19 files changed, 736 insertions, 0 deletions
diff --git a/3772/CH8/EX8.1/Ex8_1.sce b/3772/CH8/EX8.1/Ex8_1.sce
new file mode 100644
index 000000000..45c8ab5f7
--- /dev/null
+++ b/3772/CH8/EX8.1/Ex8_1.sce
@@ -0,0 +1,36 @@
+// Problem 8.1,Page no.206
+
+clc;clear;
+close;
+
+k=1 //KN/m //stiffness of spring
+P=45 //N //Maximum Load
+sigma_s=126 //MPa //Max shear stress
+L=4.5 //cm //Lenght of spring
+G=42 //GPa //Modulus of rigidity
+
+//Calculations
+
+//sigma_s_max=16*P*R*(%pi*d**3)**-1 //Max shear stress
+
+//After substituting values in above equation and simolifying we get
+//1000=42*10**9*d**4*(64*R**3*n)**-1   (//Equation 1)
+
+//R=0.175*10**6*%pi*d**3 //Radius of spring (Equation 2)
+
+//L=n*d //solid length of spring
+//Thus simplifying above equation, n=L*d**-1
+
+//substituting value of n and R in (equation 1) we get,
+
+d=(42*10**9*(1000*64*4.5*10**-2*(0.175*%pi)**3*(10**6)**3)**-1)**0.25*10**2 //cm //diameter of helical spring
+
+//substituting value d in (equation 2) we get,
+R=0.175*10**6*%pi*(d)**3*10**-6*100 //cm //Radius of coil
+D=2*R //cm //Mean diameter of coil
+n=0.045*0.00306**-1 //Number of turns
+
+
+//Result
+printf("The Diameter of wire is %.3f cm",d)
+printf("\n The Mean Diameter of coil is %.2f",D);printf(" cm")
diff --git a/3772/CH8/EX8.10/Ex8_10.sce b/3772/CH8/EX8.10/Ex8_10.sce
new file mode 100644
index 000000000..2e9e0ed21
--- /dev/null
+++ b/3772/CH8/EX8.10/Ex8_10.sce
@@ -0,0 +1,33 @@
+// Problem 8.10,Page no.213
+
+clc;clear;
+close;
+
+L=75 //cm //Legth of Leaf spring
+P=8 //KN //Load
+y_c=20 //mm //Deflection
+sigma=200 //MPa //Bending stress 
+E=200 //GPa //modulus of Elasticity
+//b=12*t
+
+//Calculation
+
+//y_c=sigma*L**2*(4*E*t)**-1 
+//After substituting values and further simplifying we get
+t=200*10**6*(75*10**-2)**2*(4*200*10**9*0.02)**-1*10**2 //Thickness of plate
+
+b=12*t //width of plate
+
+//Now using relation we get
+//sigma=3*P*L*(2*n*b*t**2)**-1 
+//After substituting values and further simplifying we get
+n=3*8*10**3*0.75*(2*200*10**6*0.084*0.007**2)**-1
+
+//Y_c=L**2*(8*R)**-1
+R=(L*10**-2)**2*(8*y_c*10**-3)**-1 //m //Radius of spring
+
+//Result
+printf("The thickness of plate is %.2f",t);printf(" cm")
+printf("\n The width of plate is %.2f",b);printf(" cm")
+printf("\n The number of plate is %d",ceil(n))
+printf("\n The Radius of plate is %.2f m",R)
diff --git a/3772/CH8/EX8.11/Ex8_11.sce b/3772/CH8/EX8.11/Ex8_11.sce
new file mode 100644
index 000000000..67671afff
--- /dev/null
+++ b/3772/CH8/EX8.11/Ex8_11.sce
@@ -0,0 +1,33 @@
+// Problem 8.11,Page no.214
+
+clc;clear;
+close;
+
+L=75 //cm //span of laminated steel spring
+P=7.5 //KN //Load
+y_c=5 //cm //Central Deflection
+sigma=400 //MPa //Bending stress
+E=200 //GPa //Modulus of Elasticity
+//b=12*t
+
+//Calculations
+
+//y_c=3*P*L**3*(8*E*n*b*t**3)**-1 //Deflection
+//After substituting values and further simplifying we get
