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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
commit7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch)
treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3131/CH11
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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initial commit / add all books
Diffstat (limited to '3131/CH11')
-rw-r--r--3131/CH11/EX11.3/11_3.sce20
-rw-r--r--3131/CH11/EX11.4/11_4.sce3
-rw-r--r--3131/CH11/EX11.5/11_5.sce21
-rw-r--r--3131/CH11/EX11.6/11_6.sce10
-rw-r--r--3131/CH11/EX11.7/11_7.sce3
5 files changed, 57 insertions, 0 deletions
diff --git a/3131/CH11/EX11.3/11_3.sce b/3131/CH11/EX11.3/11_3.sce
new file mode 100644
index 000000000..790fad652
--- /dev/null
+++ b/3131/CH11/EX11.3/11_3.sce
@@ -0,0 +1,20 @@
+clear all; clc;
+disp("Ex 11_3")
+//At required equilibrium position, theta =45 degrees
+////From virtual displacements
+//From virtual - work equation
+theta=45*%pi/180
+a=1
+b=-1.2*cos(theta)
+c=-0.13
+d=(b^2)-(4*a*c)
+e=2*a
+xc1=(-b+sqrt(d))/e
+xc2=(-b-sqrt(d))/e
+printf('\n\n The first value of xc = %0.3f m',xc1)
+printf('\n\n The second value of xc = %0.3f m \n\n',xc2)
+disp("Considering the positive value only")
+disp("xc = 0.981 m")
+//put xc in equation 4
+Cx=(-120*cos(theta)*((1.2*cos(theta))-(2*xc1)))/(1.2*sin(theta)*xc1)
+printf('\n\n Cx = %0.0f N',Cx)
diff --git a/3131/CH11/EX11.4/11_4.sce b/3131/CH11/EX11.4/11_4.sce
new file mode 100644
index 000000000..af89051eb
--- /dev/null
+++ b/3131/CH11/EX11.4/11_4.sce
@@ -0,0 +1,3 @@
+clear all; clc;
+disp("Ex 11_4")
+disp("It is a theory problem with symbolic answer. Please refer to the book for the derivation")
diff --git a/3131/CH11/EX11.5/11_5.sce b/3131/CH11/EX11.5/11_5.sce
new file mode 100644
index 000000000..4c2959531
--- /dev/null
+++ b/3131/CH11/EX11.5/11_5.sce
@@ -0,0 +1,21 @@
+clear all; clc;
+disp("Ex 11_5")
+//Using the equation for equilibrium position,
+theta=asin(0)
+printf('\n\n Theta = %0.0f degrees',theta)
+m=10//mass in kg
+W=10*9.81//weight of the mass in N
+k=200//spring constant in N/m
+l=0.6//m
+theta1=acos(1-(W/(2*k*l)))//in radian
+theta2=theta1*180/%pi
+printf('\n\n Theta = %0.1f degrees',theta2)
+//Stability: Determining the second derivative of V w.r.t theta at theta=0 and theta=53.8 degrees
+//let a=d^2V/d(theta)^2 at theta=0
+//let b=d^2V/d(theta)^2 at theta=53.8
+a=((k*l^2)*(cos(theta)-cos(2*theta)))-(((W*l)/2)*cos(theta))
+printf('\n\n Second derivative of V w.r.t theta at theta = 0 degrees is %0.1f',a)
+disp("Unstable equilibrium at theta=0 degrees")
+b=((k*l^2)*(cos(theta1)-cos(2*theta1)))-(((W*l)/2)*cos(theta1))
+printf('\n\n Second derivative of V w.r.t theta at theta = 53.8 degrees is %0.1f',b)
+disp("Stable equilibrium at theta=53.8 degrees")
diff --git a/3131/CH11/EX11.6/11_6.sce b/3131/CH11/EX11.6/11_6.sce
new file mode 100644
index 000000000..bcd7b654a
--- /dev/null
+++ b/3131/CH11/EX11.6/11_6.sce
@@ -0,0 +1,10 @@
+clear all; clc;
+disp("Ex 11_6")
+//From the potential function analysis and te equilibrium analysis
+m=69.14/10.58
+printf('\n\n m = %0.2f kg',m)
+//Second derivative of V w.r.t. theta at m=6.53 kg and theta=20 degrees is:
+theta=20*%pi/180
+a=(-73.6*sin(theta))-((m*9.81*-1*(-3.6*cos(theta))^2)/(2*2*(3.69-(3.6*sin(theta)))^(3/2)))-((m*9.81*-3.6*sin(theta))/(2*sqrt(3.69-(3.6*sin(theta)))))
+printf('\n\n The second derivative of V w.r.t. theta at m=6.53 kg and theta=20 degrees is -%0.1f',a)
+disp("Unstable equilibrium at theta = 20 degress")
diff --git a/3131/CH11/EX11.7/11_7.sce b/3131/CH11/EX11.7/11_7.sce
new file mode 100644
index 000000000..565f879e7
--- /dev/null
+++ b/3131/CH11/EX11.7/11_7.sce
@@ -0,0 +1,3 @@
+clear all; clc;
+disp("Ex 11_7")
+disp("This is a theory question. Please refer to the Book for the derivation")