3.0Solution to the Problem

3.1Analytical Problem

Force acting on dam

Force from the pressure distribution

Analytical diagram for each force (assuming breadth = 1)

At

At F1

From the hydrostatic equation, by assuming breath = 1

A

At F2

From the hydrostatic equation, by assuming breath = 1

A

At /

//

By assuming breadth = 1,

/ +

/What is the value of? Assume b = 1

For equilibrium dam, so that

(1)

and with

23.60kN/Hence, the following equation can be present in term of and

Then, substituted of these expression into equation 1

Thus, equation can be simplify into,

(2)After that, from the given , we can determine the by using and the equation (2) will be uses to determine the unknown . For the dam width, l specified, the angle,( and the maximum water depth, h are computed in the Table 1.

614.4788078275.96375653

918.3411524569.44395478

1221.5593547963.43494882

1524.3271161157.99461679

1826.7530299953.13010235

Table 1: The value for and for each given

3.2Numerical MethodNumerical method is the way to solve the mathematical problem by reformulated to be solved in arithmetic operation. From the equation (2), the following term of is uses in numerical method.

The equation can be approximated into:

The equation has rearranged by solving using Newton-Raphson Method as follow:

3.2.1Algorithm

1.Start

2.Input the value of dam width, l

3.Compute the value of theta

theta = atan(24/)

4.State the values that known as parameters, starting approximation for root, h, error, eps, tolerance, tol, maximum number of iteration, total, and first iteration, k

5.Predefined the function of f and its derivative, f1

f =(1.635*(h^3))+(3.27*(l^2)*h)-(183.895*(l^2))-(44.145/(sin(theta)^2))

f1 =(4.905*(h^2)) + (3.27*(l^2))6.Calculate for new root value, hnew

hnew = h-f/f1

7.Compute the absolute error, ea

eps = abs(hnew-h)

ea = eps/hnew

8.Show the values of the number of iteration, k, maximum height, h and absolute error, ea

9.Finish3.2.2Flowchart

3.2.3Coding Program

The maximum water height at every dam width can be compute by using Matlab coding as given below:

% Name : Muhammad Aliff Irfan Bin Mohd Suhaimi% IC Number: 940803035865% Matric Number: A12KM0249% Year/Course : 3rd year/Naval Architecture & Offshore Engineering% Project : 1% Lecturer : Dr. Yasser Mohamed Ahmed Abdel Razak% Date : 17 December 2014% to find the maximum water height, h (m) by using Newton-Raphson% equation of function in term of h,f(h) = (1.635*(h^3))+(3.27*(l^2)*h)-(183.895*(l^2))-(44.145/(sin(theta)^2))l = input('Dam Width (m) : ');theta = atan(24/l); % angleh = 24; % the value starting approximate to the rooteps = 1;tol = 10^(-14);total = 100;k = 0;format long;while ((eps > tol) & (k < total)) f = (1.635*(h^3))+(3.27*(l^2)*h)-(183.895*(l^2))-(44.145/(sin(theta)^2)); f1 = (4.905*(h^2))+(3.27*(l^2)); % First derivation at h = h_khnew = h-f/f1; % New approximation value for the rooteps = abs(hnew-h);h = hnew;k = k+1;ea = eps/hnew;fprintf('k = %2.0f, x = %12.10f, ea = %12.10f\n',k,h,ea);endF2

F1

A

W

C

F3

F1

F1

F2

C

+Type equation here.