ME 360 – H02
Name _________________________
1) Norton text problem 5-42. See Section 4.17 in Norton for help with thin wall pressure
vessels.
P = 150 psi = 1.034 MPa = 1.034 N/mm2
d = 500 mm
r = 250 mm
hoop stress
σt =
P r (1.034 MPa )(250 mm )
= 258.5 Mpa = 37.49 ksi
=
t
1 mm
axial stress
σa =
Pr
= 129.3 Mpa = 18.75 ksi
2t
σ1 = 258.5 MPa
N=
σ2 = 129.3 MPa
SY
400 MPa
=
= 1.79
σ' 223.9 MPa
N = 1.79
t = 1 mm
SY = 400 Mpa
l = 1000 mm
σ' = σ12 + σ 22 − σ1σ 2 = 223.9 Mpa = 32.47 ksi
ME 360 – H02
Name _________________________
2) Norton text problem 5-68. HOWEVER - use 1045 cold rolled steel fully keyed shaft
available from McMaster-Carr www.mcmaster.com . Do not include stress concentrations for
the keyway. What is the cost for a two foot length of this shaft? What is the standard size
keyway used for your shaft?
use 1045 cold rolled shaft
try
SY = 77 ksi
Norton Table A-9
d = 2 inch
T = 6500 in.lbf c = 1 inch J = π d4 / 32 = 1.5707 in4
τ=
T c (6500 in.lbf )(1 in )
= 4.14 ksi
=
J
1.5707 in 4
(
)
M = 9800 in.lbf c = 1 inch I = π d4 / 64 = 0.785 in4
σ=
M c (9800 in.lbf )(1 in )
= 12.48 ksi
=
I
0.785 in 4
(
)
ME 360 – H02
Name _________________________
von Mises σ' = σ 2 + 3τ 2 = 14.39 ksi
d NEW = d NEW 3
desire N = 2, may use smaller d
try
N = SY / σ’ = 5.35
2
= 1.44 inch
5.35
d = 1.5 inch
T = 6500 in.lbf c = 0.75 in J = π d4 / 32 = 0.4970 in4
τ=
T c (6500 in.lbf )(0.75 in )
= 9.81 ksi
=
J
0.4970 in 4
(
)
M = 9800 in.lbf c = 0.75 inch I = π d4 / 64 = 0.2485 in4
σ=
M c (9800 in.lbf )(0.75 in )
= 29.6 ksi
=
I
0.2485 in 4
(
)
von Mises σ' = σ 2 + 3τ 2 = 34.13 ksi
N = SY / σ’ = 2.26
for d = 1 7/16 inch = 1.4375 inch
c = 0.7188 in
J = 0.4192 in4
I = 0.2096 in4
τ = 11.14 psi
σ = 33.60 psi
σ’ = 38.75 psi
use
d = 1 1/2 inch
Norton Eq. 10.6a
d = 1.4407 inch
N = 2.26
d3 =
cost = $66.08
N = 1.99
3/8 inch wide keyway
32 N
3
32(2 )
M2 + T2 =
4
π SY
π(77,000 )
(9800)2 + 3 (6500)2
4
ME 360 – H02
Name _________________________
3) Norton text problem 5-73. HOWEVER - use 0.5 inch thick 1018 cold rolled steel square or
flat bar stock available from Online Metals www.onlinemetals.com . Do not include stress
concentrations for connections at A and B. Material properties are available at
http://www.onlinemetals.com/productguides/steelguide.cfm What is the cost and weight of one
bar?
use 1018 cold rolled bar
t = 0.5 inch
SY = 53.7 ksi
Online Metals
try w = 0.5 inch
F = 3500 in.lbf
σ=
3500 lbf
F
= 14 ksi
=
t w (0.5 in )(0.5 in )
N = SY / σ’ = 3.86
von Mises σ’ = σ = 14 ksi
too low
try w = 0.625 inch
σ=
3500 lbf
F
= 11.2 ksi
=
t w (0.5 in )(0.625 in )
density = 1.0763 lbm/ft
use
½ x 5/8 bar
N = SY / σ’ = 4.79
OK
use 7 feet = 84 inches
N = 4.79
m = 7.53 lbm
cost = $23.36
ME 360 – H02
Name _________________________
4) Norton text problem 5-73. HOWEVER - use 0.5 inch thick standard size 6061-T6511 (or
6061-T6) aluminum square or flat bar stock available from Online Metals
www.onlinemetals.com . Do not include stress concentrations for connections at A and B.
Material properties are available at
https://www.onlinemetals.com/productguides/aluminumguide.cfm What is the cost and weight
of one bar?
use 6061-T6 (6061-T6511) aluminum bar
t = 0.5 inch
σ' =
F
tw
F = 3500 in.lbf
N=
SY = 40 ksi
Online Metals
N>4
SY SY t w
=
σ'
F
w=

N F 4 (3500 lbf ) 
in 2

 = 0.70 inch
=
(0.5 in )  40,000 lbf 
SY t
try w = 0.75 inch
σ' =
(3500 lbf ) = 9.33 ksi
F
=
t w (0.5 in )(0.75 in )
N = SY / σ’ = 4.29
OK
density = 0.441 lbm/ft
use
½ x 3/4 bar
use 7 feet = 84 inches
N = 4.29
m = 3.09 lbm
cost = $12.02
5) Is steel or aluminum preferrable for the connecting rod in Norton text problem 5-73? Explain
your choice and be explicit about your decision process.
use steel but determine bar size based on infinite fatigue life
aluminum does not have infinite fatigue life