# Volume of a taper column

1. Oct 6, 2012

### bugatti79

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

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Last edited: Oct 6, 2012
2. Oct 6, 2012

### voko

It is unclear what is given and what you are supposed to find. Please formulate the problem properly.

3. Oct 6, 2012

### bugatti79

I have reformulated the question as requested. Thanks

4. Oct 6, 2012

### voko

What is the cross-section of the column? Circle? Square? Something else?

5. Oct 7, 2012

### bugatti79

I have attached the geometry of the problem. It has a front view and the rectangular side view.

Thanks

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6. Oct 7, 2012

### voko

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?

7. Oct 7, 2012

### bugatti79

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..

8. Oct 7, 2012

### voko

What is the bottom base for some arbitrary x?

9. Oct 7, 2012

### bugatti79

Not sure I follow.

The bottom base $b_1$ is 1.5m...?

10. Oct 7, 2012

### voko

It is 1.5 only for x = 2.0. For x < 2.0, it is less than that. What is it exactly?

11. Oct 7, 2012

### bugatti79

$A(x)=f(b_2, 0.5<b_1(x) \le 1.5)$

Not sure what $b_2$ would be exactly for some value of x! $b_2=0.5+x$..?

12. Oct 7, 2012

### voko

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.

13. Oct 7, 2012

### bugatti79

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....

14. Oct 7, 2012

### voko

Do you see that b(x) - b(0) is proportional to x?

15. Oct 7, 2012

### bugatti79

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!

16. Oct 7, 2012

### voko

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?

17. Oct 8, 2012

### bugatti79

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 $\displaystyle A(x)=\frac{x}{2} (b_1+b_2)=\frac{x}{2} (2(\frac{1}{4} x+0.25)+b_2)$...?

18. Oct 8, 2012

### voko

Correct.

19. Oct 9, 2012

### bugatti79

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 $W(x)= 6.25(1+.05x)x=9.375kN$..?

20. Oct 9, 2012