Bending in structure. Small vehicle frame.

  • Thread starter elsikre
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  • #1
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Hi

I need help for this structure. It is loaded as shown - point E and G is wheels added as free bearings. I need to know the bending in Beam AD and also EF.
I tried isolating beam AD and see it as an statically inderteminant beam to the second degree - but i am not sure this is the right approach ? Can anyone help !!

Thanks
 

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  • #2
nvn
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What are the dimensions of your structure, and what are the cross-sectional properties (E, A, and I) of each member? Are all joints pinned or welded? The current diagram shows all joints welded. The supports at points E and G are free to roll horizontally and free to rotate, correct? Is this a school assignment?
 
  • #3
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Hey, the dimensions are
AB = 0.6 m
BC = 0.2 m
CD = 0.2 m
AE = 0.45 m

Furthermore (for all members)
E = 200 GPa (steel)
I = 0.04 * 10^6 mm^4 (beam 30x30x3,2 mm)
A = 343.04 mm^2 (approximately)

And the beams are welded. It is for a project where we are construction a vehicle (so E and G are wheels), here is the frame shown in 3 D.
 

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  • #4
nvn
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Isn't I = 41 650 mm^4? I assumed it is. The maximum bending moment in beam AD occurs on the left-hand end of member CD and is 62 400 N*mm. Therefore, the maximum stress in beam AD is 22.47 MPa. The maximum bending moment in member EF occurs on the left-hand end of member EF and is 35 956 N*mm. The maximum stress in member EF is 13.43 MPa.
 
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  • #5
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If i take a look at Beam AD as free supported at A and D i get Maximum bending moment to 75 N*m by m.max = (1/8)*w*l but I guess its wrong because I didn't consider internal reactions at B and C - how did you find the momentum curves from A to D - could you tell me your approach ?

Thanks
 

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  • #6
nvn
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elsikre: The reactions at supports E and G are statically determinate; therefore, you can use statics to find the reactions at supports E and G. After you obtain those, make a section cut at joint C to determine the moment on the left-hand end of member CD. The other members are statically indeterminate, and therefore their forces are more difficult to obtain. You could use the finite element method, which is too complicated to explain on a forum.
 
  • #7
Reactions:
Ve+Vg=600*1+400*0.6 => Ve+Vg=840
E is a hinge so moment at E=0 => so from the balance of moments at E:
0=Me=600*1*0.5+400*0.6*0.3-Vg*1=>
Vg=372 => Ve=468

IIRC you can also cut at C, to find the internal shear force Vc and the moment Mc there:
Vc+372-600*0.2=0 => Vc = 252 downwards
Mc - 372*0.2 + 600*0.2*0.1 = 0 => Mc = 62.4 "clockwise"for the part C and rightwards.

From then on things get more complicated, as it is not statically determinate...
 
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