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Volume of a taper column

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Folks,

    Ask to calculate the weight ##W(x)## of this tapered column at a distance x from the top as shown in the sketch.

    2. Relevant equations
    Given:
    A trapezoid of bottom base 1.5m, top is 0.5m, length is 0.5m and height 2m.

    Specific weight is 25kN/m^3

    Area of trapezoid ##A=\displaystyle \frac{h(b_1+b_2)}{2}##

    3. The attempt at a solution

    I have just repeated what is in the sketch for clarity.
    Weight ##W(x)=V* \gamma## where ##\gamma## is the specific weight.

    Area ##A= \frac{x}{2}(b_1+b_2)##
    ##V= A*L##

    Thus ##W(x)=\frac{x}{2}(b_1+b_2)*L*25##

    The book answer is ##\displaystyle W(x)= 0.5\frac{0.5+(0.5+0.5x)}{2}x*25##

    I dont understand how there x is repeated twice in the above expression.

    What have I done wrong?

    Regards
     

    Attached Files:

    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2
    It is unclear what is given and what you are supposed to find. Please formulate the problem properly.
     
  4. Oct 6, 2012 #3

    I have reformulated the question as requested. Thanks
     
  5. Oct 6, 2012 #4
    What is the cross-section of the column? Circle? Square? Something else?
     
  6. Oct 7, 2012 #5
    I have attached the geometry of the problem. It has a front view and the rectangular side view.

    Thanks
     

    Attached Files:

  7. Oct 7, 2012 #6
    OK, now it is all clear.

    Let's start with the trapezoid "face" from 0 to x. What is its area? More specifically, what are its bases?
     
  8. Oct 7, 2012 #7
    Area of a quadrilateral with 1 pair of parallel sides is ##\displaystyle A(x)=x \frac{(b_1+b_2)}{2} ##

    where ##b_1## is the bottom base and ##b_2## is the top base say..
     
  9. Oct 7, 2012 #8
    What is the bottom base for some arbitrary x?
     
  10. Oct 7, 2012 #9
    Not sure I follow.

    The bottom base ##b_1## is 1.5m...?
     
  11. Oct 7, 2012 #10
    It is 1.5 only for x = 2.0. For x < 2.0, it is less than that. What is it exactly?
     
  12. Oct 7, 2012 #11
    ##A(x)=f(b_2, 0.5<b_1(x) \le 1.5)##

    Not sure what ##b_2## would be exactly for some value of x! ##b_2=0.5+x##..?
     
  13. Oct 7, 2012 #12
    As can be seen from the drawing, b(x) is a linear function of x. b(0) = 0.5, b(2) = 1.5. That is enough to determine it.
     
  14. Oct 7, 2012 #13
    Sorry, still dont see how the area can be calculated for some x because we dont know what the length of ##b_2## is apart from it being greater than 0.5 and less than 2....
     
  15. Oct 7, 2012 #14
    Do you see that b(x) - b(0) is proportional to x?
     
  16. Oct 7, 2012 #15
    No, doesnt make me see the answer! I was thinking of it another way. Just take a copy of that sketch and rotate it 180degrees and put it beside it gives a parallegrom of area ##h(b_1+b_2)##. We only need half of that so then actual area is ##h \frac{(b_1+b_2)}{2}##

    Finally, repace the h with x to give ##A(x)= \frac{x}{2} (b_1+b_2)##

    Im baffled by such a simple problem!
     
  17. Oct 7, 2012 #16
    Try something else. Rotate the column 90 degree counterclockwise, so that the bottom is on the right, and align it so that the middle of the column is on the X axis, and the top is at x = 0. What used to be the rightmost edge of the column is now a line above the X axis. It is a straight line. What is its equation?
     
  18. Oct 8, 2012 #17
    That helps. The length of the bottom base ##b_1## is given by

    ##b_1(x)=2(\frac{1}{4} x+0.25)##

    Thus the area of the trapezoid is ##\displaystyle A(x)=\frac{x}{2} (b_1+b_2)=\frac{x}{2} (2(\frac{1}{4} x+0.25)+b_2)##...?
     
  19. Oct 8, 2012 #18
    Correct.
     
  20. Oct 9, 2012 #19
    Ok, so the volume becomes ##\displaystyle A(x)*L=\frac{x}{2} (2(\frac{1}{4} x+0.25)+b_2)*L##

    The weight ##W(x)## is the volume times the specific weight ie ## \displaystyle V (m^3) *\gamma (kN/m^3)= \frac{x}{2} (2(\frac{1}{4} x+0.25)+b_2)*0.5*25kN/m^3##

    If we let x=1m we get ##W(x)=0.375kN## whereas the book has ##W(x)= 6.25(1+.05x)x=9.375kN##..?
     
  21. Oct 9, 2012 #20
    You made a mistake in arithmetic somewhere. Your formula is correct.
     
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