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Uniformly Varying Load/Uniform Load to Concentrated Point Load

 
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Sep19-09, 07:40 PM   #1
 

Uniformly Varying Load/Uniform Load to Concentrated Point Load


Hi all,
I'm taking a Structures course in University and are learning about Freebody diagrams and figuring out Reaction forces (magnitude, direction, sense etc) and have a very basic question. Can someone please tell me how to convert a uniformly varying load into a concentrated point load, and how does one figure out at which point on the member to place this concentrated point load? i've attached a scan from my book but it doesn't show how to derive the point load. appreciate thanks.

I do know how to convert a uniform load into a concentrated point load.

Also if anyone would like to tutor me in Structures, (i will reimburse in an hourly basis or we could work out a deal), please let me know? I'm having difficulty in my Structures class, and desperately need to pass it and would like to have a tutor who I can depend on to have questions that I may have answered. thanks

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Sep19-09, 08:37 PM   #2
 
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The point load is placed in line with the centroid of the right triangle. It is found by

[tex]X = \frac{2B}{3}[/tex]

where

B = The length of the total varying load.
X = distance from the large end of the load triangle (in this case, where the large dot is right below the 15 k/ft)

Look here for a reference:
http://www.ele.uri.edu/~daly/106/06/.../centroid.html
Sep20-09, 12:11 PM   #3
 
Thanks Fred... can you tell me how does one arrive at the value concentrated load of 67.5? I know with a uniform load we have to multiply the uniform load and the length it affects. eg: 2k/ft x 6ft
Sep20-09, 12:57 PM   #4
 
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Uniformly Varying Load/Uniform Load to Concentrated Point Load


Quote by accesskb View Post
Thanks Fred... can you tell me how does one arrive at the value concentrated load of 67.5? I know with a uniform load we have to multiply the uniform load and the length it affects. eg: 2k/ft x 6ft
If you think of a triangle as one half of a rectangle...
Oct26-09, 01:33 PM   #5
 
It is very simple. Just find the area of triangular load. (1/2)*(3+6)*15 = 67.5. And place the load at the centroid of the triangle (1/3)*9= 3..
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