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+clc
+//
+//
+
+//Variable declaration
+n=-1
+P1=15 // Force(kN)
+P2=18 // Force(kN)
+a=50 // Distance(mm)
+b=60 // Distance(mm)
+c=0.020 // Distance(m)
+F=P1 // Force(kN)
+V=P2 // Force(kN)
+t=0.040 // Distance(m)
+Iz=125.7*((10**-9)) // Moment of inertia(m**4)
+
+//Calculation
+//Internal Forces in Given Section
+T=P2*a // Torque(N.m)
+My=P1*a // Moment(N.m)
+Mz=P2*b // Moment(N.m)
+// Case(a) Normal and Shearing Stresses at Point K
+// Geometric Properties of the Section
+A=(%pi)*(c**2) // Area of cross section(m**2)
+Iy=(1/4.0)*(%pi)*(c**4) // Moment of inertia(m**4)
+Jc=(1/2.0)*(%pi)*(c**4) // Moment of inertia(m**4)
+Q=(A/2.0)*((4*c)/(3.0*(%pi)))
+t=2*c // Distance(m)
+// Normal Stresses
+Sx=(n*(F/A))/(1000.0) + ((My*c)/(Iy))/(1000000.0) // Normal stress(MPa)
+// Shearing Stresses
+txyV=((V*Q)/(Iz*t))/(1000.0) // Shearing stress(MPa)
+txytwist=((n*(T*c))/(Jc))/(1000000.0) // Shearing stress(MPa)
+txy=(txyV + txytwist) // Shearing stress(MPa)
+// Case(b) Principal Planes and Principal Stresses at Point K
+CD=(1/2.0)*(107.4) // Stress(MPa)
+OC=(1/2.0)*(107.4) // Stress(MPa)
+DX=52.5 // Stress(MPa)
+phyp=44.4/2.0 // Angle(degree)
+R=sqrt(53.7**2 + 52.5**2) // Stress(MPa)
+Smax=OC+R // Maximum principal stress(MPa)
+Smin=OC-R // Minimum principal stress(MPa)
+// Case(c) Maximum shearing stress at point k
+tmax=75.1 // Shearing stress(MPa)
+
+// Result
+printf("\n Case(a) Normal stress = %0.3f MPa' ,Sx)
+printf("\n Case(a) Shearing stress = %0.3f MPa' ,txy)
+printf("\n Case(b) Principal axis angle = %0.3f degree' ,phyp)
+printf("\n Case(b) Maximum principal stress at point k = %0.3f MPa' ,Smax)
+printf("\n Case(b) Minimum principal stress at point k = %0.3f MPa' ,Smin)
+printf("\n Case(c) Maximum shearing stress at point k = %0.3f MPa' ,tmax)