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.
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.