Lessons · Engineering · axial deformation, PL over AE
How much a rod stretches: PL over AE
A rod of length L and area A under axial load P, in a material of modulus E, stretches δ = P L / (A E).
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.
What it is for
A tie rod holds two walls together. If it stretches 6 mm under load the walls move 6 mm, and the plaster tells everyone. Whether it is 0.5 mm or 6 mm is one line of arithmetic, done before the rod is ordered.
How to think about it
Convert everything to newtons, metres, square metres and pascals before substituting; the formula has no unit factor in it and will not forgive one term left in millimetres. Longer, heavier-loaded, thinner or softer means more stretch, and each in simple proportion.
Worked example
δ = P L / (A E)Stretch is load times length, over area times modulus.
P = 50 kN = 50,000 N; L = 2.0 m; A = 1000 mm² = 0.001 m²; E = 200 GPa = 200e9 PaEvery term in base units first. This line is where the marks are won.
δ = (50,000 × 2.0) / (0.001 × 200e9) = 100,000 / 2e8 = 0.0005 mSubstitute, then the arithmetic.
δ = 0.5 mmHalf a millimetre on a two-metre rod, which is about what a tight steel tie does.
Your turn
P = 20 kN, L = 1.5 m, A = 0.0004 m², E = 200e9 Pa. Write the elongation.
δ = (20,000 × 1.5) / ( × 200e9) = 0.000375 m
Solve one, graded on the server
The trap
One term left in millimetres. Area as 1000 instead of 0.001 makes the answer a million times too small, and it looks plausible because a tiny stretch is what you expected to see.