Calculating Failure Angle of Silo Wedge

Question

I am working on an optimization problem, specifically calculating the lateral earth pressures on the inside of a silo wall. I have been trying to solve for the failure angle of a silo wedge based on its volume and weight using the Coloumb method. I am aware of the Jannsen method but am doing research on alternative methods specifically using the Coloumb method. There are two ways to calculate the earth pressure:

$$Pa = {\gamma*H^2 \over 2}*Ka$$

$$Pa = {Weight*sin[\rho-\phi] \over sin[\pi-\alpha-\rho+\phi+\delta]}$$

I have solved for the centroid, area, volume, and weight for the wedge of silo material. Then I calculated the weight per unit length of silo wall and used that in the Pa equation. I then set the Pa equations equal to each other and solved for Ka.

$$Ka = {2*Weight*sin[\rho-\phi] \over \gamma*H^2*sin[\pi-\alpha-\rho+\phi+\delta]}$$

Which simplified is

$$Ka = \frac{csc[\alpha-\delta+\rho-\phi]*sin[\rho-\phi]*(-4*n+tan[\beta]+tan[\rho])}{3*(8*n^2-6*n*tan[\beta]+tan^2[\beta])}$$

Where n is the aspect ratio of the height of the pile divided by the diameter of the silo. I took the derivative of this equation with respect to $\rho$ and set it equal to zero.

$$\frac{\partial Ka}{\partial \rho}=0$$

This is where Mathematica started freezing. I attached a picture of my calculations and am looking to simplify this answer to solve for $\rho$ and possibly get a closed-form solution.Treat all other variables not given as constants.

This shows the steps for calculating the centroid, area, volume, and weight for the wedge of silo material.

This shows how to get the force per unit width of the wall. It then shows Coulombs lateral earth pressure equations set equal to each other solving for the variable Ka. The derivative of Ka is taken with respect to rho and then set to zero.

Answer 1

An idea would be to expand these trigonometric expressions and replace:

$$ x = \cos\rho, \quad \quad \quad y = \sin\rho, \quad \quad \quad x^2 + y^2 = 1. $$

Now we have reduced to a rational system and Mathematica is happy:

h0 = 2 n r;
h = h0 - r Tan[β];

c = Integrate[x (h0 + Tan[β] x), {x, 0, -r}] / Integrate[h0 + Tan[β] x, {x, 0, -r}] -    
    Integrate[x (x + r) Tan[ρ], {x, 0, -r}] / Integrate[(x + r) Tan[ρ], {x, 0, -r}];

a = Integrate[h0 + Tan[β] x, {x, -r, 0}] - Integrate[(x + r) Tan[ρ], {x, -r, 0}];

v = 2 π c a;
w = γ v;
wp = w / (2 π r);
k = 2 wp Sin[ρ - ϕ] / (γ h^2 Sin[π - α - ρ + ϕ + δ]);

dk = Simplify[TrigExpand[D[k, ρ]] /. {Cos[ρ] -> x, Sin[ρ] -> y, Tan[ρ] -> y/x, 
                                      Sec[ρ] -> 1/x, Csc[ρ] -> 1/y, Cot[ρ] -> x/y}];

sol = Solve[{dk == 0, x^2 + y^2 == 1}, {x, y}];
sol = Simplify[Thread[{ρ -> ArcTan[x, y] /. sol}]];

They aren't certainly nice expressions to see, but I've seen worse!