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+// Problem no 13.5,Page No.304
+
+clc;clear;
+close;
+
+D=50 //cm //Overall depth
+b=19 //cm //width of flange
+t=2.5 //cm //Thickness of Flange
+w_t=1.5 //cm //Web thickness
+w_d=45 //cm //web thickness
+F=400*10**3 //N //Shear Force
+I=64500*10**-8 //m**4 //M.I
+
+//Calculations  (Part-1)
+ 
+a_1=b*t //cm**2 //Area of flange
+a_2=w_d*w_t //cm**2 //Area of web
+y_1=t*2**-1 //cm //C.G of flange
+y_2=t+w_d*2**-1
+
+//As section is symmetrical 
+Y=D*2**-1 //cm
+
+//Shear Stress in flange at the junction with web
+//Let tou(Shear stress)=S 
+//Change in the notifications of Shear Stress For convenience
+S_1=(F*a_1*(Y-y_1)*10**-6)*(I*b*10**-2)**-1*10**-3 
+
+//Shear Stress in web at the junction with flange 
+S_2=(F*a_1*(Y-y_1)*10**-6)*(I*w_t*10**-2)**-1*10**-3 
+
+//Max Shear Stres at N.A
+S_max=(F*(a_1*(Y-y_1)+(w_t*(Y-t))*((Y-t)*2**-1))*10**-6)*(I*w_t*10**-2)**-1*10**-3 //kPa
+
+//Calculations (Part-2)
+
+//consider a strip in the flange of thickness dy at a distance y from N.A 
+
+//S=F*(b*(Y-y)*(Y+y)*2**-1*10**-6)*(I*b*10**-2)**-1
+//after substituting values we get
+//S=625-y**2*(3225*10**-8)**-1
+
+//shear force carried by small strip
+//F_1=625-y**2*(3225*10**-8)**-1*b*dy*10**-4
+
+//Now Integrating above Equation we get
+a =625
+b =-1
+I = integrate('625-y**2','y', 22.5, 25)//, args=(a,b))
+//Shear force carried by one flange
+F_1=19*3225**-1*10**4*I
+
+//Shear force carried by two flange
+F_2=2*F_1
+
+//Shear force carried by web
+F_3=F-F_2
+
+//Result
+printf("The shear Force int the section is %.2f",S_max);printf(" kPa")
+printf("\n Total Shear Force in the web is %.2f",F_3);printf(" N")
+
+
+printf("\n Shear stress distribution Diagram")
+
+//Plotting the Shear stress distribution Diagram
+
+X_1=[0,2.5,2.5,25,47.5,47.5,50]
+Y_1=[0,S_1,S_2,S_max,S_2,S_1,0]
+Z_1=[0,0,0,0,0,0,0]
+plot(X_1,Y_1,X_1,Z_1)
+xlabel("Length x in m")
+ylabel("Shear Stress in kN/m**2")