#Given
T1 = 4250.0 #kNmm, torque
T2 = -3000.0 #kNm
T3 = T1+T2 #kNm
r = 75.0 #mm, radius
#Calculation
#Section Property
import math
J = (math.pi/2.0)*(r**4) #polar moment of inertia
#Shear Stress
c_a = 75 #mm
tou_a = (T3*c_a*1000)/J #tou = Tc/J
c_b = 15 #mm
tou_b = (T3*c_b*1000)/J #tou = Tc/J
#Display
print'The shear stress developed at A = ',round(tou_a*10,1),"ksi"
print'The shear stress developed at B = ',round(tou_b*10,2),"ksi"
#Given
di = 80 #mm, inside diameter
ri = 40/1000.0 #m, inside radius
d0 = 100 #mm, outside diameter
ro = d0/2000.0 #m outside radius
F = 80 #N, force
l1 = 0.2 #m, length
l2 = 0.3 #m
#Internal Torque
T = F*(l1+l2)
#Section Property
import math
J = (math.pi/2.0)*((ro**4)-(ri**4))
#Shear Stress
c_o = 0.05#m
tou_o = (T*c_o)/(J*10**6)
c_i = 0.04 #m
tou_i = (T*c_i)/(J*10**6)
#Display
print'The shear stress in the inner wall = ',round(tou_i,3),"MPa"
print'The shear stress in the outer wall = ',round(tou_o,3),"MPa"
#Given
P = 5 #hp
N = 175 #rpm
allow_shear = 14.5 #ksi
#Calculations
import math
P_=P*550 #ftlb/s
ang_vel = (2*math.pi*N)/60.0 # rad/s
T = P_/ang_vel #P = T*angular velocity
c = ((2*T*12)/(math.pi*allow_shear*1000))**(1/3.0)
d =2*c
#Display
print'The required diameter of the shaft = ',round(d,3),"inch"
#Given
E = 80*10**9 #N/m**2, longitudinal stress
d = 14/1000.0 #m, diameter
r = d/2.0 #m, radius
R = 100 #mm
l_ac = 0.4 #m, length
l_cd = 0.3 #m
l_de = 0.5 #m
T_c = 280 #Nm, torque
T_a = 150 #Nm
T_d = 40 #Nm
T_ac = T_a #Nm
#Calculation
T_cd = T_ac - T_c
T_de = T_cd - T_d
#Angle of Twist
import math
J = (math.pi/2.0)*(r**4)
phiA=T_ac*l_ac/(J*E)+T_cd*l_cd/(J*E)+T_de*l_de/(J*E)
Sp=phiA*R
#Display
print'The angle of twist of the shaft = ',round(phiA,3),"rad"
print'The displacement of tooth P on gear A =',round(Sp,3),"mm"
#Given
T = 45 #N, torque
G = 80 #GPa, pressure
d = 20/1000.0 #m
r = d/2.0 #m
l_dc = 1.5 #m
l_ab = 2 #m
r1 = 75/1000.0 #m
r2 = 150/1000.0 #m
#Calculation
#Internal Torque
F = T/r2
T_d_x = F*r1
#Angle of twist
import math
J = (math.pi/2)*(r**4)
phi_c = (T*l_dc)/(2*J*G*10**9)
phi_b = (phi_c*r1)/r2
phi_ab = (T*l_ab)/(J*G*10**9)
phi_a = phi_b + phi_ab
#Display
print'The angle of twist of end A of shaft AB = ',round(phi_a,3),"rad"
#Given
d = 2 #inch, diameter
r = d/2.0 #radius
c = d/2.0
l_buried = 600 #mm, buried length
G = 5500*10**3 #MPa
F = 1 #N
l_handle= 150 #mm
l_ab = 36 #inch
#Internal Torque
T_ab = F*2*l_handle
t = T_ab/l_buried
#Maximum Shear Stress
import math
J = (math.pi/2.0)*(r**4)
tou_max = (T_ab*c)/(J)
#Angle of Twist
from scipy import integrate
def f(x):
return(x)
x=integrate.quad(f,0,24) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s)
X= x[0]
phi_a = ((T_ab*l_ab)+(50*X/4.0))/(J*G)
#Display
print'The maximum shear stress in the post =',round(tou_max,1),"psi"
print'The angle of twist at the top of the post = ',round(phi_a,5),"rad"
#Given
d = 20/1000.0 #m, diameter
r = d/2.0
l_bc = 0.2
l_cd = 1.5
l_da = 0.3
T_c = 800 #Nm, torque
T_d = -500 #Nm
#Calculation
#Eqn 1 300 = T_a + T_b
#Compatibility
#Eqn 2
coeff_Tb = -l_bc
coeff_Ta = l_cd + l_da
#Solving Equations simultaneously using matrices
T_b = 645
T_a = -345
#Display
print'The reaction at A = ',T_a,"Nm"
print'The reaction at B = ',T_b,"Nm"
#Given
T = 250 #Nm, torque
G_st = 80 #GPa, pressure
G_br = 36 #GPa
ri = 0.5 #inch, inside radius
ro = 1 #inch, outside radius
l_ab = 1.2 #m
#Equilibrium
# -Tst-Tbr+250Nm = 0
coeff1_st = -1
coeff1_br = -1
b1 = -250
#Compatibility
#phi = TL/JG
import math
J1 = (math.pi/2.0)*(ro**4 - ri**4)
J2 = (math.pi/2.0)*(ri**4)
coeff2_st = 1/(J1*G_st*10**3)
coeff2_br = -1/(J2*G_br*10**3)
b2 = 0
#Solving the above two equations simultaneously using matrices
T_st = 2911.5 #lb-inch
T_br = 88.5 #lb-inch
shear_br_max = (T_br*10**3*ri)/(J2) #tou = (Tr)/J
shear_st_min = (T_st*10**3*ri)/(J1) #tou = (Tr)/J
shear_st_max = (T_st*10**3*ro)/(J1) #tou = (Tr)/J
shear_strain = shear_br_max / G_br
shear_strain = shear_strain
#Display
print'The maximum shear stress experienced by Steel =',round(shear_st_max/1000),"psi"
print'The minimum shear stress experienced by Steel =',round(shear_st_min/1000),"psi"
print'The maximum shear stress experienced by Brass ',round(shear_br_max/1000),"psi"
#Given
import math
l = 4*12 #m, length
a = 1.5 #inch
tou_allow = 8000 #lb
phi_allow = 0.02 #rad
G = 3.7*10**6 #lb/inch**2, pressure
alpha = (60*math.pi)/180.0 #degrees
#Calculations
T_shear1 = (tou_allow*a**3)/(20.0) # allowable shear stress = (20T)/(a**3)
T_twist1 = (phi_allow*a**4*G)/(46*l) #angle of twist =(46TL)/(a**4*G)
T1 = min(T_shear1, T_twist1)
#Circular Cross Section
c_ = (a*a*math.sin(alpha))/(math.pi*2)
c = math.sqrt(c_)
J = (math.pi/2.0)*(c**4)
T_shear2 = (tou_allow*J)/(c*1000)
T_twist2 = (phi_allow*J*G*10**3)/(l*10**6)
T2 = min(T_shear2, T_twist2)
#Display
print'The largest torque that applied at the end of the triangular shaft ',round(T1,0),"lb-in"
print'The largest torque that applied at the end of the circular shaft ',round(T2*1000,0),"lb-in"
#Given
#The given dimension are
l_cd = 0.5 #m
l_de = 1.5 #m
h =60/1000.0 #m
w = 40/1000.0 #m
t_h = 3/1000.0 #m
t_w = 5/1000.0 #m
T_c = 60 #Nm
T_d = 25 #Nm
G = 38*10**9 #N/m**2
T1 = T_c - T_d
#Calculation
#Average Shear Stress
area = (w-t_w)*(h-t_h)
shear_a = T1/(2*t_w*area*10**6)
shear_b = T1/(2*t_h*area*10**6)
#Angle of Twist
phi=(T_c*l_cd/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))+(T1*l_de/(4*area**2*G))*((2*57/5.0)+(2*35/3.0))
#Display
print'The average shear stress of the tube at A = ',round(shear_a,2),"MPa"
print'The average shear stress of the tube at B = ',round(shear_b,2),"MPa"
print'The angle of twist of end C = ',round(phi,5),"rad"
#Given
fillet_r = 6 #mm, fillet radius
D = 40/1000.0 #m, diameter
d = 20/1000.0 #m
T = 30 #Nm
#Calculation
D_d = D/d
r_d = fillet_r/d
k = 1.3
#Maximum Shear Stress
import math
c = D/2.0
J = (math.pi/2.0)*(c**4)
max_shear = (k*T*c)/(J*10**6) # tou = K(Tc/J)
#Display
print'The maximum shear stress in the shaft is = ',round(max_shear,1),"MPa"
#Given
ro = 50/1000.0 #m, outside radius
ri = 30/1000.0 #m inside radius
c = ro
shear = 20*10**6 #N/m**2
#Maximum Elastic Torque
import math
J = (math.pi/2.0)*((ro**4)-(ri**4))
T_y = (shear*J)/c # tou = Tc/J
T_y = T_y/1000.0 #in kN
#Plastic Torque
x0 = 0.03
x1 = 0.05
from scipy import integrate
def f(rho):
return(rho**2)
I=integrate.quad(f,x0,x1) #Strain formula for short line segment = delta(sdash) =(1+e_z)delta(s)
Tp =(2*math.pi*I[0]*shear)
Tp_= Tp/1000.0
#Outer Shear Strain
strain = (0.286*10**-3*ro)/(ri)
#Display
print'The maximum torque that can be applied to the shaft ',round(T_y,2),"kNm"
print'The plastic torque that can be applied to the shaft',round(Tp_,2),"kNm"
print'The minimum shear strain at the outer radius of the shaft ',round(strain,6),"rad"
#Given
r = 20/1000.0 #m, radius
l = 1.5 #m, length
phi = 0.6 #rad
shear_y = 75*10**6 #N/m**2
#Calculations
max_shear_strain = (phi*r)/(l) #phi = (strain*L)/r
strain_y = 0.0016
r_y = (r*strain_y)/(max_shear_strain) #by ratios
#T= (math.pi*shear_y)*(4c**3 - r_y**3)/6.0
import math
c = r
T = (math.pi*shear_y)*(4*c**3 - r_y**3)/6.0
T = T/1000.0
#Display
print'The torque needed to twist the shaft by 0.6 rad ',T,"kNm"
#Given
l = 5 #m, length
G = 12*10**3 #GPa
co = 2 #inch
ci = 1 #inch
shear_y = 12 #N/mm**2
strain_y = 0.002 #rad, strain
#Plastic Torque
import math
T_p = ((2*math.pi)*(co**3 - ci**3)*shear_y)/3.0
phi_p = (strain_y*l*shear_y)/ci
J = (math.pi/2.0)*(co**4 - ci**4)
shear_r = (T_p*co)/J
shear_i = (shear_r*ci)/(co)# shear = Tc/J
G = shear_y/strain_y
phi_dash = (T_p*l*10**3)/(J*G) #phi = TpL/JG
phi = phi_p - phi_dash
#Display
print'The plastic torque Tp = ',round(T_p,1),"kip in"
print'shear stress at inner wall is ',round(shear_i,2),"ksi"