12 files changed, 594 insertions, 0 deletions
diff --git a/3772/CH14/EX14.1/Ex14_1.sce b/3772/CH14/EX14.1/Ex14_1.sce
new file mode 100644
index 000000000..ece91e449
--- /dev/null
+++ b/3772/CH14/EX14.1/Ex14_1.sce
@@ -0,0 +1,49 @@
+// Problem no 14.1,Page No.325
+
+clc;clear;
+close;
+b=2 //m //width 
+FOS=1.5 //Factor of safety
+//rho_mason=2.5*rho_w
+mu=0.5 //coeffeicient of friction
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+//W=b*H*L*rho
+//After substituting values and Further simplifying we get
+//W=2*H*rho
+
+//Total Pressure
+//P=W*H**2*2**-1
+
+x_bar=b*2**-1 //Distance of Line of action of W from waterface
+
+//Distance of pt where resultant cuts the base measured from Line of action
+//x=P*W**-1*H*3**-1
+//After substituting values and Further simplifying we get
+//x=H**2*30**-1
+
+//x_bar+x=2*b*3**-1
+//After substituting values and Further simplifying we get
+//1+H**2*30**-1=2*b*3**-1
+H=(30*(2*b*3**-1-1))**0.5 //height of dam
+
+//Frictional Resistance offered at the base
+//F=mu*W
+//After substituting values and Further simplifying we get
+//F=3.16*rho
+
+//Total Lateral Pressure
+//P=W*H**2*2**-1
+//P=4.99*W
+
+//Factor of safety against sliding
+//FOS1=F*P**-1=3.16*4.99**-1*rho_mason*rho_w**-1
+FOS1=3.16*4.99**-1*2.5
+
+//FOS1>FOS
+
+//Result
+printf("Dam is safe against sliding  = %.2f m",FOS1)
diff --git a/3772/CH14/EX14.10/Ex14_10.sce b/3772/CH14/EX14.10/Ex14_10.sce
new file mode 100644
index 000000000..4bdf26731
--- /dev/null
+++ b/3772/CH14/EX14.10/Ex14_10.sce
@@ -0,0 +1,48 @@
+// Problem no 14.10,Page No.337
+
+clc;clear;
+close;
+H=6 //m //height of dam
+a=1 //m //top width
+b=3 //m //Bottom width
+rho_mason=22 //KN/m**3 //weight of mason
+rho_earth=16 //KN/m**3 //density of water
+phi=30 //Degree //angle of repose
+mu=0.5 //Coeffecient of friction
+
+//Calculations
+
+//Let Length of dam ,L=1 m
+L=1 //m
+
+//weight of dam
+W=(a+b)*2**-1*L*H*rho_mason
+
+//Lateral thrust
+P=rho_earth*H**2*L*2**-1*((1-sin(30*%pi*180**-1))*(1+sin(phi*%pi*180**-1))**-1)
+
+//Distance of Line of action from vertical base
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+//distance of pt where resultant cuts the base
+x=P*W**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+e_max=b*6**-1
+
+//stress at toe
+sigma2=W*10**3*b**-1*(1+6*e*b**-1)*10**-3
+
+//Safe agaainst bearing
+
+//Frictional Resistance 
+F=mu*W
+
+if F>P then
+    //it is safe against sliding
+
+//Result
+printf("Safe against bearing as well as sliding")
+
+end
diff --git a/3772/CH14/EX14.11/Ex14_11.sce b/3772/CH14/EX14.11/Ex14_11.sce
new file mode 100644
index 000000000..24ce862aa
--- /dev/null
+++ b/3772/CH14/EX14.11/Ex14_11.sce
@@ -0,0 +1,57 @@
+// Problem no 14.11,Page No.338
+
+clc;clear;
+close;
+H=8 //m //height of dam
+a=1 //m //top width
+b=4.5 //m //Bottom width
+rho_mason=24 //KN/m**3 //weight of mason
+rho_earth=20 //KN/m**3 //density of water
+phi=30 //Degree //angle of repose
+mu=0.5 //Coeffecient of friction
+BC=120 //KN/m**2
+
+
+//Calculations
+
+//Let Length of dam ,L=1 m
