1. Oct 25, 2015

goldfish9776

1. The problem statement, all variables and given/known data
i am asked to find the reaction at A and B ...
Here's my working :
15(3/2) +15(5) +15(6)+5(6/2)=RA +RB

moment about A = 15(3/2)(3x2/3) + (15)(5)(8) + (5x6/2)(8+4) + (15)(6)(14) -RB(14) = 0
so , i have my RB=148.9N , RA= 53.5N , which is wrong ? which part of my working is nt correct?
i have 15(3/2)(3x2/3) because the centroid is located 2/3 of 3m from the triangle .
I have (5x6/2)(8+4) because the centroid is located 8+(6x2/3) from A

2. Relevant equations

3. The attempt at a solution

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2. Oct 25, 2015

PhanthomJay

3. Oct 25, 2015

goldfish9776

moment about A = 15(3/2)(3x2/3) + (15)(5)(3+2.5) + (5x6/2)(8+4) + (15)(6)(12) -RB(14) = 0
now , my RB = 122.67 , which part is wrong again ?

4. Oct 25, 2015

PhanthomJay

Looks like the moment arm of the last uniformly distributed load is still wrong

5. Oct 25, 2015

Mister T

Can you tell me the correct answers? It's been a while since I've looked at this kind of stuff.

6. Oct 25, 2015

SteamKing

Staff Emeritus
The centroid of the first triangular load is located @ (2/3) * 3 m from the toe of the triangle, or point A.
This load calculation and location of the centroid from point A is correct.
This part of the UDL is OK:
(15)(5)(3+2.5)

Did you check the location of the centroid for this part of the UDL?:
(15)(6)(12)

The left end of this UDL starts at 8 m from point A and ends at 14 m from point A. Is 12 m midway between these two locations?

7. Oct 25, 2015

Mister T

Never mind. I figured it out.

8. Oct 27, 2015

goldfish9776

After changing the third part of location of centroid to11, I have (15)(3/2)(3*2/3)+ (15)(5)(5.5) + (15)(6)(11) + (5)(6/2)(11) -FB(14)=0...
FB=115.17N , RA=87.3N , which is still different from the ans given

9. Oct 27, 2015