Ironworker Salary
by City (2026)
Erect structural steel frameworks for buildings, bridges, and other structures. National median: $68K/year. See how your city compares.
Ironworker salary by city
Sorted by median salaryCOL-adjusted: where does your salary go furthest?
A $68K Ironworker salary in Houston (COL 91, no tax) has significantly more purchasing power than the same salary in San Francisco (COL 224, 9.3% state tax). Use our calculator to see the real numbers.
Try COL Calculator βIronworker Salary Guide (2026)
The national median salary for a Ironworker is $68,000 per year. Entry-level positions typically start at $52,000, while experienced professionals in the top 25% earn $88,000 or more. The highest-paid 10% of Ironworkers nationally earn $115,000+.
Education and requirements
Apprenticeship program; journeyman certification. This role is 0% remote-eligible nationally, which significantly expands the geographic job market for workers in lower cost-of-living cities.
Salary growth outlook
Ironworker compensation is growing at 5% annually, well above the national average wage growth rate.
Ironworker salary FAQ
How much does a Ironworker make?
The median Ironworker salary in the US is $68,000 per year β about $33/hour. Most earn between $52,000 (25th percentile) and $88,000 (75th percentile).
What is the starting salary for a Ironworker?
Entry-level Ironworker positions typically start around $52,000 per year. Pay rises with experience, specialization, and location.
How much do the highest-paid Ironworkers earn?
The top 10% of Ironworkers earn $115,000 or more per year β usually senior, lead, or specialized professionals in high-cost or high-demand markets.
Which city pays Ironworkers the most?
San Francisco has the highest median Ironworker salary at $102K among the cities we track. But high-cost metros like San Francisco and New York often have lower real purchasing power once cost of living and taxes are factored in.
Are Ironworker salaries going up?
Yes β Ironworker pay is growing about 5% year over year, above the US average wage-growth rate.