Upload
others
View
6
Download
0
Embed Size (px)
Citation preview
98
Chap. 4 Equilibrium Method4.1 Introduction
- Based on lower bound- Relationship between the strength of structure and
the the applied load is found by adjusting the unknown redundants
- The moment condition is not violated- The mechanism condition may or may not be satisfied
4.2 Basis of the method
- Determinate structures- Indeterminate structures
CM
BM12 BM
12P C BM M M= -
99
4.3 Moment equilibrium equations
a) Select redundants to make determinate structures
b) Draw BMD for determinate structurefor applied load
c) Draw BMD for redundants
d) Superpose them
e) Write moment equation at critical section
4.5 Strength of beams
a) Shear force effect
b) Lateral torsional buckling
4.5.1 Shear force effect
( )2
23
P p w f
y fw
V t d t
F d tt d
d
t= -
-æ ö= ç ÷
è ø
1 0.551.073
yP w y w
FV t d F t d= @
100
4.5.2 Lateral Torsional BucklingpM
pL rL
Plastic
Inelastic Elastic
3u yf f³
bL
2
cr y y wb b
EM EI GJ I CL Lp pæ ö
= + ç ÷è ø
rM
( )cr y r xM F F S= -
( ) ( )2121 1y
r y ry r
r XL X F F
F F= + + -
-
101
Residual stress in rolled section
Both flange sides get coolThen fixed
Tension
102
Equilibrium method example
A B C D E
+
=
AMCM EM
4PL
PM
PMPM PM
PMPM
PM
PM
PM
103
Equilibrium method example
A B C
D E
Primary structure and Primary moment
redundant structure and redundant moment
PMPM
PM
x
2
max
2 2
02
P
P
wl w xM x x Ml
MdM lxdx wl
= - -
= Þ = -
w
2wl
M
x
( )2
3 82PwlM = -
w
2wl
2wl PM
/PM l/PM l
104
B
E
PMPM
PM
x
D
A52.81
74.69
354.69 inP
y
MZF
= =
( )2
3 82PwlM = -
5 kips/ft15 ft
wl=ì
í =î
Try WF16 x 36
274.69 3.59 in36 / 3
Bw
y
VAt
= = =
Max. Shear @ B
3
2
64 in4.42 inw
ZA
ì =í
=î
2 2
74.69 16.89 ksi4.42
Z 1
31
Z 57.07 54.69
w
ps wy
yw
y
ps
Z Z
Z Z
t
ss
s t
s
= =
æ ö= - -ç ÷ç ÷
è øæ ö-ç ÷= - -ç ÷è ø
= >
Use WF16 x 36
ys
ts
s
fM
wsM
1
ps f ws
ws wy
ps f ws ws ws
p wy
M M M
M M
M M M M M
M M
ss
ss
= +
=
= + - +
æ ö= - -ç ÷ç ÷
è ø
1ps p wy
Z Z Z ss
æ ö= - -ç ÷ç ÷
è ø
105
Equilibrium method example
A B C D E F G
8520
PM-
¢
170
170170
85 85
2M
3M1M
3x2x
1x
by WF18x50(A36)PM
A B C D E F G
Mc and ME are not determined
by WF18x50(A36)PM
( )2 2
@ ,
31 y
PS P y WC Ey
M M Zs t
ss
é ù-ê ú= - -ê úë û
( )@ ,3267PS C E
M =
3267 3267
3267 69344DPLM = - =
Shear stress in the web @D
85 14.21w wd t
t = =
1) Cover plate for the mid-span
106
Try
required ZPL
BM3267
170 20 12 3267 85674 2BM ´ ´
= - =
Shear stress in the web @D85 14.21w wd t
t = =
2) Cover plate for the end-spans
38567 3267Z 147.23 in36PL-
=
0.75 10.54PL PLt b¢¢ ¢¢= =
2 231 3479y
