Young's Modulus Calculator: Stress, Strain and Elongation

Free Young's modulus calculator: find stress, strain, elongation or the modulus of elasticity from the force, area and length, with a live picture of the stretched bar, safety factor, buckling load and 30 materials. SI and US units, every formula shown.

Material

Units

Compare with

and

True stress–strain Shade areas Plateau serrations

Speed

Yield point or smooth yielding?

Mild and structural steels (A36, 1020) have a yield point. Carbon and nitrogen atoms pin the dislocations, so the stress climbs a little past the yield strength until a band of metal breaks free: that is the upper yield point. The force then drops to the lower yield point and the bar stretches at nearly constant stress while Lüders bands spread along it (the Lüders strain, about 1 to 3% for A36), with small serrations as each band starts. Only then does strain hardening begin, up to the tensile strength (58 to 80 ksi, 400 to 550 MPa for A36, at about 15 to 20% strain) and necking. The upper yield is usually only a few % to about 10% above the lower one, and its height depends on the test: a stiff machine, a fast pull and a smooth, well-aligned specimen give a taller, sharper peak; a soft machine or a slightly bent bar can hide it. The line also bends a little just before the peak (some grains yield early), so the top is rounded rather than a sharp corner. Codes therefore use the lower yield (or 0.2% offset) as the yield strength: 250 MPa (36 ksi) for A36.

Aluminium alloys (2014-T6, 6061-T6), copper, stainless steel and most other metals yield continuously: no peak, no drop, no plateau. The curve just bends away from the straight line, so the yield strength is defined by the 0.2% offset rule: draw a line with slope E starting at 0.2% strain, and where it meets the curve is the yield strength (414 MPa, 60 ksi, for 2014-T6).

Engineering stress\(\displaystyle \sigma = \frac{F}{A_0},\)\(\displaystyle \varepsilon = \frac{L - L_0}{L_0}\)

True stress\(\displaystyle \sigma_t = \frac{F}{A} = \sigma\,(1+\varepsilon),\)\(\displaystyle \varepsilon_t = \ln(1+\varepsilon)\)

Hardening\(\displaystyle \sigma_t = K\,\varepsilon_p^{\,n}\)

Necking starts\(\displaystyle \frac{d\sigma_t}{d\varepsilon_t} = \sigma_t \;\Rightarrow\; \varepsilon_t \approx n\)

Resilience\(\displaystyle U_r = \frac{\sigma_y^2}{2E}\)

Toughness\(\displaystyle U_t = \int_0^{\varepsilon_f} \sigma\,d\varepsilon\)

The curves are a model fitted to each material's typical Young's modulus, yield strength, tensile strength and elongation (Hollomon / Ramberg–Osgood hardening, a Lüders plateau for mild steel). Specimen: a standard round bar, 12.5 mm across with a 50 mm gauge length (0.500 in and 2.000 in in US units). Real curves also depend on the test speed, temperature and the exact alloy.

Supports

Loads drag them along the beam; drag a handle to change the size

Cross-section

Deflection limit

Bending stress (flexure formula)\(\displaystyle \sigma = \frac{M\,c}{I} = \frac{M}{S},\)\(\displaystyle S = \frac{I}{c}\)

Shear stress\(\displaystyle \tau = \frac{V\,Q}{I\,t}\)

Deflection curve\(\displaystyle E I\,\frac{d^2 y}{dx^2} = M(x),\)\(\displaystyle \frac{dM}{dx} = V\)

Rectangle\(\displaystyle I = \frac{b\,h^3}{12},\)\(\displaystyle S = \frac{b\,h^2}{6}\)

Simply supported, load P at the middle\(\displaystyle M_{max} = \frac{P L}{4},\)\(\displaystyle \delta = \frac{P L^3}{48 E I}\)

Simply supported, spread load w\(\displaystyle M_{max} = \frac{w L^2}{8},\)\(\displaystyle \delta = \frac{5 w L^4}{384 E I}\)

Cantilever, load P at the end\(\displaystyle M_{max} = P L,\)\(\displaystyle \delta = \frac{P L^3}{3 E I}\)

Fixed both ends, spread load w\(\displaystyle M_{ends} = \frac{w L^2}{12},\)\(\displaystyle \delta = \frac{w L^4}{384 E I}\)

The beam is solved exactly for any mix of loads (singularity functions: equilibrium plus the deflection conditions at the supports), so the fixed and propped cases, which statics alone cannot solve, come out right too. It assumes a straight, elastic beam with small deflections (Euler–Bernoulli), bending about the strong axis, no buckling sideways (long, slender I-beams without bracing can twist out of the way: lateral-torsional buckling) and no self-weight unless you add it as a spread load.

