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Chap 3. Tools used in Plastic Analysis and Design
1. Introduction2. Assumption of Ductility of Steel3. Small deflection4. Virtual work equations5. Fundamental theorem6. Upper and Lower Bound Solutions
71
72
3.1 introduction
- Equilibrium
- Compatibility
- Constitutive Relationship
- Equilibrium
- Failure mechanism
- Yield condition
- Virtual works
Mechanics Limit Analysis
Lower bound
Upper bound
3.2 Ductility ss
e e
d
M
f f
M
pM
ys
F
Plastic Idealization
Virtual Work Method
• The virtual work equation relates a system of forces in equilibrium to a system of compatible displacement.
• Equilibrium systems• Displacement systems
73
Assumptions
Equilibrium Compatibility
Stress-strain relationship
Stress or strain field
Internal Stress i.o.t. applied load
Straini.o.t. displacement
Dual
74
Statics ß-----------à Kinematics
Equilibriumß-----------à Compatibility
Stress ß-----------à Strain
Force ß-----------à Displacement
75
3.4 Virtual work theorem1P 2P
P
d1d 2dvirtuald
M
virtuaal lreMEI
ff = =
M
fvirtualf
, 1 1 2 2i virtual iP P Pd d d= +å
virtualM dxfò
,i virtual i virtualP M dxd f=å ò
Real displacement
Only at loading pointsare considered
VirtualWork
d
P
d
RealworkR
eal force
Real displacement
Real or virtual displacementReal or virtual force
76
y
ee
yEse
1.0-1.0
0.67
0.67-
x
y
P
1A2A 3A
1L2L 3L
2dD =
1 3 4A A A= =
1 3
2
2L L LL L
= =
=
2A A= 2113 y
E es ee
é ùæ öê ú= - ç ÷ç ÷ê úè øë û
2
1Ty
E Es ee e
é ùæ ö¶ ê ú= = - ç ÷ç ÷¶ ê úè øë û
Deformation
iyix izi x y z
i i i
LL LL L L
d d d d= + +
1
2
3
2
2
d
d
d
Dì =ïï
= Díï Dï =î
( )2 2
1 1 1 1 1 1 12*4 1 1 1 1*2 2 3 2 2 2 3y y
AE AE PL L L Le e
é ù é ùæ ö æ öD D D Dæ öê ú ê ú- + - =ç ÷ ç ÷ç ÷ç ÷ ç ÷ê ú ê úè øè ø è øë û ë û
217 1248 y
PL L AEe
é ùæ öD Dê ú- =ç ÷ç ÷ê úè øë û
Virtual displacement
Load-displacement relationship Unit displacement
Equilibrium
77
x
y xP
1A2A 3A
1L2L 3L
xu d=
iyix izi x y z
i i i
LL LL L L
d d d d= + +
The principle of virtual displacements
yv d=
q
1 1 2 2 3 3 x yN N N Pu P vd d d+ + = +
1
2
3
cos sin
cos sin
u vv
u v
d q qd
d q q
ì = +ï =íï = - +î
( )( )
1 3
1 3 2
cos cos
sin sin 0x
y
u N N P
v N N N P
q q
q q
- -
+ + + - =
1 3
1 3 2
cos cos 0sin sin 0
x
y
N N PN N N P
q qq q
- - =ìí + + - =î
The principle of virtual forces
1 1 2 2 3 3 x yN N N Pu P vd d d+ + = +
1 3
1 3 2
cos cos 0sin sin 0
x
y
N N PN N N P
q qq q
ì - - =ïí + + - =ïî
( ) ( )1 1 3 3 1 2( cos sin ) cos sin 0N u v N u v N vd q q d q q d- - + + - + - =
1
2
3
cos sin 00
cos sin 0
u vv
u v
d q qd
d q q
- - =ìï - =íï + - =î
Virtual displacement
