Leaning Tower of Pisa

  • #1
unscientific
1,734
13

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?
 

Attachments

  • pisa1.jpg
    pisa1.jpg
    51.6 KB · Views: 401
Last edited:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,657
It helps if you show us how you have been trying.
Can you not find the max and min locations without knowing C2?
 
  • #3
voko
6,054
391
What are the equilibrium conditions?
 
  • #4
unscientific
1,734
13
What are the equilibrium conditions?

1) Vector Sum of forces about any direction = 0

2)no resultant moment about any point
 
  • #5
voko
6,054
391
That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.
 
  • #6
unscientific
1,734
13
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
unscientific
1,734
13
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
voko
6,054
391
Observe how the second term of pressure behaves. Its moment about O must be non-zero.
 
  • #9
abel_ghita
9
0
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
unscientific
1,734
13
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
voko
6,054
391
∫ 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
unscientific
1,734
13
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
voko
6,054
391
Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.
 
  • #14
unscientific
1,734
13
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..
 
Last edited:
  • #15
voko
6,054
391
Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?
 
  • #16
unscientific
1,734
13
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!
 
Last edited:
  • #17
voko
6,054
391
Do the vector product :)
 
  • #18
voko
6,054
391
I am sure you can integrate that now.
 
  • #19
unscientific
1,734
13
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
voko
6,054
391
So, have you solved the entire problem?
 
  • #21
unscientific
1,734
13
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)
 
Last edited:
  • #22
voko
6,054
391
Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.
 
  • #23
unscientific
1,734
13
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
voko
6,054
391
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
unscientific
1,734
13
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
voko
6,054
391
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
unscientific
1,734
13
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
unscientific
1,734
13
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...)
 

Attachments

  • axes.jpg
    axes.jpg
    3.6 KB · Views: 342
Last edited:
  • #29
voko
6,054
391
Explain why you have cos 5.6 in front of both integrals. That does not seem correct to me.
 

Suggested for: Leaning Tower of Pisa

  • Last Post
Replies
2
Views
276
  • Last Post
Replies
1
Views
501
  • Last Post
Replies
9
Views
512
Replies
27
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
728
Top