From a669cf65ee5039c931e66bc7c63d704562d92120 Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Wed, 21 Sep 2016 23:30:25 +0600 Subject: Added(A)/Deleted(D) following books A f_by_df/chapter03_16.ipynb A f_by_df/chapter03_16_1.ipynb A f_by_df/screenshots/1.png A f_by_df/screenshots/2.png A f_by_df/screenshots/3.png A f_by_df/screenshots/image.png A f_by_df/screenshots/image_1.png A f_by_df/screenshots/image_2.png A sample_notebooks/SufiyanSiddique/Chapter2.ipynb A sample_notebooks/chayas/DimensionsSystemUnits.ipynb --- f_by_df/chapter03_16.ipynb | 864 +++++++++++++++++++++ f_by_df/chapter03_16_1.ipynb | 864 +++++++++++++++++++++ f_by_df/screenshots/1.png | Bin 0 -> 75812 bytes f_by_df/screenshots/2.png | Bin 0 -> 120130 bytes f_by_df/screenshots/3.png | Bin 0 -> 101972 bytes f_by_df/screenshots/image.png | Bin 0 -> 286668 bytes f_by_df/screenshots/image_1.png | Bin 0 -> 286668 bytes f_by_df/screenshots/image_2.png | Bin 0 -> 286668 bytes sample_notebooks/SufiyanSiddique/Chapter2.ipynb | 257 ++++++ .../chayas/DimensionsSystemUnits.ipynb | 169 ++++ 10 files changed, 2154 insertions(+) create mode 100644 f_by_df/chapter03_16.ipynb create mode 100644 f_by_df/chapter03_16_1.ipynb create mode 100644 f_by_df/screenshots/1.png create mode 100644 f_by_df/screenshots/2.png create mode 100644 f_by_df/screenshots/3.png create mode 100644 f_by_df/screenshots/image.png create mode 100644 f_by_df/screenshots/image_1.png create mode 100644 f_by_df/screenshots/image_2.png create mode 100644 sample_notebooks/SufiyanSiddique/Chapter2.ipynb create mode 100644 sample_notebooks/chayas/DimensionsSystemUnits.ipynb diff --git a/f_by_df/chapter03_16.ipynb b/f_by_df/chapter03_16.ipynb new file mode 100644 index 00000000..46ae736b --- /dev/null +++ b/f_by_df/chapter03_16.ipynb @@ -0,0 +1,864 @@ + + + + + +chapter03_16 + + + + + + + + + + + + + + + + + + + + + + + + +
+
+ +
+
+
+
+
+

Chapter 3 : Parallel Forces In A Plane

+
+
+
+
+
+
+
+
+

Example 3.3-1,Page No:64

+
+
+
+
+
+
In [1]:
+
+
+
import math
+
+#Initilization of variables
+
+W=1000 #N
+Lab=1 #m
+Lac=0.6 #m
+theta=60 #degree #angle made by the beam with the horizontal
+
+#Calculations
+
+Q=(W*Lac*cos(theta*(pi/180)))/(Lab*cos(theta*(pi/180))) #N # from eq'n 2
+P=W-Q #N # from eq'n 1
+
+#Results
+
+print"The load taken by man P is ",round(P),"N"
+print"The load taken by man Q is ",round(Q),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The load taken by man P is  400.0 N
+The load taken by man Q is  600.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-2,Page No:64

+
+
+
+
+
+
In [2]:
+
+
+
import math
+
+#Initilization of variables
+
+F=1000 #N
+Lab=1 #m
+Lbc=0.25 #m
+Lac=1.25 #m
+
+#Calculations
+
+Rb=(F*Lac)/Lab #N # from eq'n 2
+Ra=Rb-F #N # fom eq'n 1
+
+#Results
+
+print"The reaction (downwards)at support A is ",round(Ra),"N"
+print"The reaction (upwards)at support B is ",round(Rb),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction (downwards)at support A is  250.0 N
+The reaction (upwards)at support B is  1250.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-3,Page No:65

+
+
+
+
+
+
In [7]:
+
+
+
import math
+
+#Inilitization of variables
+
+Lab=12 #m
+Mc=40 #kN-m 
+Md=10 #kN-m
+Me=20 #kN-m
+Fe=20 #kN #force acting at point E
+
+#Calculations
+
+Xa=-(Fe) #kN #take sum Fx=0
+a=Me+Md-Mc #N #take moment at A
+Rb=a*(Lab)**-1
+Ya=-Rb #N #take sum Fy=0
+
+#Results
+
+print"The vertical reaction (upwards) at A is ",round(Ya,3),"kN"
+print"The horizontal reaction (towards A) is ",round(Xa,2),"kN"
+print"The reaction (downwards) at B is ",round(Rb,3),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The vertical reaction (upwards) at A is  0.833 kN
+The horizontal reaction (towards A) is  -20.0 kN
+The reaction (downwards) at B is  -0.833 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-5,Page No:66