+//nt**4=9.887*10**-9   (Equation 1)
+
+//We Know sigma=3*P*L*(2*n*b*t**3)**-1 //bending stress
+//Again after substituting values and further simplifying we get
+//nt**3=1.757*10**-6   (Equation 2)
+
+//After Divviding (Equation 1) by (Equation 2) we have
+t=9.887*10**-9*(1.757*10**-6)**-1*10**2 //cm 
+
+//substituting value of t in Equation 2) we get
+n=1.757*10**-6*((t*10**-2)**3)**-1 //Number of plates
+R=(L*10**-2)**2*(8*y_c*10**-2)**-1 //Radius of curvature
+
+//Result
+printf("The thickness of Plates is %.2f",t);printf(" cm")
+printf("\n The Number of Plates is %d",ceil(n))
+printf("\n The Radius of Curvature of Plates is %.2f",R);printf(" m")
diff --git a/3772/CH8/EX8.12/Ex8_12.sce b/3772/CH8/EX8.12/Ex8_12.sce
new file mode 100644
index 000000000..0615f73a7
--- /dev/null
+++ b/3772/CH8/EX8.12/Ex8_12.sce
@@ -0,0 +1,26 @@
+// Problem 8.12,Page no.214
+
+clc;clear;
+close;
+
+L=1.3 //m //Length of carriage spring
+b=10 //cm //width of spring
+t=12 //mm //thickness of spring
+sigma=150 //MPa //Bending stresses
+E=200 //GPa //Modulus of Elasticity
+U=120 //N*m //Strain Energy
+
+//Calculation
+
+//V=n*b*t*L //Volume of carriage spring
+//U=sigma**2*(6*E)**-1*V
+//After substituting values in above equation and further simplifying we get
+n=120*6*200*10**9*2*((150*10**6)**2*10*10**-2*12*10**-3*1.3)**-1
+
+sigma_1=(120*6*200*10**9*2*(9*0.1*0.012*1.3)**-1)**0.5*10**-6 //MPa //Actual Bending stress
+
+R=E*t*(2*sigma_1)**-1 //m 
+
+//Result
+printf("The number of plates is %d",ceil(n))
+printf("\n Radius of curvature is %.3f m",R)
diff --git a/3772/CH8/EX8.13/Ex8_13.sce b/3772/CH8/EX8.13/Ex8_13.sce
new file mode 100644
index 000000000..866448121
--- /dev/null
+++ b/3772/CH8/EX8.13/Ex8_13.sce
@@ -0,0 +1,48 @@
+// 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")
diff --git a/3772/CH8/EX8.14/Ex8_14.sce b/3772/CH8/EX8.14/Ex8_14.sce
new file mode 100644
index 000000000..db5cda509
--- /dev/null
+++ b/3772/CH8/EX8.14/Ex8_14.sce
@@ -0,0 +1,45 @@
+// Problem 8.14,Page no.215
+
+clc;clear;
+close;
+
+L=70 //cm //Length of Longest plate
+n=10 //Number of turns
+b_1=5 //cm //width of plates
+P=3.5 //KN //central Load
+t=6 //mm //thickness of plates
+L=75 //cm //Length of plates
+y_c=1.8 //cm //central deflection
+sigma=190 //MPa //allowable bending stress
+//b=12*t
+E=200 //GPa //Modulus of Elasticity
+
+//Calculation
+//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
+
+y_c=3*P*L**3*(8*n*E*b*t**3)**-1 //Deflection (//Equation 1)
+sigma=3*P*L*(2*n*b*t**2)**-1 //stress
+//Dividing Equation 1 by Equation 2 we get
+//y_c*sigma**-1=L**2*(4*E*t)**-1
+//After substituting values in above equation and further simplifying we get
+t=190*10**6*0.7**2*(1.8*10**-2*4*200*10**9)**-1*10**3 //thickness of plate
+b=12*t //Width of plates
+
+//sigma=3*2**-1*P*L*(n*b*t**2)**-1 //stress
