Stress

A tension of 128 kN in a bridge chord is not yet a statement about safety, because the same 128 kN would part a wire rope and would be nothing at all in a metre of solid steel. Everything so far has produced internal forces; this lesson turns them into the quantity a material property is written in.

Force per unit area

Cut a bar of uniform cross section $A$ carrying an axial force $N$ , and the cut face has to transmit that force. If the force is shared evenly over the face, the intensity of the transmission is

σ = \frac{N}{A}

which is the normal stress, normal because it acts perpendicular to the cut face. Its unit is the pascal, one newton per square metre, and the pascal is uselessly small for structures: atmospheric pressure is about $10^{5}$ Pa and steel yields somewhere near $2.5 \times 10^{8}$ Pa. So structural work is done in megapascals, and there is an identity that makes the arithmetic almost free:

1 MPa = 1 {N mm}^{- 2}

Newtons and millimetres are the units drawings are dimensioned in, so a force in newtons divided by an area in square millimetres comes out directly in megapascals with no conversion at all. This is worth adopting as a habit rather than a trick.

Tension is positive and compression negative, carrying over the convention from the truss lesson. A bar in compression has exactly the same $σ = N / A$ with a negative sign, and everything in this lesson applies to it, with one large caveat about slender members that the buckling lesson deals with.

Example. The midspan bottom chord of the truss analysed two lessons ago carries $128$ kN of tension. It is made from a flat steel bar $25$ mm by $50$ mm. What is the stress, and how does it compare with the yield strength of ordinary structural steel, $275$ MPa?

The area is $25 \times 50 = 1250$ mm², so $σ = 128 000 / 1250 = 102.4$ MPa, about $37$ per cent of yield. That is a sensible working figure for a member sized with a margin.

Now you. The same truss has an end diagonal carrying $100$ kN of compression. Suppose it is a solid circular rod of $40$ mm diameter. What is the stress?

Answer

The area is $π \times 20^{2} = 1257$ mm², so $σ = 100 000 / 1257 = 79.6$ MPa of compression.

Where the uniform assumption comes from

Dividing by the area assumes the force spreads itself evenly, and near the ends of a bar it does not. Grip a strip in a testing machine and the stress just under the jaws is concentrated wherever the grip bites, varying wildly across the width.

Saint-Venant's principle, stated by Adhémar Barré de Saint-Venant in 1855, is the rescue: two different load distributions with the same resultant force and moment produce essentially the same stress field at distances greater than about the width of the member. A strip $50$ mm wide loaded through a pin, a weld, or a clamp all look identical $50$ mm in from the end. So the uniform formula is a statement about the middle of a member, and the ends are detailing problems solved by rules of thumb, testing and generous local material.

The principle is not a theorem and it fails where the geometry changes abruptly, which is the subject of the last section here.

Shear stress, and where it is the governing case

If the force on the cut face lies in the plane of the face rather than perpendicular to it, the intensity is a shear stress

τ = \frac{V}{A}

again as an average over the face. Shear is what a bolt, a rivet, a weld, a pin and a glue line carry, and connections fail in shear far more often than members fail in tension.

The count of shear planes matters and is easy to get wrong. A bolt joining two overlapping plates has one plane of material sliding over another and is in single shear, so the whole force crosses one bolt cross section. A bolt through a central plate sandwiched between two outer plates is in double shear, with two cross sections sharing the load, so the stress halves for the same force.

Example. A $20$ mm diameter bolt in double shear transmits $60$ kN. Find the shear stress.

The bolt has area $π \times 10^{2} = 314.2$ mm² per plane, and two planes share the load, so the effective area is $628.3$ mm² and $τ = 60 000 / 628.3 = 95.5$ MPa.

Now you. A $16$ mm bolt in single shear transmits $25$ kN. Find the shear stress.

Answer

The area is $π \times 8^{2} = 201.1$ mm² on the single plane, so $τ = 25 000 / 201.1 = 124.3$ MPa.

A third kind appears at the same connection. The bolt presses sideways on the hole it sits in, and the plate can be crushed or torn out even if the bolt itself is fine. The bearing stress is conventionally taken as the force divided by the projected area of contact, the bolt diameter times the plate thickness, which for the $20$ mm bolt above through a $12$ mm plate gives $60 000 / (20 \times 12) = 250$ MPa. That is a large number by design: the convention deliberately understates the true contact area, and the allowable bearing stress is correspondingly set high, often above the material's yield stress, because a little local crushing at a hole is harmless.

A pressure vessel, cut two ways

The neatest use of stress as a bookkeeping quantity is a thin-walled cylinder under internal pressure, because equilibrium alone gives the answer with no material properties involved at all. Take a cylinder of internal radius $r$ and wall thickness $t$ , with $t$ small compared with $r$ , under gauge pressure $p$ .

