Hi, I have an equation which is sh

[aj_k]

Hi,
I have an equation which is shown below, which says qc (cone resistance) = .....
Now i need to calculate the effective angle of internal friction from the below formula. So i need to make a new formula
which starts with φ'= ......
q[sub]c[/sub]= 1.3 × exp[2π tanφ'] × tan[sup]2[/sup][(π/4)+ fracφ'2] × σ[sub]v0[/sub]
And i have no idea what is this fracφ'2.
I am stuck with this value.
I cannot use any other formula to calculate φ'. It is that i have to use this formula./
Does anyone know how to reverse the equation so that i can calculate effective angle of internal friction?
The qc value and effective stress are given.

 

[BigH]

From where does this equation of yours come? Source?

 

[aj_k]

This is the equation for calculating φ' (effective friction angle) by "De Beer" method. In Belgium practice, they follow this method for calculating the pile bearing capacity. And i am in the starting stage of the calculation. I cannot go any further if i dont have an equation for φ'. I have to follow this method, because the project is in Belgium.
I have tried to simplify this equation and now i am getting like

Let θ be φ' in the above equation and x is a constant.
Now i need to solve for θ in terms of x.
Does anyone know how to do it?

 

[IRstuff]

Unless you luck out with Wolfram Alpha or some other symbolic solver, I would use plug into Excel or MAthcad and numerically solve for it.

It seems odd that an equation in general use for this application wouldn't already have a symbolic solution or approximation.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[HTURKAK]

This is one of the means to estimate the friction angle Ø using cone bearing.
qc= 1.3 × exp[2π tanØ] × tan2[(45 + Ø/2] × σv0)
σv0 = ɣ'z (overburden pressure) and q'c are both "effective" stresses and you must know them..

You can simplfy the equation (σv0)/(1.3*σv0)= exp[2π tanØ] × tan2[(45 + Ø/2]

Now, the solve the implicit function with numerical analysis.. One of the methods, trial ,..

In your final formula, take natural log of both sides ln x = 2π tanØ + 2ln ( π/4+ Ø/2)

tanØ = (ln x-2ln ( π/4+ Ø/2))/( 2π ) now start with trial Ø=30 degr but in radians (π /6) you will get another Ø
and try with new Ø.. The formula will converge after 4-5 trials..

Or you may look to the Figure 3-22 in the famous book Foundation Analysis and Design Joseph E. Bowles page 180.

I have attached the graph..

 

[aj_k]

Thank you so much for your help.
But in the above comment, when we take log on both sides, tan2 is missing from your equation.

And also in the attached graph, according to De Beer, the equation is used on the condition that : when Ø´ is greater than Ø(provided). Inorder to find out this Ø´ i have use the below equation.

And for calculating the effective friction angle, i need to convert the above formula into Ø´= ....
Where the values of Ø , Ckd(=qc) and σv0 will be provided.

 

[HTURKAK]

Typing mistakes...yes tan**2 is forgotten...

qc= 1.3 × exp[2π tanØ] × tan2[(45 + Ø/2] × σv0)

qc/(1.3σv0 ) = exp[2π tanØ] × tan2[(45 + Ø/2]

C= exp[2π tanØ] × tan2[(45 + Ø/2]

Now take the natural log of both sides,

ln C= 2π tanØ + 2*ln tan [(45 + Ø/2] another way, define function Ω = 2π tanØ + 2*ln tan [(45 + Ø/2] -Ln C

The value of Ø which the function Ω = 0 will be the answer..

Regarding your quote: Vb" = Ckd /Pb where Pb = y'z (overburden pressure) and Ckd (cone resistance) are both "effective" stresses ..

 

[aj_k]

Thank you so much for your help.

But i am sorry beacuse still i couldnt get an equation for Ø
When Ω = 0 ,
0= 2π tanØ + 2tan [(45 + Ø/2] - LnC
In this above equation , Suppose C= 20, How to get a value for Ø
I need to get the Ø value in terms of C. There are still two Ø terms in this equation.
I cannot go for derivation again, because if derivate again, this C will become zero.
I need to calculate the values of Ø for different values of C.
Thank you so much.

 

[HTURKAK]

C is a constant used for simplification of the formula...In this case , C=qc/(1.3σv0 )..You must know the σv0 (overburden pressure) and qc (cone resistance) both effective stress...

0= 2π tanØ + 2tan [(45 + Ø/2] - LnC

This nonlinear eq. can be solved with iterative methods. For example, Bisection method,

Give arbitrary value to Ø .. Consider two points, (a, f (a)), and (b, f (b)) which the function having (+ and - ) opposite signs .Thus, we know that there is a root of f in the interval (a, b). Let c be the midpoint of the interval [a, b], so that c = (a+b)/2, and evaluate f (c). One of the following is true: Either f (c) = 0, f (a) and
f (c) have opposite signs, or f (c) and f (b) have opposite signs. Now try d=(a+c)/2 or d=(b+c)/2 ..a or b will be decided so that they must opposite sign with c...After a few iteration , the diffrence will be negligible and you will find Ø .

 

[aj_k]

Thank you so much for your help.
I will try this method