What is Unit Converter used for in structural engineering?

Unit Converter provides values required for steel design calculations including capacity checks, connection design, and detailing. It is referenced in structural design codes and used by practicing engineers during the design of buildings, bridges, and industrial structures.

Can I use these reference values directly in my design?

Yes, provided the values match your governing code edition and project specifications. Always cross-reference against the primary source (code document, manufacturer data, or published standard). The information on this page is for educational use — final verification by a licensed engineer is required.

How often are these reference values updated?

Reference content is maintained to align with current code editions. When a new edition of AISC 360, AS 4100, EN 1993, or CSA S16 is published, values are reviewed and updated. Check the last-modified date at the bottom of the page and verify against the current edition for your jurisdiction.

Unit Converter | Steel Calculator

-------- | ---- | ----------- | ------------------- | | kips (kip) | kN | 4.448 | 50 kip = 222.4 kN | | kN | kips | 0.2248 | 200 kN = 44.96 kip | | kips | lbf | 1,000 | 10 kip = 10,000 lbf | | kN | N | 1,000 | 25 kN = 25,000 N | | metric tons | kN | 9.807 | 10 t = 98.07 kN | | US tons | kips | 2.0 | 5 ton = 10 kip |

Stress and pressure conversions

From	To	Multiply By	Example
ksi	MPa	6.895	50 ksi = 344.7 MPa
MPa	ksi	0.1450	250 MPa = 36.26 ksi
psi	MPa	0.006895	36,000 psi = 248.2 MPa
psf	kPa	0.04788	100 psf = 4.788 kPa
ksf	kPa	47.88	2 ksf = 95.76 kPa
MPa	N/mmÃÂÃÂ²	1.0	345 MPa = 345 N/mmÃÂÃÂ²

Length and area conversions

From	To	Multiply By	Example
ft	m	0.3048	30 ft = 9.144 m
m	ft	3.2808	10 m = 32.81 ft
in	mm	25.4	12 in = 304.8 mm
mm	in	0.03937	200 mm = 7.874 in
inÃÂÃÂ²	mmÃÂÃÂ²	645.2	10 inÃÂÃÂ² = 6,452 mmÃÂÃÂ²
inÃÂ¢ÃÂÃÂ´	mmÃÂ¢ÃÂÃÂ´	416,231	448 inÃÂ¢ÃÂÃÂ´ = 186,471,000 mmÃÂ¢ÃÂÃÂ´
inÃÂÃÂ³	mmÃÂÃÂ³	16,387	64 inÃÂÃÂ³ = 1,048,768 mmÃÂÃÂ³
ftÃÂÃÂ²	mÃÂÃÂ²	0.09290	1,000 ftÃÂÃÂ² = 92.90 mÃÂÃÂ²

Moment and distributed load conversions

From	To	Multiply By	Example
kip-ft	kN-m	1.3558	200 kip-ft = 271.2 kN-m
kN-m	kip-ft	0.7376	300 kN-m = 221.3 kip-ft
kip-in	N-mm	112,984	12,000 kip-in = 1,355,818 N-mm
klf	kN/m	14.59	2.0 klf = 29.19 kN/m
kN/m	klf	0.06852	30 kN/m = 2.056 klf
psf	kN/mÃÂÃÂ²	0.04788	50 psf = 2.394 kN/mÃÂÃÂ²
kN/mÃÂÃÂ²	psf	20.88	5.0 kN/mÃÂÃÂ² = 104.4 psf

Temperature conversions

From	To	Formula	Example
Fahrenheit	Celsius	C = (F - 32) ÃÂÃÂ 5/9	70ÃÂÃÂ°F = 21.1ÃÂÃÂ°C
Celsius	Fahrenheit	F = C ÃÂÃÂ 9/5 + 32	20ÃÂÃÂ°C = 68ÃÂÃÂ°F
Fahrenheit	Kelvin	K = (F - 32) ÃÂÃÂ 5/9 + 273.15	70ÃÂÃÂ°F = 294.3 K

Density and mass conversions

From	To	Multiply By	Example
lb/ftÃÂÃÂ³	kN/mÃÂÃÂ³	0.1571	490 lb/ftÃÂÃÂ³ = 76.98 kN/mÃÂÃÂ³
lb/ftÃÂÃÂ³	kg/mÃÂÃÂ³	16.018	490 lb/ftÃÂÃÂ³ = 7,849 kg/mÃÂÃÂ³
lb/ft	kg/m	1.488	35 lb/ft = 52.09 kg/m

Common Steel Design Values — Quick Reference

Material properties in both unit systems

Property	Imperial	Metric
Steel density	490 lb/ftÃÂÃÂ³	7,850 kg/mÃÂÃÂ³
E (carbon steel)	29,000 ksi	200,000 MPa
Fy (A36)	36 ksi	248 MPa
Fy (A572 Gr 50 / A992)	50 ksi	345 MPa
Fu (A992)	65 ksi	450 MPa
Fu (A325 bolt)	120 ksi	827 MPa
Fu (A490 bolt)	150 ksi	1,034 MPa
Concrete f'c (typical)	4,000 psi	27.6 MPa
Concrete f'c (high)	8,000 psi	55.2 MPa

