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NIST SP 811 factors: stress in a 0.5 in rod under 1,000 lbf, shown in ksi and MPa
| Tool | Formula with Units |
| Source | Thompson, A., and Taylor, B. N. (2008). Guide for the Use of the International System of Units (SI). NIST Special Publication 811, National Institute of Standards and Technology, Gaithersburg, MD. Appendix B.8 (lbf, in, psi). |
| Level | Undergraduate |
Problem
A round rod 0.5 in in diameter carries an axial force of 1,000 lbf. Evaluate σ = F/(π d²/4), check that the result is a stress, and express it in psi, ksi and MPa using the exact SP 811 factors for the inch and the pound-force.
Inputs to Enter
- Variables F = 1000 lbf, d = 0.5 in (loaded by default, also the 'Stress in a rod' example)
- Formula F / (pi * d^2 / 4), result name sigma, result unit ksi
Answers
- σ = 5,092.958 psi = 5.0930 ksi (1 lbf = 4.4482216152605 N, 1 in = 0.0254 m exactly)
- σ = 35.1147 MPa (1 psi = 6,894.757293168 Pa)
- Base-unit form kg·m⁻¹·s⁻², reconstructed as Pa
Notes
The tool's default inputs are this case; press Submit. Every factor is applied through the site's base-unit engine, not typed.