# Homework Help: Maximum Bending Stress in Beam

1. May 2, 2010

### joemama69

1. The problem statement, all variables and given/known data
Note the attachment. I scanned it out of my book and I drew a FBD on it.

2. Relevant equations

3. The attempt at a solution

Ok so i havent gotten very far at all with this one because it just doesnt seem to have enough information.

So first I summed the forces in the y direction.

Fy = 0 = 9k - RB.... RB = 9k

Then I found the Moment @ B

MB = 9k(192) = 1728k

and thats about as far as I got. It seemed odd that there was no information given about the forces or distances at point A.

so I know that the Bending stress = -My/I

but how do I find the y. Im assuming it would come from k = 1/p but I do not know any of those quantities. Same with I = integral y2dA. Can I get a hint as to my next step

File size:
487.3 KB
Views:
576
2. May 2, 2010

### PhanthomJay

There is a force at A (and B), but the max bending moment at B (in inch-kips) is the same on either side of support B, so you don't need the force at A to calculate it. You should calculate the moment of Inertia , I, of the wide flanged beam, using the parallel axis theorem. Ignore the small fillets where the web connects to the flanges.

3. May 3, 2010

### joemama69

Ok so find I i used an equation from wikipedia

http://en.wikipedia.org/wiki/Second_moment_of_area#Parallel_axis_theorem

I calculated it for Ix & Iy, but it my book it doesnt specify which ones to use. do i add them together or did i use the wrong formulas

Ix = 1995 & Iy = 75

my book says for doubly symmetric shapes... $$\sigma$$max = M/S where S = I/c

where do I go from here

4. May 3, 2010

### PhanthomJay

The beam is bending about the strong axis of the beam, not the weak axis, so you just need to use the I that applies, not both. You don't add the moments of inertias. Don't forget your units of I (in^4). I assume you understand the value to use for 'c' ?

5. May 3, 2010

### joemama69

Ok so i used the Iy because that is the direction of movment = 75 in^4

and c would just be half of the height because it is a symetric beam

S = I/c = 75in^4/11in = 6.81in^3

$$\sigma$$ = M/S = 1725kip*in/6.81in^3 = 254 kips/in^2

is this correct

6. May 3, 2010

### PhanthomJay

You've got your Ix and Iy mixed up. But you managed to get the correct c distance.

7. May 3, 2010

### joemama69

ok I missunderstood

S = 1995/11 = 181.4 in^3

Stress max = M/S = 1728kips*in / 181.4 in^3 = 9.5 is kips/in^2 = psi

8. May 3, 2010

### PhanthomJay

yes, that's 9.5 kips/in^2, or 9.5 ksi, or 9500 psi.