+
+
+
+
+
+
In [11]:
+
+
+
import math
+import numpy as np
+
+#Initilization of variables
+
+W=1000 #N
+Lad=7.5 #m
+Lae=1.5 #m
+La1=3.75 #m #distance of 1st 1000N load from pt A
+La2=5 #m #distance of 2nd 1000N load from pt A
+La3=6 #m # distance of 3rd 1000N load from pt A
+
+# Calculations (part1)
+
+#using matrix to solve the given eqn's 1 & 2
+
+A=np.array([[1 ,-2.5],[3.5 ,-5]])
+B=np.array([1000,7250])
+C=np.linalg.solve(A,B)
+
+#Resuts
+
+print"The reaction at F i.e Rf is ",round(C[0]),"N"
+print"The reaction at D i.e Rd is ",round(C[1]),"N"
+
+#Calculations (part 2)
+#Consider combined F.B.D of beams AB,BC &CD. Take moment at A
+
+Re=((W*La1)+(W*La2)+(W*La3)+(C[1]*Lad)-(C[0]*La3))/Lae #N
+Ra=C[1]-Re-C[0]+(3*W) #N #Taking sum of forces in Y direction
+
+#Results
+
+print"The reaction at pt E i.e Re is ",round(Re),"N"
+print"The reaction at pt A i.e Ra is ",round(Ra),"N" #acting vertically downwards
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction at F i.e Rf is  3500.0 N
+The reaction at D i.e Rd is  1000.0 N
+The reaction at pt E i.e Re is  833.0 N
+The reaction at pt A i.e Ra is  -333.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-7,Page No:69

+
+
+
+
+
+
In [14]:
+
+
+
import math
+
+#Initilization of variables
+
+Ws=2 #kN #weight of scooter
+Wd=0.5 #kN #weight of driver
+Lab=1 #m
+Led=0.8 #m
+Leg=0.1 #m
+
+#Calculations
+
+Rc=((2*Leg)+(Wd*Led))/Lab #kN #take moment at E
+Ra=(2+Wd-Rc)/2 #kN # as Ra=Rb,(Ra+Rb=2*Ra)
+Rb=Ra # kN
+
+#Results
+
+print"The reaction at wheel A is ",round(Ra,2),"kN"
+print"The reaction at wheel B is ",round(Rb,2),"kN"
+print"The reaction at wheel C is ",round(Rc,2),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction at wheel A is  0.95 kN
+The reaction at wheel B is  0.95 kN
+The reaction at wheel C is  0.6 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-8,Page No:69

+
+
+
+
+
+
In [25]:
+
+
+
import math
+
+#Initilization of variables
+
+W1=15 #N #up
+W2=60 #N #down
+W3=10 #N #up
+W4=25 #N #down
+Lab=1.2 #m
+Lac=0.4 #m
+Lcd=0.3 #m
+Ldb=0.5 #m
+Lad=0.7 #m
+Leb=0.417 #m #Leb=Lab-x
+
+#Calculations
+
+#(a) A single force
+
+Ry=W1-W2+W3-W4 #N #take sum Fy=0
+x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) #m
+
+#(b) Single force moment at A
+
+Ma=(Ry*x) #N-m
+
+# Single force moment at B
+
+Mb=W2*Leb #N-m
+
+#Results
+
+print"The reaction for single force (a) is ",round(Ry,2),"N"
+print"The distance of Ry from A is ",round(x,3),"m"
+print"The moment at A is ",round(Ma,2),"N-m"
+print"The moment at B is ",round(Mb,2),"N-m"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction for single force (a) is  -60.0 N
+The distance of Ry from A is  0.783 m
+The moment at A is  -47.0 N-m
+The moment at B is  25.02 N-m
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-9,Page No:71