Standard deflection limits

Member	Limit	Metric Equivalent
Floor beam (live load)	L/360	L/360 (same ratio)
Roof beam (live load)	L/240	L/240
Floor beam (total load)	L/240	L/240
Cantilever (live load)	L/180	L/180
Brittle finishes	L/480	L/480

Worked Example — Converting a Beam Design from Imperial to Metric

Problem: A W16x36 floor beam was designed using AISC 360 in imperial units. Convert the key design values to SI for comparison with an AS 4100 check.

Step 1 — Section properties

W16x36 (imperial):  Ix = 448 inÃÂ¢ÃÂÃÂ´,  Sx = 56.5 inÃÂÃÂ³,  Zx = 64.0 inÃÂÃÂ³
                    d = 15.86 in,  bf = 6.985 in,  tw = 0.295 in

Convert to metric:
  Ix = 448 ÃÂÃÂ 416,231 = 186,471,488 mmÃÂ¢ÃÂÃÂ´ = 186.5 ÃÂÃÂ 10ÃÂ¢ÃÂÃÂ¶ mmÃÂ¢ÃÂÃÂ´
  Sx = 56.5 ÃÂÃÂ 16,387 = 925,866 mmÃÂÃÂ³ = 926 ÃÂÃÂ 10ÃÂÃÂ³ mmÃÂÃÂ³
  Zx = 64.0 ÃÂÃÂ 16,387 = 1,048,768 mmÃÂÃÂ³ = 1,049 ÃÂÃÂ 10ÃÂÃÂ³ mmÃÂÃÂ³
  d = 15.86 ÃÂÃÂ 25.4 = 402.8 mm
  bf = 6.985 ÃÂÃÂ 25.4 = 177.4 mm
  tw = 0.295 ÃÂÃÂ 25.4 = 7.49 mm

Step 2 — Loading and moment

Service load: w = 2.0 klf = 2.0 ÃÂÃÂ 14.59 = 29.18 kN/m
Span: L = 20 ft = 20 ÃÂÃÂ 0.3048 = 6.096 m

Service moment: M = wLÃÂÃÂ²/8 = 29.18 ÃÂÃÂ 6.096ÃÂÃÂ² / 8 = 29.18 ÃÂÃÂ 37.16 / 8 = 135.6 kN-m
  Check: 100 kip-ft ÃÂÃÂ 1.3558 = 135.6 kN-m ÃÂ¢ÃÂÃÂ

LRFD factored load: wu = 3.2 klf = 3.2 ÃÂÃÂ 14.59 = 46.69 kN/m
LRFD moment: Mu = 46.69 ÃÂÃÂ 37.16 / 8 = 216.9 kN-m
  Check: 160 kip-ft ÃÂÃÂ 1.3558 = 216.9 kN-m ÃÂ¢ÃÂÃÂ

Step 3 — Capacity

PhiMn = 240 kip-ft = 240 ÃÂÃÂ 1.3558 = 325.4 kN-m
Steel: E = 200,000 MPa, Fy = 345 MPa
Section: d = 402.8 mm, Zx = 1,049 ÃÂÃÂ 10ÃÂÃÂ³ mmÃÂÃÂ³
phiMn = 0.90 ÃÂÃÂ 345 ÃÂÃÂ 1,049,000 = 325,701,000 N-mm = 325.7 kN-m ÃÂ¢ÃÂÃÂ

The conversion checks out: both unit systems give the same result when the conversion factors are applied correctly.

Common pitfalls and how to avoid confusion

Hidden assumptions: some checks require assumptions that are not explicit in the UI (e.g., end restraint idealization, load distribution, slip requirements). If you canÃÂ¢ÃÂÃÂt state the assumption, do not treat the result as verified.
Standard mismatch: names like ÃÂ¢ÃÂÃÂyield strengthÃÂ¢ÃÂÃÂ and ÃÂ¢ÃÂÃÂultimate strengthÃÂ¢ÃÂÃÂ are universal, but how they are used in a resistance model is standard-specific.
Axis confusion: major/minor axis properties, sign conventions, and local coordinate systems can flip a result.
Detailing constraints: minimum edge distances, minimum weld sizes, and installation constraints often govern before a strength limit state does.
Over-trusting a single ratio: a utilization < 1.0 does not prove the detail is acceptable; it only indicates the evaluated checks passed under the toolÃÂ¢ÃÂÃÂs assumptions.

