part i - design of steel members

7/29/2019 Part I - Design of Steel Members

1/91

Diseo de Estructuras de Acero - Parte I

CIMEPI - Agosto 2013

Instructor: Ing. Telmo A. Snchez, Ph.D. 1

1

Part I Design of Steel

Members

2

In this course we will be studying the design of the

following components, according to the provisions of the

2010 American Institute of Steel Construction (AISC)

Specification for Structural Steel Buildings:

Introduction to Steel Design

1.1 Basics of Structural Steel Construction:

Tension Members

Bearing and Column Base Plates

ColumnsBeams

Beam - Columns

Steel - to - Steel Connections

3


1.1 Basics of Structural Steel Construction:

Typical framing:

2 3

A

B

Spandrel beam

Beam

Deck direction

Definition:

Bay is the area bounded

by 3 or 4 columns

4


1.2 Design Specifications:

ASCE-7: This specification

provides the criteria for the

calculation of design loads in

buildings in general (steel

buildings, concrete buildings, etc.)(ASCE, 2005, 2010).


2/91




5



AISC: This specification provides

the criteria for the design of

structural steel buildings and their

connections (AISC, 2005, 2010).

6



CEC: Este cdigo contiene

normativas locales para el diseo

de estructuras de acero (CEC,

2012).

7



AISI: This specification dealswith the design of cold-formed

steel structures (AISI, 2007).

8



AASHTO: This specification

covers the design of highway

bridges and related structures

(AASHTO, 2012).


3/91




9



Steel Construction Manual: Themanual is a book that contains

design aids and other valuable

information for the designer. The

specification and its commentary

are included in the manual inSection 16.

AISCM Manual

AISCS Specification

10


Design Philosophies:

There are two basic specification philosophies:

ASD Allowable Stress Design

LRFD Load and Resistance Factor Design

* LRFD is sometimes referred to as Limit States Design


11


Define:

Limit State: Condition in which a structure or

component becomes unfit for service and is

judged either to be no longer useful for itsintended function (serviceability limit state) or to

have reached its ultimate load-carrying capacity

(strength limit state).


12


For all specifications:


Loading d Resistance

Define:

Loading

ResistanceSafetyofFactor

= accounts for under-strengthand over loads

Basic Factor of Safety = 1.67 (determined by society)


4/91


CIMEPI - Agosto 2013Instructor: Ing. Telmo A. Snchez, Ph.D. 4

13



Define:

Loading

ResistanceSafetyofFactor

Basic Factor of Safety = 1.67 (determined by society)

Factor of Safety is direct in ASD

Factor of Safety is in two parts in LRFD:

Resistance Factor

Load Factor

14


1.3 2005 and 2010 AISC Specifications

The 2005 AISCS was the first that combines ASD and

LRFD procedures. Previous to this specification, there

were separate ASD and LRFD specifications.

ASD (1989) LRFD (1999) ASD,LRFD ( , 2010 )

15


1.3 2010 AISC Specification

ASD: General format for ASD

Required strength referred to as

a loading combination

where:

:n

i

RQ d

iQ

:nR Allowable strength

i Load typeiQ Nominal (service or working) load

nR Nominal resistance

: Factor of safety16



LRFD: General format for LRFD

Required strength, ,referred to

as a loading combination

where:

nunii RRRQ dd IIJ iiQJ

nRI Allowable strength

i Load type

iQ Nominal (service or working) load

nR Nominal resistanceResistance factor corresponding to nR

iJ Load factor corresponding to iQ

uR

or


5/91




17


I


LRFD Specification Basis:

LRFD is based on a probabilistic model. The load

factor, , and the resistance factor, , reflect the fact

that loads, load effects and resistances can be

determined only to imperfect degrees of accuracy.

takes into account variations in loading over the life

structuretakes into account:

Ii

J

iJ

1. unavoidable design model

2. variations in dimensions

3. variations in erected positions

inaccuracies

18




Q,R

FrequencyQ: loading R: resistance

failure

19

Frequency

mean




VE

E = number of std. deviations= reliability index= 3.0 for members, 4.0 for conn.

V = standard deviation

Q

Rln

Q

Rln

20



Load Factors and Load Combinations for LRFD:

The load combinations for LRFD recognize that, when

several transient loads act in combination, only one

assumes its maximum lifetime value, based on a 50-

year recurrence, while the others are at their

arbitrary-point-in-time values.