Then
Equilibrium
Virtual forces
Compatibility equation
yP
u
v
q
q
1vd3vd
3ud
1ud
78
mw
/ 2l / 2l,z w
x
2
8ql
-
/ 2l
/ 2l
/ 2l
/ 2l
/ 2l/ 2l
/ 2l
1
1
1
/ 4l
/ 2l
2
8ql
2
8ql
2
16ql
1A
2A3A
/ 4l
1A 1A
2A3A 4A
1A
2A
The principle of virtual forces
79
( )f x
( )g x
( ) ( ) ?f x g x dx =ò
( )1g x
1x
fgdx =ò ( )1g x´
Area Centroid
12bh
23b
23bh 5
8b
35b
( )3
2 2n bn++
34bh
1n bh
n +
2ax
3ax
nax
bh
2
8ql
2
8ql
2
16ql
1A
2A3A
2 2 2
2 2 2 2 4
1 2 3 1 2 1 13 2 8 8 2 2 2 8 3 2 2 2 16 3 2
1 1 1 1 1 12 8 8 6 24 2 8 12 192
ml ql l l ql l l ql lw
EI
l ql l ql qlEI EI EI
é ù= - + +ê ú
ë ûé ùæ ö= - + + = =ç ÷ê úè øë û
/ 4l
1A 1A
2A3A
1A 2A 3A
2 2 2
2 2 2 2 4
1 2 5 1 1 1 22* 2*3 2 8 8 4 2 2 8 3 4 2 2 16 3 4
1 5 1 1 1 12 8 24 24 12 2 8 12 192
ml ql l l ql l l ql lw
EI
l ql l ql qlEI EI EI
é ù= - + +ê ú
ë ûé ùæ ö= - + + = =ç ÷ê úè øë û
3A
1A
2A
2
8ql
2
16ql
/ 2l
Bq
2 2
2 2 4
4 4
1 2 3 1 12 3 2 8 8 2 2 2 16 3 2
1 1 12 8 8 24 2 48
1 1 116 12 6 192
m B
m
l l ql l l ql lwEI
l ql l qlwEI EI
ql qlEI EI
qé ù
- = - +ê úë û
é ùæ ö= - + +ç ÷ê úè øë û
æ ö= - + =ç ÷è ø
80
1cj
2cj
3cj
4cj
5cj
6cj
1cj¢
2cj¢
3cj¢
4cj¢
5cj¢
6cj¢
cij¢
cijcl
cl
i clb
11 ici
i
bjb
æ ö-¢ +ç ÷è ø
1 21 1 2 2
1 2
2
2 3 51 11 12 2 3 2 2 3
11 112 2 3 2
c c c c c cy c c c c c c c c c c
i c cci c c c ci c c cr
i
l l l l l ll rl rl l rl rl
l ll rl i l rl i l
b bj j j jb b
bj j j jb
æ ö æ ö- -æ ö æ ö æ ö æ öD = - - + - + - - + - + ×××ç ÷ ç ÷ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è øè ø è ø
æ ö-é ù é ùæ ö æ ö+ - - - + - - + ×××+ -ç ÷ç ÷ ç ÷ê ú ê úè ø è øë û ë ûè ø
( ) ( )
2
2
2
116
6 0.5 3 16
crc
r
c cii
i r i
l
l r i r i
bb
j bb=
æ ö-+ç ÷
è ø
= - + - - -é ùë ûå
81
11 1
1
22 2
2
2
2112 2 3
3 5112 2 3
11 112 2 3
2
c c cy c c c c c
c c cc c c c c
i cci c c c ci c c
i
ccr
l l ll rl rl
l l ll rl rl
ll rl i l rl i l
l
bj jb
bj jb
bj jb
j j
æ ö-æ ö æ öD = - - + -ç ÷ç ÷ ç ÷è ø è øè ø
æ ö-æ ö æ ö+ - - + - + ×××ç ÷ç ÷ ç ÷è ø è øè ø
æ ö-é ù é ùæ ö æ ö+ - - - + - - + ×××ç ÷ç ÷ ç ÷ê ú ê úè ø è øë û ë ûè ø
+ -
( ) ( )
2
2
2
116
6 0.5 3 16
crc
r
c cii
i r i
l
l r i r i
bb
j bb=
æ ö-+ç ÷
è ø
= - + - - -é ùë ûå
cij¢
cij
cl
i clb
11 ici
i
bjb
æ ö-¢ +ç ÷è ø
Application of virtual work method for plastic analysis
1.Obtain the geometric relationships of mechanism motion by assuming appropriate equilibrium.2.Make a moment check for given mechanism by assuming appropriate displacement sets3. Prove the uniqueness, unsafe, and safe theorem4. Obtain bounding solutions5. Calculate deflections at collapse load
82
83
1.Obtain the geometric relationships of mechanism motion by assuming appropriate equilibrium.