+
+
+
+
+
+
In [6]:
+
+
+
import math
+import numpy as np
+
+#Initilization of variables
+
+Ra=5000 #N
+Ma=10000 #Nm
+alpha=60 #degree #angle made by T1 with the pole
+beta=45 #degree #angle made by T2 with the pole
+theta=30 #degree #angle made by T3 with the pole
+Lab=6 #m
+Lac=1.5 #m
+Lcb=4.5 #m
+
+#Calculations
+
+T3=Ma/(4.5*sin(theta*(pi/180))) #N #take moment at B
+
+# Now we use matrix to solve eqn's 1 & 2 simultaneously,
+
+A=np.array([[-0.707, 0.866],[0.707, 0.5]])
+B=np.array([2222.2,8848.8])
+C=np.linalg.solve(A,B)
+
+#Results
+
+print"Tension in wire 1 i.e T1 is ",round(C[1],1),"N" #answer may vary due to decimal variance
+print"Tension in wire 2 i.e T2 is ",round(C[0],1),"N"
+print"Tension in wire 3 i.e T3 is ",round(T3,1),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
Tension in wire 1 i.e T1 is  8104.7 N
+Tension in wire 2 i.e T2 is  6784.2 N
+Tension in wire 3 i.e T3 is  4444.4 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-10,Page No:76

+
+
+
+
+
+
In [32]:
+
+
+
import math
+
+#Initilization of variables
+
+w=2000 #N/m
+Lab=3 #m
+
+#Calculations
+
+W=w*Lab/2 #N# Area under the curve
+Lac=(0.6666)*Lab #m#centroid of the triangular load system
+Rb=(W*Lac)/Lab #N #sum of moment at A
+Ra=W-Rb #N
+
+#Results
+
+print"The resultant of the distibuted load lies at ",round(Lac),"m"
+print"The reaction at support A is ",round(Ra),"N"
+print"The reaction at support B is ",round(Rb),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The resultant of the distibuted load lies at  2.0 m
+The reaction at support A is  1000.0 N
+The reaction at support B is  2000.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-12,Page No:78

+
+
+
+
+
+
In [33]:
+
+
+
import math
+
+# Initilization of variables
+
+w1=1.5 #kN/m # intensity of varying load at the starting point of the beam
+w2=4.5 #kN/m # intensity of varying load at the end of the beam
+l=6 #m # ength of the beam
+
+# Calculations
+
+# The varying load distribution is divided into a rectangle and a right angled triangle
+
+W1=w1*l #kN # where W1 is the area of the load diagram(rectangle ABED)
+x1=l/2 #m # centroid of the rectangular load system
+W2=(w2-w1)*l/2 #kN # where W1 is the area of the load diagram(triangle DCE)
+x2=2*l/3 #m # centroid of the triangular load system
+W=W1+W2 #kN # W is the resultant
+x=((W1*x1)+(W2*x2))/W #m # where x is the distance where the resultant lies
+
+#Results
+
+print"The resultant of the distributed load system is ",round(W),"kN"
+print"The line of action of the resulting load is ",round(x,1),"m"
+
+ +
+
+
+ +
+
+ + +
+
+
The resultant of the distributed load system is  18.0 kN
+The line of action of the resulting load is  3.5 m
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-13,Page No:78

+
+
+
+
+
+
In [35]:
+
+
+
import math
+
+# Initiization of variables
+
+W1=10 #kN #point load acting at D
+W2=20 #kN # point load acting at C at an angle of 30 degree
+W3=5 #kN/m # intensity of udl acting on span EB of 4m
+W4=10 #kN/m # intensity of varying load acting on span BC of 3m
+M=25 #kN-m # moment acting at E
+theta=30 #degree # angle made by 20 kN load with the beam
+Lad=2 #m
+Leb=4 #m
+Laf=6 #m #distance between the resultant of W3 & point A
+Lac=11 #m
+Lag=9 #m #distance between the resultant of W4 and point A
+Lbc=3 #m
+Lab=8 #m
+
+# Calculations
+
+Xa=20*cos(theta*(pi/180)) #kN # sum Fx=0
+Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*sin(theta*(pi/180))*Lac)+((W4*Lbc*Lag)/2))/Lab #kN # taking moment at A
+Ya=W1+(W2*sin(theta*(pi/180)))+(W3*Leb)+(W4*Lbc/2)-Rb #kN # sum Fy=0
+Ra=(Xa**2+Ya**2)**0.5 #kN # resultant at A
+
+#Results
+
+print"The horizontal reaction at A i.e Xa is ",round(Xa,2),"kN"
+print"The vertical reaction at A i.e Ya is ",round(Ya),"kN"
+print"The reaction at A i.e Ra is ",round(Ra),"kN"
+print"The reaction at B i.e Rb is ",round(Rb),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The horizontal reaction at A i.e Xa is  17.32 kN
+The vertical reaction at A i.e Ya is  10.0 kN
+The reaction at A i.e Ra is  20.0 kN
+The reaction at B i.e Rb is  45.0 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-14,Page No:79

