How Tall Should Stiffeners Be to Meet Allowable Stress on a Baseplate?

[mr_dude32]

A ¾” by 21” wide base plate is required to resist an applied moment of 250 in-kips,
(bent about the weak axis). The allowable stress for the plate is 0.75*fy and fy = 36 ksi.
The plate may not work and may require stiffeners. Three stiffeners can be welded onto
the plate and used to increase the moment capacity. The stiffened plate has an allowable
bending stress of 0.66*fy and of course fy still = 36 ksi. Determine the moment capacity
of the un-stiffened plate, find the stress on the un-stiffened plate due to the moment.
Lastly, (and most importantly), find out how tall the stiffeners need to be to satisfy the
allowable stress requirements.

- I know i have to find the moment around the weak axis using Iz=(b)(h^3)/12 where b=21 and h=3/4, and i also know for the siffeners I=sum{(bh^3)/12 + Ad^2}

 

[nvn]

mr_dude32: OK. So list the other relevant equations. And show your work. We do not tell you how to solve your homework here. We just check math.

 

[mr_dude32]

okay so here is what i have so far:

the moment cap. of the un-stiffened plate i used Iz=(bh^3)/12= (21)(.75^3)/12= .73828
then knowing \sigmaallow=Mc/Iz i solved for M where c=(3/4)/2 resulting in M equaling roughly 53.15 kip*in.

my next step is finding the centroid of the stiffened base plate using the variable X for the height of ea. extension so far resulting in Y(bar)= (1.125x^2+1.6875x+5.90625)/(2.25x+15.75)

then to find the MOI using Sum{ (bh^3)/12 + Ad^2 } where d is going to be the |Y(bar)-centroid for each rectagle|
Note: stiffened base plate is broken into 4 parts 3 of which are .75*x and the 4th is 21*.75

Is this the right track for solving?

 

[nvn]

mr_dude32: Your answer for M is currently incorrect. Try again. If you still get your current answer, include how you computed M. (You can hit the Edit button to correct your post, if you wish.)

 

[mr_dude32]

thanks for catching that, i came out with 53.156 kip-in when i did it again!

 

[nvn]

Excellent. I have not yet been able to obtain your ybar function. I need to know where your datum is located for your ybar derivation.

 

[mr_dude32]

To solve for my Ybar i broke the region into 4 parts:

Part_______ Area______Ybar_______Ybar*Area
1:________.75*x_____.75+x/2____.5625x+.375x^2
2:________.75*x_____.75+x/2____.5625x+.375x^2
3:________.75*x_____.75+x/2____.5625x+.375x^2
4:________.75*21______.375________5.90625

Area Total: 2.25x+15.75 (Ybar*Area Total): 1.125x^2+ 1.6875x+ 5.90625

Where (Ybar)\sum(Area)= \sum(Ybar*Area Total)
therefor Ybar=(\sum(Ybar*Area Total))/\sumArea)

Ybar=(1.125x^2+1.6875x+5.90625)/(2.25x+15.75 )
or in another form Ybar=.5x-2.75+49.21875/(2.25x+15.75)

 

[nvn]

Excellent work, mr_dude32. Your ybar is correct. I.e., the second to last line of post 7, and ybar in post 3, are correct. (The last line of post 7 looks incorrect; but we can ignore it.)

Your procedure in post 3 is correct. Do you plan on solving this problem analytically (closed-form); i.e., using algebra and hand calculations? Or do you plan on solving it numerically, using a computer?

 

[mr_dude32]

I just tried solving it by hand and got an X value of -5.65 inches so i know its wrong. do you know of any "safer" means of solving it. Thank you so much for helping check my work.

The next step i took was to sum the for the MOI for all 4 parts:

Iz=\sum(bh^3)/12 + Ad^2 , which resulted in a 3rd degree polynomial but i must have made a mistake somewhere when solving by hand. any recomendations for how you would sum all the MOI?

 

[nvn]

I could be wrong, but my current impression is that it might be too complicated to solve analytically. If so, other solution methods would be (a) numerically, using a computer, or (b) trial and error, by hand.

 

[nvn]

I just tried solving it by hand, and got an x value of -5.65 inches, so I know it's wrong. ... The next step I took was to sum the MOI.

mr_dude32: As you know, you cannot compute the MOI until you obtain a valid value for x. I do not yet understand how you solved for x by hand, because it involves a fourth-degree polynomial. How do you plan to solve for x? Try again.

 

[pongo38]

Trial and error is good for a student because it makes you repeat and learn a procedure. It also gives you an opportunity to detect and correct errors.