# Chapter 1:Equilibrium of a deformable body¶

### Example 1.1 Page no 11¶

In :
#Given
w_varying = 270.0     #N/m, torque
l_crossection = 9.0   #m , length
l_cb = 6.0            #m length
l_ac = 2.0             #m length

#Calculation
w_c = (w_varying/l_crossection) * l_cb #By proportion, load at C is found.
f_resultant_c = 0.5* w_c *l_cb
#Balancing forces in the x direction:
n_c = 0
#Balncing forces in the y direction:
v_c = f_resultant_c
# Balncing the moments about C:
m_c = - (f_resultant_c*l_ac)

#Results
print"The resultant force at C  is  ",f_resultant_c,"N"
print"The horizontal force at C is  ",n_c,"N"
print"The vertical force at C   is  ",v_c,"N"
print"The moment about C        is  ",m_c,"Nm"

The resultant force at C  is   540.0 N
The horizontal force at C is   0 N
The vertical force at C   is   540.0 N
The moment about C        is   -1080.0 Nm


### Example 1.2 Page no 12¶

In :
#Given
f_d = 225         #N
w_uniform = 800   # N/m
l_ac = 0.200      #m
l_cb = 0.05+0.1   #m
l_bd = 0.100      #m
l_bearing = 0.05    #m
f_resultant = w_uniform*l_cb #120N
l_f_resultant_b = (l_cb/2)+ l_bearing #0.125m
l = l_ac + l_cb + l_bearing + l_bd

#Calculation
m_b = 0                                                        # Net moment about B is zero for equilibrium
a_y = -((f_d*l_bd) - (f_resultant*l_f_resultant_b))/ (l - l_bd) # finding the reaction force at A

# Refer to the free body diagram in Fig.1-5c.
f_c = 40                        #N
#Balancing forces in the x direction
n_c = 0
#Balncing forces in the y direction
v_c = a_y - f_c            #-18.75N - 40N-Vc = 0
# Balncing the moments about C
m_c =  ((a_y * (l_ac + 0.05)) - f_c*(0.025) ) # Mc+40N(0.025m)+ 18.75N(0.250m) = 0

# Result
print'The horizontal force at C =',n_c,"N"
print'The vertical force at C   =',v_c,"N"

The horizontal force at C = 0 N
The vertical force at C   = -58.75 N
The moment about C        = -5.69 NM


### Example 1.3 Page no 13¶

In :
# Given
l_ac = 2     #m.,length
l_cb = 1     #m.
#r_a = 0.125  #m.
#r_d = 0.125  #m.
W = 500      # N, force
g=9.81

#Calculation
#Balancing forces in the x direction
n_c = (W*g*(l_ac+l_cb))/(3*2/5.0) # N
#Balncing forces in the y direction
v_c = n_c*(4/5.0)                 #N
# Balncing the moments about C
m_c = n_c*(3/5.0)-(W*g)

# Result
print'The horizontal force at C   =  ',-v_c/1000,"KN"
print'The vertical force at C     =  ',-m_c/1000,"KN"
print'The moment about C          =  ',-m_c/1000,"KNm"

The horizontal force at C   =   -9.81 KN
The vertical force at C     =   -2.4525 KN
The moment about C          =   -2.4525 KNm


### Example 1.4 Page no 14¶

In :
# Given
l_ag = 2           #Length of AG is 1m.
l_gd = 1         #Length of GD is 1m.
l_de = 3         #Length of DE is 1m.
f_a = 1500       #Force at A is 1500N.
l_ec = 1.5       #Length of EC is 1m.
l = l_ag +l_gd +l_de
w_uniform_varying = 600 #Nm.
f_ba = 7750      #N
f_bc = 6200      #N
f_bd = 4650      #N

#Calculation
w_resultant = 0.5*l_de*w_uniform_varying
# calling point of action of resultant as P
l_ep = (2/3.0)*l_de
l_ap = l - l_ep

#Free Body Diagram Using the result for Fba, the left section AG of the beam is shown in Fig 1-7d.
# Equations of equilibrium
#Balancing forces in the x direction
n_g = -f_ba * (4/5.0)    # N
#Balncing forces in the y direction
v_g = -f_a + f_ba*(3/5.0) #N