20
105
25
10
1q
3q
2q
1-
1-
1-
ApproachMake three set of equilibrium system
Displacement system
Equilibrium system
84
1-
1
1 20
1
125
20
105
25
10
1q
3q
2q
2 210 25+
2510
325
25
10
250
5
20
2 25 20+
3255
20
5
20
3255
205
20
200
3 1250 325 0q q- + =1 220 25 0q q- =
85
2.Make a moment check for given mechanism by assuming appropriate displacement sets
pM-
pM
?BM =+
+
-
-
qq-
2q
q- q-
P/ 2P
-+
Beam mechanism Side-sway mechanism
L
/ 2L 2q
2q-
q- q
/ 2Lq
/ 2Lq
/ 2 / 22ePW L L Pq q= ´ + ´
( ) ( ) ( )( )2 2d p pW M Mq q- = + ´ + + - -
0e dW W- =3 44 pP L Mq q=
163
pu
MP
L=
Virtual displacement
86
pM-
pM
?BM =+
+
-
-
q+
q-
2q+
q- q-
Equilibrium set
Displacement set
Displacement set 1 Displacement set 2
ii i p iP Md q´ = ´å å
( ) ( )( ) ( )( )22
3
B p p
pB
LP M M M
MM
q q q q= - + + + + - -
=
( ) ( )( )2
3
B p
pB
LP M M
MM
q q q= + + - -
=
163
pu
MP
L=
P/ 2P
87
3. Prove the uniqueness, unsafe, and safe theorem
uP
D
P
Upper bound
Lower bound
The ultimate load must exist between the lower and upper bounds
88
A B C D
BWl CWl
Aq ¢- Bq ¢ Dq ¢-Aq ¢¢- Cq ¢¢ Dq ¢¢-
Bd ¢ Cd ¢Bd ¢¢ Cd ¢¢
BWl¢CWl¢ BWl¢¢
CWl¢¢
PM-PM- PM-
PM-PM PM
Displacement sys 1 Displacement sys 2
Equilibrium sys 1 Equilibrium sys 2
Displacement sys 1 Equilibrium sys 1
Equilibrium sys 2
B B C C P A P B P DW W M M Ml d l d q q q¢ ¢ ¢ ¢ ¢ ¢ ¢+ = + +
B B C C P A B B P DW W M M Ml d l d q q q¢¢ ¢ ¢¢ ¢ ¢ ¢ ¢+ = + +
( )( ) ( )B B C C B P BW W M Ml l d d q¢ ¢¢ ¢ ¢ ¢- + = -
0l l¢ ¢¢- ³
( ) 0B P BM Mq ¢ - ³
XDisplacement sys 1
X
---- (A)
---- (B)
(A) – (B)
Then
so
Uniqueness theorem
89
Aq ¢- Bq ¢ Dq ¢-Aq ¢¢- Cq ¢¢ Dq ¢¢-
Bd ¢ Cd ¢Bd ¢¢ Cd ¢¢
PM-PM- PM-
PM-PM PM
Displacement sys 1 Displacement sys 2
Equilibrium sys 1 Equilibrium sys 2
Displacement sys 2 Equilibrium sys 1
B B C C P A C C P DW W M M Ml d l d q q q¢ ¢¢ ¢ ¢¢ ¢¢ ¢¢ ¢¢+ = + +
B B C C P A P C P DW W M M Ml d l d q q q¢¢ ¢¢ ¢¢ ¢¢ ¢¢ ¢¢ ¢¢+ = + +
( )( ) ( )B B C C C P CW W M Ml l d d q¢¢ ¢ ¢¢ ¢¢ ¢¢- + = -
0l l¢¢ ¢- ³
( ) 0C P CM Mq ¢¢ - ³
0l l¢ ¢¢- ³l l¢ ¢¢=
X
Displacement sys 2 X Equilibrium sys 2
---- (A)
---- (B)
(B)- (A)
BWl¢CWl¢ BWl¢¢ CWl¢¢
90
A B C D
P
Actual Displacement Assumed Displacement
cAq- c
Bq cDq-
Aq- Cq Dq-
cBd
Bd Cd
cP
PM-PM- PM-
PM-PM PM
Actual Equilibrium Assumed EquilibriumuP
Actual Displacement sys
cAq- c
Bq cDq-
cBd
cP
PM-PM- AM-