+//After substituting values in above equation and further simplifying we get
+n=3*3.5*10**3*0.7*(2*190*10**6*0.077583*(6.465*10**-3)**2)**-1
+
+//Now sigma*y**-1=E*R**-1 
+//simplifying above equationwe get
+R=200*10**9*6.465*10**-3*(2*190*10**6)**-1 //Radius of Curvature
+a=L*10**-2*(2*n)**-1*10**3 //Overlap
+
+//Result
+printf("size of the plate is: %.2f",b);printf(" mm")
+printf("\n                     : %.2f",t);printf(" mm")
+printf("\n Overlap of plates is %.2f",a);printf(" mm")
+printf("\n Number of plates is %d",ceil(n))
+printf("\n The Radius of curvature is %.3f m",R)
diff --git a/3772/CH8/EX8.15/Ex8_15.sce b/3772/CH8/EX8.15/Ex8_15.sce
new file mode 100644
index 000000000..d4da33452
--- /dev/null
+++ b/3772/CH8/EX8.15/Ex8_15.sce
@@ -0,0 +1,34 @@
+// Problem 8.15,Page no.216
+
+clc;clear;
+close;
+
+alpha=30 //degree //helix angle
+dell=2.3*10**-2 //m //Vertical displacement
+W=40 //N //Axial Load
+d=6*10**-3 //steel rod diameter
+E=200*10**9 //Pa 
+G=80*10**9 //Pa 
+
+//Calculations
+
+//from equation of deflection of the spring under the Load we get
+//R**3*n=8.49*10**-4 
+
+//Let R**3*n=X
+X=8.49*10**-4  //Equation 1
+
+//from equation of angular rotation
+//R**2*n=8.1*10**-3
+
+//Let R**2*n=Y
+Y=8.1*10**-3   //Equation 2
+
+//After dividing equation 1 by equation 2 we get R 
+//Let Z=R
+
+Z=X*Y**-1
+R=Z*10**2 //cm //Mean Radius
+
+//Result
+printf("Mean Radius of Open coiled spring of helix angle is %.2f",R);printf(" cm")
diff --git a/3772/CH8/EX8.16/Ex8_16.sce b/3772/CH8/EX8.16/Ex8_16.sce
new file mode 100644
index 000000000..791095d23
--- /dev/null
+++ b/3772/CH8/EX8.16/Ex8_16.sce
@@ -0,0 +1,48 @@
+// Problem 8.16,Page no.217
+
+clc;clear;
+close;
+
+n=10 //Number of coils
+sigma=100 //MPa //Bending stress
+sigma_s=110 //MPa //Twisting stress
+//D=8*d
+dell=1.8 //cm //Max extension of of wire
+E=200 //GPa //Modulus of Elasticity
+G=80 //GPa //Modulus od Rigidity
+
+//Calculation
+
+//M=W*R*sin_alpha=%pi*d**3*sigma_1*32**-1 //(Equation 1) //Bending moment
+//As D=8*d 
+//then R=D*2**-1
+//Therefore, R=4*d
+
+//Now substituting values in equation 1 we get
+//W*sin_alpha=2454369.3*d**2  (Equation 2)
+
+//T=W*R*cos_alpha=%pi*d**3*sigma_s //Torque  (Equation 3)  
+//Now substituting values in equation 3 we get
+//W*cos_alpha=5399612.4*d**2    (Equation 4)
+
+//Dividing Equation 2 by Equation 4 we get,
+//tan_alpha=0.4545
+alpha=atan(0.4545)*180*%pi**-1
+
+//From Equation 2 we get
+//W=2454369.3*d**2*(sin24.443)**-1
+//W=5931241.1*d**2   (Equation 5)
+
+//dell=64*W*R**3*n*sec_alpha*(d**4)**-1*((cos_alpha)**2*G**-1+2*sin_alpha**2*E**-1)
+//Now substituting values in above equation  we get
+//W=33140.016*d  (Equation 6)
+
+//From Equation 5 and Equation 6 we get
+//5931241.1*d**2=33140.016*d
+//After simplifying above equation we get
+d=33140.016*5931241.1**-1 //m //Diameter of wire
+W=33140.016*d //N //MAx Permissible Load
+