+
+
+
+
+
+
In [37]:
+
+
+
import math
+
+# Initilization of variables
+
+h=4 #m #height of the dam wall
+rho_w=1000 # kg/m^3 # density of water
+rho_c=2400 # kg/m^3 # density of concrete
+g=9.81 # m/s^2
+
+# Calculations
+
+P=(rho_w*g*h**2)/2 # The resultant force due to water pressure per unit length of the dam
+x=(0.6666)*h #m # distance at which the resutant of the triangular load acts 
+b=((2*P*h)/(3*h*rho_c*g))**0.5 # m # eq'n required to find the minimum width of the dam
+
+# Results
+
+print"The minimum width which is to be provided to the dam to prevent overturning about point B is ",round(b,3),"m" 
+
+ +
+
+
+ +
+
+ + +
+
+
The minimum width which is to be provided to the dam to prevent overturning about point B is  1.491 m
+
+
+
+ +
+
+ +
+
+
+ + diff --git a/f_by_df/chapter03_16_1.ipynb b/f_by_df/chapter03_16_1.ipynb new file mode 100644 index 00000000..46ae736b --- /dev/null +++ b/f_by_df/chapter03_16_1.ipynb @@ -0,0 +1,864 @@ + + + + + +chapter03_16 + + + + + + + + + + + + + + + + + + + + + + + + +
+
+ +
+
+
+
+
+

Chapter 3 : Parallel Forces In A Plane

+
+
+
+
+
+
+
+
+

Example 3.3-1,Page No:64

+
+
+
+
+
+
In [1]:
+
+
+
import math
+
+#Initilization of variables
+
+W=1000 #N
+Lab=1 #m
+Lac=0.6 #m
+theta=60 #degree #angle made by the beam with the horizontal
+
+#Calculations
+
+Q=(W*Lac*cos(theta*(pi/180)))/(Lab*cos(theta*(pi/180))) #N # from eq'n 2
+P=W-Q #N # from eq'n 1
+
+#Results
+
+print"The load taken by man P is ",round(P),"N"
+print"The load taken by man Q is ",round(Q),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The load taken by man P is  400.0 N
+The load taken by man Q is  600.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-2,Page No:64

+
+
+
+
+
+
In [2]:
+
+
+
import math
+
+#Initilization of variables
+
+F=1000 #N
+Lab=1 #m
+Lbc=0.25 #m
+Lac=1.25 #m
+
+#Calculations
+
+Rb=(F*Lac)/Lab #N # from eq'n 2
+Ra=Rb-F #N # fom eq'n 1
+
+#Results
+
+print"The reaction (downwards)at support A is ",round(Ra),"N"
+print"The reaction (upwards)at support B is ",round(Rb),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction (downwards)at support A is  250.0 N
+The reaction (upwards)at support B is  1250.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-3,Page No:65

+
+
+
+
+
+
In [7]:
+
+
+
import math
+
+#Inilitization of variables
+
+Lab=12 #m
+Mc=40 #kN-m 
+Md=10 #kN-m
+Me=20 #kN-m
+Fe=20 #kN #force acting at point E
+
+#Calculations
+
+Xa=-(Fe) #kN #take sum Fx=0
+a=Me+Md-Mc #N #take moment at A
+Rb=a*(Lab)**-1
+Ya=-Rb #N #take sum Fy=0
+
+#Results
+
+print"The vertical reaction (upwards) at A is ",round(Ya,3),"kN"
+print"The horizontal reaction (towards A) is ",round(Xa,2),"kN"
+print"The reaction (downwards) at B is ",round(Rb,3),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The vertical reaction (upwards) at A is  0.833 kN
+The horizontal reaction (towards A) is  -20.0 kN
+The reaction (downwards) at B is  -0.833 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-5,Page No:66

+
+
+
+
+
+
In [11]:
+
+
+
import math
+import numpy as np
+
+#Initilization of variables
+
+W=1000 #N
+Lad=7.5 #m
+Lae=1.5 #m
+La1=3.75 #m #distance of 1st 1000N load from pt A
+La2=5 #m #distance of 2nd 1000N load from pt A
+La3=6 #m # distance of 3rd 1000N load from pt A
+
+# Calculations (part1)
+
+#using matrix to solve the given eqn's 1 & 2
+
+A=np.array([[1 ,-2.5],[3.5 ,-5]])
+B=np.array([1000,7250])
+C=np.linalg.solve(A,B)
+
+#Resuts
+
+print"The reaction at F i.e Rf is ",round(C[0]),"N"
+print"The reaction at D i.e Rd is ",round(C[1]),"N"
+
+#Calculations (part 2)
+#Consider combined F.B.D of beams AB,BC &CD. Take moment at A
+
+Re=((W*La1)+(W*La2)+(W*La3)+(C[1]*Lad)-(C[0]*La3))/Lae #N
+Ra=C[1]-Re-C[0]+(3*W) #N #Taking sum of forces in Y direction
+
+#Results
+
+print"The reaction at pt E i.e Re is ",round(Re),"N"
+print"The reaction at pt A i.e Ra is ",round(Ra),"N" #acting vertically downwards
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction at F i.e Rf is  3500.0 N
+The reaction at D i.e Rd is  1000.0 N
+The reaction at pt E i.e Re is  833.0 N
+The reaction at pt A i.e Ra is  -333.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-7,Page No:69