# Balncing the moments about C
m_g = (f_ba * (3/5.0)*l_ag) - (f_a * l_ag) #Nm

# Result
print'The horizontal force at G =  ',n_g,"lb"
print'The vertical force at G   =  ',v_g,"lb"
print'The moment about G        =  ',m_g,"lb-ft"

The horizontal force at G =   -6200.0 lb
The vertical force at G   =   3150.0 lb
The moment about G        =   6300.0 lb-ft


### Example 1.5 Page no 15¶

In :
# Given
f_a = 50       #N, force
m_a = 70       # Moment at A in Nm
l_bd = 0.5     #Length of BD in m.
l_cb = 0.75    #Length of BC in m.
w_l  = 2       #Kg/m
g = 9.81       #N/kg- acceleration due to gravity

#Free Body Diagram
import math
w_bd = w_l*l_bd*g #in N.
# Equations of Equilibrium
#Balancing forces in the x, y and z direction
f_b_x = 0     #  N
f_b_y = 0     #N
f_b_z = g + w_ad + f_a      #N
# Balancing Moments in the x,y and z direction
m_b_x = - m_a + (f_a*l_bd) + (w_ad*l_bd) + (l_bd/2.0)*g #Nm
m_b_z = 0                       #Nm
v_b_shear = sqrt(f_b_z **2 + 0) #Shear Force in N
t_b = - m_b_y                   #Torsional Moment in Nm
m_b = math.sqrt(m_b_x **2+ 0)   # Bending moment in Nm

# Result
print' The weight of segment BD                =',w_bd,"N"
print' The force at B in the Z direction       =',f_b_z,"N"
print' The moment about B in the X direction   =',m_b_x,"Nm"
print' The moment about G in the Y direction   =',m_b_y,"Nm"
print' The Shear Force at B                    =',v_b_shear,"N"
print' The Torsional Moment at B               =',t_b,"Nm"
print' The Bending Moment at B                 =',m_b,"Nm"

 The weight of segment BD                = 9.81 N
The weight of segment AD                = 24.525 N
The force at B in the Z direction       = 84.335 N
The moment about B in the X direction   = -30.285 Nm
The moment about G in the Y direction   = -77.828125 Nm
The Shear Force at B                    = 84.335 N
The Torsional Moment at B               = 77.828125 Nm
The Bending Moment at B                 = 30.285 Nm


### Example 1.6 Page no 28¶

In :
#Given
netf_b = 18*(10 **3)      #N Net force at B.
netf_c = 8*(10**3)        #N Net force at C.
f_a = 12 *(10**3)         #N Force at A.
f_d = 22* (10**3)         #N Force at D.
w = 35.0                    #mm Width.
t = 10.0                    #mm Thickness.

#calculations
p_bc = netf_b + f_a          #N Net force in region BC.
a = w*t                      #m**2 The area of the cross section.
avg_normal_stress = p_bc/a   #Average Normal Stress.

# Result
print'The Average Normal Stress in the bar when subjected to load = ',round(avg_normal_stress,1),"MPa"

The Average Normal Stress in the bar when subjected to load =  85.7 MPa


### Example 1.7 Page no: 29¶

In :
#Given :
m_lamp = 80   #Mass of lamp in Kg.
d_ab = 10     # Diameter of AB in mm.
d_bc = 8      # Diameter of BC in mm.
ab_h = 60 *(math.pi/180.0) # In degrees - Angle made by AB with the horizontal.
w = m_lamp*9.81 #N
a_bc = (math.pi/4.0)*(d_bc**2) #m**2 Area of cross section of rod BC
a_ab = (math.pi/4.0)*(d_ab**2) #m**2 Area of cross section of rod AB

#calculation
import math   # Equations of equilibrium: Solving equilibrium equations simultaneously ,
f_bc=395
f_ab=f_bc*(4/5.0)/(math.cos(60*3.14/180.0))
avg_normal_stress_a = f_ab / a_ab #Mpa Average Normal Stress in AB
avg_normal_stress_c = f_bc/ a_bc# Mpa   Average Normal Stress in BC

# Displaying results:
print"The Average Normal Stress in AB when subjected to load = MPa",round(avg_normal_stress_a,2),"MPa"
print"The Average Normal Stress in BC when subjected to load = MPa",round(avg_normal_stress_c,2),"MPa"

The Average Normal Stress in AB when subjected to load = MPa 8.04 MPa
The Average Normal Stress in BC when subjected to load = MPa 7.86 MPa