+//Result
+printf("The Max Permissible Load is %.2f",W);printf(" N")
+printf("\n The Wire Diameter is %.6f m",d)
diff --git a/3772/CH8/EX8.18/Ex8_18.sce b/3772/CH8/EX8.18/Ex8_18.sce
new file mode 100644
index 000000000..73c4da4bb
--- /dev/null
+++ b/3772/CH8/EX8.18/Ex8_18.sce
@@ -0,0 +1,36 @@
+// Problem 8.18,Page no.218
+
+clc;clear;
+close;
+
+//Calculation
+
+n=10 //number of coils
+d=2*10**-2 //m //Diameter of wire
+D=12*10**-2 //m //Diameter of coiled spring
+R=0.06 //m //Radius of coiled spring
+dell=0.5*10**-2 //Deflection
+E=200*10**9 //Pa 
+G=80*10**9 //Pa 
+alpha=30 //degree
+
+//Calculations
+
+//beta=64*W*R**2*n*sinalpha*(d**4)**-1*(1*G**-1-2*E**-1)+64*T*R*n*secalpha*(d**4)**-1*(sin**2alpha*G**-1+2*cos**2alpha*E**-1)=0
+//From above equation anf simplifying we get
+
+//T=-6.11*10**-3*W
+
+//dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*[(cos(alpha))**2*G**-1+2*(sin(alpha))**2*E**-1]+64*T*R**2*n*sin(alpha)*(d**4)**-1*[1*G**-1+2*E**-1]
+
+//After substituting Values and further simplifying we get
+//1.1847*10**-5*W+1.62*10**-4*T=0.005
+
+//Now substituting value of T in above equation we get
+//1.1847*10**-5*W-9.8982*10**-7*W=0.005
+W=0.005*(1.1847*10**-5-9.8982*10**-7)**-1 //N
+T=-6.11*10**-3*W //N*m
+
+//Result
+printf("The axial Load is %.2f",W);printf(" N")
+printf("\n Necesscary torque is %.2f",T);printf(" N*m")
diff --git a/3772/CH8/EX8.19/Ex8_19.sce b/3772/CH8/EX8.19/Ex8_19.sce
new file mode 100644
index 000000000..d8812b713
--- /dev/null
+++ b/3772/CH8/EX8.19/Ex8_19.sce
@@ -0,0 +1,34 @@
+// Problem 8.19,Page no.219
+
+clc;clear;
+close;
+
+d=6 //mm //Diameter of steel wire
+n=1 //number of turns
+D=6.5 //cm //Mean of diameter
+G=80 //GPa //modulus of rigidity
+P_1=150 //Load
+p=1.5 //cm //%pitch of coil
+
+//Calculation 
+
+R=D*2**-1
+//For one turn deflection is
+dell=p-d*10**-1 //cm 
+
+//dell=64*P*R**3*n*(G*d**4)**-1 
+//Now, after simplifying further we get,
+P=dell*10**-2*G*10**9*(d*10**-3)**4*(64*(R*10**-2)**3*n)**-1 //N //Axial Load
+
+dell_2=dell*8 //cm //Total Displacement //Notification has been changed
+U=P*dell_2*10**-2*2**-1 //N-m //Strain Energy
+
+//Potential Energy given by 150N Load is
+//U=150*(h+0.072)
+
+//After simplifying above equation we get
+h=(U*P_1**-1-0.072)*10**2 //cm //Height from which 150 N load falls
+
+//Result
+printf("Axial Load is %.2f",P);printf(" N")
+printf("\n Height from which 150 N load falls is %.2f",h);printf(" cm" )
diff --git a/3772/CH8/EX8.2/Ex8_2.sce b/3772/CH8/EX8.2/Ex8_2.sce
new file mode 100644
index 000000000..0bd89e308
--- /dev/null
+++ b/3772/CH8/EX8.2/Ex8_2.sce
@@ -0,0 +1,47 @@
+// Problem 8.2,Page no.207
+
+
+clc;clear;
+close;
+
+L=15 //cm //Length of close coiled helical spring
+U=50 //N*m //Strain energy
+sigma_s=140 //MPa //Shear stress
+D=10 //cm //Mean coil diameter
+G=80 //GPa //Modulus of rigidity
+
+R=D*2**-1 //cm //Mean coil Radius
