
#1
Aug2211, 01:21 AM

P: 3

Hi Guys,
I'm having real difficulty trying to understand shear force and bending moment diagrams  both cantilever beams and supported ones like below. I've given it my best shot to solve the reactions at the points and would greatly appreciate any help on how to form the diagram and the equations after using method of sections. Thanks 1. The problem statement, all variables and given/known data Create SFD and BFD of following beam. 2. Relevant equations Fy=0 Fx=0 M=0 3. The attempt at a solution Ma=0 Rb(6) +30(4.5)+[(10x4.5)/3.75]20=0 Rb=21.2kN Mb=0 (30)(1.5)+[(10X4.5)/2.25]206Ra=0 Ra=7.5kN http://imageshack.us/photo/myimages/710/unledosv.jpg/ 



#2
Aug2211, 01:26 AM

P: 3

Heres my attempt at using method of sections to write the equations:
I 0<x<a v=7.5kN m=20kNM II 0<x<0 v=7.5kN + 10x m=20kNm + (10x^2)/2 III v=7.5 kN + 10x + 30kN  21.2kN m=20kN + (10x^2)/2 + 30kN.x 21.2kN(6) As you can probably tell I'm struggling pretty bad on this lol. Any help is appreciated! 



#3
Aug2211, 01:51 AM

HW Helper
Thanks
P: 5,580

You have not written the correct equilibrium equations for this beam. The sum of the forces must equal zero, and the sum of the moments, taken about a single reference point, also must equal zero. You have written moment equations about two different points. The expression for the moment of the distributed load is also incorrect. Until you calculate the reactions, you will not be able to construct the SF and BM diagrams for this beam.




#4
Aug2211, 02:51 AM

P: 3

SFD & BMD of beam with point force through distributed load
oh ok thanks. so you're not meant to take the moment of the two points? An example i saw did it that way i think so i thought i'd give it a go. oh well, back to the drawing board!
ps. you calculate the moment of the distributed load by multiplying the magnitude by its length and then placing a point force at the middle of the distributed load, don't you? Sorry i'm struggling big time 


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