Problem 5-15 & 16 Extra Part of Problem 5-15pkwon/me471/Lect 5.2.pdf · 5! Brittle Materials!...
Transcript of Problem 5-15 & 16 Extra Part of Problem 5-15pkwon/me471/Lect 5.2.pdf · 5! Brittle Materials!...
5
Brittle Materials Modified I Mohr
( )
BAuc
B
A
BBA
uc
B
utuc
Autuc
A
BBA
BAut
A
nS
nSSSSS
nS
σσσ
σσ
σσσσ
σσ
σσ
σσσ
≥≥−=
>≥≥=−−
≤≥≥
≥≥=
0
1and01
1and0
0
5-11 Selection of Failure Criteria
Problem 5-15 & 16 ( )
( )( )
( )ksi175.9
2375.0375.0)4(190
ksi52.274375.0
375.0)6(190
4
4
===
===
πτ
πσ
JTcIMc
xz
x
967.0076.3332
32076.3354.16175.976.13
76.13252.27
22
==
=>=
=+=
==
n
SDR
C
yA
D
A σx
τxz
( ) ( )
007.178.3132
78.31175.9352.273 222/122
==
=+=+=
n
xyx τσσ
Extra Part of Problem 5-15
( ) ksi94.4547.11328.41:Misesvon
ksi23.4747.1164.2022;64.20228.41:Tresca
ksi47.11175.925.1175.9ksi28.4152.275.152.27
133.0/;33.1/
22
22
'
=+=
=+×=×===
=×==
=×==
==
req
reqy
txz
tsx
yS
RSC
KK
drdD
τ
σ
¾” 1”
0.1” Determine the required yield strength of a material after considering the stress concentration factors.
6
21
Problem 5-5
3/2
)3/4)(2/(3
2max
rrryAQ
=
== ππ
( )ksi21.1630.1609.04/)5.0()5.0(1600
5.075
42
=+−=
+−=
+−=
ππ
σIcM
AR
x
c
yy
Mz=125(8)-75(3) =775 lb-in
Mx=200(8) =1600 lb-in
A
B
Ty=200(3) =600 lb-in
A
( )( ) ( ) ksi84.2
2/5.0)5.0(600
)1(4/5.03/)5.0(2125
44
3
max
−=−
+=
+=
ππ
τJcM
ItQR yx
xy
B ( )ksi78.7
4/)5.0()5.0(775
5.075
42
=
+−=
+−=
ππ
σIcM
AR z
c
yy
( )( ) ( )ksi4.3
2/5.0)5.0(600
)1(4/5.03/)5.0(2200
44
3
max
=
+=
+=
ππ
τJcM
ItQR yz
yz
A
B
y
x
z
3’
8’
Fy=75lb Fx=125lb
Fz=200lb
d = 1”
π34r
Problem 5-5 Mz=125(8)-75(3) =775 lb-in
Mx=200(8) =1600 lb-in
A
B
Ty=200(3) =600 lb-in
75lb 125lb
200lb
775 lb-in
1600 lb-in 125lb
200lb inlb82.1777
1600775 22
−=
+=RM
( ) lb9.2484.2532sin85.235lb85.235200125 22
=−⋅=
=+=
c
R
VV
32o
25.84o
( ) ( )ksi195.18095.01.18
5.075
45.05.082.1777
24
−=−−=
−⋅
−=−−=ππ
σAR
IcM yR
( )( )ksi095.3055.304.0
2600
24329.24
44
3
=+=
+⋅
=+=cc
ccc
JTc
IbVQ
ππτ
ksi34.214.52.922 22 =+== RD
Problem 5-5
1’ 1.1’
0.05’
D / d =1.1; r / d = 0.05σ x' = Kts27.52+Kt 0.095=1.85×18.1+1.87×0.095= 33.66ksi
τ xz = Kt9.175=1.25×3.055= 3.82ksi
Tresca :C = 33.662
=16.85; Syreq = 2×R = 2× 16.852 +3.822 = 34.555ksi
vonMises : Sy
req = 33.662 +3 3.82( )2 = 34.3ksi
Determine the required yield strength of a material after considering the stress concentration factors.
Rectangular plate with hole subjected to axial load. (a) Plate with cross-sectional plane; (b) one-half of plate with stress distribution; (c) plate with elliptical hole subjected to axial load.
Rectangular Plate with Hole
5-12 Introduction of Fracture Mechanics
7
(a) Mode I, opening; (b) mode II, sliding; (c) Mode III, tearing.
Three Modes of Fracture Infinite Plate with Crack
( )⎩⎨⎧
+=
=
⎟⎠
⎞⎜⎝
⎛ +=
⎟⎠
⎞⎜⎝
⎛ −=
yxz
Ixy
Iy
Ix
rKr
Kr
K
σσνσ
θθθπ
τ
θθθπ
σ
θθθπ
σ
023cos
2cos2
sin2
23sin
2sin1
2cos
2
23sin
2sin1
2cos
2
ToughnessFracture:ICI KaK ≥= πσ
Crack Propagate when
Yield stress and fracture toughness data for selected engineering materials at room temperature.
Fracture Toughness propagateCrackICI KaK ≥= πβσ
aKI πβσ=Other geometries
I
IC
KKn =Factor of Safety