Analyzing the Leaning Tower of Pisa: Solving for C1 and C2

[unscientific]

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?

 

[Simon Bridge]

It helps if you show us how you have been trying.
Can you not find the max and min locations without knowing C₂?

 

[voko]

What are the equilibrium conditions?

 

[unscientific]

What are the equilibrium conditions?

1) Vector Sum of forces about any direction = 0

2)no resultant moment about any point

 

[voko]

That should give you two somewhat different integrals, hence you should be able to determine the two constants from them.

 

[unscientific]

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..

 

[unscientific]

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..

 

[voko]

Observe how the second term of pressure behaves. Its moment about O must be non-zero.

 

[abel_ghita]

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
\partial^2p/\partialr^2 * dr+ \partial^p/\partialθ^2 * dθ
and your only variable is C2

 

[unscientific]

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
\partial^2p/\partialr^2 * dr+ \partial^p/\partialθ^2 * dθ
and your only variable is C2

And with that, you only get r = 0... when u equate each = 0 ..

 

[voko]

∫ 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.

 

[unscientific]

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...

 

[voko]

Draw the resultant moment in a few different places and you will see different directions. You can't add them together as scalars.

 

[unscientific]

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..

 

[voko]

Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?

 

[unscientific]

Expand the vectors involved in terms of the coordinate unit vectors i, j, k. Do the vector product. What do you get?

d\overline{M}
= \overline{r} x d\overline{F}
= (\overline{r} x \overline{p}) dA
= (r cosθ \widehat{i} + r sinθ \widehat{j}) x (C₁ + C₂ r^0.625 cosθ)\widehat{k}
= -( C₁ cosθ + C₂ r^1.625 cos²θ) \widehat{j} + (C₁ r sinθ + C₂ r^1.625 sinθcosθ ) \widehat{i}


Along j

\int d\overline{M}
= - \int^{9.8}_{0} dr \int^{2∏}_{0} dθ [C₁ r cosθ + C₂ r^1.625 cos²θ]
= -478.9 C₂

Along j
Everything cancels out to be zero!

 

[voko]

Do the vector product :)

 

[voko]

I am sure you can integrate that now.

 

[unscientific]

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

 

[voko]

So, have you solved the entire problem?

 

[unscientific]

So, have you solved the entire problem?

I found C₁ and C₂;

C₁ = 477 300
C₂ = 799 832

which gives:

p_min = -2853243
p_max = 3807843

How can there be negative pressure? It doesn't make any sense... (Given the initial assumption that all forces act upwards)

 

[voko]

Your answers are obviously incorrect, pressure cannot be negative. I guess the equation of forces or torques is incorrect.

 

[unscientific]

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

\int dF
= \int\int pr dr dθ
= \int^{9.8}_{0} dr \int^{2∏}_{0} dθ [C₁r + C₂r^1.625 cosθ ]
= 2∏ \int^{9.8}_{0} 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)

 

[voko]

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 :)

 

[unscientific]

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:

= (\bar{r} x \bar{F})
= (\bar{r} x \bar{p}) dA

 

[voko]

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.

 

[unscientific]

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:

P_min = 8109

P_max = 951 891

 

[unscientific]

Assume: \bar{P} acts along tower

\bar{P} = ( C₁ + C₂r^0.625 cosθ )cos5.6^o \hat{k} + ( C₁ +C₂r^0.625 cosθ )sin 5.6^o ) \hat{i}We know vector sum of forces along \hat{k} must be equal to 0.

We first find the contribution by the soil:

Along k

dF = ( C₁ + C₂r^0.625 cosθ)cos 5.6^o r*dθ dr\int dF = cos 5.6^o \int^{9.8}_{0} dr \int^{2π}_{0} dθ [C₁ r + C₂r^1.625 cosθ ]
= 300C₁

300C₁ = 144*10⁶
C₁ = 480000
Finding C₂

d\bar{M}
= \bar{r} x d\bar{F}
= \bar{r} x \bar{P}dA
= [r cosθ \hat{i} + r sinθ \hat{j}] x [ (C₁+C₂r^0.625cosθ)cos5.6^o \hat{k} + (C₁+C₂r^0.625cosθ)sin5.6^o \hat{i}] (r*dr dθ)

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

Thus the only contribution is along \hat{j}

-cos 5.6^o \int^{9.8}_{0} dr \int^{2π}_{0} dθ [C₁r²cosθ + C₂r^2.625cos²θ]

=-3380 C₂

Thus,

3380 C₂ = (144*10⁶)(27.1 * sin 5.6^o)
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...)

 

[voko]

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