Calculate elongation ∆𝑙 of tendon with initial length 5 cm and diameter 4 mm due to the action of internal force 31.4 N. Young’s modulus of tendon is 2·109 N/m2 .

Answer 1

Assuming circular cross-section of the tendon and 2*109 is 2*10^9:

Givens:

L = 5 cm = .05 m

d = 4 mm = 0.004 m

E = 2 * 10^9 N/m^2

F = 31.4 N

Equations:

σ = E * ε:

σ = F / A = F / (π * (d^2) / 4)

ε = ΔL / L

Substituting:

F / (π * (d^2) / 4) = E * ΔL / L

Solving:

ΔL = (F / (π * (d^2) / 4) * L) / E

Plugging in Numbers:

ΔL = (31.4 N) / (π * (0.004 m^2) / 4) * 0.05 m ) / 2 * 10^9 N/m^2

Solution:

ΔL = 0.0000625 m = 0.0625 mm