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Lesson 34 of 78 · Structures & Material Response

Stress, Strain & Safety Factor

Stress converts internal force into intensity over an area; strain measures deformation relative to original size. For a uniform axial member,

\[ \sigma=\frac{N}{A}, \qquad \epsilon=\frac{\Delta L}{L}. \]

In the linear elastic range, \(\sigma=E\epsilon\), where \(E\) is Young’s modulus. MIT’s material courses emphasize that modulus, yield, plasticity, creep, and fracture describe different parts of a material’s response—not one ranking called “strength” 1.

Worked tie-rod calculation

A 10 mm diameter steel tie rod carries 12 kN in tension.

\[ A=\frac{\pi d^2}{4}=78.5\ \text{mm}^2, \qquad \sigma=\frac{12{,}000}{78.5}=153\ \text{MPa}. \]

If \(E=200\) GPa, elastic strain is \(153/200{,}000=0.000765\). Over 400 mm, predicted elongation is 0.306 mm. If the material’s specified minimum yield strength is 350 MPa, a simple yield-based factor is \(350/153=2.29\).

That number is incomplete. Threads reduce area and concentrate stress; load may be eccentric; fatigue may govern; corrosion can remove section; temperature can change properties. A safety factor belongs to a named limit state and load case.

Concentrations and local stress

Holes, grooves, sharp shoulders, keyways, and thread roots disturb nominal stress. A geometric stress-concentration factor \(K_t\) estimates peak elastic stress: \(\sigma_{max}=K_t\sigma_{nom}\). Do not apply a factor from memory to an unmatched geometry. Better design often comes from increasing fillet radius, smoothing load flow, moving holes away from high moment, or adding section locally instead of thickening everything.

A disciplined margin statement

Write:

For load case LC-3, the predicted bracket-root stress is 82 MPa. Against a documented 205 MPa minimum yield strength, the nominal yield margin is 2.5 before the listed corrections for weld geometry, residual stress, and fatigue.

This is stronger than “factor of safety equals 2.5” because it exposes demand, resistance, failure mode, and exclusions.

Practice

Build a three-column table for one component: limit state, demand model, resistance evidence. Include yielding, excessive deflection, buckling, fatigue, wear, fastener slip, and loss of alignment even if several are later screened out. Safety is not obtained by multiplying one stress by one customary number; it comes from finding plausible ways the function can be lost.

Source trail

References

  1. 1
    Mechanical Behavior of Materials. MIT OpenCourseWare. verifiedUndergraduate treatment of elastic and plastic deformation, creep, fracture, and the processing-structure-property relationship. Cited at: elasticity through fracture.
Further reading
  • Mechanics of Materials. MIT OpenCourseWare. verifiedOpen modules on stress, strain, trusses, torsion, bending, deflection, yielding, fracture, fatigue, and material properties.

Check your understanding

  1. Normal stress in a uniformly loaded axial member is estimated by which expression?
  2. Why is factor of safety not a universal constant?