+
+//Calculations
+
+//Let dell be the deflection of the spring when fully compressed
+// 0.15-dell=n*d (Equation 1)
+ 
+//U=(sigma_s)**2*V*(4*G)**-1 //Strain energy
+
+//After substituting values in above equation and simolifying we get
+V=50*4*80*10**9*((140*10**6)**2)**-1 //m**3 //Volume of spring
+
+//But V=%pi*4**-1*d**2*2*%pi*R*n
+//After substituting values in above equation and simolifying we get
+//n=3.308*10**-3*(d**2)**-1 //Number of turns
+
+//We know, T=P*R 
+//Now substituting values in T and simolifying we get
+//P=549.7787*10**6*d**3  //Load
+
+//U=P*dell*2**-1
+//After substituting values in above equation and simolifying we get
+//dell=0.18189*10**-6*d**3 //Deflection 
+
+//After substituting values in above equation and simolifying we get
+
+//d**3-22.0533*10**-3*d**2-1.21261*10**-6=0
+
+Coeff=[1 -22.0533*10**-3 0 -1.21261*10**-6]
+d=roots(Coeff)  //Diameter of steel wire
+n=3.308*10**-3*((d(1)**2)**-1) //no.of coils
+
+//Result
+printf("Diameter of steel wire is %.5f m",d(1))
+printf("\n number of coils  = %d",ceil(n))
diff --git a/3772/CH8/EX8.20/Ex8_20.sce b/3772/CH8/EX8.20/Ex8_20.sce
new file mode 100644
index 000000000..86efd46ff
--- /dev/null
+++ b/3772/CH8/EX8.20/Ex8_20.sce
@@ -0,0 +1,32 @@
+// Problem 8.20,Page no.219
+
+clc;clear;
+close;
+
+alpha=30 //degree 
+E=200*10**9 //Pa
+G=80*10**9 //pa 
+
+//Calculations
+
+//For alpha=30 //Degree
+//dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)
+//Now substituting values in above equation we get
+
+//dell_1=64*W*R**3*n*(d**4)**-1*1330*(10**9)**-1  (equation 1)
+
+//For alpha=0 //Degree
+//dell=64*W*R**3*n*sec(alpha)*(d**4)**-1*(cos(alpha)**2*G**-1+2*sin(alpha)**2*E**-1)
+//Now substituting values in above equation we get
+
+//dell_2=64*W*R**3*n*(d**4)**-1*1250*(10**9)**-1  (equation 2)
+
+//subtracting equation 1 and equation 2 we get
+//Let dell_1-dell_2=X
+//X=64*W*R**3*n*(d**4)**-1*80*(10**9)
+
+//Let Y=X*dell_1**-1*100
+Y=80*1330**-1*100 //% under estimation of axial extension
+
+//Result
+printf("%% under estimation of axial extension is %.2f",Y)
diff --git a/3772/CH8/EX8.3/Ex8_3.sce b/3772/CH8/EX8.3/Ex8_3.sce
new file mode 100644
index 000000000..d7cd05c91
--- /dev/null
+++ b/3772/CH8/EX8.3/Ex8_3.sce
@@ -0,0 +1,40 @@
+// Problem 8.3,Page no.208
+
+clc;clear;
+close;
+
+k=10 //KN/m //stiffness
+L=40 //cm //Length of coil when adjascent coil touch each other
+G=80 //GPa //Modulus of rigidity
+//dell=0.002*n //Max compression
+
+//Calculation
+
+//k=G*d**4*(8*D**3*n)**-1 //Stiffness
+//After substituting values in above equation and simolifying we get
+//d**4=D**3*n*10**-6     (Equation 1)
+
+//L=n*d, //After substituting values we get
+//n=0.4*d**-1            (Equation 2)
+
+//Again, d*D**-1=1*10**-1
+//After solving above ratios we get,D=10*d
+
+//After substituting values in Equation 1 And Equation 2 we get
+d=(10**3*0.4*10**-6)**0.5*100 //cm
+D=10*d //cm //Mean Diameter 
+R=D*2**-1 //cm //Mean Radius
+n=0.4*(d*10**-2)**-1 //Number of turns
+dell=0.002*n*100 //Deflection
+
+//k=P*dell**-1 
+//after solving above equation we get
+P=k*10**3*dell*10**-2 //N //Load
+
+sigma_s_max=16*P*R*10**-2*(%pi*(d*10**-2)**3)**-1 //N/m**2 //Max shear stress
+
+//Result
+printf("The wire diameter is %.2f",d);printf(" cm")
+printf("\n The Mean diameter is %.2f",D);printf(" cm")
+printf("\n Max Load applied is %.2f",P);printf(" N")
+printf("\n Max shear stress is %.f N/m^2",sigma_s_max)
diff --git a/3772/CH8/EX8.4/Ex8_4.sce b/3772/CH8/EX8.4/Ex8_4.sce
new file mode 100644
index 000000000..239803ffd
--- /dev/null
+++ b/3772/CH8/EX8.4/Ex8_4.sce
@@ -0,0 +1,35 @@
+// Problem 8.4,Page no.209
+
+clc;clear;
+close;
+
+G=80 //GPa //Modulus of rigidity
+P=1 //KN //Load
+dell=10 //cm //Deflection
+sigma_s=350 //MPa //Max shear stress
+rho=78000 //N/m**3 //Density of materials
+
+//Calculations
+
+U=P*1000*dell*10**-2*2**-1 //N*m //Energy stored
+
+//Again U=sigma_s**2*V*(4*G)**-1 
+//After substituting values in above equation and further simplifying we get
+V=50*4*80*10**9*((350*10**6)**2)**-1 //m**3 //Volume
+
+W=V*rho //N //Weight
+
+//Now T=P*R=%pi*d**3*sigma_s*16**-1
+//After substituting values in above equation and further simplifying
+D=(10**6*16*(2*%pi*350*10**6)**-1)**0.5*10**2 //cm //Mean diameter of coil
+
+k=P*10**3*(dell*10**-2)**-1 //stiffness
+
+//Also k=D*n**-1*10**6
+//After substituting values in above equation and further simplifying
+n=10**6*D*10**-2*k**-1 //number of turns
+
+//Result
+printf("The Value of weight is %.3f N",W)
+printf("\n Mean coil diameter is %.2f",D);printf(" cm")
+printf("\n The number of Turns is %.d",ceil(n))
diff --git a/3772/CH8/EX8.5/Ex8_5.sce b/3772/CH8/EX8.5/Ex8_5.sce
new file mode 100644
index 000000000..90ac3d864
--- /dev/null
+++ b/3772/CH8/EX8.5/Ex8_5.sce
@@ -0,0 +1,44 @@
+// Problem 8.5,Page no.210
+
+clc;clear;
+close;
+
+d=6 //mm //Diameter of steel wire
+n=50 //number of turns
+D=5 //cm //Mean Diameter
+R=D*2**-1 //cm //Radius of coil
+G=80 //GPa //Modulus of Rigidity
+P=150 //KN //Load
+
+//Calculation
+
+//Dell=64*P*R**3*n*(G*d**4)**-1 //Deflection
+//After substituting values in above equation and simplifying we get
+//P=2073.6*dell //Gradually applied equivalent Load
+
+//loss of potential Energy of the weight=Gain of strain Energy of the spring
+//150*(0.05+dell)=P*dell*2**-1
+//After substituting values in above equation we get
+
+//dell**2-0.1446*dell-0.00723=0
+//Above Equation is in the form of ax^2+bx+c=0
+
+a=1
+b=-0.1446
+c=-0.00723
+
+//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*10**2
+dell_2=(-b-X**0.5)*(2*a)**-1*10**2
+
+P=2073.6*dell_1*10**-2 //N 
+
+sigma=16*P*R*10**-2*(%pi*(d*10**-3)**3)**-1 //N/m**2 //Max stress
+
+//Result
+printf("The Max Extension of the Spring is %.2f",dell_1);printf(" cm")
