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//
//
//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)
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