers $FS = 2.01$ , as the worked example shows next.

Worked example

A steel rod of circular cross-section, 16 mm diameter, carries an axial tensile load of $P = 40$ kN, and the steel has ultimate tensile stress $\sigma_{ult} = 400$ MPa. Find the average normal stress and the resulting factor of safety, then check a bolted lap joint that must carry the load onward.

Step 1. Cross-sectional area of the rod: $A = \dfrac{\pi d^2}{4} = \dfrac{\pi(0.016)^2}{4} = \dfrac{\pi\times2.56\times10^{-4}}{4} = 2.011\times10^{-4}\text{ m}^2$ .

Step 2. Normal stress: $\sigma = \dfrac{P}{A} = \dfrac{40\times10^3}{2.011\times10^{-4}} = 1.989\times10^{8}\text{ Pa} = 198.9$ MPa.

Step 3. Factor of safety: $FS = \dfrac{\sigma_{ult}}{\sigma} = \dfrac{400}{198.9} = 2.01$ . The rod runs at roughly half of $\sigma_{ult}$ , a typical margin for a ductile steel member.

Step 4. Now the same 40 kN must pass through a bolted lap joint using a single 12 mm bolt in single shear (the shear plane is the bolt's cross-section), but the load there is a transmitted shear force $V = 25$ kN (only part of the total load routes through this particular bolt). Shear area: $A = \dfrac{\pi(0.012)^2}{4} = 1.131\times10^{-4}\text{ m}^2$ . Shear stress: $\tau = \dfrac{V}{A} = \dfrac{25\times10^3}{1.131\times10^{-4}} = 2.211\times10^{8}\text{ Pa} = 221.1$ MPa.

Step 5. With an allowable shear stress $\tau_{allow} = 120$ MPa, $\tau = 221.1$ MPa exceeds it, so the single-shear bolt is not safe. Switching to a double-shear connection (two shear planes carrying the same $V$ ) doubles the shear area to $2.262\times10^{-4}\text{ m}^2$ , giving $\tau = 25\times10^3/2.262\times10^{-4} = 1.105\times10^{8}\text{ Pa} = 110.5$ MPa — now below $\tau_{allow}$ , so the double-shear joint is safe.

Final answer: the 16 mm rod has $\sigma = 198.9$ MPa and $FS = 2.01$ ; the bolt must be arranged in double shear ( $\tau = 110.5$ MPa) rather than single shear ( $\tau = 221.1$ MPa) to be safe.

Common traps

Using the nominal or original cross-sectional area for $\sigma = P/A$ even when the member has a hole or reduced (net) section at the cut of interest — always use the actual area at the plane being checked.
Confusing single shear (one shear plane, as in a simple lap joint) with double shear (two shear planes) — double shear doubles the area and halves the stress for the same load; forgetting this is the single most common bolt-sizing error.
Treating $\sigma = P/A$ as the exact stress at every point rather than an average — near load-application points, fillets, or holes the local stress is higher (stress concentration), a topic outside this course's scope but worth knowing exists.
Swapping $\sigma_{ult}$ and $\sigma_{allow}$ in the factor-of-safety formula, or quoting $FS$ with units attached — it is a pure, dimensionless ratio.

Ask the tutor: if you're unsure whether a bolted or riveted joint is in single or double shear, or how to size a member from a target factor of safety, ask the AI tutor to walk through the free-body diagram with you.

Why is stress used instead of just comparing forces directly?

Because whether a member fails depends on force relative to its size. A thick rod and a thin rod can carry very different loads before breaking even though both are "steel," because the thin rod concentrates the same force into less area. Stress, force divided by area, normalizes for size so that a material property like ultimate stress can be compared directly against the computed stress in any member, regardless of its cross-section.

Is stress really uniform across the whole cross-section?

Only approximately, and only away from complications. Saint-Venant's principle says that far enough from where a load is actually applied, or from a hole or sudden change in shape, the stress distribution smooths out to the uniform average $\sigma = P/A$ . Near those disruptions the true local stress can be significantly higher (stress concentration). This course treats the average value as the working answer; the localized picture comes later.

Why do engineers use an allowable stress instead of just staying under the ultimate stress?

Because the ultimate stress is where the material actually breaks, and no sensible design operates right at the edge of failure. Loads are estimated with some uncertainty, material properties vary from sample to sample, and the consequences of failure differ by application. The factor of safety builds in a deliberate margin: $\sigma_{allow} = \sigma_{ult}/FS$ keeps the design stress well below the breaking point, with the margin sized to how much uncertainty and risk the situation carries.

Stress at a point

The core idea

Worked example

Common traps

Before this

Where this leads