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// Problem 11.5,Page no.277
clc;clear;
close;
b=4 //m //width of %pier
d=3 //m //depth of %pier
e_x=1 //m //distance from y axis
e_y=0.5 //m //distance from x axis
P=80 //KN //Load
//Calculations
A=b*d //m**2 //Area of %pier
I_x_x=b*d**3*12**-1 //m**4 //M.I about x-x axis
I_y_y=d*b**3*12**-1 //m**4 //M.I about y-y axis
M_x=P*e_y //KN*m //Moment about x-x axis
M_y=P*e_x //KN*m //Moment about y-y axis
x=2 //m //Distance between y-y axis and corners A and B
y=1.5 //m ////Distance between x-x axis and corners A and D
//Part-1
//Stress developed at each corner
sigma_A=P*A**-1+M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 //KN/m**2 //stress at A
sigma_B=P*A**-1+M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 //KN/m**2 //stress at B
sigma_C=P*A**-1-M_x*y*I_x_x**-1+M_y*x*I_y_y**-1 //KN/m**2 //stress at C
sigma_D=P*A**-1-M_x*y*I_x_x**-1-M_y*x*I_y_y**-1 //KN/m**2 //stress at D
//Part-2
//Let f be the additional load that should be placed at centre
//sigma_c=F*A**-1 //KN/m**2 //compressive stress
//For no tension in %pier section, compressive stress is equal to tensile stress
sigma_c=10 //KN/m**2
F=sigma_c*A //KN
//Part-3
sigma=F*A**-1 //KN/m**2 //stress due to additional load of 120 KN
sigma_A_1=sigma_A+10 //stress at A
sigma_B_1=sigma_B+10 //stress at B
sigma_C_1=sigma_C+10 //stress at C
sigma_D_1=sigma_D+10 //stress at D
//Result
printf("Stress at each corner are as follows:stress_A %.2f",sigma_A);printf(" KN/m**2")
printf("\n :stress_B %.2f",sigma_B);printf(" KN/m**2")
printf("\n :stress_C %.2f",sigma_C);printf(" KN/m**2")
printf("\n :stress_D %.2f",sigma_D);printf(" KN/m**2(tensile)")
printf("\n\n Additional load that should be placed at centre is %.2f",F);printf(" KN")
printf("\n\n Stresses at the corners with the additional load in centre are as follows:Stress_A_1 %.2f",sigma_A_1);printf(" KN/m**2")
printf("\n :Stress_B_1 %.2f",sigma_B_1);printf(" KN/m**2")
printf("\n :Stress_C_1 %.2f",sigma_C_1);printf(" KN/m**2")
printf("\n :Stress_D_1 %.2f",sigma_D_1);printf(" KN/m**2")
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