Shear Force and Bending Moment Diagram

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 7K views
jasonbot
Messages
16
Reaction score
0

Homework Statement



I have to find the Shear Force and Bending Moment diagrams for this problem. The lengths in the picture are 1.2m each and the peak force is 4kN/m and I calculated the support reactions to be:

Ay=4,8kN (up)
Ma= 5.76kN.m (anticlockwise)

Homework Equations



These are what I'm looking for but failed to do...

The Attempt at a Solution

I first section the beam 0<x<4

Efy=0

+4.8 -.5x*3.33x-Vx=0

I got the 3.33x by using similar triangles; if triangle 1 is (1,2;4) then sectioned triangle should be (x;y) so therefore 4/1.2=y/x x=3.33xthis gave me Vx=-1.67x^3+4.8 which is wrong.

Then EMo=0

+Mx +.5x*3.33x*(x/3) -4.8x +5.76=0

which gets me Mx=-0.56x^3 +4.8x +5.76The correct answers for this question are:

Vx=4.8 -(x^3/6)
Mx=4.8x -(x^3/18) -5.76I have a feeling I'm not using the triangle force correctly. Please advise.
 

Attachments

  • Screen shot 2009-11-29 at 12.33.32 PM.png
    Screen shot 2009-11-29 at 12.33.32 PM.png
    5.7 KB · Views: 663
Physics news on Phys.org
jasonbot, welcome to PF!

jasonbot said:

Homework Statement



I have to find the Shear Force and Bending Moment diagrams for this problem. The lengths in the picture are 1.2m each
Is it 12 m or 1.2 m?
and the peak force is 4kN/m and I calculated the support reactions to be:

Ay=4,8kN (up)
Ma= 5.76kN.m (anticlockwise)
yes, looks good
I first section the beam 0<x<4
0<x<12 m
Efy=0

+4.8 -.5x*3.33x-Vx=0

I got the 3.33x by using similar triangles; if triangle 1 is (1,2;4) then sectioned triangle should be (x;y) so therefore 4/1.2=y/x x=3.33x


this gave me Vx=-1.67x^3+4.8 which is wrong.
if you correct the 1.2 m to 12m , then Vx = -.167x^2[/color] +4.8
Then EMo=0

+Mx +.5x*3.33x*(x/3) -4.8x +5.76=0

which gets me Mx=-0.56x^3 +4.8x +5.76
something is afoul between 1.2 m and 12 m, the decimal point is messed.., but you are still not quite right, don't let the plus and minus signs wear you down...
`
The correct answers for this question are:

Vx=4.8 -(x^2[/color]/6)
typo error
Mx=4.8x -(x^3/18) -5.76
taking clockwise moments as plus, I don't know about that decimal point, however


I have a feeling I'm not using the triangle force correctly. Please advise.
Looks like you did well, but check out your plus and minus signs, they'll bite you every time...
 
Last edited: