Leaning Tower of Pisa

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In summary, the problem is about finding the equilibrium conditions for a tower with two unknown constants, C1 and C2. The problem involves performing double integration and using the equilibrium condition that the vector sum of forces in any direction must be zero. The solution involves expanding the vectors involved in terms of the coordinate unit vectors and performing a vector product. The final answer depends on the correct expression for the torque equation, which must include the area element. The assumptions made include assuming all forces act upwards and ignoring the inclination of the base slab.
  • #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 I am assuming all forces from the soil act along the tower (no friction) and managed to solve C1.

But I am not sure how to solve C2?
 

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  • #2
It helps if you show us how you have been trying.
Can you not find the max and min locations without knowing C2?
 
  • #3
What are the equilibrium conditions?
 
  • #4
voko said:
What are the equilibrium conditions?

1) Vector Sum of forces about any direction = 0

2)no resultant moment about any point
 
  • #5
That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
 
  • #6
voko said:
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
voko said:
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
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
abel_ghita said:
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
unscientific said:
∫ 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
voko said:
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
Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.
 
  • #14
voko said:
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
Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?
 
  • #16
voko said:
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
Do the vector product :)
 
  • #18
I am sure you can integrate that now.
 
  • #19
voko said:
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
So, have you solved the entire problem?
 
  • #21
voko said:
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
Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.
 
  • #23
voko said:
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 C1Sum of forces in z-direction must be zero:

301.7 C1 = 144*106
C1 = 477 300Not sure what's wrong here...

My assumptions are:
1) All forces ground acting on base are upwards (z-direction)
 
  • #24
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
voko said:
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
 
  • #26
The area element must include r, which it does not. That makes the integral on order of magnitude smaller than it should be, and the constant, consequently, an order of magnitude greater.
 
  • #27
Bearing in mind what you said, I reworked everything out and here's what I got:

C1 = 480 000 (Very nice number, I know)

C2 = 113 325

This gives:

Pmin = 8109

Pmax = 951 891
 
  • #28
Assume: [itex]\bar{P}[/itex] acts along tower

[itex]\bar{P}[/itex] = ( C1 + C2r0.625 cosθ )cos5.6o [itex]\hat{k}[/itex] + ( C1 +C2r0.625 cosθ )sin 5.6o ) [itex]\hat{i}[/itex]We know vector sum of forces along [itex]\hat{k}[/itex] must be equal to 0.

We first find the contribution by the soil:

Along k

dF = ( C1 + C2r0.625 cosθ)cos 5.6o r*dθ dr[itex]\int[/itex] dF = cos 5.6o [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2π}_{0}[/itex] dθ [C1 r + C2r1.625 cosθ ]
= 300C1

300C1 = 144*106
C1 = 480000
Finding C2

d[itex]\bar{M}[/itex]
= [itex]\bar{r}[/itex] x d[itex]\bar{F}[/itex]
= [itex]\bar{r}[/itex] x [itex]\bar{P}[/itex]dA
= [r cosθ [itex]\hat{i}[/itex] + r sinθ [itex]\hat{j}[/itex]] x [ (C1+C2r0.625cosθ)cos5.6o [itex]\hat{k}[/itex] + (C1+C2r0.625cosθ)sin5.6o [itex]\hat{i}[/itex]] (r*dr dθ)

Then you find when you integrate dθ from 0 to 2∏ components along [itex]\hat{i}[/itex] and [itex]\hat{k}[/itex] turn out to be 0.

Thus the only contribution is along [itex]\hat{j}[/itex]

-cos 5.6o [itex]\int[/itex][itex]^{9.8}_{0}[/itex] dr [itex]\int[/itex][itex]^{2π}_{0}[/itex] dθ [C1r2cosθ + C2r2.625cos2θ]

=-3380 C2

Thus,

3380 C2 = (144*106)(27.1 * sin 5.6o)
C2 = 113 325

I just realized my answer is more accurate than the one provided by the tutor! (The tutor assumed P acts only directly upwards in the z-direction...)
 

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  • #29
Explain why you have cos 5.6 in front of both integrals. That does not seem correct to me.
 

What is the Leaning Tower of Pisa?

The Leaning Tower of Pisa is a famous bell tower located in Pisa, Italy. It is known for its distinctive tilt, caused by an architectural flaw during its construction in the 12th century.

Why does the Leaning Tower of Pisa lean?

The tower's lean is due to an uneven settling of the ground it was built on, combined with its soft foundation. This caused the tower to lean to one side, giving it its iconic tilt.

Is the Leaning Tower of Pisa safe to visit?

Yes, the tower is safe to visit. In the late 20th century, engineers worked to stabilize the tower and prevent it from leaning any further. It is now open to tourists and has been declared safe for at least the next 200 years.

How tall is the Leaning Tower of Pisa?

The tower stands at 183.3 feet (55.86 meters) tall on the lowest side and 186.02 feet (56.71 meters) tall on the highest side. It has eight levels and a total of 294 steps to reach the top.

Can you go inside the Leaning Tower of Pisa?

Yes, you can go inside the tower and climb to the top. Tickets must be purchased in advance and there is a limit to the number of visitors allowed inside at one time. The climb can be challenging, with a steep incline and narrow steps, but the view from the top is worth it.

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