+printf("\n The Max stress is %.3e N/m^2",sigma)
diff --git a/3772/CH8/EX8.6/Ex8_6.sce b/3772/CH8/EX8.6/Ex8_6.sce
new file mode 100644
index 000000000..b0c9e6436
--- /dev/null
+++ b/3772/CH8/EX8.6/Ex8_6.sce
@@ -0,0 +1,40 @@
+// Problem 8.6,Page no.209
+
+clc;clear;
+close;
+
+W=200 //N //weight 
+v=4 //m/s //velocity of spring
+sigma=600 //MPa //max allowable stress in spring
+G=80 //GPa //Modulus of rigidity
+rho=78000 //N/m**3 //density
+d=8 //mm //diameter of spring
+D=5 //cm //Mean Diameter of coil
+
+
+//Calculation 
+
+E=W*v**2*(2*9.81)**-1 //N*m //Kinetic Energy //Notification has been changed
+
+//U=sigma_s**2*V*(4*G)**-1 //Strain Energy stored inthe spring
+
+//After substituting values in above equation and simplifying we get
+V=163.1*4*80*10**9*((600*10**6)**2)**-1  //Volume 
+
+W=rho*V //N //Weight of spring
+
+//Now V=%pi*4**-1*d**2*%pi*D*n
+//After substituting values in above equation and simplifying we get
+n=0.000145*4*(%pi**2*0.008**2*0.05)**-1 //number of turns of spring
+
+//T=P*R=%pi*16**-1*d**3*sigma_s //Torsion
+//After substituting values in above equation and simplifying we get
+P=%pi*0.008**3*600*10**6*(0.025*16)**-1 //N
+
+//Now U=P*dell*2**-1 
+//Again,After substituting values in above equation and simplifying we get
+dell=163.1*2*(2412.743)**-1*10**2 //cm
+
+//Result
+printf("The Max Deflection Produced is %.2f",dell);printf(" cm")
+printf("\n Number of coil are %d",ceil(n))
diff --git a/3772/CH8/EX8.7/Ex8_7.sce b/3772/CH8/EX8.7/Ex8_7.sce
new file mode 100644
index 000000000..51ccd3456
--- /dev/null
+++ b/3772/CH8/EX8.7/Ex8_7.sce
@@ -0,0 +1,43 @@
+// Problem 8.7,Page no.211
+
+clc;clear;
+close;
+
+n=12 //number of coils
+d=3 //cm //mean diameter
+k=720 //N/m //stiffness of spring
+sigma_s=190 //MPa //Max shear stress
+G=80 //GPa //Modulus of rigidity
+D=3 //mm //Diameter of outer spring
+
+//Calculations
+R=D*2**-1 //mm //Radius of outer spring
+
+//Dell_1=64*P*(R*10**-3**3*n*(G*10**9*(d*10**-3)**4)**-1 //m //Extension of first spring
+//After substituting values and further simplifying we get
+//Dell_1=0.0004*P //m
+
+//Dell_2=64*P*(R*10**-3**3*n*(G*10**9*(d_1)**4)**-1 //m //Extension of first spring
+//After substituting values and further simplifying we get
+//Dell_2=3.24*10**-14*P*(d_1**4)**-1 //m //where d_1 is diameter of inner spring
+
+//Dell=Dell_1+Dell_2
+//After substituting values and further simplifying we get
+//dell=0.0004*P+3.24*10**-14*P*((d)**4)**-1
+
+//But dell=P*k**-1=P*720**-1
+
+//Now substituting value of dell in above equation we get
+d_1=(3.24*10**-14*(1*720**-1-0.0004)**-1)**0.25 //cm //diameter of inner spring
+
+//Now T=P*R=%pi*d_1**3*dell_s*sigma_s*16**-1
+//simplifying above equation further 
+//P=%pi*d**3*sigma_s*(16*R)**-1 
+//Now substituting values and further simplifying we get
+P=%pi*d_1**3*sigma_s*10**6*(16*R*10**-2)**-1 //N //Limiting Load