+L=1 //m
+
+//weight of dam
+W=(a+b)*2**-1*L*H*rho_mason
+
+//Rankine's coeff earth pressure
+K=((1-sin(30*%pi*180**-1))*(1+sin(phi*%pi*180**-1))**-1)
+
+//Lateral thrust
+P=rho_earth*H**2*L*2**-1*K
+
+//Distance of Line of action from vertical base
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+//distance of pt where resultant cuts the base
+x=P*W**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Pressure at heel B
+sigma1=W*b**-1*(1-6*e*b**-1)
+
+//Pressure at heel C
+sigma2=W*b**-1*(1+6*e*b**-1)
+
+//sigma2>120 //KN/m**2,so it is unsafe against bearing capacity of the soil
+
+//result
+printf("Unsafe against the bearing capacity of soil")
+
+//Plotting the Shear Force Diagram
+
+X1=[0,L,L]
+Y1=[sigma2,sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
diff --git a/3772/CH14/EX14.12/Ex14_12.sce b/3772/CH14/EX14.12/Ex14_12.sce
new file mode 100644
index 000000000..a3c5c5884
--- /dev/null
+++ b/3772/CH14/EX14.12/Ex14_12.sce
@@ -0,0 +1,52 @@
+// Problem no 14.12,Page No.340
+
+clc;clear;
+close;
+H=6 //m //height of dam
+a=1.5 //m //top width
+b=3.5 //m //Bottom width
+rho_s=16 //KN/m**3 //density of soil
+rho_mason=22.5 //KN/m**3 //density of mason
+phi=30 //Degree //angle of repose
+
+//Calculations
+
+//Let Length of dam ,L=1 m
+L=1 //m
+
+//weight of dam
+W=(a+b)*2**-1*L*H*rho_mason
+
+//Rankine's coeff earth pressure
+K=((1-sin(30*%pi*180**-1))*(1+sin(phi*%pi*180**-1))**-1)
+
+//Lateral thrust
+P=rho_s*H**2*L*2**-1*K
+
+//Distance of Line of action from vertical base
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+//distance of pt where resultant cuts the base
+x=P*W**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Pressure at heel B
+sigma1=W*b**-1*(1-6*e*b**-1)
+
+//Pressure at heel C
+sigma2=W*b**-1*(1+6*e*b**-1)
+
+//Result
+printf("The Max Intensities of soil at the wall is %.2f",sigma2);printf(" KN/m**2")
+printf("\n The Min Intensities of soil at the wall is %.2f",sigma1);printf(" KN/m**2")
+
+//Plotting the Shear Force Diagram
+X1=[0,L,L]
+Y1=[sigma2,sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
diff --git a/3772/CH14/EX14.13/Ex14_13.sce b/3772/CH14/EX14.13/Ex14_13.sce
new file mode 100644
index 000000000..64e5d892c
--- /dev/null
+++ b/3772/CH14/EX14.13/Ex14_13.sce
@@ -0,0 +1,55 @@
+// Problem no 14.13,Page No.341
+
+clc;clear;
+close;
+H=6 //m //height of dam
+a=1 //m //top width
+b=3 //m //Bottom width
+rho_s=18 //KN/m**3 //density of soil
+rho_mason=24 //KN/m**3 //density of mason
+alpha=20
+phi=30
+
+//Calculations
+
+//Let Length of dam ,L=1 m
+L=1 //m
+
+a2=cos(alpha*%pi*180**-1)
+b2=(cos(alpha*%pi*180**-1)-((cos(alpha*%pi*180**-1)**2-cos(phi*%pi*180**-1)**2))**0.5)
+c2=(cos(alpha*%pi*180**-1)+((cos(alpha*%pi*180**-1)**2-cos(phi*%pi*180**-1)**2)**0.5))
+
+X=a2*b2*c2**-1
+
+//Total Pressue on the wall
+P=rho_s*H**2*2**-1*X
+
+//The Horizontal component of pressure
+P_H=P*cos(20*%pi*180**-1)
+
+//The Vertical component of pressure
+P_V=P*sin(20*%pi*180**-1)
+
+//weight of wall
+W=(a+b)*H*rho_mason*2**-1
+