+
+
+
+
+
+
In [14]:
+
+
+
import math
+
+#Initilization of variables
+
+Ws=2 #kN #weight of scooter
+Wd=0.5 #kN #weight of driver
+Lab=1 #m
+Led=0.8 #m
+Leg=0.1 #m
+
+#Calculations
+
+Rc=((2*Leg)+(Wd*Led))/Lab #kN #take moment at E
+Ra=(2+Wd-Rc)/2 #kN # as Ra=Rb,(Ra+Rb=2*Ra)
+Rb=Ra # kN
+
+#Results
+
+print"The reaction at wheel A is ",round(Ra,2),"kN"
+print"The reaction at wheel B is ",round(Rb,2),"kN"
+print"The reaction at wheel C is ",round(Rc,2),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction at wheel A is  0.95 kN
+The reaction at wheel B is  0.95 kN
+The reaction at wheel C is  0.6 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-8,Page No:69

+
+
+
+
+
+
In [25]:
+
+
+
import math
+
+#Initilization of variables
+
+W1=15 #N #up
+W2=60 #N #down
+W3=10 #N #up
+W4=25 #N #down
+Lab=1.2 #m
+Lac=0.4 #m
+Lcd=0.3 #m
+Ldb=0.5 #m
+Lad=0.7 #m
+Leb=0.417 #m #Leb=Lab-x
+
+#Calculations
+
+#(a) A single force
+
+Ry=W1-W2+W3-W4 #N #take sum Fy=0
+x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) #m
+
+#(b) Single force moment at A
+
+Ma=(Ry*x) #N-m
+
+# Single force moment at B
+
+Mb=W2*Leb #N-m
+
+#Results
+
+print"The reaction for single force (a) is ",round(Ry,2),"N"
+print"The distance of Ry from A is ",round(x,3),"m"
+print"The moment at A is ",round(Ma,2),"N-m"
+print"The moment at B is ",round(Mb,2),"N-m"
+
+ +
+
+
+ +
+
+ + +
+
+
The reaction for single force (a) is  -60.0 N
+The distance of Ry from A is  0.783 m
+The moment at A is  -47.0 N-m
+The moment at B is  25.02 N-m
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-9,Page No:71

+
+
+
+
+
+
In [6]:
+
+
+
import math
+import numpy as np
+
+#Initilization of variables
+
+Ra=5000 #N
+Ma=10000 #Nm
+alpha=60 #degree #angle made by T1 with the pole
+beta=45 #degree #angle made by T2 with the pole
+theta=30 #degree #angle made by T3 with the pole
+Lab=6 #m
+Lac=1.5 #m
+Lcb=4.5 #m
+
+#Calculations
+
+T3=Ma/(4.5*sin(theta*(pi/180))) #N #take moment at B
+
+# Now we use matrix to solve eqn's 1 & 2 simultaneously,
+
+A=np.array([[-0.707, 0.866],[0.707, 0.5]])
+B=np.array([2222.2,8848.8])
+C=np.linalg.solve(A,B)
+
+#Results
+
+print"Tension in wire 1 i.e T1 is ",round(C[1],1),"N" #answer may vary due to decimal variance
+print"Tension in wire 2 i.e T2 is ",round(C[0],1),"N"
+print"Tension in wire 3 i.e T3 is ",round(T3,1),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
Tension in wire 1 i.e T1 is  8104.7 N
+Tension in wire 2 i.e T2 is  6784.2 N
+Tension in wire 3 i.e T3 is  4444.4 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-10,Page No:76