Words used here

Bending moment M: How much the loads try to bend the beam at a point: force × distance, summed over everything on one side. Sagging (smiling, top squeezed) is drawn upward here; hogging (frowning, top stretched, over supports) downward.
Shear force V: The force trying to slide one part of the beam past the next. It jumps at every point load and support, and the slope of the moment diagram equals it, so M is largest where V crosses zero.
Neutral axis: The line through the section that is neither stretched nor squeezed. For these symmetric sections it sits at mid-depth; the stress grows in a straight line away from it.
Second moment of area I: How the area is spread away from the neutral axis (m⁴, in⁴). Depth counts cubed, which is why I-beams put their metal in the flanges and why a plank is far stiffer on edge.
c and section modulus S = I/c: c is the distance from the neutral axis to the outermost fibre, where the stress is largest. S rolls I and c into one number: σmax = M/S.
Flexure formula σ = Mc/I: The largest bending stress in the section, valid while the material stays elastic.
Q and τ = VQ/(It): Q is the first moment of the area above the neutral axis and t the width there. The shear stress is largest at the neutral axis; in an I-beam the thin web carries nearly all of it.
Deflection δ and L/360: How far the beam sags. Codes limit it as a fraction of the span (L/360 for floors under live load, L/240 or L/180 for roofs) so floors do not feel bouncy and plaster does not crack.
Simply supported, cantilever, fixed, propped, overhang: A pin and a roller at the ends; built into a wall at one end only; built in at both ends; built in at one end and resting on a support at the other; resting on two supports with part of the beam sticking out past one of them.
Point load, spread load (UDL), moment: A force at one spot (kN, kip); a force spread evenly along a length (kN/m, kip/ft), such as a floor's weight; a twisting couple applied at one spot (kN·m, kip·ft).
EI (flexural rigidity): Young's modulus times I: how hard the beam is to bend. Deflection is inversely proportional to it.

For learning and first estimates. Real parts need certified material values, the design code that applies (AISC, Eurocode, ASME…) and a check by a qualified engineer.

Young's modulus, stress and strain calculator

Type any four of the force, cross-section, length, change in length, stress, strain and Young's modulus, and the calculator works out the other three. It shows which values are still missing, warns when the bar would yield or break, gives a safety factor against the material's yield strength and, for a push, the load at which a long bar buckles. The picture is live: drag the red force arrow and the bar stretches (the stretch is drawn larger than real, and the drawing says by how much). It works in SI and US units, and every box takes its own unit, so you can mix kN with inches if your data is like that.

The three formulas

Stress is force per area: \(\sigma = F / A\). One megapascal (MPa) is one newton on every square millimetre; 1 ksi = 6.895 MPa.
Strain is the stretch per unit length: \(\varepsilon = \Delta L / L\). It has no unit; 0.001 is 0.1% or 1,000 microstrain (µε).
Young's modulus (the modulus of elasticity) is stress divided by strain while the material is elastic: \(E = \sigma / \varepsilon\). Put together, the stretch of a bar is \(\Delta L = F L / (A E)\).

Worked examples

A 10 mm steel rod, 1 m long, pulled with 10 kN. The area is π × 10² ÷ 4 = 78.54 mm², so the stress is 10,000 N ÷ 78.54 mm² = 127.3 MPa. With E = 200 GPa the strain is 127.3 ÷ 200,000 = 0.000637, so the rod stretches 0.637 mm. A36 steel yields at 250 MPa, so the safety factor is about 1.96. Pull the same rod with 50 kN and the stress is 636.6 MPa: past the tensile strength, so it breaks.

Finding E from a test. A 12.5 mm bar with a 50 mm gauge length stretches 0.0245 mm under 12 kN. The stress is 12,000 ÷ 122.7 = 97.8 MPa and the strain 0.0245 ÷ 50 = 0.00049, so E = 97.8 MPa ÷ 0.00049 = 199.6 GPa: a steel. Type F, the diameter, L and ΔL and the calculator does this for you (the "Find E from a test" example).

In US units. A 1/2 in A36 rod (0.1963 in²) carrying 2 kip has a stress of 10.19 ksi. With E = 29,000 ksi the strain is 0.000351, so a 40 in rod stretches 0.0140 in.

Sizing a rod. Type the force, pick the material and press "Size it for a safety factor of 2": the stress is set to the yield strength ÷ 2 and the diameter you need is worked out. For 25 kN in A36 that is 125 MPa, 200 mm², a 15.96 mm bar: round up to 16 mm.

Questions

What does Young's modulus tell you?

How stiff a material is: how much stress it takes to stretch it by a given strain. Steel (about 200 GPa) is three times stiffer than aluminium (about 70 GPa), so the same bar in aluminium stretches three times as much under the same load. Stiffness is not strength: a stiff material can still be weak, and glass is stiffer than aluminium but breaks at a far lower stress.

Does Young's modulus depend on the size of the bar?

No. It is a property of the material. A thicker bar is stiffer because it has more area, and a longer bar stretches more because there is more of it, but E stays the same. That is why the stretch formula has F, L and A in it as well as E.

When does Hooke's law stop working?

At the proportional limit, close to the yield strength. Beyond it the bar stretches much more for each extra newton and stays longer when unloaded. The calculator warns you, and the tensile test simulator shows the whole stress-strain curve, including what happens when you unload past yield.

Is E the same in compression?

For metals, yes. Concrete, cast iron and wood behave differently in tension and compression, and a long thin bar in compression usually buckles sideways long before its stress gets high. The calculator gives the Euler buckling load for pinned ends, \(P_{cr} = \pi^2 E I / L^2\), when you choose "Push".

Why is my measured E different from the table?

Table values are typical. Real E varies by a few percent with the alloy and its condition, and a test measures a little low if the grips slip or the strain is taken from the machine's crosshead instead of an extensometer on the bar. Very small strains are also hard to measure: 0.025 mm is a thin hair.

What are the units of Young's modulus?

The same as stress: pascals. Engineers use GPa (10⁹ Pa) in SI and ksi or Msi (10⁶ psi) in US units. 200 GPa = 29,000 ksi = 29 Msi.

Comments