+//TotaL Weight 
+W1=W+P_V
+
+//Taking moment of vertical Loads about B,M_B=0
+x_bar=(rho_mason*a*H*0.5+rho_mason*H*2)*W1**-1
+
+x=P_H*W1**-1*H*3**-1
+
+//eccentricity
+e=x_bar+x-b*2**-1
+
+//Stress at the toe at C
+sigma_max=W1*b**-1*(1+6*e*b**-1)
+
+//Stress at the heel at B
+sigma_min=W1*b**-1*(1-6*e*b**-1)
+
+//Result
+printf("Pressure at the base of the wall:Pressure at the heel %.2f",sigma_min);printf(" KN/m**2")
+printf("\n                                 :Pressure at the toe %.2f",sigma_max);printf(" KN/m**2")
diff --git a/3772/CH14/EX14.2/Ex14_2.sce b/3772/CH14/EX14.2/Ex14_2.sce
new file mode 100644
index 000000000..614d5841f
--- /dev/null
+++ b/3772/CH14/EX14.2/Ex14_2.sce
@@ -0,0 +1,28 @@
+// Problem no 14.2,Page No.327
+
+clc;clear;
+close;
+D=2 //m //External Diameter
+d=1.5 //m //Internal Diameter
+P=1600 //N/m**2 //N/m**2 //Wind Pressure
+W=19200 //N/m**2 //Weight of masonry
+
+//Calculations
+
+//Let H be max height of dam
+
+//W2=%pi*4**-1*(D**2-d**2)*H*W //weight of chimney
+//W2=26400*H
+
+//Eccentricrty
+x=(D**2+d**2)*(8*D)**-1
+
+//P2=H*D*P //Lateral thrust of wind on chimney
+//P2=3200*H
+
+//Now by using the relation we get P*W**-1=x*(H*2**-1)**-1
+//After substituting values and Further simplifying we get
+H=0.39*2*26400*3200**-1
+
+//result
+printf("The Height of Dam is %.2f",H);printf(" m")
diff --git a/3772/CH14/EX14.3/Ex14_3.sce b/3772/CH14/EX14.3/Ex14_3.sce
new file mode 100644
index 000000000..ce0752bb6
--- /dev/null
+++ b/3772/CH14/EX14.3/Ex14_3.sce
@@ -0,0 +1,50 @@
+// Problem no 14.3,Page No.327
+
+clc;clear;
+close;
+rho_w=10 //KN/m**3 //Density of water
+rho_mason=22.4 //KN/m**3 //Density of mason
+H=6 //m //height of dam
+a=1 //m //width of top
+b=4 //m //bottom width
+h=5.5 //m //Weight of water depth 
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+
+//weight of dam
+W=(a+b)*2**-1*H*a*rho_mason
+
+//Lateral thrust
+P=rho_w*h**2*a*2**-1
+
+//distance of Line of action of W measured from vertical face
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+//distance of pt where resultant cuts the base
+x=P*W**-1*h*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Stress at Pt B
+sigma1=W*b**-1*(1-6*e*b**-1)
+
+//stress at Pt C
+sigma2=W*b**-1*(1+6*e*b**-1)
+
+//Result
+printf("Max stress intensities at the base is %.2f",sigma2);printf(" KN/m**2")
+printf("\n Min stress intensities at the base is %.2f",sigma1);printf(" KN/m**2")
+
+//Plotting the Shear Force Diagram
+
+X1=[0,L,L]
+Y1=[sigma2,sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
diff --git a/3772/CH14/EX14.4/Ex14_4.sce b/3772/CH14/EX14.4/Ex14_4.sce
new file mode 100644
index 000000000..bfe3387d7
--- /dev/null
+++ b/3772/CH14/EX14.4/Ex14_4.sce
@@ -0,0 +1,50 @@
+// Problem no 14.4,Page No.329
+
+clc;clear;
+close;
+H=10 //m //height od dam
+a=2 //m //top width
+b=5 //m //bottom width
+W=25 //KN/m**3 //weight of mason
+rho_w=10 //KN/m**3 //density of water
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+
+//weight of dam
+W2=(b+a)*H*L*W*2**-1
+
+////Lateral thrust
