Moment about distributed load (trapezium)

[goldfish9776]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[PhanthomJay]

Your errors are not with the triangular load but with the uniform loads. Your moment arms from the cg of the uniform loads are incorrect.

 

[goldfish9776]

Your errors are not with the triangular load but with the uniform loads. Your moment arms from the cg of the uniform loads are incorrect.

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 ?

 

[PhanthomJay]

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

 

[Mister T]

so , i have my RB=148.9N , RA= 53.5N , which is wrong ?

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

 

[SteamKing]

Homework Statement

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 .

The centroid of the first triangular load is located @ (2/3) * 3 m from the toe of the triangle, or point A.

I have (5x6/2)(8+4) because the centroid is located 8+(6x2/3) from A

This load calculation and location of the centroid from point A is correct.

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 ?

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?

 

[Mister T]

Never mind. I figured it out.

 

[goldfish9776]

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?

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

 

[Mister T]

My answers are slightly different.

You had this:

15(3/2)(3x2/3) + (15)(5)(3+2.5) + (5x6/2)(8+4) + (15)(6)(12) -RB(14) = 0

And SteamKing told you you had only one error. Now you have this:

(15)(3/2)(3*2/3)+ (15)(5)(5.5) + (15)(6)(11) + (5)(6/2)(11) -FB(14)=0

You have changed more than one thing! I'm not sure if I'm using the right terminology here, but between the positions of 8 m and 14 m you have a UDL of 15 kN, and above it a triangular load. Note that the centroid of the rectangle and the centroid of the triangle have different positions.