Project Problem 3

here is my current code (unfinished)

% Analysis of the Structure using the SLope Deflection Method

% Inputs Values 

w= 30 % KN/m

I = 1 % Moment of Interia not given so value of one is used 

E = 1 % Youngs Modulus   not given so value of one is used

% Fixed End Moments (FEM) 

(FEM)ad = -(w*Lad^2)/20 

(FEM)da = (w*Lad^2)/20 

(FEM)dc = 0

(FEM)cd = 0

(FEM)cb = 0 

(FEM)bc = 0  

%Length of each member 

Lad = 3 

Ldc = 3 

Lcb = 3  

%Moment of Inertia in the sections of the beams 

Iad = 1

Idc = 1 

Icb = 1

% Stiffness factor in each member 

Kad = (Iad)/Lad 

Kdc = Idc/Ldc 

Kcb = Icb/Lcb 

% Angular disp calculation 

ThetaA = [2*E*Kad,0,1,zeros(1,4);-2*E*Kda,-2*E*Kda,zeros(1,3);-2*E*Kda,-4*E*Kbd,0,0,1,0,0;-2*E*Kcb,zeros(1,4),1,0;-2*E*Kbc,zeros(1,5),1;0,0,1,1,0,1,0; zeros(1,4),1,0,0]  

    ThetaB = [(FEM)ad;(FEM)da;(FEM)dc,(FEM)cd;(FEM)cb;(FEM)bc]

    F = ThetaA/ThetaB

% Equalibrum Equations are 

% Mdc + Mda = 0

% Mcd + Mcb = 0 

% Use above Equilibrium Equations to find the angles and the delta 

ThetaC= 261/(56*E*I) 

ThetaD = 9/(56*E*I) 

Delta = 153/(56*E*I)  

%Angular disp and moments 

fprintf('Angular displacement at C

% Equations for End moments are then calculated, and substitution of Mdc,Mda,Mcd, and Mcb into equations it was done on paper

%Mad = -25.93 

%Mda = -3.321 

%Mdc = 3.321 

%Mcd = 6.321

%Mbc = 9.43 

%Mcb = 6.321