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Frequently Asked Questions

What are the most common unit pitfalls when mixing metric and imperial structural calculations? The most frequently confused pairs are MPa vs ksi (stress), kN vs kip (force), and kNÃÂÃÂ·m vs kipÃÂÃÂ·ft (moment). A MPa is a megapascal (10ÃÂ¢ÃÂÃÂ¶ N/mÃÂÃÂ²) while ksi is kips per square inch; the conversion is 1 ksi = 6.895 MPa. Mixing these without conversion in a formula produces errors of approximately 7ÃÂÃÂ in stress results. Similarly, 1 kip = 4.448 kN, so substituting kN values into a kip-based formula without conversion inflates forces by a factor of roughly 4.5.

How do I convert between kNÃÂÃÂ·m and kipÃÂÃÂ·ft for bending moments? The exact conversion is 1 kipÃÂÃÂ·ft = 1.3558 kNÃÂÃÂ·m, or equivalently 1 kNÃÂÃÂ·m = 0.7376 kipÃÂÃÂ·ft. A bending moment of 100 kipÃÂÃÂ·ft equals 135.6 kNÃÂÃÂ·m; a design moment of 200 kNÃÂÃÂ·m equals 147.5 kipÃÂÃÂ·ft. This conversion is needed constantly when comparing AISC (imperial) design aids against AS 4100 or EN 1993 (SI) calculations, or when checking a supplierÃÂ¢ÃÂÃÂs load table against a design performed in different units.

What is the exact conversion between ksi and MPa? 1 ksi (kip per square inch) = 6.8948 MPa exactly (based on 1 lbf = 4.44822 N and 1 inch = 25.4 mm). In practice 6.895 MPa is used. Common structural values: A36 yield strength Fy = 36 ksi = 248 MPa; A572 Gr.50 Fy = 50 ksi = 345 MPa; A325 bolt Fnt = 90 ksi = 620 MPa. Memorising a few anchor values makes it easy to sanity-check converted stress figures in the field.

When should I use imperial units versus SI units on a project? The governing standard and the project specification dictate which unit system to use — AISC 360 is published in both customary (kip, in) and SI (kN, mm) editions, while AS 4100 and EN 1993 are SI-only. On international projects or joint ventures, explicitly agreeing on a single unit system at the outset prevents mixed-unit errors in transmitted calculations. When a calculation package spans both systems (for example, US-sourced material specs combined with SI drawings), maintain a dedicated conversion table at the front of the calculation and always label every numeric value with its unit.

How do unit consistency errors cause structural calculation failures? Unit errors typically manifest as results that are off by a fixed factor — 1000ÃÂÃÂ (kN vs N), 6.895ÃÂÃÂ (ksi vs MPa), or 304.8ÃÂÃÂ (ft vs mm) — which can make an under-designed member appear over-capacity or vice versa. A classic case is entering a force in kN into a formula that expects kips: the result is approximately 4.4ÃÂÃÂ smaller than the correct answer, making the connection appear to have 4ÃÂÃÂ more capacity than it actually does. Dimensional analysis (checking that numerator and denominator units cancel correctly) is the single most effective technique for catching these errors before they reach design decisions.

What are the key area and section modulus unit conversions for steel design? 1 inÃÂÃÂ² = 645.2 mmÃÂÃÂ²; 1 inÃÂ¢ÃÂÃÂ´ = 416,231 mmÃÂ¢ÃÂÃÂ´; 1 inÃÂÃÂ³ = 16,387 mmÃÂÃÂ³. These conversions matter when using tabulated AISC section properties (in inÃÂÃÂ² and inÃÂ¢ÃÂÃÂ´) in SI-based capacity formulas. For example, the elastic section modulus S of a W18ÃÂÃÂ35 is 57.6 inÃÂÃÂ³ = 944,000 mmÃÂÃÂ³. A factor of 10ÃÂ¢ÃÂÃÂ¶ difference between mmÃÂ¢ÃÂÃÂ´ and cmÃÂ¢ÃÂÃÂ´ (1 cmÃÂ¢ÃÂÃÂ´ = 10ÃÂ¢ÃÂÃÂ´ mmÃÂ¢ÃÂÃÂ´) is another frequent source of error when comparing European and Australian section tables, which often list I in cmÃÂ¢ÃÂÃÂ´, against North American tables listed in inÃÂ¢ÃÂÃÂ´ or mmÃÂ¢ÃÂÃÂ´.

Disclaimer (educational use only)

This page is provided for general technical information and educational use only. It does not constitute professional engineering advice, a design service, or a substitute for an independent review by a qualified structural engineer. Any calculations, outputs, examples, and workflows discussed here are simplified descriptions intended to support understanding and preliminary estimation.

All real-world structural design depends on project-specific factors (loads, combinations, stability, detailing, fabrication, erection, tolerances, site conditions, and the governing standard and project specification). You are responsible for verifying inputs, validating results with an independent method, checking constructability and code compliance, and obtaining professional sign-off where required.

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