Beam Analysis Help - Get Expert Guidance Now!

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The discussion focuses on beam analysis, where the original poster expresses uncertainty about their calculations for reactions, shear force diagrams (SFD), and bending moment diagrams (BMD). Several users point out errors in signage and equations, emphasizing the importance of correctly summing moments and forces. After multiple attempts, the poster revises their equations but continues to struggle with achieving equilibrium in vertical forces. Additionally, they seek assistance with drawing shear and bending moment diagrams and inquire about other methods like Macaulay's and the moment area method. The conversation highlights the learning curve in structural engineering and the value of foundational knowledge in practical applications.
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Beam analysis Help!

I have attached the question and my attempt at a solution, although I am pretty sure I have calculated the reactions wrong and therefore my sfd and bmd are wrong as well. Any help to solve this would be great.
Thanks.
 

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At quick glance, you have a lot of errors, including signage errors, and incorrect equations. When you sum moments, the 20 kN-m couple must be included..you seem to have crossed it out. When you sum forces in the vertical direction, don't include moments or couples. Watch your plus and minus signs...V_A is 5 kN down. Correct typo errors. Calculate the proper moment for a distributed load. You are on the right track...keep on pluggin'...
 


Hi.
I have redone my first equation and got:
BMc=4(Va)+20=0, therefore Va=5 kn acting upwards
Then for my second equation I have:
BMe=-8(5)+20+15(4)-2(Vd)=0, therefore Vd=20kN acting upwards
Then the sum of the vertical forces:
5+20-15-15(4)+Vf=0, therefore Vf=50kN acting upwards
Sum of the moment about F +clockwise:
Mf-15(4)(2)+10(6)-15(8)+20+5(12)=0, therefore Mf=20kNm

Thanks for your help. I have redone the equations but I am sure that there is still something wrong as my vertical forces don't sum to give zero.
 


mariechap89 said:
Hi.
I have redone my first equation and got:
BMc=4(Va)+20=0, therefore Va=5 kn acting upwards
When you sum moments about C = 0, sum of clockwise moments must equal sum of counterclockwise moments. Since the 20 kN-m couple is clockwise, then Va must be 5 kN downwards to produce the counterclockwise moment of 20 kN-m.
Then for my second equation I have:
BMe=-8(5)+20+15(4)-2(Vd)=0, therefore Vd=20kN acting upwards
Again, you are getting careless with signs. If you consider ccw moments as plus, the cw moments must be minus. Correct the first term signage due to your initial error in calculating the direction of Va, then check that sign again for the couple.
Then the sum of the vertical forces:
5+20-15-15(4)+Vf=0, therefore Vf=50kN acting upwards
This equation will be OK once you make the necessary numerical value and signage corrections
Sum of the moment about F +clockwise:
Mf-15(4)(2)+10(6)-15(8)+20+5(12)=0, therefore Mf=20kNm
Again, make the necessary numerical corrections...I think you will have a good handle on the signage on this one...
Thanks for your help. I have redone the equations but I am sure that there is still something wrong as my vertical forces don't sum to give zero.
your free body diagrams are good...just watch your signage...sum of clockwise moments = sum of counterclockwise moments..
 


Hi. Does this look any better??
BMe would be 8(5)-20+15(4)-2(Vd)=0 therefore Vd=40
Sum of vertical forces:
5+40-15-15(4)+VF=0 therefore Vf=30
Sum of moment about F + clockwise: Mf-15(4)(2)+40(6)-15(8)-5(12)=0, therefore Mf=60kNm

I am also having trouble drawing the shear force and bending moment diagrams for this.
If possible would you also be able to look at my posts about macaulays method and moment area method ? If you could that would be great, Thanks for your help so far.
 


mariechap89 said:
Hi. Does this look any better??
BMe would be 8(5)-20+15(4)-2(Vd)=0 therefore Vd=40
, upwards, looks good!
Sum of vertical forces:
5+40-15-15(4)+VF=0 therefore Vf=30
That first term ,5, should be -5; Solve Vf = 40 upwards
Sum of moment about F + clockwise: Mf-15(4)(2)+40(6)-15(8)-5(12)=0, therefore Mf=60kNm
you forgot the 20 kn-m couple...Mf = 40 kn-m clockwise
I am also having trouble drawing the shear force and bending moment diagrams for this.
Correct the values...your approach on the shear diagram is good. Your M diagram is a little shaky at the end...moments are max or min at points of zero shear... the slope of the M diagram is the shear at that point
If possible would you also be able to look at my posts about macaulays method and moment area method ? If you could that would be great, Thanks for your help so far.
Yow, I haven't used those methods in 40 years. Funny, all this stuff you learn in college is essential to structural engineering..but when you get out into the work force, you start from scratch...using the basics to guide you...but learning from others as you go. Every job has different requirements. I do a lot of catenary design..never even studied that in school ...but you pick it up as you learn, with the basics as your guide. After 10 years or so, you become the expert, and then everyone asks YOU the questions. Good luck.:wink:
 

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