Thank you! It's been 10 years since I learned Statics in College. I am studying for my FE and I am finding these practice problems quite helpful in providing me with a quick refresher on the subject.
No actually it is in units of kN. We integrate to get the area under the curve, but the units on the y axis are kN/m, and the units on the x axis are m. If it was a rectangular shape, base * height... the units would be kN/m * m = kN. Same for this parabolic shape.
This example is confusing for students because the 16kN never appears to come into play. Or doesn't seem to in this example. I feel like the example is confusing and unclear because the beam is 4m and 4^2 is 16m. Therefore the bounds of integration appears to be based upon the beam length, not the value of x which in the function produces the output of 16kN. So it might cause students confusion if they are unaware were the 4 came from. They might think it is distance and ask... Does that mean that the moment and force created by 70000kN over 4m if it is distributed parabolically is the same as 16kN over 4m? That seems to violate real life expectations...and on a test or assignment use the distance as the bounds of integration. For clarity maybe change 16 to 25 or 36 or change the distance to 8m.
Hey thanks for commenting! Watch this video: www.engineer4free.com/4/how-to-find-the-centroid-of-simple-shapes I explain in it why the x appears in the numerator. Hope that helps!
Thank you! It's been 10 years since I learned Statics in College. I am studying for my FE and I am finding these practice problems quite helpful in providing me with a quick refresher on the subject.
Thanks greatly appreciated!
No problem!!! More distributed loading problems @ engineer4free.com/statics =)
Thank you so much for this explanation !!
You’re welcome!!! 🙂
Isn’t 21.33 in units of meters? The force should be that times 16 kN correct?
No actually it is in units of kN. We integrate to get the area under the curve, but the units on the y axis are kN/m, and the units on the x axis are m. If it was a rectangular shape, base * height... the units would be kN/m * m = kN. Same for this parabolic shape.
This example is confusing for students because the 16kN never appears to come into play. Or doesn't seem to in this example. I feel like the example is confusing and unclear because the beam is 4m and 4^2 is 16m. Therefore the bounds of integration appears to be based upon the beam length, not the value of x which in the function produces the output of 16kN. So it might cause students confusion if they are unaware were the 4 came from. They might think it is distance and ask...
Does that mean that the moment and force created by 70000kN over 4m if it is distributed parabolically is the same as 16kN over 4m? That seems to violate real life expectations...and on a test or assignment use the distance as the bounds of integration. For clarity maybe change 16 to 25 or 36 or change the distance to 8m.
@@MathTutorVideos omg that's really what I was going to ask. Where did 4 came from?
At 2:22 when you're showing the location of x bar formula, for what reason does the top integral require multiplying y by x? Thanks
Hey thanks for commenting! Watch this video: www.engineer4free.com/4/how-to-find-the-centroid-of-simple-shapes I explain in it why the x appears in the numerator. Hope that helps!