+
+
+
+
+
+
In [32]:
+
+
+
import math
+
+#Initilization of variables
+
+w=2000 #N/m
+Lab=3 #m
+
+#Calculations
+
+W=w*Lab/2 #N# Area under the curve
+Lac=(0.6666)*Lab #m#centroid of the triangular load system
+Rb=(W*Lac)/Lab #N #sum of moment at A
+Ra=W-Rb #N
+
+#Results
+
+print"The resultant of the distibuted load lies at ",round(Lac),"m"
+print"The reaction at support A is ",round(Ra),"N"
+print"The reaction at support B is ",round(Rb),"N"
+
+ +
+
+
+ +
+
+ + +
+
+
The resultant of the distibuted load lies at  2.0 m
+The reaction at support A is  1000.0 N
+The reaction at support B is  2000.0 N
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-12,Page No:78

+
+
+
+
+
+
In [33]:
+
+
+
import math
+
+# Initilization of variables
+
+w1=1.5 #kN/m # intensity of varying load at the starting point of the beam
+w2=4.5 #kN/m # intensity of varying load at the end of the beam
+l=6 #m # ength of the beam
+
+# Calculations
+
+# The varying load distribution is divided into a rectangle and a right angled triangle
+
+W1=w1*l #kN # where W1 is the area of the load diagram(rectangle ABED)
+x1=l/2 #m # centroid of the rectangular load system
+W2=(w2-w1)*l/2 #kN # where W1 is the area of the load diagram(triangle DCE)
+x2=2*l/3 #m # centroid of the triangular load system
+W=W1+W2 #kN # W is the resultant
+x=((W1*x1)+(W2*x2))/W #m # where x is the distance where the resultant lies
+
+#Results
+
+print"The resultant of the distributed load system is ",round(W),"kN"
+print"The line of action of the resulting load is ",round(x,1),"m"
+
+ +
+
+
+ +
+
+ + +
+
+
The resultant of the distributed load system is  18.0 kN
+The line of action of the resulting load is  3.5 m
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-13,Page No:78

+
+
+
+
+
+
In [35]:
+
+
+
import math
+
+# Initiization of variables
+
+W1=10 #kN #point load acting at D
+W2=20 #kN # point load acting at C at an angle of 30 degree
+W3=5 #kN/m # intensity of udl acting on span EB of 4m
+W4=10 #kN/m # intensity of varying load acting on span BC of 3m
+M=25 #kN-m # moment acting at E
+theta=30 #degree # angle made by 20 kN load with the beam
+Lad=2 #m
+Leb=4 #m
+Laf=6 #m #distance between the resultant of W3 & point A
+Lac=11 #m
+Lag=9 #m #distance between the resultant of W4 and point A
+Lbc=3 #m
+Lab=8 #m
+
+# Calculations
+
+Xa=20*cos(theta*(pi/180)) #kN # sum Fx=0
+Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*sin(theta*(pi/180))*Lac)+((W4*Lbc*Lag)/2))/Lab #kN # taking moment at A
+Ya=W1+(W2*sin(theta*(pi/180)))+(W3*Leb)+(W4*Lbc/2)-Rb #kN # sum Fy=0
+Ra=(Xa**2+Ya**2)**0.5 #kN # resultant at A
+
+#Results
+
+print"The horizontal reaction at A i.e Xa is ",round(Xa,2),"kN"
+print"The vertical reaction at A i.e Ya is ",round(Ya),"kN"
+print"The reaction at A i.e Ra is ",round(Ra),"kN"
+print"The reaction at B i.e Rb is ",round(Rb),"kN"
+
+ +
+
+
+ +
+
+ + +
+
+
The horizontal reaction at A i.e Xa is  17.32 kN
+The vertical reaction at A i.e Ya is  10.0 kN
+The reaction at A i.e Ra is  20.0 kN
+The reaction at B i.e Rb is  45.0 kN
+
+
+
+ +
+
+ +
+
+
+
+
+
+

