Calculate forces on a fixed beam

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SUMMARY

The discussion centers on calculating the forces on a fixed beam ABC subjected to an evenly distributed load q and a diagonal load F at point C. Participants emphasize the importance of correctly summing forces and moments, particularly noting that the resultant force of the distributed load is calculated as q multiplied by the length of the beam segment affected. Additionally, the moment caused by the diagonal load F must account for its angle and distance from point A, specifically using the formula F*sin(54.4°)*6.3. Accurate calculations hinge on correctly identifying the resultant load and its application point.

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Homework Statement



A beam ABC is fixed in point A. There is an evenly distributed load q working on top of the beam. In addition there is a diagonal load F working at point C.

Calculate force Ax, Ay and the reaction couple MA.




The Attempt at a Solution



Please see attached file for figures and calculations.
I believe I'm on the right path, but I fear there might be operator mistakes(+ and -) somewhere.
(Ironically that's what I find to be the most difficult thing with statics. Get a + wrong and all your calculations are off...)
 

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dreamliner said:

Homework Statement



A beam ABC is fixed in point A. There is an evenly distributed load q working on top of the beam. In addition there is a diagonal load F working at point C.

Calculate force Ax, Ay and the reaction couple MA.




The Attempt at a Solution



Please see attached file for figures and calculations.
I believe I'm on the right path, but I fear there might be operator mistakes(+ and -) somewhere.
(Ironically that's what I find to be the most difficult thing with statics. Get a + wrong and all your calculations are off...)
Oh those pesty plus and minus signs are always cause for concern in Physics and Engineering, but that is not your problem here in your incorrect solution. You are mixing up forces and moment calculations. When you sum forces in the y direction, there are no moment arm distances to consider. Check your value for the resultant force of the distributed load. And when you sum moments, check your value for the moment about A of the force F applied at C.
 
PhanthomJay said:
When you sum forces in the y direction, there are no moment arm distances to consider.

So in this case it should be only q*4,1(the length of the beam force q is working on) even though force q isn't distributed across the entire beam? (I was under the assuption you had to consider cg of force q in such cases)

PhanthomJay said:
And when you sum moments, check your value for the moment about A of the force F applied at C.

Should have been a moment arm distance there, yes. So the correct value would be F*sin 54,4*6,3
 
dreamliner said:
So in this case it should be only q*4,1(the length of the beam force q is working on) even though force q isn't distributed across the entire beam?
Correct. The load distribution is in units of force per unit length, and thus the resultant load, which must be in force units, is ql, where l is the length of the beam over which q is applied.
(I was under the assuption you had to consider cg of force q in such cases)
Once you calculate the resultant load of the force distribution, ql, the resultant acts at the cg of the load, and the magnitude and location of that resultant is used to determine end reactions when summing moments about any point.
Should have been a moment arm distance there, yes. So the correct value would be F*sin 54,4*6,3
yes.
 

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