Cut it lengthways along a diameter and take half the cylinder over a length $L$ as the free body. The pressure acts on the projected area $2 r L$ , pushing the half shell outwards with a force $2 p r L$ . The only thing resisting is the wall, cut in two places, each of area $t L$ , carrying a circumferential or hoop stress $σ_{θ}$ . Equilibrium gives $2 σ_{θ} t L = 2 p r L$ , so

σ_{θ} = \frac{p r}{t}

Now cut it across instead, perpendicular to the axis. The pressure acts on the end area $π r^{2}$ with force $p π r^{2}$ , and the wall resists over a ring of area $2 π r t$ , carrying a longitudinal stress $σ_{z}$ . So $σ_{z} 2 π r t = p π r^{2}$ and

σ_{z} = \frac{p r}{2 t}

The hoop stress is twice the longitudinal stress, which is why a cylindrical vessel that bursts splits along a line parallel to its axis rather than snapping in two, and why a sausage splits lengthways when it is grilled. Both stresses are derived from equilibrium only, so they hold for steel, aluminium, carbon fibre and cardboard alike.

Example. A compressed air receiver has an internal diameter of $1.5$ m, a wall thickness of $12$ mm, and works at $1.8$ MPa gauge. Find both stresses and check the thin-wall assumption.

The radius is $750$ mm, so $r / t = 62.5$ , comfortably above the usual threshold of $10$ for calling a shell thin. Then $σ_{θ} = 1.8 \times 750 / 12 = 112.5$ MPa and $σ_{z} = 56.25$ MPa. Against a yield of $275$ MPa the hoop stress carries a factor of about $2.4$ , which is on the low side for a pressure vessel and is why real codes also demand a corrosion allowance and a weld efficiency factor.

Now you. A pipeline of $600$ mm internal diameter and $8$ mm wall runs at $2.5$ MPa. Find the hoop and longitudinal stresses.

Answer

$r = 300$ mm and $r / t = 37.5$ , so the shell is thin. $σ_{θ} = 2.5 \times 300 / 8 = 93.75$ MPa and $σ_{z} = 46.875$ MPa.

Stress concentration

Saint-Venant's principle promises uniformity away from the ends. It says nothing about a hole, a notch, a sharp internal corner or an abrupt change of width, and near those the stress can be several times the average.

The reference case has an exact elastic solution, published by Gustav Kirsch in 1898: a small circular hole in a wide plate under uniform tension $σ$ produces a stress of $3 σ$ at the two edges of the hole on the diameter perpendicular to the load. The stress concentration factor $K_{t} = 3$ is independent of the size of the hole, which is the counterintuitive part. A pinhole raises the local stress by the same factor as a large one, provided the plate is wide compared with the hole.

Two consequences follow, in opposite directions.

For a ductile material loaded steadily, the concentration matters less than it looks. The material at the hole yields locally, stops taking more stress, and passes the excess to its neighbours, so a mild steel plate with a hole fails at close to the load its net section predicts. This is precisely why bolted steelwork is possible at all.

For a brittle material, or for any material under fluctuating load, it matters enormously. Cast iron and glass have no yielding mechanism to blunt the peak. Fatigue cracks start at concentrations essentially always, which is why the design rule is to round every internal corner generously: the factor for a sharp corner rises without bound as the radius goes to zero. The square windows of the de Havilland Comet, whose corners cracked in 1954 after repeated pressurisation, are the standard illustration, and the aircraft that followed have rounded windows for exactly this reason.

Example. A steel plate $100$ mm wide and $10$ mm thick carries $40$ kN in tension and has a $20$ mm hole drilled on its centre line. Estimate the net section stress, and say what the peak stress near the hole would be if $K_{t}$ were $3$ .

The net area at the hole is $(100 - 20) \times 10 = 800$ mm², so the net section stress is $40 000 / 800 = 50$ MPa. A factor of three on that would give $150$ MPa. In practice the hole is a fifth of the width rather than tiny, so the true factor is nearer $2.5$ and the peak nearer $125$ MPa, but either way the plate is comfortable in mild steel and would be a serious concern in cast iron.

Now you. The same plate carries $60$ kN instead. What is the net section stress, and does the plate reach a yield of $275$ MPa at the hole under a factor of $2.5$ ?

Answer

The net section stress is $60 000 / 800 = 75$ MPa, and $2.5 \times 75 = 187.5$ MPa, still below $275$ MPa. Local yielding would not begin, though under a fluctuating load the hole would still be where a crack eventually started.

What one number cannot say

Two limits are worth stating before moving on, and both will be taken up later.

The first is that stress at a point is not one number. A cut through the same point at a different angle exposes a different combination of normal and shear stress, and a bar in pure tension of $100$ MPa has a shear stress of $50$ MPa on planes at $45$ degrees to its axis, which is why a ductile tensile specimen fails on a slanted surface rather than a square one. Describing the state of stress at a point properly needs three numbers in a plane, and the lesson on combined stress does it.

The second is that stress alone cannot solve a structure. Nothing in this lesson has used a material property, which is exactly why the results are so general and also why they cannot supply the missing equations for the indeterminate structures set aside in the first lesson. For that, the material has to be allowed to deform, and how much it deforms under a given stress is a measured property of the substance. That measurement is the next lesson.