+P=rho_w*H**2*L*2**-1 
+
+//Resultant thrust
+R=(P**2+W**2)**0.5
+
+//Distance of Line of action from vertical base
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+////distance of pt where resultant cuts the base
+x=P*W2**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Stress at Pt B
+sigma1=W2*b**-1*(1-6*e*b**-1)
+
+//stress at Pt C
+sigma2=W2*b**-1*(1+6*e*b**-1)
+
+//Result
+printf("The Resultant Thrust on the base is %.2f",R);printf(" KN")
+
+//Plotting the Shear Force Diagram
+X1=[0,L,L]
+Y1=[-sigma2,-sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
diff --git a/3772/CH14/EX14.5/Ex14_5.sce b/3772/CH14/EX14.5/Ex14_5.sce
new file mode 100644
index 000000000..26e70beca
--- /dev/null
+++ b/3772/CH14/EX14.5/Ex14_5.sce
@@ -0,0 +1,65 @@
+// Problem no 14.5,Page No.330
+
+clc;clear;
+close;
+H=30 //m //height of Dam
+a=2 //m //top width
+b=5 //m //bottom width
+
+h=29 //m //height of water
+rho_w=9810 //N/m**3 
+rho_mason=22560 //N/m**3
+sigma1=0 //KN/m**3
+sigma2=880 //KN/m**3
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+
+//weight of dam
+//W=(a+b)*2**-1*L*H*rho_mason*10**-3
+//After substituting values and Further simplifying we get
+//W=338.4*(a+b) //equation1
+
+//Pressure at B=0, Sinc etension at base has just been avoided
+
+//Eccentricity
+e=b*6**-1  //as sigma1=0
+
+//Pressure at C
+//sigma2=W2*b**-1*(1+6*e*b**-1)
+//After substituting values and Further simplifying we get
+//W=440*b
+
+//From equation1,440*b=338*(a+b)
+//b=3.33*a
+
+//the distance of C.Gof dam
+//x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+//After substituting values and Further simplifying we get
+//x_bar=1.187*a
+
+//distance of pt where resultant cuts the base
+//x=P*W2**-1*H*3**-1
+//After substituting values and Further simplifying we get
+//x=27.214*a**-1
+
+//Now x_bar+x=2*3**-1*b
+//After substituting values and Further simplifying we get
+a=(27.17*(2.22-1.187)**-1)**0.5
+b=3.33*a
+
+//Result
+printf("The top width dam is %.2f",a);printf(" m")
+printf("\n The bottom width dam is %.2f",b);printf(" m")
+
+
+//Plotting the Shear Force Diagram
+X1=[0,L,L]
+Y1=[sigma2,sigma1,0]
+Z1=[0,0,0]
+plot(X1,Y1,X1,Z1)
+xlabel("Length x in m")
+ylabel("Shear Force in kN")
+title("the Shear Force Diagram")
diff --git a/3772/CH14/EX14.6/Ex14_6.sce b/3772/CH14/EX14.6/Ex14_6.sce
new file mode 100644
index 000000000..254272dc2
--- /dev/null
+++ b/3772/CH14/EX14.6/Ex14_6.sce
@@ -0,0 +1,55 @@
+// Problem no 14.6,Page No.332
+
+clc;clear;
+close;
+H= 4 //m //height of dam
+a=1 //m //Top width
+b=3 //m //bottom width
+rho1=9810 //N/m**3 //weight of water
+rho2=19620 //n/m**3 //Weight of mason
+
+//Calculations
+
+//Let L=1 m (length of dam)
+L=1
+
+//weight of dam
+W=(a+b)*2**-1*L*H*rho2*10**-3
+
+////Lateral thrust
+P=rho1*H**2*L*2**-1*10**-3 
+
+//Distance of Line of action from vertical base
+x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+
+//distance of pt where resultant cuts the base
+x=P*W**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Stress at Pt B