### Example 1.8 Page no 30¶

In :
#Given
h_above_ab = 2.75        #ft , height
h_below_ab = 0.2
d_a = 0.75               #diameter ft
d_b = 0.1
sp_w = 490

# Equation of Equilibrium
import math
a = math.pi* (d_a**2)  # Area of cross section in m**2
p = sp_w * h_above_ab * a
avg_comp_stress = p/a # The average compressive stress in kN/m**2

#Display
print'The average compressive stress = ',avg_comp_stress,"lb/ft**2"

The average compressive stress =  1347.5 lb/ft**2


### Example 1.9 Page no 31¶

In :
#Determine Distance of force from AB

#Given
f = 3000 #N Force acting at distance x from AB.
l_ac = 200 #Length of AC in mm.
a_ab = 400 #Cross sectional area of AB in mm**2.
a_c = 650 # area of C in mm**2.

#Calculation
#Equations are
#Fab+Fc=3000
#-3000*x+Fc*200=0
Fc=1857
Fab=3000-Fc
x=Fc*200/3000

#Display
print'Distance of force from AB =  ',x,"mm"

Distance of force from AB =   123 mm


### Example 1.10 Page no 35¶

In :
#Given
af =30		 #kN Axial force along centroidal axis
t1 = 2 		#m thickness of square cross section
t2=4
import math
Fb=af*t1/((t1+t2)*(4/5.0))
Ax=Fb*(3/5.0)
Ay=af-(Fb*(4/5.0))
Fa=math.sqrt(Ax**2+Ay**2)
Va=Fa/t1
Vb=Fb
Aa=(math.pi/4.0)*((t1/100.0)**2)
Ab=(math.pi/4.0)*(((t1+t2)/200.0)**2)
Ta=Va/Aa
Tb=Vb/Ab

#Display
print'The Average Normal Stress for section b-b   =  ',round(Ta/1000,0),"Mpa"
print'The Average Shear Stress for section b-b    =  ',round(Tb/1000,1),"Mpa"

The Average Normal Stress for section b-b   =   34.0 Mpa
The Average Shear Stress for section b-b    =   17.7 Mpa


### Example 1.11 Page no 36¶

In :
#Given
f = 6          #kN
d_rod = 0.1        #Diameter of steel rod in mm.
l_bc = 0.125        #Length of side bc in mm.
l_ac=0.15
#Calculation
F=f/2.0
#Appling Force balance
Va=F
Vb=F
a_rod=d_rod*l_ac
a_strut=l_bc*l_ac
#Average shear stress
avg_shear_rod = F/a_rod #for rod in Mpa
avg_shear_strut = (f/2)/a_strut #for strut

#Display
print'The average shear stress for the rod    =  ',avg_shear_rod,"Kpa"
print'The average shear stress for the strut  =  ',avg_shear_strut,"Kpa"

The average shear stress for the rod    =   200.0 Kpa
The average shear stress for the strut  =   160.0 Kpa


### Example 1.12 Page no 37¶

In :
#Given
l_bc = 1 	#Length of BC in inc.
l_db = 3 	# in.
l_ed = 2 	# in.
l_ab = 1.5 	# in.
f_diagonal = 600 	#lb

#Calculation
a1 = l_bc*l_ab 		#Area of face AB in mm**2.
a2 = l_ab*l_ed  	#mm**2.
a3 = l_db*l_ab  	# mm**2.
#Balancing forces along the x- direction.
f_ab = f_diagonal*(3/5.0) #Force on segment AB in N
V = f_ab

#Balancing forces along the Y direction.
f_bc = f_diagonal*(4/5.0) #Force on segment BC in N.
avg_comp_ab = f_ab/a1 # N/mm**2
avg_comp_bc = f_bc/a2 # N/mm**2
avg_shear = f_ab/a3 # N/mm**2

#Display
print a1
print'The average compressive stress along AB  =  ',avg_comp_ab,"psi"
print'The average compressive stress along BC  =  ',avg_comp_bc,"psi"
print'The average shear stress along EDB       =  ',avg_shear,"psi"

1.5
The average compressive stress along AB  =   240.0 psi
The average compressive stress along BC  =   160.0 psi
The average shear stress along EDB       =   80.0 psi


### Example 1.13 Page no 49¶

In :
#Given
shear_allow = 3 	#ksi, stress
tensile_allow = 5 	#ksi, stress
l_C1 = 3		 #in
l_C2 = 2 		 #in
l_ac=8            #inch
Tauallow=8       #ksi

#Calculation
import math
Fab=(tensile_allow*(3/5.0)*(l_C1+l_C2)+shear_allow*l_C1)/(l_ac)
Cx=-shear_allow+tensile_allow*(4/5.0)
Cy=tensile_allow*(3/5.0)+l_C1
Fc=math.sqrt(Cx**2+Cy**2)
A=(Fc/2.0)/Tauallow
d=2*math.sqrt(A/math.pi)

#Result
print" The diameter of rod is =  ",round(d,3),"inch"
print"Wse a pin of diameter =",3/4.0,"inch"

 The diameter of rod is =   0.696 inch
Wse a pin of diameter = 0.75 inch


### Example 1.14 Page no 50¶

In :
#Given
P= 20.0		      #kN, force
d_hole = 40.0 	  #mm
normal_allow = 60.0 #MPa, pressure
shear_allow = 35.0  #MPa, pressure

#Diameter of Rod
import math
area1 = (P*10**3)/(normal_allow*10**6) #Area in m**2
d = ((math.sqrt((4*area1)/math.pi))*1000) # Area = (math.pi\4)d**2
#Thickness of disc
V = P
area2 = (V*10**3)/(shear_allow*10**6) #Area in m**2
thickness = (area2*10**6)/(d_hole*math.pi)# A = pi*d*t

#Result
print"The diameter of rode   = ",round(d,2),"mm"
print"The thickness of disc  = ",round(thickness,2),"mm"

The diameter of rode   =  20.6 mm
The thickness of disc  =  4.55 mm


### Example 1.15 Page no 51¶

In :
#Given
import math
bearing_allow = 75.0	 #MPa, pressure
tensile_allow = 55.0 	#MPa, pressure
d_shaft = 60.0		 #mm
r_shaft = d_shaft/2.0 	#mm
area_shaft = math.pi*(r_shaft**2) #Area = pi*r**2
d_collar = 80 #mm
thick_collar = 20 #mm

#Calculation
r_collar = d_collar/2.0 #mm
area_collar = math.pi*(r_collar**2) #Area = pi*r**2
#Normal Stress
P1 = (tensile_allow* area_shaft)/3.0 #Tensile stress = 3P/A.
P1_kN = P1/1000.0
#Bearing Stress
bearing_area = area_collar-area_shaft
P2 = (bearing_allow*bearing_area)/3
P2_kN= P2/1000.0

#Result
print"The load calculated by Normal Stress  ",round(P1_kN,1),"kN"
print"The load calculated by Bearing Stress ",round(P2_kN,0),"kN"

The load calculated by Normal Stress   51.8 kN
The load calculated by Bearing Stress  55.0 kN


### Example 1.16 Page no 52¶

In :
#Given
d_ac= 20.0            #mm
area_al = 1800.0       #mm**2, area
d_pins = 18.0          #mm
st_fail_stress = 680.0 #MPa, pressure
al_fail_stress = 70.0 #MPa, pressure
shear_fail_pin = 900.0 #MPa
fos = 2.0             #Factor of safety
l_ab = 2.0            #m
l_ap = 0.75           #m

#Calculation
import math
area_pins = math.pi*(d_pins/2.0)**2
area_ac = math.pi*(d_ac/2.0)**2 #Area = (math.pi\4)d**2
st_allow= st_fail_stress /fos #MPa
al_allow = al_fail_stress/fos #MPa
pin_allow_shear = shear_fail_pin/fos #MPa

#Rod AC
f_ac = (st_allow*area_ac)/1000.0
P1 = ((f_ac*l_ab)/(l_ab-l_ap))

#Block B
f_b =(al_allow*area_al)/1000.0
P2 = ((f_b*l_ab)/l_ap)

#Pin A or C
V = (pin_allow_shear*area_pins)/1000.0
P3 = (V*l_ab)/(l_ab-l_ap)

#Result
print"The load allowed on rod AC   ",round(P1,0),"kN"
print"The load allowed on block B  ",P2,"kN"
print"The load allowed on pins A or C ",round(P3,0),"kN"

The load allowed on rod AC    171.0 kN
The load allowed on block B   168.0 kN
The load allowed on pins A or C  183.0 kN