PS P y Wy
M M Zs t
ss
é ù-ê ú= - - =ê úë û
2 231 3391y
PS P y Wy
M M Zs t
ss
é ù-ê ú= - - =ê úë û
required ZPL
36934 3391Z in36PL-
=
32673391
[ ] 3Z 17.99 99.17 inPL PL PL PLb t t= + =
3391
( ) 10200 3267 339110
M x x= - Þ
1110 6.95
2xx x ¢= - Þ =
1 / 2xDM
107
32673479
2x
BM
1x
8567
2
2 2
3
3 3
8567 347910
10 5.948567 3267 3267 3267
1010 4.48
M x
x x x
M x
x x x
= Þ
¢= - Þ =+
= - Þ
¢= - Þ =
Lateral support spacing
13600 200P
pd yy
MML r
F
+=
For W 16 x 36 1.52yr ¢¢=
12.67pdL ¢¢=
108
Design of columnsnP
cl
( )
2
2crEIP
klp
=
1 yc
Fklr E
lp
=
2 II Ar rA
= Þ =
2
2cr
crP E FA kl
r
p= Þ
æ öç ÷è ø
2
2 2
1 1cr cr
y y y c
P P EF A P F kl
r
pl
= = =æ öç ÷è ø
: slenderness parametercl
2
1cr
y c
PP l
=
79 0.87790
=
For Design of columns
2
2
0.8771.5
1.5 0.658 c
c n yc
c n y
P P
P Pl
ll
l
³ =
£ =
109
h
V
l
H
V
H
free redundant
V
H
2V Hh
l+2
V Hhl
-
H
A CB D E
A
B C D
E
/ 1 and / 1/ 3/ 1 and / 3
/ 3 and / 1/ 3/ 3 and / 3
l h V Hl h V H
l h V Hl h V H
= =ìï = =ïí = =ïï = =î
110
A CB D E
BMCM
DM
BMCM
DM
2 4
D P
B P
C P
M M ShM Hh M
Hh VlM M
ìï = =ï
= -íïï = + -î
Case I) / 1 and / 1/ 3l h V H= =
Then7
2 12 12Sincethe second PH occurs @B
2 =
B P
C P P
B C
P
M Hh MHh Hh HhM M M
M M
MHh
= -
= + - = -
>
\
Case II) / 1 and / 3l h V H= =
Then3 5
2 4 4Sincethe second PH occurs @C
1.6 =
B P
C P P
C B
P
M Hh MHh Hh HhM M M
M M
MHh
= -
= + - = -
>
\
111
Case III) / 3 and / 1/ 3l h V H= =
Then3
2 4 4Sincethe second PH occurs @B
2 =
B P
C P P
B C
P
M Hh MHh Hh HhM M M
M M
MHh
= -
= + - = -
>
\
Case IV) / 3 and / 3l h V H= =
Then3 3 2.75
2 4Sincethe second PH occurs @C
0.727 =
B P
C P P
C B
P
M Hh MHh H hM M Hh M
M M
MHh
= -´
= + - = -
>
\
112
h
V
l
H
A
B C D
E
Ex. 4.6.2
20/ 3
all WF16 45
l hV H
¢= ==
´
Determine the limit values V and Ha) w/o considering the effect of axial forceFrom the previous solution
1.6 PMHh
=
From AISC
( )( )
2963 kip-in
=1.6 19.75 kip-in20 12
3 59.25 kip
Px x y
P
M Z FMH
V H s
= =
=
= =
h
V
H
A
B C D
E
V
H
b) w/ the effect of axial forceFor Member BD
0.6 7.41 kipsPMT H Sh
= - = =
WF16 45 P 13.3 36 478.8 kipsy´ Þ = ´ =
=0.018 0.2t y
PPf
<
Reduced Mpc + 1.0
0.0181 0.9912
pc
t y b p
PC b p b p
MPP M
M M M
f f
f f
£ Þ
æ ö= - =ç ÷è ø
113
b) w/ the effect of axial forceFor Member DE
V
HB C
EA
D
49.382V HhP
l= + =
( )
( )
2
1 0.843
1 0.398
0.658 355.6 kipsc
y ycy
y
yxcx
x
n y
kL Fr E
kL Fr E
P Pl
lp
lp
= = Ú
= =
= =
49.38The ratio 0.1630.85 355.6
The reduced
12
0.163 1.02 0.9
0.827
1.6 16.34kips
3 49kips
c n
pc
pc
c n b nx
pc
nx
pc p
pc
PP
MMP
P MMM
M MNow
MH
hV H
f
f f
= =´
+ £
+ =
=
= =
= =
114
Ex. 4.6.3
