def Problem3Test():
""" Problem 8-33"""
steel = 20 # mm
brass = 20 # mm
d = 12 # in
D = 1.5 * d
l = steel + brass #in grip length
ksteel = Screw.SpringRate(207, d, D, steel)
kbrass = Screw.SpringRate(100, d, D, brass)
k = 1 / (1 / ksteel + 1 / kbrass)
print(k)
def Problem8_4():
d = 25 #mm
p = 5 #mm
F = 5 #Kn
fc = 0.06
ft = 0.09
dc = 45 #mm
dm = d - p / 2
l = p * 1 # Single threaded, n = 1
Tr = Screw.TorqueToRaiseWCollar(F, ft, fc, dm, dc, l)
Tl = Screw.TorqueToLowerWCollar(F, ft, fc, dm, dc, l)
e = Screw.Efficiency(F, l, Tr)
print(Tr, Tl, e)
def Problem3():
"""WRONG
Supposing that you need to clamp a 1.3" thick steel
plate with a 0.5" thick brass plate using a 1/4"-20
through-hole bolt and nut, what is the stiffness of the
members in Mlbf/in? Assume there isn't a need for a washer."""
steel = 1.3 # in
brass = 0.5 # in
d = .25 # in
D = 1.5 * d
l = steel + brass #in grip length
ksteel = Screw.SpringRate(30, d, D, steel)
kbrass = Screw.SpringRate(15.4, d, D, brass)
k = 1 / (1 / ksteel + 1 / kbrass)
print(k)
def Problem3Test():
"""Supposing that you need to clamp a 0.4"
thick steel plate with a 0.4" thick brass
plate using a 1/4"-20 through-hole bolt and
nut, what is the stiffness of the members
in Mlbf/in? Assume there isn't a need for
a washer."""
tan = 0.5773502692
t1 = 20 #in
t2 = 20 #in
d = 12 #in
D2 = 1.5 * d
l = t1 + d / 2
D1 = D2 + l * tan
a = t1 - l / 2
b = l - t1
D3 = D1 - 2 * a * 0 * tan
k1 = Screw.SpringRate(207, d, D2, l / 2)
k2 = Screw.SpringRate(207, d, D3, a)
k3 = Screw.SpringRate(100, d, D2, b)
ans = 1 / (1 / k1 + 1 / k2 + 1 / k3)
print(ans)
def Problem8_8():
d = 0.75 #in
ft = 0.15
fc = ft
p = 1 / 6
dc = 1 #in
F = 8 #lbf
r = 3.5 #in
T = F * r
dm = d - p / 2
l = p
F = Screw.ForceFromTorqueRaiseCollar(T, dm, dc, ft, fc, l)
print(F)
def Problem3():
"""Supposing that you need to clamp a 0.6"
thick steel plate with a 0.6" thick brass
plate using a 1/4"-20 through-hole bolt and
nut, what is the stiffness of the members
in Mlbf/in? Assume there isn't a need for a
washer."""
tan = 0.5773502692
t1 = 0.6 #in steel plate
t2 = 0.6 #in brass plate
# lg = A + B #grip length
d = 1 / 4
D2 = 7 / 16 # Table A-29
l = t1 + d / 2
D1 = D2 + l * tan
a = t1 - l / 2
b = l - t1
D3 = D1 - 2 * a * tan
k1 = Screw.SpringRate(30, d, D2, l / 2)
k2 = Screw.SpringRate(30, d, D3, a)
k3 = Screw.SpringRate(15.4, d, D2, b)
ans = 1 / (1 / k1 + 1 / k2 + 1 / k3)
print(ans)
def Problem2():
""" CORRECT
What is the torque, in pound-inches, required
to raise a 7 kip load using a square thread power
screw with a major diameter of 1.5 inches and a
pitch of 0.2 inches, and a collar with mean diameter
of 2.8 inches? Assume the screw to be single threaded
and the friction on the thread and collar to be 0.14."""
