Home / Strength of materials

Tension, compression and shear

Enter force, section and material to get stress, factor of safety on yield and ultimate strength, elongation and minimum area. For pins, check single or double shear and plate bearing.

Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.

Check
Load
Section
Material and criterion

For ductile materials the factor applies to yield; for cast iron, to ultimate strength.

—Working stress
—

Schematic stress-strain curve

Typical shape for the chosen material using minimum strengths; the diamond is the operating point.

Tap the chart to see the values.

Elastic range enlarged

Hooke’s line up to yield, allowable stress for the target factor and operating point.

Tap the chart to see the values.

Minimum material properties

Minimum yield and tensile strength from the material specs; typical E, G and ν. Elongation in 2 in.

Next step

Take this page’s results to another calculator with the fields already filled in.

Properties of this section (inertia, radius of gyration) →Hydraulic cylinder force and buckling →

The report gathers this page's inputs, results, equations, assumptions and references, with fields for project, engineer and revision. In the print dialog, choose Save as PDF.

Method, assumptions and references

Equations used

Normal stress: σ = F ÷ A; shear: τ = F ÷ (n · A), n = shear planes; bearing: σb = F ÷ (d · t)
Hooke: ε = σ ÷ E; δ = F · L ÷ (A · E); γ = τ ÷ G; lateral strain = ν · ε
Factor of safety: n = Sy ÷ σ (ductile), Su ÷ σ (brittle); shear: Ssy = 0.577 Sy (von Mises), Ssu ≈ 0.67 Su (steels and aluminum); minimum area = n · F ÷ S
Compression (K = 1): λ = L ÷ r; λc = π √(2E ÷ Sy); λ ≥ λc: σcr = π² E ÷ λ² (Euler); λ < λc: σcr = Sy − (Sy λ ÷ 2π)² ÷ E (Johnson)

Assumptions and limits

  • Static, axial, centered load with uniform stress. Stress concentrations at holes, threads or notches, fatigue and high temperature are not included.
  • Minimum properties: ASTM A36, A240 for 304, B221 for 6061-T6, A48 class 30 for cast iron; SAE 1020 and 1045 per Shigley Table A-20. Cast iron compressive strength is typical.
  • For pin shear the average stress is used; the peak in a solid round section is 4/3 of the average.

Standards and technical documents

  • Budynas, R. G.; Nisbett, J. K.: Shigley’s Mechanical Engineering Design, 11th ed., McGraw-Hill (Tables A-5, A-20, A-24)
  • Beer, F. P.; Johnston, E. R.: Mechanics of Materials, McGraw-Hill
  • ASTM A36/A36M; ASTM A240/A240M; ASTM B221; ASTM A48/A48M
  • ISO 6892-1: Metallic materials — Tensile testing — Method of test at room temperature

U.S. regulations

  • ANSI/AISC 360-22: Specification for Structural Steel Buildings (Chapters D, E and J)
  • ASME BTH-1: Design of Below-the-Hook Lifting Devices
  • OSHA 29 CFR 1910.184: slings and rigging hardware

Frequently asked questions

How do I calculate tensile stress in a bar?

Divide force by area. A 3/4 in round bar (0.442 in²) carrying 7,000 lbf sees 15.8 ksi; in A36 steel (Fy = 36 ksi) the factor of safety is 2.3.

How much does a steel bar stretch?

δ = F L ÷ (A E). The same 3/4 in bar with 7,000 lbf over 6 ft stretches about 0.039 in, with E = 29,000 ksi.

What is the difference between single and double shear?

In single shear the pin is cut on one plane; in double shear, as in a clevis, on two. With 7,000 lbf on a 3/4 in pin, stress is 15.8 ksi in single and 7.9 ksi in double shear.

What factor of safety should I use?

For well-known static loads on steel, 1.5 to 2 on yield. With impact or uncertain loads, 3 or more. For cast iron, 3 to 5 on ultimate strength.