V
H
V
H
H
1
2HhVL
+1
2HhVL
-
SSA
B
C
D
E
+
115
A B C D E
1Hh
1
2Hh
4VL
1Sh1Sh( )1 2S h h+
BM
CM
DM
( )
11
1 1
11 2
12.35 kips
20 247
2 4
pp
B
C
MSh M S
hM Hh Sh H
Hh VLM S h h
= Þ = =
= - = -
= + - +
55 370.5assume
24.7 kips
C
B p
M HM M
H
= -
=
=
55 24.7 370.5 988
So the second PH occurs @C2962.8 55 370.5
1224.7 kips
3 33.69 kips
C
C p
C p
MM M
M M H
HV H
= ´ - =
>
= = = -
== =
116
4.8 Practical procedure for large structures
a) Select the redundantsb) Obtain the primary momentsc) Obtain BMD for redundantsd) Assume a failure mechanisme) Combine b)+c) and equate the sum of moment to the Mpf) Solve the equations to determine the redundants and the limitg) Check the yield conditionsh) If the solution is exact, If not proceed further to determine the upper and lower solutions
0max
pt up
MP P
M=
117
2 24 4 4 4 4 4 4
12 12
A
B C D
E
A1
B1 B2 B3 D3 D2 D1
E1
2 4 4 4 2
12
A
B C
A1
B1 B2 B3
24 4 4
12
C D
E
D3 D2 D1
E1
2
T
S
SM
M
Primary Redundant
118
joint A A1 A2 B B1 B2 B3 C D3 D2 D1 D E1 Efree -384 -240 -192 -108 -48 -12 0 -12 -48 -108 -192 -192 -144 0
T 16T 8T 0 0 0 0 0 0 0 0 0 0 8T 16T
S 24S 24S 24S 18S 12S 6S 0 -6S -12S -18S -24S -24S -24S -24S
M M M M M M M M M M M M M M M
Total 1 -96 48 96 156 192 204 192 156 96 12 -96 -96 -48 96
Total 2 -128 -16 0 68 112 132 128 100 48 -28 -128 -128 -48 128
Total 3 -129.2 -18.4 -3.6 64.8 108.8 129.3 125.6 98 46 -29.6 -12.96 -129.2 -48 129.6
119
A
BC
D
EA
B D
E
384 16 24192 16
A p
B
M T S M MM T
= - + + + = -= - + 24
192 16p
D
S M MM T
+ + == - + 24
384p
E
S M MM
+ + = -= - 16 24
96 kip-ft192 kip-ft
4 kips0
p
p
T S M M
MMST
ìïïíïï + + + = -î
=ìï =ïí =ïï =î
max 20496 204p
MM=
£ £
Mechanism 1
384 16 24192
A p
B
M T S M MM
= - + + + = -= - 16T+ 24S+
192 16p
D
M MM T
+ == - + 24
384p
E
S M MM
- + = -= - 16 24
128 kip-ft128 kip-ft2.67 kips
4 kips
p
p
T S M M
MMST
ìïïíïï + - + =î
=ìï =ïí =ïï =î
max 132128 132p
MM=
£ £
Mechanism 2
120
A
BB3
D
E
384 16 2412 16
A p
B
M T S M MM T
= - + + + = -= - + 6
192 16p
D
S M MM T
+ + == - + 24
384p
E
S M MM
- + = -= - 16 24
129.2 kip-ft125.6 kip-ft2.615 kips4.15 kips
p
p
T S M M
MMST
ìïïíïï + - + =î
=ìï =ïí =ïï =î
max 129.2129.2 p
MM
=
=
Mechanism 3
121
A
B
C
D
E
A
B
C
D
E
A
BC
D
E
Mechanism 1Mechanism 3
EX. 4.7.2
122
A
B
C
D
E
VHM MV
joint A B C D Eprimary -274.66 -261.03 0 -222.07 -153.33
M M M M M MV -12V -12V 0 12V 12VH 12.97H 4.97H 0 4.97H 12.97H
261.03 12 4.97
222.07 12 4.97153.33 12 12.97
B p
C p
D p
E p
M M V H MM M M
M M V H MM M V H M
= - + - + = -ìï = =ïí = - + + + = -ïï = - + + + = -î
A
BC
D
E
Mechanism 1
87.681.62
