Shear Force and Bending Moment Diagram

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SUMMARY

The discussion focuses on calculating Shear Force and Bending Moment diagrams for a beam with a length of 1.2m and a peak force of 4kN/m. The support reactions were determined to be Ay=4.8kN (up) and Ma=5.76kN.m (anticlockwise). The user initially attempted to derive the equations for Vx and Mx but encountered errors related to the beam length and the application of similar triangles. The correct equations are Vx=4.8-(x^2/6) and Mx=4.8x-(x^3/18)-5.76, emphasizing the importance of accurate decimal placement and sign conventions.

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  • Understanding of Shear Force and Bending Moment concepts
  • Familiarity with static equilibrium equations
  • Knowledge of similar triangles in structural analysis
  • Proficiency in algebraic manipulation of equations
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  • Learn about static equilibrium and its application in beam analysis
  • Explore the use of similar triangles for calculating forces in beams
  • Review sign conventions in structural engineering to avoid common mistakes
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Engineering students, structural analysts, and professionals involved in beam design and analysis will benefit from this discussion, particularly those focusing on Shear Force and Bending Moment calculations.

jasonbot
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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.
 

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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:

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