Strain and stiffness

Statics ran out of equations in the first lesson, at the beam propped in three places, and stress in the previous lesson did nothing to help because it used no material property at all. The missing conditions are geometric: after the structure deforms, the parts still have to fit together. Writing that down needs a measure of deformation.

Strain

A bar of original length $L$ stretched by an amount $δ$ has normal strain

ε = \frac{δ}{L}

Strain is a ratio of lengths, so it is dimensionless, and structural strains are small: a steel member working at a sensible stress is strained by roughly one part in a thousand. Because the numbers are inconveniently small they are often quoted in microstrain, one part per million, so a strain of $0.000842$ is $842$ microstrain. Electrical resistance strain gauges, which are the standard way of measuring it on a real structure, read directly in those units.

Defining strain as the extension divided by the original length is the engineering convention and it is an approximation. For strains beyond a few per cent the length is changing while the stretching happens, and the honest definition integrates $d L / L$ to give true strain $\ln (L / L_{0})$ . At structural strains the two agree to better than a tenth of a per cent, so this course uses the engineering definition and the next lesson says where it breaks.

Shear has its own measure. A block whose top face slides sideways relative to its bottom face by $Δ$ over a height $h$ has shear strain $γ = Δ / h$ , which for small distortions is the angle in radians by which a right angle in the material has closed.

The tensile test

Everything about a material's response comes from one experiment. A round bar of known diameter is pulled in a machine that records force and extension, over a marked gauge length so the ends are excluded and Saint-Venant's principle applies. Dividing force by original area and extension by gauge length gives a stress-strain curve, and because both axes are normalised, the curve is a property of the material rather than of the specimen.

The first part of the curve is straight, and its slope is the material's stiffness:

E = \frac{σ}{ε}

called Young's modulus. Robert Hooke published the proportionality in 1676 as an anagram and unscrambled it two years later as ut tensio, sic vis, as the extension so the force; Thomas Young turned it into a material constant in 1807 by dividing out the geometry. Because strain is dimensionless, $E$ has the units of stress and is huge: it is the stress that would double the length of a bar if the material stayed linear that far, which no metal does.

The numbers are worth memorising, because they are the ones that decide how a structure behaves rather than how strong it is. Steel is $210$ GPa, and remarkably it is $210$ GPa whether the steel is soft mild steel or a high strength alloy: heat treatment changes the strength enormously and the stiffness hardly at all. Aluminium is $70$ GPa, one third of steel, which is close to the ratio of their densities too, and a fact the last lesson of this course makes a great deal of. Titanium is $110$ , glass about $70$ , ordinary concrete around $30$ , timber along the grain around $10$ , nylon about $3$ , and natural rubber about $0.01$ , some twenty thousand times less stiff than steel.

Example. A steel rod of $12$ mm diameter and $2.5$ m length carries $20$ kN. Find the stress, the strain and the extension, taking $E = 210$ GPa.

The area is $π \times 12^{2} / 4 = 113.1$ mm², so $σ = 20 000 / 113.1 = 176.8$ MPa. Then $ε = σ / E = 176.8 / 210 000 = 8.42 \times 10^{- 4}$ , or $842$ microstrain. The extension is $ε L = 8.42 \times 10^{- 4} \times 2500 = 2.11$ mm.

Now you. An aluminium rod of $16$ mm diameter and $3$ m length carries $30$ kN, with $E = 70$ GPa. Find the stress and the extension.

Answer

The area is $201.1$ mm², so $σ = 30 000 / 201.1 = 149.2$ MPa. The strain is $149.2 / 70 000 = 2.13 \times 10^{- 3}$ and the extension is $2.13 \times 10^{- 3} \times 3000 = 6.39$ mm, three times what a steel rod of the same stress would give.

Combining the definitions in one step gives the formula that does most of the work in this course:

δ = \frac{N L}{A E}

Read as a spring, this says the axial stiffness of a bar is $k = A E / L$ newtons per millimetre. A structure is a set of springs, and that is what makes the indeterminate cases solvable.

Poisson's ratio, and strain energy

Stretch a bar and it also gets thinner. The lateral strain is proportional to the axial one, and the constant is Poisson's ratio

ν = - \frac{ε_{lateral}}{ε_{axial}}

defined with a minus sign so that it comes out positive. For most metals $ν$ is close to $0.3$ , for concrete about $0.2$ , for rubber almost exactly $0.5$ , and for cork nearly zero, which is why a cork can be pushed into a bottle neck and a rubber bung fights back. A value of $0.5$ means the volume does not change at all under load, which is why rubber has to be given somewhere to bulge sideways or it behaves as though it were nearly incompressible. Thermodynamic stability limits $ν$ for an ordinary isotropic material to between $- 1$ and $0.5$ , and engineered foams with re-entrant cells do achieve negative values.

The work done stretching an elastic bar is stored and recoverable. Load rises linearly from zero to $N$ while the extension rises from zero to $δ$ , so the work is the area under that line:

U = \frac{1}{2} N δ = \frac{N^{2} L}{2 A E}

For the steel rod above, $U = 0.5 \times 20 000 \times 0.00211 = 21.1$ J. That looks small until it is compared with what happens when it is released suddenly. Strain energy is why a snapped cable whips, why a bolt tightened into a long grip is a better fastener than a short one, and why toughness matters more than strength for anything that might be hit.

Thermal strain

A material expands when heated, by $ε = α Δ T$ , where $α$ is the coefficient of linear thermal expansion. For steel $α \approx 12 \times 10^{- 6}$ K⁻¹, for aluminium $23 \times 10^{- 6}$ , for concrete about $10 \times 10^{- 6}$ , which is the happy accident that makes reinforced concrete possible, and for Invar, a nickel-iron alloy discovered by Charles Guillaume in 1896, about $1 \times 10^{- 6}$ .

If the member is free to move, thermal strain produces no stress at all: it simply gets longer. If it is restrained, the restraint has to squeeze it back to its original length, and the stress required is

σ = E α Δ T

which is independent of the length of the member. That is the important and counterintuitive part. A restrained bar 1 metre long and one 100 metres long develop exactly the same thermal stress.

Example. A steel bar is held between two immovable walls and its temperature rises by $40$ K. What stress does it develop?

$σ = 210 000 \times 12 \times 10^{- 6} \times 40 = 100.8$ MPa of compression, over a third of the yield strength of mild steel, from a temperature change a sunny afternoon supplies.

Now you. Repeat for an aluminium bar, with $E = 70$ GPa and $α = 23 \times 10^{- 6}$ K⁻¹.

Answer

$σ = 70 000 \times 23 \times 10^{- 6} \times 40 = 64.4$ MPa. Aluminium expands twice as much but is three times less stiff, so the stress is lower.

Structures are therefore detailed to let temperature happen. Bridges get sliding bearings and expansion joints, long pipe runs get loops, and rails are either gapped or, in continuous welded track, deliberately stressed at installation so that the working range never reaches buckling. Continuous track that is laid at the wrong temperature buckles sideways in summer, which is a real and regular failure mode.

Compatibility, and the indeterminate structure solved

Now the promise of the first lesson can be kept.

Example. A short reinforced concrete column is a $300$ mm square of concrete with four $20$ mm steel bars cast into it, carrying $900$ kN in compression through a rigid plate at the top. How does the load divide between the two materials?

Equilibrium gives one equation, that the two materials share the load: $P = σ_{c} A_{c} + σ_{s} A_{s}$ . Two unknown stresses, one equation. The missing equation is compatibility: the plate is rigid and the bars are bonded to the concrete, so both materials shorten by the same amount over the same length, and therefore they have the same strain.

ε_{c} = ε_{s} \Rightarrow \frac{σ_{c}}{E_{c}} = \frac{σ_{s}}{E_{s}} \Rightarrow σ_{s} = \frac{E_{s}}{E_{c}} σ_{c}

Taking $E_{s} = 200$ GPa and $E_{c} = 25$ GPa, the modular ratio $E_{s} / E_{c}$ is $8$ , so the steel carries eight times the stress of the concrete beside it. The steel area is $4 \times π \times 20^{2} / 4 = 1257$ mm² and the concrete area is $90 000 - 1257 = 88 743$ mm². Substituting,

P = σ_{c} (A_{c} + 8 A_{s}) = σ_{c} (88 743 + 10 053) = 98 796 σ_{c}

so $σ_{c} = 900 000 / 98 796 = 9.11$ MPa and $σ_{s} = 72.9$ MPa. The steel carries $72.9 \times 1257 = 91.6$ kN, which is $10.2$ per cent of the load on $1.4$ per cent of the area. Stiffness attracts load: in an indeterminate structure, the stiffer path takes the larger share, and that principle is worth more than the arithmetic.

Now you. The same column is reinforced with four $25$ mm bars instead. Find the stress in each material and the fraction of the load the steel carries.

Answer

$A_{s} = 4 \times π \times 25^{2} / 4 = 1963$ mm² and $A_{c} = 90 000 - 1963 = 88 037$ mm². Then $A_{c} + 8 A_{s} = 103 744$ , so $σ_{c} = 900 000 / 103 744 = 8.68$ MPa and $σ_{s} = 69.4$ MPa. The steel force is $69.4 \times 1963 = 136$ kN, or $15.1$ per cent of the total.

The same two steps settle the other classic case, a steel bar of uniform section fixed rigidly at both ends with a load $P$ applied at a point dividing it into lengths $a$ and $b$ . Equilibrium gives $R_{1} + R_{2} = P$ and nothing more. Compatibility says the total length is unchanged, so the extension of one portion equals the shortening of the other: $R_{1} a / (A E) = R_{2} b / (A E)$ , giving $R_{1} a = R_{2} b$ . Solving, $R_{1} = P b / (a + b)$ and $R_{2} = P a / (a + b)$ . The shorter portion, being the stiffer spring, takes the larger force, and $A$ and $E$ cancel out entirely because both portions are the same material and section.

The recipe generalises. Write every equilibrium equation available, count the shortfall, and supply exactly that many statements about how the deformed parts must fit. Each one turns into an equation in the unknown forces through $δ = N L / A E$ . This is the whole method for indeterminate structures, and the deflection lesson will use it again on beams.

What the straight line does not cover

Two limits are set by the curve itself.

Hooke's law is the first part of a measured curve, not a law of nature. It holds up to a proportional limit and no further, and past that a structure analysed by these methods is being analysed with the wrong constitutive relation. Where that limit is, and what happens beyond it, is the next lesson.

And even inside the elastic range, $E$ is a single number only for a material that behaves the same in every direction. Timber along the grain is roughly twenty times stiffer than across it. A carbon fibre laminate can be tuned to almost any stiffness in any direction by choosing the ply angles. Concrete is stiff in compression and effectively has no tensile strength at all, which is why it is reinforced. Assuming one isotropic $E$ is correct for steel and for most metals, and it is an approximation that has to be checked for everything else.