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+// Problem 8.13,Page no.215
+
+clc;clear;
+close;
+
+P=200 //N //Load
+h=10 //cm //Height of Load dropped
+n=10 //Number of turns
+b_1=5 //cm //width of plates
+t=6 //mm //thickness of plates
+L=75 //cm //Length of plates
+E=200 //GPa //Modulus of Elasticity
+
+//Calculaion
+
+//Let P be the equivalent gradually applied load whuch would cause the same stress as is caused by the impact Load
+//200(0.1+dell)=P*dell*2**-1 (Equation 1)
+
+//dell=3*P*L**3*(8*E*n*b*t**3)**-1 //Deflection
+//After substituting values in above equation and further simplifying we get
+//P=136533.33*dell
+
+//After substituting values of P in (equation 1) and further simplifying we get
+//200(0.1+dell)=136533.33*dell**2*2**-1
+
+//simplifying above equation we get
+//dell**2-2.93*10**-3*dell-2.93*10**-4=0
+//The Above Equation is in the form of ax**2+bx+c=0
+a=1
+b=-2.93*10**-3
+c=-2.93*10**-4
+
+//First computing value of  b^2-4ac and store it in a variable say X
+X=b**2-(4*a*c)
+//now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X
+
+dell_1=(-b+X**0.5)*(2*a)**-1
+dell_2=(-b-X**0.5)*(2*a)**-1
+
+//Now deflection cannot be negative so consider value of dell_1
+
+P=136533.33*dell_1
+sigma=3*P*L*10**-2*(2*n*b_1*10**-2*(t*10**-3)**2)**-1*10**-6 //MPa //Max instantaneous stress
+R=(L*10**-2)**2*(8*dell_1)**-1 //Radius of curvature
+ 
+//Result
+printf("Max instantaneous stress in plates is %.2f",sigma);printf(" MPa")
+printf("\n Radius of curvature of spring is %.2f",R);printf(" m")