Example 3.3-14,Page No:79

+
+
+
+
+
+
In [37]:
+
+
+
import math
+
+# Initilization of variables
+
+h=4 #m #height of the dam wall
+rho_w=1000 # kg/m^3 # density of water
+rho_c=2400 # kg/m^3 # density of concrete
+g=9.81 # m/s^2
+
+# Calculations
+
+P=(rho_w*g*h**2)/2 # The resultant force due to water pressure per unit length of the dam
+x=(0.6666)*h #m # distance at which the resutant of the triangular load acts 
+b=((2*P*h)/(3*h*rho_c*g))**0.5 # m # eq'n required to find the minimum width of the dam
+
+# Results
+
+print"The minimum width which is to be provided to the dam to prevent overturning about point B is ",round(b,3),"m" 
+
+ +
+
+
+ +
+
+ + +
+
+
The minimum width which is to be provided to the dam to prevent overturning about point B is  1.491 m
+
+
+
+ +
+
+ +
+
+
+ + diff --git a/f_by_df/screenshots/1.png b/f_by_df/screenshots/1.png new file mode 100644 index 00000000..04dc7a94 Binary files /dev/null and b/f_by_df/screenshots/1.png differ diff --git a/f_by_df/screenshots/2.png b/f_by_df/screenshots/2.png new file mode 100644 index 00000000..3d55a09c Binary files /dev/null and b/f_by_df/screenshots/2.png differ diff --git a/f_by_df/screenshots/3.png b/f_by_df/screenshots/3.png new file mode 100644 index 00000000..f00ffd91 Binary files /dev/null and b/f_by_df/screenshots/3.png differ diff --git a/f_by_df/screenshots/image.png b/f_by_df/screenshots/image.png new file mode 100644 index 00000000..c1fd6fdb Binary files /dev/null and b/f_by_df/screenshots/image.png differ diff --git a/f_by_df/screenshots/image_1.png b/f_by_df/screenshots/image_1.png new file mode 100644 index 00000000..c1fd6fdb Binary files /dev/null and b/f_by_df/screenshots/image_1.png differ diff --git a/f_by_df/screenshots/image_2.png b/f_by_df/screenshots/image_2.png new file mode 100644 index 00000000..c1fd6fdb Binary files /dev/null and b/f_by_df/screenshots/image_2.png differ diff --git a/sample_notebooks/SufiyanSiddique/Chapter2.ipynb b/sample_notebooks/SufiyanSiddique/Chapter2.ipynb new file mode 100644 index 00000000..7a4bda7a --- /dev/null +++ b/sample_notebooks/SufiyanSiddique/Chapter2.ipynb @@ -0,0 +1,257 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 : Amplification" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.1, page 92" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ouput voltage of and amplifier = 15.2 V\n" + ] + } + ], + "source": [ + "#Initialisation\n", + "Ri=1000 #Input Resistance of amplifier in Ohm\n", + "Rs=100 #Output Resistance of sensor in Ohm\n", + "Rl=50 #Load Resistance\n", + "Ro=10 #Output Resistance of amplifier in Ohm\n", + "Av=10 #Voltage gain\n", + "Vs=2 #Sensor voltage\n", + "\n", + "#Calculation\n", + "Vi=(Ri*Vs)*(Rs+Ri)**-1 #Input Voltage of Amplifier\n", + "Vo=Av*Vi*Rl/(Ro+Rl) #Output Voltage of Amplifier\n", + "\n", + "#Result\n", + "print'Ouput voltage of and amplifier = %.1f V'%Vo\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.2, page 93" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage Gain, Av = 8.35\n" + ] + } + ], + "source": [ + "#Initialisation\n", + "Vo=15.2 #Output Voltage of Amplifier\n", + "Vi=1.82 #Input Voltage of Amplifier\n", + "\n", + "#Calculation\n", + "Av=Vo/Vi #Voltage gain\n", + "\n", + "#Result\n", + "print'Voltage Gain, Av = %.2f'%Av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.3, page 94" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ouput voltage of and amplifier = 20.0 V\n" + ] + } + ], + "source": [ + "#Initialisation\n", + "Av=10 #Voltage gain\n", + "Vi=2 #Input Voltage of Amplifier\n", + "Rl=50 #Load Resistance\n", + "Ro=0 #Output Resistance of amplifier in Ohm\n", + "\n", + "\n", + "#Calculation\n", + "Vo=Av*Vi*Rl/(Ro+Rl) #Output Voltage of Amplifier\n", + "\n", + "#Result\n", + "print'Ouput voltage of and amplifier = %.1f V'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.4, page 96" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output Power, Po = 4.6 W\n" + ] + } + ], + "source": [ + "#Initialisation\n", + "Vo=15.2 #Output Voltage\n", + "Rl=50 #Load Resistance\n", + "\n", + "#Calculation \n", + "Po=(Vo**2)/Rl #Output Power\n", + "\n", + "#Result\n", + "print'Output Power, Po = %.1f W'%Po" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.5, page 98" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power Gain, Ap = 1395\n" + ] + } + ], + "source": [ + "#Initialisation\n", + "Vi=1.82 #Input Voltage of Amplifier\n", + "Ri=1000 #Input Resistance of amplifier in Ohm\n", + "Vo=15.2 #Output Voltage of Amplifier\n", + "Rl=50 #Load