# Chapter 7 Internal Forces¶

## Ex 7.1 Page no 261¶

In [16]:
# Example 7.1

# Calculation
# Applying +ΣF_y(upward) = 0 Refer fig 7-4b
A_y = 16+4-12  #[kilo Newton]

# Applying +ΣF_y(upward) = 0 to segment AB
NB = 8  #[kilo Newton]

# Applying +ΣF_y(upward) = 0 to segment DC
NC = 4  #[kilo Newton]

# Result
print"A_y = ",(A_y),"kN"
print"NC = ",(NC),"kN"

A_y =  8 kN
NC =  4 kN


## Ex 7.2 Page no 262¶

In [5]:
# Example 7.2

# Calculation
# Applying +ΣM_x = 0 Refer fig 7-5b
TD = -10+15+20  #[Newton meter]

# Applying +ΣM_x = 0 to segment AB
TB = -10+15  #[Newton meter]

# Applying +ΣM_x = 0 to segment CD
TC = 25  #[Newton meter]

# Result
print"TB = ",(TB),"N.m"
print"TC = ",(TC),"N.m"

TB =  5 N.m
TC =  25 N.m


## Ex 7.3 Page no 263¶

In [4]:
# Example 7.3
from __future__ import division

# Calculation
# Applying +ΣMD(counterclockwise) = 0 Refer fig 7-6b
A_y = (9+6*6)/9

# Consider segment AB
# Applying +ΣF_x(right) = 0
NB = 0  #[kilo Newton]
# Applying +ΣF_y(upward) = 0
VB = 5  #[kilo Newton]
# Applying +ΣMB(counterclockwise) = 0
MB = 5*3  #[kilo Newton meter]

# Consider segment AC
# Applying +ΣF_x(right) = 0
NC = 0  #[kilo Newton]
# Applying +ΣF_y(upward) = 0
VC = 6-5  #[kilo Newton]
# Applying +ΣMC(counterclockwise) = 0
MC = 5*3  #[kilo Newton meter]

# Result
print"NB = ",(NB),"kN"
print"VB = ",(VB),"kN"
print"MB = ",(MB),"kN.m"
print"NC = ",(NB),"kN"
print"VC = ",(VC),"kN"
print"MC = ",(MC),"kN.m"

NB =  0 kN
VB =  5 kN
MB =  15 kN.m
NC =  0 kN
VC =  1 kN
MC =  15 kN.m


## Ex 7.4 Page no 264¶

In [15]:
# Example 7.4
from __future__ import division

# Calculation
# Refer fig 7-7b
# Applying +ΣMA(counterclockwise) = 0
FDC = (400*4)/((3/5)*8)  #[Newton]
# Applying +ΣF_x(right) = 0
A_x = (4/5)*333.3   #[Newton]
# Applying +ΣF_y(upward) = 0
A_y = 400-((3/5)*333.3)   #[Newton]

# Applying equations of equilibrium to segment AB
# Applying +ΣF_x(right) = 0
NB = 266.7   #[Newton]
# Applying +ΣF_y(upward) = 0
VB = 200-200   #[Newton]
# Applying +ΣMB(counterclockwise) = 0
MB = 200*4-200*2   #[Newton meter]

# Result
print"NB = ",(NB),"N"
print"VB = ",(VB),"N"
print"MB = ",(MB),"N.m"

NB =  266.7 N
VB =  0 N
MB =  400 N.m


## Ex 7.5 Page no 265¶

In [14]:
# Example 7.5
import math
from __future__ import division

# Calculation
# Using  +ΣF_y(upward) = 0 Refer fig 7-8b
R = round(600/math.sin(math.pi*45/180),1)  #[Newton]
# Applying equations of equilibrium
# Applying +ΣF_x(right) = 0
VE = round(R*math.cos(math.pi*45/180),1)  #[Newton]
# Applying +ΣF_y(upward) = 0
NE = round(R*math.sin(math.pi*45/180),1)  #[Newton]
# Applying +ΣME(counterclockwise) = 0
ME = round(R*math.cos(math.pi*45/180)*0.5,1)  #[Newton meter]

# Result
print"VE = ",(VE),"N"
print"NE = ",(NE),"N"
print"ME = ",(ME),"N.m"

VE =  600.0 N
NE =  600.0 N
ME =  300.0 N.m


## Ex 7.6 Page no 266¶

In [13]:
# Example 7.6
from __future__ import division
import numpy as np

# Calculation
W = 650*9.81  #[kilo Newton]
Fw = 900/(6*2.5)  #[kilo Newton]
# Using +ΣF_x(right) = 0
FA_x = 13.5  #[kilo Newton]
FA_y  = 0  #[kilo Newton]
FA_z = 6.376  #[kilo Newton]
# Using ΣMA = 0, MA + r X (Fw + W)
a = np.cross([0,3,5.25],[-13.5,0,6.376])
MA_x = round(-a[0],1)  #[kilo Newton meter]
MA_y = round(-a[1],1)  #[kilo Newton meter]
MA_z = round(-a[2],1)  #[kilo Newton meter]

# Result
print"MA_x = ",(MA_x),"kN.m"
print"MA_y = ",(MA_y),"kN.m"
print"MA_z = ",(MA_z),"kN.m"  # Correction in MA_z

MA_x =  -19.1 kN.m
MA_y =  70.9 kN.m
MA_z =  -40.5 kN.m


## Ex 7.13 Page no 286¶

In [30]:
# Example 7.13
from __future__ import division
import math

# Calculation
# Consider free body diagram Refer fig 7-21b
# Using +ΣME(counterclockwise) = 0
A_y = (4*15+15*10+3*2)/18   #[kilo Newton]
# Using +ΣF_y(upward) = 0
E_y = -12+4+15+3  #[kilo Newton]

# Consider leftmost section which cuts cable BC Refer fig 7-21b
# Using +ΣMC(counterclockwise) = 0
A_x = (12*8-4*5)/12  #[kilo Newton]
thetaBC = round(math.degrees(math.atan((12-4)/6.33)),1)  #[Degrees]
# Using +ΣF_x(right) = 0
TBC = round(6.33/math.cos(math.pi*thetaBC/180),1)  #[kilo Newton]
thetaAB = round(math.degrees(math.atan(12/6.33)),1)  #[Degrees]

# Applying +ΣF_x(right) = 0 at point A
TAB = round(12/math.sin(math.pi*thetaAB/180),1)  #[kilo Newton]
thetaCD = round(math.degrees(math.atan((-10.2*math.sin(math.pi*51.6/180)+15)/(10.2*math.cos(math.pi*51.6/180)))),1)  #[Degrees]

# Applying +ΣF_x(right) = 0 at point C
TCD = round((10.2*math.cos(math.pi*51.6/180))/math.cos(math.pi*thetaCD/180),1)  #[kilo Newton]
thetaED = round(math.degrees(math.atan(10/6.33)),1)  #[Degrees]

# Applying +ΣF_x(right) = 0 at point E
TED = round(10/math.sin(math.pi*thetaED/180),1)  #[kilo Newton]

print"TBC = ",(TBC),"kN"
print"TAB = ",(TAB),"kN"
print"TCD = ",(TCD),"kN"
print"TED = ",(TED),"kN"

TBC =  10.2 kN
TAB =  13.6 kN
TCD =  9.5 kN
TED =  11.8 kN

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