DM-PM BM
Actual Equilibrium sys Assumed Equilibrium sysLP
CM
uB P A P C P DP M M Md q q q= + +
cB P A C C P DP M M Md q q q= + +
CM
Assum Dis Assum EquilX
Assum Dis Act EquilX
( ) ( )u cB P C CP P M Md q- = -
u cP P³
L c c c cB A A B B D DP M M Md q q q= + +
c c c c cB P A P B P DP M M Md q q q= + +
Act Dis sys Ass Equil sysX
Act Dis sys Act Equil sysX
( ) ( ) ( ) ( )L c c c c cB A P A B P B D P DP P M M M M M Md q q q- = - + - + -
L cP P£
Unsafe theorem Safe theorem
91
Obtain bounding solutions
3.5.4 Corollaries
cl l=
cl l³
cl l£
Mechanism
Equilibrium
Moment condition
3.6 Upper-and-Lower Bound Solutions
720
24024040l
60l
40l
60l5q
2q+
2q-10q
ii i p iP Md q´ = ´å å
( ) ( ) ( )( ) ( )( ) ( )( )40 5 60 10 720 2 240 2 240l q l q q q q+ = + + + - - + + +
2.7l =60 2.7´
40 2.7´
240PM =
240PM =
720PM =
?CM =
?BM =
92
60 2.7´
40 2.7´
240PM =
240PM =
720PM =
?CM =
?BM =
Equilibrium System
Displacement (virtual) system 1
Displacement (virtual) system 2
q- 2q
q-
q-
q-
2q
( )( ) ( )( ) ( )( )60 2.7 10 720 2 24060
c
c
MM
q q q q´ ´ = + - + + + + - -
Þ =
60CM =
40 2.7´
Equilibrium system
( )( ) ( )( )40 2.7 5 60 2300
B
B
MM
q q q´ ´ = + - + +
Þ =240P
B P
MM M
=>
240 0.8 2.7 0.8 2.16300
= Þ ´ =60 2.16´
40 2.16´
0.8 PM
0.8 640PM =
0.8 60 48CM = ´ =
240B PM M= =
0.8 PM2.16 2.7cl£ £
93
3.6.2 design
3 PM
PM96
144
PM
Find Mp for given load
96
1445q
2q2q-10q
ii i p iP Md q´ = ´å å
( ) ( ) ( )( ) ( )( ) ( )( )96 5 144 10 3 2 2213
P P P
P
M M MM
q q q q q+ = + +
=144
96
213PM =
213PM =
639PM =
?CM =
?BM =
Displacement system 1
q- 2q
q-( )( ) ( )( ) ( )( )639 2 213
144 1053
c
c
M
M
q q qq
+ - + + - -
= ´Þ =
94
Displacement system 2
q-
q-
2q
53CM =
96
Equilibrium system
( )( ) ( )( )53 2 96 5267
B
B
MM
q q q+ - + = ´
Þ =
213 267PM£ £144
96
0.8 PM
0.8 640PM =
53CM =
267BM =
0.8 PM
BM
3.7 Example
0.1H ¢¢D =
0.5V ¢¢D =
0.004 radianq =
Vd
1
ii i p iP Md q´ = ´å å
( )( ) ( )( ) ( )( )1 1 0.5 5 12 0.0040.74
v
v
dd
+ - = - ´ -
¢¢Þ = -
95
A
B
C
D
A
B
C
D
A
B
C
D
A
yxx y
LLL L
d = D + D
/ 2 / 2 102x
L LL L
d -= D + = D
D
DD
Tension failure
Compression failure
96
B
C
D
yxx y
LLL L
d = D + D
/ 2 / 2 102x
L LL L
d -= D + = D
D
A
B
C
D
DD
A
2 412
uyP P d
d
D =
= D
2uyP P=
( )2 2uyP P d
d
D =
= D
uyP P=
yP yP
yP
2 yP
yP
/ 2yP
Mechanism 1
Mechanism 2
97
Example of highly redundant system
D
1
N
N