Resistance\n", + "\n", + "\n", + "#Calculation\n", + "Pi=(Vi**2)*Ri**-1 #Input Power in Watt\n", + "Po=(Vo**2)*Rl**-1 #Output Power in Watt\n", + "Ap=Po*Pi**-1 #Power Gain\n", + "\n", + "\n", + "#Result\n", + "print'Power Gain, Ap = %d'%Ap\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.6, page 99" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power Gain (dB) = 31.5 dB\n" + ] + } + ], + "source": [ + "import math \n", + "\n", + "#Initialisation\n", + "P=1400 #Power gain\n", + "\n", + "#Calculation\n", + "pdb=10*math.log10(P) #Power Gain in dB\n", + "\n", + "#Result\n", + "print'Power Gain (dB) = %.1f dB'%pdb\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [] + } + ], + "metadata": { + "anaconda-cloud": {}, + "kernelspec": { + "display_name": "Python [Root]", + "language": "python", + "name": "Python [Root]" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.12" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/chayas/DimensionsSystemUnits.ipynb b/sample_notebooks/chayas/DimensionsSystemUnits.ipynb new file mode 100644 index 00000000..cdb3b1c1 --- /dev/null +++ b/sample_notebooks/chayas/DimensionsSystemUnits.ipynb @@ -0,0 +1,169 @@ +{ + "metadata": { + "name": "DimensionsSystemUnits" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Dimensions and systems of units" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Ex.1.1" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Ex1.1 pageno 12\npg=40 #negative pressure\npatom=47.2 #atom pressure\npa=patom-pg #absolute pressure of 6000\nprint \" pa = \",round(pa,1), \"kPa\" ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " pa = 7.2 kPa\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Ex 1.2" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Ex 1.2 pageno 13\nsigma=0.073 #N/m\nga= 9800 #density\nD= 2*10**-3\nh=(4*sigma)/(ga*D)*10**3\nprint '%s % 2g' %(\"\\n h(meter)= \",h), \"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "\n h(meter)= 14.898 mm\n" + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Ex 1.3" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Ex 1.3 pageno 13 \n# from table given in textbook\nimport math\nz=10000 #meter\nT=223.3 #kelvin\nr=287\nk=1.4\nc=math.sqrt(k*r*T)\nprint \" velocity of sound C= \",round(c,1), \"m/s\" \nv=(800*1000) # given speed of aircraft flies 800 km/hr\nt=3600\nspeed= v/t\nprint '%s %.1f %s' %(\"speed of aircraft v= \",speed,\"m/s\" )\nm=speed/c\nprint '%s %.2f'%(\"mach number of aircraft M= \",m)", + "language": "python", + "metadata": {}, + "outputs": [], + "prompt_number": "*" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "EX 1.4" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# pageno 13\ns=0.91 #specific gravity of water\nd=1000 #density of water\np=s*d # fluid density of water\nprint \"fluid density\",p,\"kg/m3\"\nd1=25*10**-3\nv1=2.6\nmu=0.38\nre=(p*d1*v1)/mu\nprint '%s %.1f' %(\"Reynolds number Re is \",re)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "fluid density 910.0 kg/m3\nReynolds number Re is 155.7\n" + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Ex 1.5" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# page no 14\nb=3 #pressure bar of gauge\nv2=101.3 #conversion value \nap=(b*v2)+(v2) # absolute pressure in the tank in kpa\nprint '%s %.1f %s' %(\"Absolute pressure in the tank in kpa =\",ap,\"kPa\")\np=ap*10**3\nR=287 # R Radius of Surface tension\nT=288 # Temperature in kelvin\np1=p/(R*T)\nprint '%s %.1f %s' %(\"density p= \",p1,\"Kg/m3\")\nv=0.85 # volume of airtank in m3\ng= 9.8 # gravational force \nw=p1*v*g\nprint '%s %.2f %s'%(\"Weight of air W = pvg= \",w,\"Kg\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Absolute pressure in the tank in kpa = 405.2 kPa\ndensity p= 4.9 Kg/m3\nWeight of air W = pvg= 40.84 Kg\n" + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Ex 1.6" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#page no 14 \nR=1*10**-3 # radius in meter\nsigma1=72.7*10**-3 # N/M\nE=(2*sigma1)/R\nprint '%s %.1f %s' %(\" Excess pressure p= \",E,\"N/m2\")", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Excess pressure p= 145.4 N/m2\n" + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Ex 1.7\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#page no 15\n# derivation of equation by substuting so no calculations \nprint \"shear stress t=-BD/4\"\nprint \" Also at position r=D/4;t=-BD/8\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "shear stress t=-BD/4\n Also at position r=D/4;t=-BD/8\n" + } + ], + "prompt_number": 63 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit