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Formula for cantilever beam with distributed equal force

Hey, I make a project in which I use a cantilever beam with distributed equal load across the whole beam and I'm using the formulas above to calculate the displacement, moment...

My specifications for the beam are: 4.5m long aluminum 2024 beam, 4x3 cm walls (the force is applied on the 4cm side), 2mm thickness. and a young's modulus of 70 GPa. a force of 40 N/m is applied

from my calculations the moment of inertia is 37 mm^4, meaning the maximum displacement should be 0.6m at X=4.5 and the maximum moment is 400 N-m at X=0.

When I put the results in the formula for bending stress, the maximum bending stress is 162 MPa, while the yield stress for this type of aluminum is around 300 MPa. Bending stress formula

something here doesn't seems right for me. the beam is bent 0.6m but does not exceed the yield point, maybe I did a mistake somewhere?

george456
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  • I can see it bending that much - even a 3m length of copper pipe will form a nice curve if you load it gently. – Solar Mike Jan 06 '20 at 09:11
  • I think you mean" deflection" . Al has a modulus 1/3 of steel so it deflects more than steel components . It is not affected by yield strength until you exceed the yield. – blacksmith37 Jan 06 '20 at 15:37
  • You are right, but it just seems really strange for me, the beam is bent 0.6 meters from it's original position but does not exceed the yield point? – george456 Jan 06 '20 at 17:26
  • Steel will tolerate 0.3 % elastic strain before yielding ( up to 0.6 % for high yield steel) . I don't know the number for Al. – blacksmith37 Jan 06 '20 at 21:23

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