Vertical reaction at supports.
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blamocur
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Me too
Can you post an explicit statement of the problem, i.e., what is given, and what you are trying to compute.
Thanks.
?? I’m looking to find the vertical reactions at the supports.Me too
Can you post an explicit statement of the problem, i.e., what is given, and what you are trying to compute.
Thanks.
blamocur
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I still don't understand your drawings. Where does your problem come from? If you cannot describe it in more detail I don't see how I can be of any help, but maybe someone else can be more successful in deciphering your post.
topsquark
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I'm guessing that there are two supports? Is A one of them? Is B the other? Which is which?
And that's the weirdest Extended FBD I've ever seen! It's certainly not standard in Physics.
Honestly, I'm guessing because this isn't labelled. But in the first line (under moments about A) I note two things:
1. None of your torques are negative. At least one of them has to be for the net torque to be 0.
2. There isn't any force a distance of 3 m from the center line (which is where I presume the axis of rotation is for this line).
Please show a better labelled diagram (in fact, the whole original problem would be nice.) We can help you better if we know what the problem is stating.
-Dan
One problem with the FBD & analysis is the absence of the loading due to the weight of the beam.
This problem is probably from Engineering Statics.
blamocur
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Only slightly. I still don't understand what "Total U.D.L" (or is it "U.O.L.") ? What it means that it is 3m from R1?
The uniformly distributed load can be substituted by a concentrated load acting in the centre of gravity of the UDL. The total load on beam is the UDL multiplied by the length of the beam, i.e. 5 kN/m × 10.00 m = 50 kN.
Looks good - except the final answer for RA & RB should include units (kN), and,
If I were solve this problem, I would have left the answers in fraction - i.e. as............. 53\(\displaystyle \frac{1}{3}\) kN .............and.................. 26\(\displaystyle \frac{2}{3}\) kN