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+// Problem no 4.4.15,Page No.105
+
+clc;clear;
+close;
+L=8 //m //Length of beam
+L_AD=4 //m //Length of AD
+w=300 //KN //u.d.l
+
+//Calculations
+
+//Let R_A and R_C be the reactions at A and C
+//R_A+R_C=300
+
+//Taking moment at A
+//LEt x be the distance from Pt B L_CB=x
+//R_C*(L-L_CB)=300*L*2**-1
+//R_C=1200*(8-x)**-1
+//After substituting values and further simplifying we get
+//R_A=300-R_C
+//R_A=1200-300*x*(8-x)**-1
+
+//B.M at D
+//M_D=R_A*L_AD-w*2**-1*2=0
+
+//Now substituting value of R_A we get
+//M_D=4*1200-300*x*(8-x)**-1-300=0
+
+//Further on simplification we get
+L_CB=600*225**-1
+x=L_CB;
+R_C=1200*(8-x)**-1
+R_A=(1200-300*x)*(8-x)**-1
+
+//Pt of contraflexure
+//Let E be the pt and BE=y
+//V_E=0=-R_A*2**-1*L_BE+R_C
+L_BE=R_C*(R_A*2**-1)**-1
+L_AE=L-L_BE
+L_AC=L-L_CB
+L_EC=L_BE-L_CB
+
+//Shear Force at B
+V_B=0
+
+//Shear Force at C
+V_C1=-w
+V_C2=-V_C1+R_C
+
+//Shear Force at A
+V_A=-w+R_C
+
+//B.M at C
+M_C=-w*L_CB
+
+//B.M at E
+M_E=-R_A*L_AE+w*L_AE
+
+//B.M at A
+M_A=0
+
+//B.M at B
+M_B=0
+
+//Result
+printf("The Shear Force and Bending Moment Diagrams are the results")
+
+//Plotting the Shear Force Diagram
+subplot(2,1,1)
+X1=[0,L_CB,L_CB,L_CB+L_AC,L_CB+L_AC]                   
+Y1=[V_B,V_C1,V_C2,V_A,0]
+Z1=[0,0,0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
+
+//Plotting the Bending Moment Diagram
+subplot(2,1,2)
+X2=[0,L_CB,L_CB+L_EC,L_CB+L_AC]
+Y2=[M_B,M_C,M_E,M_A]          
+Z2=[0,0,0,0]
+plot(X2,Y2,X2,Z2)
+xlabel("Length in m")
+ylabel("Bending Moment in kN.m")
+title("the Bending Moment Diagram")