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diff --git a/3864/CH6/EX6.17/Ex6_17.sce b/3864/CH6/EX6.17/Ex6_17.sce
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+clear
+//
+//
+
+//Initilization of Variables
+
+d1=80 //mm //External Diameter of Brass tube
+d2=50 //mm //Internal Diameter of Brass tube
+d=50  //mm //Diameter of steel Tube
+G_b=40*10**3 //N/mm**2 //Modulus of Rigidity of brass tube
+G_s=80*10**3 //N/mm**2 //Modulus of rigidity of steel tube
+T=6*10**6 //N-mm //Torque
+L=2000 //mm //Length of Tube
+
+//Calculations
+
+//Polar Modulus of brass tube
+J1=%pi*32**-1*(d1**4-d2**4) //mm**4 
+
+//Polar modulus of steel Tube
+J2=%pi*32**-1*d**4 //mm**4
+
+//Let T_s & T_b be the torque resisted by steel and brass respectively
+//Then, T_b+T_s=T   ............................................(1)
+
+//Since the angle of twist will be the same
+//Theta1=Theta2
+//After substituting values and further simplifying we get 
+//Ts=0.360*Tb     ...........................................(2)
+
+//After substituting value of Ts in eqn 1 and further simplifying we get 
+T_b=T*(0.36+1)**-1 //N-mm
+T_s=0.360*T_b
+
+//Let q_s and q_b be the max stress in steel and brass respectively
+q_b=T_b*(d1*2**-1)*J1**-1 //N/mm**2
+q_s=T_s*(d2*2**-1)*J2**-1 //N/mm**2
+
+//Since angle of twist in brass=angle of twist in steel
+theta_s=T_s*L*(J2*G_s)**-1
+
+//Result
+printf("\n Stresses Developed in Materials are:Brass %0.2f  N/mm**2",q_b)
+printf("\n                                    :Steel %0.2f  N/mm**2",q_s)
+printf("\n Angle of Twist in 2m Length %0.3f  Radians",theta_s)