+sigma1=W*10**3*b**-1*(1-6*e*b**-1)
+
+//stress at Pt C
+sigma2=W*10**3*b**-1*(1+6*e*b**-1)
+
+//Stresses at the base when resorvoir is empty
+
+e2=x_bar-b*2**-1
+
+//Minus sign indicates sigma_b>sigma_c
+
+//Stress at C
+sigma2_2=W*10**3*b**-1*(1+6*e2*b**-1)
+
+//Stress at Pt B
+sigma1_2=W*10**3*b**-1*(1-6*e2*b**-1)
+
+//result
+printf("When the Reservoir is full :sigma1 %.2f",sigma1);printf(" N/m**2")
+printf("\n                            :sigma2 %.2f",sigma2);printf(" N/m**2")
+printf("\n When the Reservoir is empty:sigma1_2 %.2f",sigma1_2);printf(" N/m**2")
+printf("\n                            :sigma2_2 %.2f",sigma2_2);printf(" N/m**2")
+
+//Answer is wrong in the textbook.//////
diff --git a/3772/CH14/EX14.7/Ex14_7.sce b/3772/CH14/EX14.7/Ex14_7.sce
new file mode 100644
index 000000000..4fae09483
--- /dev/null
+++ b/3772/CH14/EX14.7/Ex14_7.sce
@@ -0,0 +1,49 @@
+// Problem no 14.7,Page No.333
+
+clc;clear;
+close;
+H=8 //m //height of dam
+h=7.5 //m //Height of water
+a=1 //m //top width
+mu=0.6 //Coeffeicient of friction
+rho_mason=22.4 //KN/m**3 //weight of mason
+rho_w=9.81 //KN/m**3 //density of water
+
+//Calculations
+
+//weight of dam
+//W=(a+b)*2**-1*L*H*rho2*10**-3
+//After substituting values and further simplifying we get
+//W=89600*(b+1)
+
+//Distance of Line of action from vertical base
+//x_bar=(b**2+b*a+a**2)*(3*(b+a))**-1
+//After substituting values and further simplifying we get
+//x_bar=(1+b+b**2)*(3*(1+b))**-1
+
+//Lateral thrust
+P=rho_w*h**2*2**-1
+
+//Min width to avoid tension at base
+//Z=x_bar+P*W**-1*h*3**-1
+//Z=2*3**-1*b
+//After substituting values and further simplifying we get
+//b**2+b-24.09=0
+a=1
+b=1
+c=-24.09
+
+X=b**2-4*a*c
+
+b1=(-b+X**0.5)*(2*a)**-1
+b2=(-b-X**0.5)*(2*a)**-1
+
+//Thus width cannot be negative,b1 is considered
+
+//Min width to avoid sliding
+//mu*W>P
+//After substituting values and further simplifying we get
+b=P*10**3*(mu*89600)**-1-1
+
+//Result
+printf("The Min bottom width is %.2f",b);printf(" m")
diff --git a/3772/CH14/EX14.8/Ex14_8.sce b/3772/CH14/EX14.8/Ex14_8.sce
new file mode 100644
index 000000000..551430d71
--- /dev/null
+++ b/3772/CH14/EX14.8/Ex14_8.sce
@@ -0,0 +1,36 @@
+// Problem no 14.8,Page No.334
+
+clc;clear;
+close;
+H=10 //m //height of dam
+a=1 //m //top width
+b=7 //m //Bottom width
+rho_mason=19620 //N/m**3 //weight of mason
+rho_w=9810 //N/m**3 //density of water
+
+//Calculations
+
+//Lateral thrust
+P=rho_w*H**2*2**-1
+
+//weight of dam
+W=(rho_w*H*2**-1*a)+(rho_mason*(a+b)*2**-1*H)
+
+//Taking Moment at B,M_B=0
+x_bar=((rho_w*H*2**-1*1*3**-1)+(rho_mason*H*2**-1*2*3**-1)+(rho_mason*H*1.5)+(rho_mason*H*5*11*2**-1*3**-1))*W**-1
+
+//Now using relation we get
+x=P*W**-1*H*3**-1
+
+//Eccentricity
+e=x_bar+x-b*2**-1
+
+//Max stress
+sigma_max=W*b**-1*(1+6*e*b**-1)
+
+//Min stress
+sigma_min=W*b**-1*(1-6*e*b**-1)
+
+//Result
+printf("The Max stresses on the base is %.2f",sigma_max);printf(" N/m**2")
+printf("\n The Min stresses on the base is %.2f",sigma_min);printf(" N/m**2")