Two points Catenary equation [2/2]

 

In previous post we obtain catenary equation.

 

 

Following is a example.

 

D = 500m h = 100m q = 10kN/m H = 3000kN (Horizontal force shall also be assumed) Boundary condition A(0,0) B(500,100) → z(0)=0 and z(500)=100 In first condition \[z(0)=\frac{H}{q}sinh(c_1)sinh(\frac{q}{H}0)+\frac{H}{q}cosh(c_1)cosh(\frac{q}{H}0)+c_2=0\] sinh(0)=0, cosh(0)=1 \[c_2=-\frac{H}{q}cosh(c_1)\] The equation is changed \[z=\frac{H}{q}sinh(c_1)sinh(\frac{q}{H}x)+\frac{H}{q}cosh(c_1)cosh(\frac{q}{H}x)-\frac{H}{q}cosh(c_1)\]

 

Second condition \[z(D)=\frac{H}{q}sinh(c_1)sinh(\frac{q}{H}D)+\frac{H}{q}cosh(c_1)cosh(\frac{q}{H}D)-\frac{H}{q}cosh(c_1)= h\] Solving c1 using Newton Raphson's method

\[f(c_1)=f(x)=z(D) - h = 0\] x₁(initial c1) = 1 

then c2 = - 3000 / 10 cosh(1) = -462.9241904

f(x₁) = f(1) = z(500) - 100 = 1606.285742 x₁ + dx(0.00001) = 1.00001, c2 = -462.9277161

f(1.00001) = 1606.3036996

 

\[f'(x)=\frac{f(x+dx)-f(x)}{dx}\]

 

f'(1) = [f(1.00001) - f(1)] / 0.00001 = 1795.81307

 

\[x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]

 

x₂ = 1 - 1606.285742 / 1795.81307 = 0.10553845 (second c1)

f(x₂) = 506.39333983 (not 0)

In the same way

f(x₃) = 85.38694039 (not 0)

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f(x₆) = 4.0165e-10 (same as 0)

c1 = -0.65567036 , c2 = -366.8291184

 

The equation

 

 

Now, you will solve z value when putting x value.

 

and you can check following Excel Link Excel