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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Initiization of variables
+W1=100 // N // Pt load at C
+W2=150 // N // Pt load at D
+W3=200 // N // Pt load at E
+l=1 // m // l=Lac=Lcd=Lde=Leb
+h=2 // m // dist between Rb & top
+Xa=200 // N
+Xb=200 // N
+// Calculations
+// consider the F.B.D of entire cable
+// Take moment at A
+Yb=((W1*l)+(W2*2*l)+(W3*3*l)-(Xb*h))/(4*l) // N
+Ya=W1+W2+W3-Yb // N // sum Fy=0
+// Now consider the F.B.D of AC
+// Take moment at C,
+y_c=(Ya*l)/Xa // m
+theta_1=atand(y_c/l) // degree
+T_AC=Xa/cosd(theta_1) // N // T_AC*cosd(theta_1)=horizontal component of tension in the cable
+// here, T_AC=T_max
+T_max=T_AC // N
+// Now consider the F.B.D of portion ACD
+y_d=((Ya*2*l)-(W1*l))/(Xa) // m // taking moment at D
+theta_2=atand(((y_d)-(y_c))/(l)) // degree
+T_CD=Xa/(cosd(theta_2)) // N
+// Results
+clc
+printf('(i) The component of support reaction at A (Ya) is %f N \n',Ya)
+printf('(i) The component of support reaction at B (Yb) is %f N \n',Yb)
+printf('(ii) The tension in portion AC (T_AC) of the cable is %f N \n',T_AC)
+printf('(ii) The tension in portion CD (T_CD) of the cable is %f N \n',T_CD)
+printf('(iii) The max tension in the cable is %f N \n',T_max)