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Leaning Tower of Pisa

  • #1
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Homework Statement



The problem is as attached.


The Attempt at a Solution



I tried doing double integration of r from 0 to 9.8 and θ from 0 to 2pi. But here im assuming all forces from the soil act along the tower (no friction) and managed to solve C1.

But im not sure how to solve C2?
 

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Answers and Replies

  • #2
Simon Bridge
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It helps if you show us how you have been trying.
Can you not find the max and min locations without knowing C2?
 
  • #3
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What are the equilibrium conditions?
 
  • #4
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What are the equilibrium conditions?
1) Vector Sum of forces about any direction = 0

2)no resultant moment about any point
 
  • #5
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That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
 
  • #6
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That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
Yes, but the cos θ becomes sin θ and then from 0 to 2∏ it cancels out to be zero..
 
  • #7
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That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
Here's what I got:

Let infinitesimal moment be dM.

∫ dM = ∫ r dF = ∫ rP dA = ∫∫ r2P dr dθ...

Then when ∫cosθ becomes sin θ the limits going from 0 to 2∏ it becomes 0, so C2 cannot be found..
 
  • #8
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Observe how the second term of pressure behaves. Its moment about O must be non-zero.
 
  • #9
what about..
you know that Pmax is right underneeth the center of gravity which is r=2.67 and θ=0
and also Pmax/min is the second derivative of function p..which i suppose is
[itex]\partial[/itex]^2p/[itex]\partial[/itex]r^2 * dr+ [itex]\partial[/itex]^p/[itex]\partial[/itex]θ^2 * dθ
and your only variable is C2
 
  • #10
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what about..
you know that Pmax is right underneeth the center of gravity which is r=2.67 and θ=0
and also Pmax/min is the second derivative of function p..which i suppose is
[itex]\partial[/itex]^2p/[itex]\partial[/itex]r^2 * dr+ [itex]\partial[/itex]^p/[itex]\partial[/itex]θ^2 * dθ
and your only variable is C2
And with that, you only get r = 0... when u equate each = 0 ..
 
  • #11
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∫ dM = ∫ r dF = ∫ rP dA = ∫∫ r2P dr dθ...
Like I said, this cannot be zero. The specific mistake you are making is that of ignoring the vector nature of r and F. What you denote as dM is a vector product of r and dF. Work it out.
 
  • #12
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Like I said, this cannot be zero. The specific mistake you are making is that of ignoring the vector nature of r and F. What you denote as dM is a vector product of r and dF. Work it out.
yes, dM = r x F

but I assumed all F acts upwards, so vectors r and F are perpendicular so it simply becomes r*F...
 
  • #13
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Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.
 
  • #14
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Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.
That is true...I realized that only for the same angle they are pointing in the same direction

So how do we find the net moment? It's like trying to sum up all the vectors in different directions..
 
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  • #15
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Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?
 
  • #16
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Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?

d[itex]\overline{M}[/itex]
= [itex]\overline{r}[/itex] x d[itex]\overline{F}[/itex]
= ([itex]\overline{r}[/itex] x [itex]\overline{p}[/itex]) dA
= (r cosθ [itex]\widehat{i}[/itex] + r sinθ [itex]\widehat{j}[/itex]) x (C1 + C2 r0.625 cosθ)[itex]\widehat{k}[/itex]
= -( C1 cosθ + C2 r1.625 cos2θ) [itex]\widehat{j}[/itex] + (C1 r sinθ + C2 r1.625 sinθcosθ ) [itex]\widehat{i}[/itex]


Along j

[itex]\int[/itex] d[itex]\overline{M}[/itex]
= - [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2∏}_{0}[/itex] dθ [C1 r cosθ + C2 r1.625 cos2θ]
= -478.9 C2

Along j
Everything cancels out to be zero!
 
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  • #17
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Do the vector product :)
 
  • #18
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I am sure you can integrate that now.
 
  • #19
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I am sure you can integrate that now.
Already did! (Page 1 please) Sorry for the long edit....was trying to figure out how latex works
 
  • #20
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So, have you solved the entire problem?
 
  • #21
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So, have you solved the entire problem?
I found C1 and C2;

C1 = 477 300
C2 = 799 832

which gives:

pmin = -2853243
pmax = 3807843

How can there be negative pressure? It doesn't make any sense... (Given the initial assumption that all forces act upwards)
 
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  • #22
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Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.
 
  • #23
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Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.
I can't see what's wrong with my equations for torque?

But here's what I used for my force equations:

dF = p dA

dA = r dr dθ

[itex]\int[/itex] dF
= [itex]\int[/itex][itex]\int[/itex] pr dr dθ
= [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2∏}_{0}[/itex] dθ [C1r + C2r1.625 cosθ ]
= 2∏ [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [C1r]
= (9.8)2 ∏ C1
= 301.7 C1


Sum of forces in z-direction must be zero:

301.7 C1 = 144*106
C1 = 477 300


Not sure what's wrong here...

My assumptions are:
1) All forces ground acting on base are upwards (z-direction)
 
  • #24
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First of all, you are ignoring that the base slab is inclined. So the total force of pressure is different from the total weight.

Secondly, in the torque integral I think you did not express the area element correctly.

Thirdly, since you have not shown the torque equation, I can't say much else :)
 
  • #25
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First of all, you are ignoring that the base slab is inclined. So the total force of pressure is different from the total weight.

Secondly, in the torque integral I think you did not express the area element correctly.

Thirdly, since you have not shown the torque equation, I can't say much else :)
Ok, given that the slab is inclined at 5.6 degrees, we can simply take the answer * cos 5.6 but it would roughly give the same answer..

I'm not sure where I went wrong for my torque equation:

= ([itex]\bar{r}[/itex] x [itex]\bar{F}[/itex])
= ([itex]\bar{r}[/itex] x [itex]\bar{p}[/itex]) dA
 

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