Help: Calculate steel rod strength

The usual design stress for common A36 (36,000 psi yield) structural steel is 0.6 * yield = 22,000 psi. You do need to account for reduction of area due to threaded ends, combined loads, etc. Note that this is based on yield, not ultimate tensile.

strain = stress/modulus of elasticity modulus of elasticity = 30e6 lbs/in^2 for steel

For our purposes, strain is elastic deformation per unit length, a unitless ratio, often expressed as inches/inch.

Length * strain = delta Length

or for your example

12 inches * (1000 lb / .2 in^2) / (30e6 lb/in^2) = .002 in

Ned Simmons

Whoever built my neighbor's barn around the turn of the century used discarded haywire, probly salvaged from logging operations......likely was cheaper and easier to install.

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A steel cable or wire rope would have the load ratings specified and would be much easier to install.

Have your library get a copy of Harry Parker's "Simplified Design of Structural Timber". It covers metal connectors and reinforcements.

Kevin Gallimore

Good point. While the strain would be the same the load capacity could easily be 5 time greater. Typically the cables are made from small diameter wires having ultimate tensile strengths from 200,000 to 300,000 psi.

The side forces will be larger than the actual snow load due to a "toggle effect", which depends on the angle of the roof relative to horizontal. For a shallow angle these forces get huge: side force = (vertical load) / sin(theta) where theta is the angle relative to horizontal.

Yield stress for mild steel ranges from 36-50kpsi, with ultimate stress around 60-70. The effective area depends on whether the ends will be threaded or not (use the root diameter if so). Stress concentrations almost always cause the failure, e.g., transition to threads, a hole for a clevis pin (which also reduces the area), transitions from stiff to flexible components. Stress concentrations of 2 to 4+ are common. Even accounting for these, a factor of safety of 2 to 3+ is usually used to account for the unknowns.

The formula is

elongation = P * L / (A * E)

where P = Load (lbs) L = Length (in) A = Cross sectional area (in^2) E = Modulus of Elasticity (which for steel (all types except stainless) is ~30*10^6 psi)

So for your data, elongation = 1000 lb * 12 in / (3.14*.25^2 * 30*10^6psi) = 0.002 inches.

Yup, not much. Tensile loads rarely cause much deflection. Bending, torsion, and buckling are very different stories, however.

When all is said and done, tho, the failure will probably be from the rod pulling through the wood--much harder to analyze.

These questions usually fall under text titles of "strength of materials" or "mechanics of materials". "Statics" is the foundation for this, which would cover things like the "toggle effect". "Machine Design" is the course after strength of materials, and will have much more about stress concentrations and design info. Check out half.com for some older editions--this stuff hasn't changed in 50 years. For SoM, I like Beer and Johnston for its accessibility. Timoshenko is popular but more theoretical. For Statics, Hibbeler and B&J are good.

Here's some links i found:

careful w/ bending formulas unless you have a pure cantlever load).

whole text online)

David, thanks VERY much; this is just what I need.. and I appreciate your comments and suggestions.

The down/side forces from the roof rafters are known from previous design work.

I plan to use a steel plate to distribute load, based on Hemlock rated at

360 PSI for side grain, 950 PSI with the grain. So, um, a 4"x4" plate would be good for 5760 pounds working with a 4:1 safety factor.

I have a few photos of the work so far at:

Again thanks to the many people who offered info and suggestions. I'm not alone out here in the Woods!