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+clear
+//
+
+P1=(20) //vertical loading from A at distance of 1m,KN.
+P2=(20) //vertical loading from A at distance of 2m,KN.
+P3=(20) //vertical loading from A at distance of 3m,KN.
+Ra=(P1+P2+P3)/2 //Due to symmetry
+
+Rb=Ra
+//At section 1.5 m from A
+F=(Ra-P1)*1000
+M=((Ra*1.5-P1*0.5)*1000000)
+b=(100)
+h=(180)
+
+I=((b*(h**3))/12)
+
+// Bending stress
+//f=M*y/I
+y11=0
+f1=(-1)*M*y11/I
+y22=45
+f2=(-1)*M*y22/I
+y33=90
+f3=(-1)*M*y33/I
+//Shearing stress at a fibre ‘y’ above N–A is
+//q=(F/(b*I))*(A*y1)
+//at y=0,
+y1=45
+A1=b*90
+q1=(F/(b*I))*(A1*y1)
+//at y=45
+y2=(90-45/2)
+A2=b*45
+q2=(F/(b*I))*(A2*y2)
+//at y=90
+q3=0
+
+//(a) At neutral axis (y = 0) : The element is under pure shear
+
+py=0
+
+p1=(f1+py)/2+sqrt((((f1-py)/2)**2)+(q1**2))
+
+p2=(f1+py)/2-sqrt((((f1-py)/2)**2)+(q1**2))
+printf("\n (i) p1= %0.4f N/mm^2",p1)
+printf("\n p2= %0.4f N/mm^2",p2)
+
+theta1=45
+theta2=theta1+90
+printf("\n theta= %0.0f ° and %0.0f °",theta1,theta2)
+
+//(b) At (y = 45)
+py=0
+
+p1=(f2+py)/2+sqrt((((f2-py)/2)**2)+(q2**2))
+
+p2=(f2+py)/2-sqrt((((f2-py)/2)**2)+(q2**2))
+printf("\n (ii) p1= %0.4f N/mm^2",p1)
+printf("\n p2= %0.4f N/mm^2",p2)
+
+thetab1=(atan((2*q2)/(f2-py))*180)/(%pi*2)
+thetab2=thetab1+90
+printf("\n theta= %0.0f ° and %0.0f °",thetab1,thetab2)
+//mistake in book
+printf("\n mistake in book")
+
+//(c) At Y=90
+
+py=0
+
+p1=(f3+py)/2+sqrt((((f3-py)/2)**2)+(q3**2))
+
+p2=(f3+py)/2-sqrt((((f3-py)/2)**2)+(q3**2))
+printf("\n (iii) p1= %e N/mm^2",p1)
+printf("\n p2= %0.4f N/mm^2",p2)
+
+thetac1=(atan((2*q3)/(f3-py))*180)/(%pi*2)
+thetac2=thetac1+90
+printf("\n theta= %0.0f ° and %0.0f °",thetac1,thetac2)