Tools › Structures › Stress Concentrations › Elliptic Hole Uniaxial Stress › Example

Inglis (1913): elliptic hole under uniaxial tension, K_t = 1 + 2a/b = 5 for a/b = 2

Tool	Elliptic Hole Uniaxial Stress
Source	Inglis, C. E. (1913). "Stresses in a Plate Due to the Presence of Cracks and Sharp Corners." Transactions of the Institution of Naval Architects, Vol. 55, pp. 219–241.
Level	Undergraduate

Problem

An infinite plate under a uniaxial stress of 150 ksi contains an elliptic hole with semi-axis 0.25 in normal to the load and 0.125 in along it. Find the peak tangential stress at the ends of the major axis, the stress concentration factor and the stress at the ends of the minor axis.

Inputs to Enter

σ_x = 150 ksi
a = 0.25 in (semi-axis normal to the load), b = 0.125 in (along the load), β = 0
r = 0.25 in (end of a), θ = 90° from the load axis

Open the Elliptic Hole Uniaxial Stress tool (loaded with this example)

Answers

K_t = 1 + 2a/b = 5; σ_A = 750 ksi at the ends of a (Inglis)
Equivalent form K_t = 1 + 2√(a/ρ_a) with ρ_a = b²/a = 0.0625 in
σ_B = −σ_x = −150 ksi at the ends of b, K_t,B = −1
At the probe point: σ_yy = σ_hoop = σ_max = 750 ksi, τ_max = 375 ksi, φ_p = 90°
Other ratios checked in the build: a/b = 0.5, 1, 2, 10 give K_t = 2, 3, 5, 21

Notes

The tool's default inputs are this case. a = b reproduces the Kirsch round hole (K_t = 3); the field is the Kolosov–Muskhelishvili solution on z = R(ζ + m/ζ).