I'm considering attaching a stainless steel bar on a sailboat masthead.

My initial thought was to use a bar with a cross section of 3/8x2" and 1/4x2". The bar will be supported for 3" of its 5" length.

What is the breaking strength if a load is applied 1" past the support?
What kind of safety margin would I have if the downward load is 2500 lbs?

Or would I be better off using rectangular channel 1" x 2" with a 1/8" or 1/4" wall thickness?

  • $\begingroup$ If your state university has an ASTM testing lab they may be able to answer your question $\endgroup$
    – Duckweather
    Commented Nov 12, 2015 at 2:45
  • 2
    $\begingroup$ Can you provide a sketch of the bar, how it will be mounted, and how the load will be applied? $\endgroup$
    – Carlton
    Commented Nov 12, 2015 at 16:41

1 Answer 1


There are a number of unknowns in the problem statement:

  1. The geometry of the bar
  2. The location of the supports
  3. Whether the beam is simply supported or clamped at these supports
  4. Whether the load is a point load or distributed
  5. Whether the load is in lbf or lbm
  6. What type of stainless steel is being used, etc.

However, a rough estimate can be found if we make a number of assumptions. Assume that the beam is simply supported with asymmetric point load as shown in the Wikipedia page on beam bending.

The maximum moment is $$ M_{\max} = \frac{P\,a\,b}{L} $$ where $P$ is the applied load, $a,b$ are the distances of the load point from the supports, and $L$ is the distance between the supports. For this situation the maximum stress is $$ \sigma_{\max} = \frac{M\,h}{I} = \frac{P\,a\,b\,h}{L\,I} $$ where $h$ is the height of the beam cross-section and $I$ is the area moment of inertia. For a beam with a square cross-section, $$ I = \frac{1}{3} w h^3 $$ where $w$ is the width of the beam cross-section. So $$ \sigma_\max = \frac{3\,P\,a\,b\,h}{L\,w\,h^3} = \frac{3\,P\,a\,b}{L\,w\,h^2} $$ Plugging in some numbers, $P$ = 2500 lbf, $a$ = 2 in, $b$ = 1 in, $h$ = 2 in, $L$ = 3 in, $w$ = 3/8 in, we get $$ \sigma_\max = \frac{3 \times 2500 \times 2 \times 1 }{3 \times 3/8 \times 2^2} = 3333\,\, \text{lbf/sq.in.} \approx 3500\,\text{psi} $$ Note that if the load is in lbm (or more accurately in "slugs"), then $$ P = 2500 \times 32 $$ and we have $$ \sigma_\max \approx 106000 \,\text{psi} $$ Now we have to compare this stress with the yield stress of stainless steels. If you look at the MatWeb page on stainless steels you will find that the yield strength can vary between 24,700 psi and 160,000 psi.

You will have to pick your material such that even if there are microscopic cracks in the material (and cyclic loading condition), the beam will not fail by direct fracture or fatigue fracture.

The increase in stress due to the presence of a small flaw can be estimated using a Stress intensity factor.

Suppose that you have a tiny edge crack in the beam. We can approximate the stress near the crack as uniform and uniaxial. This allows us to use the stress intensity factor for an edge crack for a rough estimate of the stress intensity factor: $$ K_{\rm I} = \sigma_0\sqrt{\pi c}\left[1.12 - 0.23\left(\frac{c}{h}\right) + 10.6\left(\frac{c}{h}\right)^2 - 21.7\left(\frac{c}{h}\right)^3 + 30.4\left(\frac{c}{h}\right)^4\right] \,. $$ where $\sigma_0$ is a uniform applied stress, $c$ is the crack length and $h$ is the height of the beam cross-section. If the crack length is $c$ = 0.01 in, $h$ = 2 in, and $\sigma_0$ = 3500 psi, we get $$ K_{\rm I} = 695 \,\,\rm{psi.in}^{1/2} $$ The stress near the crack can be found using the relation $$ \sigma = \frac{K_{\rm I}}{\sqrt{2\pi \,r}} $$ but we assume that there exists a material property called the fracture toughness ($K_{\rm{Ic}}$) that allows us to compare $K_{\rm{I}}$ with experimental data. For some stainless steels, $K_{\rm{Ic}}$ = 100,000 psi-in$^{1/2}$ (which is much larger than our calculated value of $K_{\rm I}$).

Fracture toughness values can be found in the ASM Metals Handbooks and a large list of publications can be found in the review articles by Hudson, CM and Seward, SK: A Compendium of sources of fracture toughness and fatigue-crack growth data for metallic alloys, International Journal of Fracture, vol. 14, R151-R184, 1978; vol. 20, R58-R117, 1982; vol. 39, R43-R63, 1989 vol. 48, R19-R43, 1991.

  • $\begingroup$ Just to add to this solid answer: steels exhibit a fatigue limit on the Wohler curve, meaning that for any number of cycles there is minimum cyclic fatigue failure stress. If the part is designed so that stress is always below this level, then the part should not fail in fatigue, provided all else is steady. $\endgroup$ Commented Nov 13, 2015 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.