+
+dell=P*k**-1*10**2 //cm //Total Elongation
+
+//Result
+printf("Greatest Load that can be carried by composite spring is %.f N",P)
+printf("\n Extension in spring is %.2f",dell);printf(" cm")
diff --git a/3772/CH8/EX8.8/Ex8_8.sce b/3772/CH8/EX8.8/Ex8_8.sce
new file mode 100644
index 000000000..e9f5ec01b
--- /dev/null
+++ b/3772/CH8/EX8.8/Ex8_8.sce
@@ -0,0 +1,45 @@
+// Problem 8.8,Page no.212
+
+clc;clear;
+close;
+
+//Outer spring
+n_1=10 //number of coils
+D_1=3 //cm //Diameter of coil
+d_1=3 //mm //diameter of wire
+dell_1=2 //cm //deflection of spring
+
+//Inner spring
+n_2=8 //number of coils
+
+G=80 //GPa //Modulus of rigidity
+
+//Calculation
+
+R_1=D_1*2**-1
+P_1=G*10**9*dell_1*10**-2*(d_1*10**-3)**4*(64*(R_1*10**-2)**3*n_1)**-1 //Load carried outer spring for compression of 2 cm
+
+P_2=100-P_1 //N //Load carried by inner spring
+k_2=P_2*0.01**-1 //N/m //stiffness of inner spring
+
+//D_2=D_1*10**-2-d_1*10**-3-2*dell_1*10**-2-d_2 //Diameter of inner spring
+//Further simplifying above equation we get
+//D_2=0.023-d_2
+
+//Now from stiffness equation of inner spring
+//k=G*d_2**4*(8*D_2**3*n_2)**-1
+//Now substituting values and further simplifying we get
+//d**4=(0.023-d)**3*312500**-1
+
+//As d is small compared with 0.023,as a first appromixation
+d_2_1=(0.023**3*312500**-1)**0.25 //m
+
+//Second Approximation
+d_2_2=((0.023-d_2_1)**3*312500**-1)**0.25 //m
+
+//Final approximation
+d_2_3=((0.023-d_2_2)**3*312500**-1)**0.25*100 //cm
+
+//Result
+printf("Stiffness of inner spring is %.2f",k_2);printf(" N/m")
+printf("\n Wire Diameter of inner spring is %.3f cm",d_2_3)
diff --git a/3772/CH8/EX8.9/Ex8_9.sce b/3772/CH8/EX8.9/Ex8_9.sce
new file mode 100644
index 000000000..99e641ca8
--- /dev/null
+++ b/3772/CH8/EX8.9/Ex8_9.sce
@@ -0,0 +1,37 @@
+// Problem 8.9,Page no.212
+
+clc;clear;
+close;
+
+L= 3 //m //Length of rod
+d_1=25*10**-3 //m //Diameter of rod
+n= 5 //no. of coils
+sigma=70*10**6 //MPa  //instantaneous stress
+E=70*10**9 //Pa 
+G=80*10**9 //Pa
+D=24*10**-2 //Spring diameter
+R=d_1*2**-1 //spring radius
+d=4*10**-2 //diameter of steel
+
+//Calculations
+
+dell_1=sigma*L*(E)**-1
+
+//Let P be the equivalent applied Load which will produce same stress of 70 MPa
+P=%pi*4**-1*(d_1)**2*E*10**-3 //KN
+
+//deflection of the spring is given by
+dell_2=P*64*R**3*n*(G*d**4)**-1 
+
+//Now Loss of Potential Energy of the weight=strain energy stored in the rod and the spring
+//Height measured from top of uncompressed  spring
+h=((P*dell_1*2**-1+P*dell_2*2**-1)*(5.5*10**3)**-1-(dell_1+dell_2))*10**2
+
+//Shear stress in the spring is given by
+sigma_s=16*P*R*(%pi*d**3)**-1*10**-6 //MPa 
+
+//Result
+printf("Height measured from top of uncompressed  spring %.2f",h);printf(" cm")
+printf("\n max shearing stress is %.2f",sigma_s);printf(" MPa")
+
+// Answer is wrong in the textbook.