13.32
C pM M MVH
= = =ìï = -íï =î
123
A
B
C
D
E
maxMpM
pM
pM
pM?AM =
87.681.62
13.32
C pM M MVH
= = =ìï = -íï =î
A
BC
D
E
274.66 12 12.97 5.22A p pM M V H M= - + - + = <
pM
V ¢
x q
1.62V =
13.32H =
13.32sin 1.62cosV q q¢ = +
2
2x pwxM M V x¢= + -
where6.6 kips
50cos 22.5 6 3.1 kips/ft12.98
V
w
¢ =ìïí -
= =ïî
o
( )
2
2
1
3.187.68 6.62
6.6 3.1 0 2.13
2.1387.68 6.6 2.13 3.1 94.71 kips-ft
287.68 M 94.71
x
x
C
p
xM x
dM x xdx
M
= + -
¢= - = Þ =
= + ´ - =
£ £
1
261.03 12 4.977.03 1.97 0.82
222.07 12 4.97153.33 12 12.97
B p
C p
D p
E p
M M V H MM M V H MM M V H MM M V H M
= - + - + = -ìï = - + - + =ïí = - + + + = -ïï = - + + + =î
82.4813.91.62489.96p
MH
VM
=ìï =ïí = -ïï =î
Try exact mechanism
Mechanism 3
124
A
BC
D
E
VHoM V
A
B
C
D
E
A
BC
D
E
W W
W W
WaWa
Wa Wa
oM
Mechanism 1
Mechanism 2
Mechanism 3
Mechanism 4
Mechanism 5
Primary system
Redundant system
Ex. 4.7.3
125
A
BC
D
E
VH
M MV
A
BC
D
E
W W
Wa Wa
joint A B C D Eprimary -WL(1/4+2a/5) -WL/4 0 -WL/4 -WL(1/4+2a/5)
Mo Mo Mo Mo Mo MoV 1/2VL 1/2VL 0 -
1/2VL-1/2VL
H 3/5HL 1/5HL 0 1/5HL 3/5HL
Mechanism 1
BD1B1 C
D
E
B p
C p
D p
E p
M MM MM MM M
= -ìï =ïí = -ïï =î
0 12 1505 104
p
p
WL WLM M
VMWHL
a= = +
=
= -
Now we must ascertain that
1 1
1 0
1 0
@ B &D , A1 104 101 104 10
0.625
p
B p
D p
M M
M M VL HL M
M M VL HL M
a
£
= + + + £
= + - + £
Þ ³
At A we have
01 2 1 34 5 2 5
0.25
A pM WL M VL HL Ma
a
æ ö= - + + + + ³ -ç ÷è ø
Þ £
Not possible mechanism
126
A
BC
D
E
A p
C p
D p
E p
M MM MM MM M
= -ìï =ïí = -ïï =î
03
16 20
8 2516 4
pWLM M WL
W WV
W WH
a
a
a
æ ö= = + ç ÷è ø
= - +
= -
Now we must ascertain that
1 1
1 0
@ B &D , B1 1 04 10
p
B p
M M
M M VL HL M a
£
= + + £ Þ £
For moment @D1 &B
1 0
0
14 10
14 2 5
50.417,4
D p
B p
VLM M HL M
WL VLM M HL M
a a
= - + £
= - + + + £
Þ ³ £
Not possible mechanism
Mechanism 2 Mechanism 3
1
B p
D p
D p
E p
M MM MM MM M
= -ìï =ïí = -ïï =î
0 0.1220
04
2 51
10 25p
WLM WL
VWH W
WLM WL
a
a
a
= +
=
= -
= +
Now we must ascertain that
1
0
@ A&C, D
1 2 1 34 5 2 5
00.278
p
A p
A p
M M
M WL M VL HL M
M M
a
aa
£
æ ö= - + + + + £ç ÷è ø
Þ ³³ - Þ £
For moment @C
( )
0
0
1 0.1220
0.625
0 0.278
0.1 0.04
C p
C p
p
M M WL WL M
M M M
MW
L
a
a
a
a
= = + £
Þ £= ³ -
£ £
=+
127
Mechanism 4
A p
B p
D p
E p
M MM MM MM M
= -ìï =ïí = -ïï =î
0 4250
15p
WLM
V W
H
M WL
a
a
=
=
=
=
Now we must ascertain that1 1
1
1
1
1
@ &B , D1 1 1.254 5
1 1 1.254 5
2.50.833
0.8332.5
p
C p
C p
p
B p
B p
D p
D p
M M C
M WL M WL
M M
WL M WL
M MM MM MM M
a a
a a
a
a
a
a
£
= £ = Þ ³
³ -
³ - = - Þ ³
£ Þ ³
³ - Þ ³ -
£ Þ ³
³ - Þ ³ -
52.5 pM
WL
aa
\ ³ Þ =
Mechanism 5
1
