Ae/AM/CE/ME 102c
Mid-Term Exam Formula Sheet
Plasticity
Yield Criteria:
1
3
Deviatoric stress, sij = ij kk ij
1 > 2 > 3 are the principal stresses of the stress tensor ij
1/ 2
1
Von Mises, ( J 2 )1 / 2 = sij sij
2
Tresca, max =
= o =
o
3
max min
=o = o
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Problem 1
i - 4,
-;ion
-,
a du?le cairn\~cr a m
h
$1
7
5.i.
faiio
ac Crack
mm(n
I t, i;
;
,I,
.
J
L 1 I 0 i~
GC"
cq T r
-,
r
7
Q
is pre;cribed
~
h=gx,J
load, i s i ndnpondar\+
3
-assume
E,Q
C UnS*r,t
I
!
NC
7
1 4 1/ Q
ienosh O-.
<*:
<fh
,
,
- L- .,
,
.
.
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close all
clear all
ac_w = linspace(0,0.5,10000); % generate vector for crack length divided by
width
f = (1.12-.23.*ac_w +10.55.*ac_w.
1
Problem 2.
Part a. Stress distribution in the initial elastic state and in the subsequent elastic-plastic state?
Starting with Airys stress function for an axi-symmetric problem
(r ) = A ln r + Cr 2 + D
1 2 1 A
+
= + 2C
r 2 2 r r r 2
2
A
= 2 = 2 + 2C
Final Examination
Ae/AM/CE/ME 102c
Mechanics of Structures and Solids - Spring 2012
Due June 6, 12:00 PM, 101 Guggenheim
Note:
i) You are not allowed to consult any material other than the formula sheet handed in class.
ii) Calculators are allowed.
Page 1
2007-08 Ae102b HW5 Solution
Problem 1
Problem 2
Problem 3
Problem 4
Nm N
+
=p
Rm R
with p = gt cos 0 ; Rm = ; R =
Force Balance:
r
sin 0
(R2 r2 )
Also, Nm (2r) sin 0 = g
t
cos 0
Figure 1: Schematic of conical shell
Nm = gt
R2 r 2
and N = gt cot 0
2 sin