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

= == ´ =

=

=

= =

= £

+ =