From 64d949698432e05f2a372d9edc859c5b9df1f438 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- backup/f_by_df_version_backup/chapter03_16.ipynb | 864 +++++++++++++++++++++++ 1 file changed, 864 insertions(+) create mode 100755 backup/f_by_df_version_backup/chapter03_16.ipynb (limited to 'backup/f_by_df_version_backup/chapter03_16.ipynb') diff --git a/backup/f_by_df_version_backup/chapter03_16.ipynb b/backup/f_by_df_version_backup/chapter03_16.ipynb new file mode 100755 index 00000000..46ae736b --- /dev/null +++ b/backup/f_by_df_version_backup/chapter03_16.ipynb @@ -0,0 +1,864 @@ + + + + + +chapter03_16 + + + + + + + + + + + + + + + + + + + + + + + + +
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Chapter 3 : Parallel Forces In A Plane

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Example 3.3-1,Page No:64

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In [1]:
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import math
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+#Initilization of variables
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+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"
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The load taken by man P is  400.0 N
+The load taken by man Q is  600.0 N
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Example 3.3-2,Page No:64

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In [2]:
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import math
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+#Initilization of variables
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+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"
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The reaction (downwards)at support A is  250.0 N
+The reaction (upwards)at support B is  1250.0 N
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Example 3.3-3,Page No:65

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In [7]:
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import math
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+#Inilitization of variables
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+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"
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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
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Example 3.3-5,Page No:66

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In [11]:
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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
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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
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Example 3.3-7,Page No:69

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In [14]:
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import math
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+#Initilization of variables
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+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"
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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
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Example 3.3-8,Page No:69

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In [25]:
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import math
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+#Initilization of variables
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+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
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+#(b) Single force moment at A
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+Ma=(Ry*x) #N-m
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+# Single force moment at B
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+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"
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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
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Example 3.3-9,Page No:71

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In [6]:
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import math
+import numpy as np
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+#Initilization of variables
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+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
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+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"
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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
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Example 3.3-10,Page No:76

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In [32]:
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import math
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+#Initilization of variables
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+w=2000 #N/m
+Lab=3 #m
+
+#Calculations
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+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"
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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
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Example 3.3-12,Page No:78

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In [33]:
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import math
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+# Initilization of variables
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+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
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+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"
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The resultant of the distributed load system is  18.0 kN
+The line of action of the resulting load is  3.5 m
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Example 3.3-13,Page No:78

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In [35]:
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import math
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+# Initiization of variables
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+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"
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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
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Example 3.3-14,Page No:79

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In [37]:
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import math
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+# 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" 
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The minimum width which is to be provided to the dam to prevent overturning about point B is  1.491 m
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+ +
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+ +
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+ + -- cgit