F = 7000 #lb
dc = 2.8 #in
d = 1.5 #in (Major diameter)
p = 0.2 #in
dm = d - p / 2 # (Median diameter)
ft = 0.14
l = p #Single threaded
ans = Screw.TorqueToRaiseWCollar(F, ft, ft, dm, dc, l)
print(ans)
def Problem2():
"""What is the torque, in pound-inches,
required to raise a 6 kip load using a
square thread power screw with a major
diameter of 2.3 inches and a pitch of
0.2 inches? Assume the screw to be single
threaded and the friction on the thread
to be 0.15. There is no collar."""
F = 6000 #lb
# dc = 2.8 #in
d = 2.3 #in (Major diameter)
p = 0.2 #in
dm = d - p / 2 # (Median diameter)
ft = 0.14
l = p #Single threaded
ans = Screw.TorqueRaiseThread(F, dm, ft, l)
print(ans)
def Problem2():
"""What is the torque, in pound-inches,
required to raise a 6 kip load using a
square thread power screw with a major
diameter of 1.7 inches and a pitch of
0.6 inches? Assume the screw to be single
threaded and the friction on the thread
to be 0.10. There is no collar."""
F = 6000 #kips
p = 0.6 #in
d = 1.7 #in
dm = d - p / 2
f = 0.1
l = p * 1 # lead for single threaded
Tr = F * dm / 2 * (l + np.pi * f * dm) / (np.pi * dm - f * l)
Tr2 = Screw.TorqueRaiseThread(F, dm, f, l)
print(Tr, Tr2)
def Problem5():
"""WRONG
Given a non-permanent bolted joint with a bolt stiffness
of 2.4 Mlbf/in and a member stiffness of 3.5 Mlbf/in that is
supporting a tensile load that fluctuates repeatedly between
0 and 2.1 kip, what is the factor of safety against fatigue
failure if one, 5/16”-18, SAE grade 8 bolt is used? Use Goodman
theory if needed."""
kb = 2400 #klbf/in
km = 3500 #klbf/in
P = 2.1 #kip
Sy = 130 #kpsi
d = 5 / 16
At = 0.0524 #in^2
Fi = 0.75 * At * Sy
C = kb / (kb + km)
sigmaI = Fi / At
sigmaA = C * P / (2 * At)
sigmaM = sigmaA + sigmaI
Sut = 150 #kpsi Table 8-9
Se = 23.2 #kpsi Table 8-17
n = Screw.Goodman(Se, Sut, sigmaI, sigmaA, sigmaM)
print(n)
def Problem8_51():
Pg = 6 # MPa
A = 20
B = 20
d = 12
Sp = 650
D = 1.5*d
l = A + B
H = 10.8
# L = H + l + 2*p
L = 60
Lt = 2*d + 6
ld = L - Lt
lt = l - ld
Ad = np.pi/4*d**2 #Table 8-7
At = 84.3 #Table 8-1
kb = Ad*At*207/(Ad*lt + At*ld)
kSteel = Screw.SpringRate(207, d, D, A)
kCastIron = Screw.SpringRate(100, d, D, B)
kM = 1/(1/kSteel + 1/kCastIron)
C = kb/(kb + kM)
Fi = 0.75*At*Sp
P = Pg*np.pi/4*150**2/10/1000
sigmaI = Fi/At
sigmaA = C*P*10**3/(2*At)
sigmaM = sigmaA + sigmaI
Sut = 900
Se = 140
print(Screw.Goodman(Se, Sut, sigmaI, sigmaA, 0))
print(Screw.Gerber(Se, Sut, sigmaI, sigmaA))
print(Screw.ASMEElliptic(Se, Sut, Sp, sigmaI, sigmaA))
# Begin problem 8 - 55
sigmaI = Fi / At
sigmaA = C*(P - P/2)*10**3/(2*At)
sigmaM = sigmaA + sigmaI
print('Test', Screw.Goodman(Se, Sut, 487.5, 3.675, 498.5))
print(Screw.Goodman(Se, Sut, sigmaI, sigmaA, sigmaM))