A p
B p
D p
E p
M MM MM MM M
= -ìï =ïí = -ïï =î
0 0.06251 98 20
0.3125 0.1251 7
16 40p
M WL
V W W
H W W
M WL WL
a
a
a
=
= - +
= -
= +
Now we must ascertain that
( )
@ &2.5
0.2780
0.5So 0.278 2.5
1/16 7 / 40
p
B p
B p
C p
C p
p
M M B CM MM MM MM M
MW
L
a
a
a
a
a
a
£
£ Þ ³
³ - Þ ³
£ Þ ³
³ - Þ ³ -
£ £
=+
128
( )
0 0.278
0.1 0.04pM
WL
a
a
£ £
=+
2.55 pM
WL
a
a
³
=
( )
0.278 2.5
1/16 7 / 40pM
WL
a
a
£ £
=+
W
a
Mechanism 3 Mechanism 4Mechanism 5
0.278 2.5
Mechanism 1
Mechanism 2
Mechanism 3
Mechanism 4
Mechanism 5
129
30l
A
B C D
G
E
H
VM
M
F
30l
30l 30lV
15l
15l
Example 4.8joint A B C D E F Gprimary -525l -525l 0 0 0 -525l -525l
M M M M M M M MV 15V 15V 7.5V 0 -7.5V -15V -15V
H 20H 0 0 0 0 0 20H
130
A
B C D
G
E F
525 12 207.5
225 15225 15 20
A p
C p
D p
E p
M M V H MM M V M
M M V MM M V H M
l
ll
= - + + + = -ìï = + =ïí = - + - = -ïï = - + - + =î
1107
9001901112
p
p
p
p
H M
M
V M
M M
l
ì =ïïï =ïíï =ïïï =î
A
B
C D
G
E F
2225 153
1112
107.512
B p
D p
E p
M M V M
M M M
M M V M
lì = - + + = -ïïï = =íïï = - =ïî
24.7
54.9
60.354.9
2.7 60.3
24.7
4.1
-
- -
+
+ +
- -
AFD SFD
131
247kip-ftp pc y xM M F Z= = =
Try W16X45
Since
2
60.3 kips478.8
1 0.398
1.686 control
0.877 147.7
FG
y
ycx
x
cy
n yc
PP
FkLr E
P P
lp
l
l
=
=
= =
= Ú
é ù= =ê ú
ë û
1.686 1.54cyl¢ = <
2
0.658 444.460.6 0.16 0.2
0.85 444.4
cn y
c n
P PPP
l
f
= =
= = <´
Member FG is critical
Interaction Eq.1.0
2x
c n b nx
MPP Mf f
+ £
0.1610.9 2
0.92 0.9 0.828
x
nx
pc p p
MM
M M M
æ ö£ -ç ÷´ è ø= ´ =
We reduce the load factor
71.59
900pcM
l = =
Lateral torsional buckling
13600 2200p
pd yy
MM
L rF
æ ö+ç ÷ç ÷
è ø=
1M
pM
pM
23 pM
1
1
23600 22003negative 1.57 93
36
23600 22003postive 1.57 221
36
pdp
pdp
M LM
M LM
æ ö-ç ÷è ø ¢¢Þ = =
æ ö+ç ÷è ø ¢¢Þ = =
Portion CE
103600 220012 1.57 77 180
36pdL
æ ö-ç ÷è ø ¢¢ ¢¢= = <
Portion EFPortion FG
OK
Shear force
0.55 110 kipsp y wV F t d= =
132
A
BC
D
E
A
BC
D
E
4.8.2
joint A B C D Eprimary 0 400 2495 0 0
S 0 -20S -35S -20S 0
133
35 249520
C p
D p
M S MM S M
= - + =ìí = - = -î
BMpM
pM
907pM =
400 20 505B pM S M= - = - £
x
94.9
( ) ( ) ( )
( ) ( )( )
2
2
1.8 5094.9 50
20.3
0.8 35 0.3 45.42
xM x x
xx
+= + -
- - -
50
45.4
20
1510
3
1.8 kips/ft
0.8 kips/ft
0 9.9 ftdM xdx
= Þ =
max 997907 997p
MM=
£ £
( ) ( ) ( )2
21
20
1.8 59.994.9 59.9 0.8 2.97 32
2
D p
C p
M S M
M S M
= - = -
= - - - =
134
1 2448 32
942 kip-ft47.1 kips
C p
p
M S M
MS
= - =
=ìí =î
Try W30X116Axial force: Member DE
2
94.934.2 36 12310.2210.302
0.658 1185
0.094 0.2
1.02
c
DE E
y
cx
cy
n y
c n
x
c n b nx
P VP
P PPP
MPP M
l
ll
f
f f
= == ´ =
=
=
= =
= £
+ =