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1 files changed, 52 insertions, 0 deletions

diff --git a/3864/CH6/EX6.16/Ex6_16.sce b/3864/CH6/EX6.16/Ex6_16.sce
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+clear
+//
+//
+
+//Initilization of Variables
+
+L=1000 //mm //Length of bar
+L1=600 //mm //Length of Bar AB
+L2=400 //mm //Length of Bar BC
+d1=60  //mm //Outer Diameter of bar BC
+d2=30  //mm //Inner Diameter of bar BC
+d=60   //mm //Diameter of bar AB
+T=2*10**6 //N-mm //Total Torque
+
+//Calculations
+
+//Polar Modulus of Portion AB
+J1=%pi*32**-1*d**4 //mm*4
+
+//Polar Modulus of Portion BC
+J2=%pi*32**-1*(d1**4-d2**4) //mm**4
+
+//Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC
+//Let theta1 and theta2 be the rotation of shaft in portion AB & BC
+
+//theta1=T1*L1*(G*J1)**-1 //radians
+//After substituting values and further simplifying we get 
+//theta1=32*600*T1*(%pi*60**4*G)**-1
+
+//theta2=T2*L*(J2*G)**-1 //Radians
+//After substituting values and further simplifying we get 
+//theta2=32*400*T2*(%pi*60**4*(1-0.5**4)*G)**-1 
+
+//Now For consistency of Deformation,theta1=theta2
+//After substituting values and further simplifying we get 
+//T1=0.7111*T2  ..................................................(1)
+
+//But T1+T2=T=2*10**6   ...........................................(2)
+//Substituting value of T1 in above equation
+
+T2=T*(0.7111+1)**-1
+T1=0.71111*T2
+
+//Max stress in Portion AB
+q_s1=T1*(d*2**-1)*(J1)**-1 //N/mm**2
+
+//Max stress in Portion BC
+q_s2=T2*(d1*2**-1)*J2**-1 //N/mm**2 
+
+//Result
+printf("\n Stresses Developed in Portion:AB %0.2f  N/mm**2",q_s1)
+printf("\n                              :BC %0.2f  N/mm**2",q_s2)