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%%刚体系统受力分析+符号推导  %%%
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clc;          %清除屏幕
clear all;    %清除内存

P=10;Q=20;    %载荷

theta15=15*pi/180;theta45=45*pi/180;   %角度
theta60=60*pi/180;theta75=75*pi/180;   %角度

A=zeros(12,12);                        %A矩阵清零
B=[0;0;0;0;-P;0;0;0;0;Q;0;0];          %B矩阵

%----下面是A矩阵中非0元素
A(1,1)=cos(theta15);A(1,7)=1;A(1,9)=cos(theta60);A(1,10)=1;
A(2,1)=sin(theta15);A(2,9)=sin(theta60);A(2,11)=1;
A(3,2)=-cos(theta45);A(3,7)=-1;A(3,8)=-cos(theta60);
A(4,2)=sin(theta45);A(4,8)=sin(theta60);A(4,12)=1;
A(5,3)=-cos(theta75);A(5,8)=cos(theta60);A(5,9)=-cos(theta60);
A(6,3)=-sin(theta75);A(6,8)=-sin(theta60);A(6,9)=-sin(theta60);
A(7,3)=cos(theta75);A(7,4)=cos(theta60);A(7,6)=1;
A(8,3)=sin(theta75);A(8,4)=-sin(theta60);
A(9,1)=-cos(theta15);A(9,4)=-cos(theta60);A(9,5)=cos(theta60);
A(10,1)=-sin(theta15);A(10,4)=sin(theta60);A(10,5)=sin(theta60);
A(11,2)=cos(theta45);A(11,5)=-cos(theta60);A(11,6)=-1;
A(12,2)=-sin(theta45);A(12,5)=-sin(theta60);

X=inv(A)*B;    %求解

disp(['S1=',num2str(X(1)),'N'])%在屏幕上显示结果
disp(['S2=',num2str(X(2)),'N'])%在屏幕上显示结果
disp(['S3=',num2str(X(3)),'N'])%在屏幕上显示结果
disp(['S4=',num2str(X(4)),'N'])%在屏幕上显示结果
disp(['S5=',num2str(X(5)),'N'])%在屏幕上显示结果
disp(['S6=',num2str(X(6)),'N'])%在屏幕上显示结果
disp(['S7=',num2str(X(7)),'N'])%在屏幕上显示结果
disp(['S8=',num2str(X(8)),'N'])%在屏幕上显示结果
disp(['S9=',num2str(X(9)),'N'])%在屏幕上显示结果
disp(['XA=',num2str(X(10)),'N'])%在屏幕上显示结果
disp(['YA=',num2str(X(11)),'N'])%在屏幕上显示结果
disp(['YB=',num2str(X(12)),'N'])%在屏幕上显示结果