Sparse Gerber and ASME Elliptic Criteria
Monday, February 06, 2017
5:29 AM
Ssu 0.67Sut
Ssy
Sse
Page 1
891 1.2
0.087514
0.175
0.5in
Ke 1.2
Eshear
0.1033mn
1
0.899
Tsin
1 0.899 1.2 11 1.18
becausethe arm is Horizontal
in startingposas sothearclength
isjusttheheightchange
0 54
12
T 18in
11.5 10apsi1210.5in4
120
2
8
Tmax 1017 in lb
Tmin 821inlb
Page 2
T 18in
11.5100
psi 1210.5in
arclength is thedistance between two points along a section of a curve
I Emin 39.5Ksi
Emin 1.1811 82112
Tmax 118 1101712
bg gmIemax
48.9Ksi 39.5Ksi 44.2kg
2
usaksttshdsinuahf
ya 48.9Ksi 39.5Ksi 4.7kg
265
Ka 2.7 13050
Kp
0.59
8
Kp 0.947
2155
0.897
Sse bc of Kc
Sse 0.51130Ksi 0.743 0.947 0.5910.897 24.2Ksi
Page 3
0.743
Pg315
li ii
V 8 tiisiitiei e
d' Nf
1.629
i
yieldingstrength
in shear
EE
io.skYsasis
Rf 1.024
10.577 80ksi
my Streetsville
48.9Ksi
Nf 0.944
Page 4
steps to solve problem above
1.) use Neuber Constant, notch sensitivity, and theoretical stress concentration factor to nd fatigue
stress concentration factor.
2.) Find max and min torque, then max and min stress to nd midrange and alternating stress.
3.) Use Marin Factor to nd fully corrected endurance limit.
4.) apply Gerber and ASME Elliptic equations (modi ed for shearing stress/ strength) to nd Factor of
Safety.
Some Definitions
0min is the minimum stress
Omax is the maximum stress
Oa is the amplitude component
Om is the midrange component
Is mkanpltludeeduttgpsentcommf.ge Eyeteheofmgaqag.im
deviation ofstressfrom
Them rashes MPFattesfurshegssonerepgfents.ptheadangrage or mean
Then
ashmp.pt
ggyalue
stress around
can lead to a greater overallrisk of failure even if
stress
time
MFL CW a ew 2b a Rfr Zat2b 0
Rp
cw
EY
630.18510
EFy RFL CW EW RAR
RFL CW EWRFR
1459.815lb