Shear stress distribution in triangular steel profile

AI Thread Summary
Calculating maximum shear stress in a triangular thin wall steel beam presents challenges due to the unique geometry compared to rectangular shapes. The standard formula for shear stress, tau=Vy*Sza/(b*Iz), requires adaptation for triangular profiles, particularly in determining the statistical moment. The triangular shape lacks a point of zero shear stress and has unequal leg directions, complicating calculations. A suggested approach for maximum shear stress is tau = 2.0*V/A, where A is the area based on the triangle's perimeter and wall thickness. Understanding the neutral axis properties is crucial for accurate shear stress distribution analysis.
Ieliepielie
Messages
1
Reaction score
0
For my job I have to calculate the maximum shear stress in a triangular thin wall steel beam. I know how to calculate this for a retangular shape.
The general formula is:
tau=Vy*Sza/(b*Iz)
with:
tau= shear stress
Vy=force
Sza = statistical moment
b= width
Iz = moment of Inertia

I just don't see how I could calculate the statistical moment for the triangle. It is a closed profile, therefore there is no point where the shear stress is zero and in contrary to the retangular shape the 'legs' don't have equal direction so it can not just be divided by two. I hope somebody can help me.
Thank you!
 
Engineering news on Phys.org
What do you know about the neutral axis? its properties?
 
Ieliepielie: My best guess, for maximum shear stress, is currently, tau = 2.0*V/A, where A = s*t, s = triangle perimeter, and t = wall thickness.
 
no idea knowing neutral axis and no properties. Its a random question. Also how to graph the distribution?
Thanks.
 
I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them. The math I'm using is: FA = MA / 2 = ? t-force MB * SIN(of the incline degree) = ? If MB is greater then FA, it pulls FA up as MB moves down the incline. BUT... If I reverse the 2:1 pulley. Then the math changes to... FA = MA * 2 = ? t-force MB * SIN(of the incline degree) = ? If FA is greater then MB, it pulls MB up the incline as FA moves down. It's confusing...
Hi. I noticed that all electronic devices in my household that also tell time eventually lag behind, except the ones that get synchronized by radio signal or internet. Most of them are battery-powered, except my alarm clock (which runs slow as well). Why does none of them run too fast? Deliberate design (why)? Wrong temperature for quartz crystal? Decreasing battery voltage? Or just a coincidence?
Back
Top