What is the maximum normal stress in a W18x46 beam?

AI Thread Summary
The discussion revolves around calculating the maximum normal stress in a W18x46 beam, focusing on part C of the homework problem. The user has drawn shear and moment diagrams and is determining the maximum normal stress, which they believe occurs at the end of the web, not the flange. They calculated the moment to be 152.3 kip*ft at 14.25 ft from the origin and are using the formula σ=Mc/I to find the stress, arriving at approximately 22 ksi. The user is also considering the appropriate moment of inertia (I) to use, debating between the x-x and y-y axes. Overall, the calculations seem to align with their expectations, indicating they are on the right track.
SteelDirigibl
Messages
40
Reaction score
0

Homework Statement



attachment.php?attachmentid=46421&stc=1&d=1334881670.png


Homework Equations



σ=P/A

σ=My/I

τ=VQ/It

The Attempt at a Solution



I'm on part C, pretty sure I have A and B, so I'm using a W18x46 Beam. I have drawn shear and moment diagrams. I just want to be clear on what part C is asking...
I'm assuming this point is at the end of the web, wither top of bottom, just not part of the flange. So if that's correct... it is wanting the max normal stress. Is this ging to be where moment is at it's maximum? I have determined this to be at 14.25 ft (using A as the origin), with the moment being 152.3 kip*ft.

V is at a max at C, this seems it might make more sense. but P/A just gets me 1.75 and that's the same as τ that I already found.

Reaction force at A is 21.375 kip and at C is 23.625.

Am I on the right track? Can I use the equations above, find principle stresses, and make a mohr's circle? because at this point, I have V=0, does that seem right?

I'm also not sure which I to use... the x-x axis or y-y? the I for these beams in the apendix shows x-x being 712 in4 and 22.5 in4 going by the y-y axis

Any more of my previous calculations that might be neccessary, let me know, I'll post them...
 

Attachments

  • Screen shot 2012-04-19 at 7.23.30 PM.png
    Screen shot 2012-04-19 at 7.23.30 PM.png
    11 KB · Views: 752
  • Screen shot 2012-04-19 at 8.24.21 PM.png
    Screen shot 2012-04-19 at 8.24.21 PM.png
    21 KB · Views: 2,937
Last edited:
Physics news on Phys.org
I think I may have figured it out. I just used Mc/I where c is the distance from the neutral axis.

Beam is 18.06 in deep, minus 0.625in per each flange, is 16.85, divided by two is 8.425. I used that as C, and 712 for I. Its about 22 ksi. seems reasonable to me.
 
Back
Top