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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")
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