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+// Problem no 9.2,Page no.233
+
+clc;clear;
+close;
+
+
+L=1.5 //m //Length of steelbar
+b=2 //cm //bredth of steelbar
+d=0.5 //cm //depth of steelbar
+sigma=320 //MPa //Yield point
+E=210 //GPa //modulus of Elasticity of steelbar
+
+//Calculations
+
+I_min=b*d**3*12**-1*10**-8 //m**4 //Moment of Inertia
+P=%pi**2*E*10**9*I_min*(L**2)**-1 //N //N //Crippling Load
+
+//Let dell=Central Deflection
+
+//M=P*dell //Max Bending moment
+//After substituting value in above equation we get
+//M=191.9*dell
+
+A=b*d*10**-4 //m**2 //Area of steel bar
+sigma_1=P*A**-1*10**-6 //Mpa //Direct stress
+
+Z=b*d**3*10**-6 //Section modulus
+//sigma_2=M*Z**-1 //N/m**2 //Bending stress
+//After substituting value in above equation we get
+//sigma_2=dell*2302.8*10**6 //N/m**2
+
+//sigma=sigma_1+sigma_2
+//Now substituting value of Bending stress and direct stress in above equation we get
+
+//320*10**6=1.919*10**6+2302.8*10**6*dell
+dell=((320*10**6-1.919*10**6)*(2302.8*10**6)**-1)*10**2 //cm //Central Deflection
+
+//Result
+printf("Maximum Central Deflection is %.2f",dell);printf(" cm")