6/91



21




uRFor LRFD, the required strength, , is determinedfrom the following factored combinations, which are

on p.2-8 of the Manual, and are based on ASCE-7

Section 2.3:

E.orW.D.

S.L.E.D.

RorSorL.L.W.D.W.orL.RorSorL.D.

orRSorL.L.D.

D.

r

r

r

016190

20500121

50506121

80506121

506121

41

r

r

22



Where:

loadearthquake

loadwind

rainwatertodueload

Ecuador)inapplicable(notsnow

loadliveroof

occupacytodueloadlive

loaddead

E

W

R

S

L

L

D

r


23


Example 1:

The compression member shown in the

figure is subject to the following service

loads. Determine the controlling load

combination, and the corresponding

required strength.

D=115 kipsL= 60 kipsLr= 23 kips


24


Example 1:

kips...

RorSorL.L.W.D.

kips...

W.orL.RorSorL.D.

kips...orRSorL.L.D.

kips.

D.

r

r

r

1802350605011521

50506121

2056050236111521

80506121

2462350606111521

506121

16111541

41



7/91



25


kipsRu 246

Example 1:

kips.

E.orW.D.

kips..

S.L.E.D.

10411590

016190

168605011521

20500121

r

r

Load combination 2 controls

Note that some of the load combinations can

be eliminated by inspection.


26


Example 1:Ru= 246 kips


27


1.4 Structural Shapes:

Types of shapes:

Hot-rolledBuilt-up

Types of hot-rolled shapes:

a) H-shapes: could be W,M,S,HP

flange

web

28



plf.076weight

.in.d 923



i) W-shapes, p. 1-10+

Example: W24x76 (W610x113)

Note that nominal depth and

actual depth are not the

same.


8/91



29





ii) M-shapes, p.1-28+

Example: M4x6

iii) S-shapes, p.1-30+

Example: S12x40.8 plf.840weight .ind 12

30





iv) HP-shapes, p.1-32+

Example: HP12x84

b) Channels or C - or MC sections, p. 1-34+

Example: C12x50MC 18x58

31




c) Angles, p.1-40+

Example: L8x8x5/8 equal-leg angle

L8x6x3/4 unequal-leg angle

toe

leg

toeheel

32




d) Tee-sections, p.1-48+

Example: WT20x196 cut from a W40x392

stem

Also cut from M - and S - Shapes


9/91



33




e) Hollow Structural Sections (HSS), p.1-72+

Example: HSS 20x12x5/8 rectangular HSS

HSS 10.000x0.635 round HSS

34




f) Pipe, p.1-99

Three types: strong, extra strong, and double-

extra strong

.ind 12|

Example: Pipe 12 x-Strong

35

LLBB SLBB

Gusset

Plate



Types of built-up shapes:

a) Double angles, p.1-100+

LLBB: Long leg back to backSLBB: Short leg back to back

2L8x6x LLBB

Example:

36




b) Double channels, p1-108+

x

y


10/91




37




c) I Sections

Weld (316 -1

4 in.)

38


1.5 Structural Steels:

High strength steel

Mild steel F

F

V

yVuV

H

39

yield plateau

Estrain

hardening

regionelastic

region



yH1512

yV

uuyy F,F VV

yH

In the AISC Specification:

40



ksiFy ksiFu

Common structural steels used in buildings:

ASTM A36 36 58-65ASTM A992 50 65

ASTM A572 Gr. 42 42 60

ASTM A572 Gr. 50 50 65

ASTM A572 Gr. 60 60 75

Steel


11/91




41



Most common structural steels:

ASTM A36 for connection plates and angles

ASTM A992 for hot-rolled members

ASTM A572 Gr. 50 for plates used in built-up members

42

43 44


12/91




45

Tension Member Design

Types: a) Wire rope, bridge strand

b) Rods

Non-threaded (welded)

Threaded

Non-upsetUpset

2.1 Introduction:

46

c) Plates primarily splice plates

d) Single rolled members

splice plates


47

e) Built-up members

truss bottom chord


48

Limit States (Failure Modes):

Strength:

T

Serviceability:

rupture (Fu)

T

yielding (Fy)

i) slendernessi) yielding

ii) deflection at service loadsii) rupture



13/91




49

2.2 Design Criteria:

D1. Slenderness Limitations

D2. Tensile Strength

D3. Area DeterminationD4. Built-up Members

D5. Pin-connected Members

D6. Eyebars

Chapter D applies to members subject to axial

tension caused by static forces acting through

the centroidal axis.


50

D1. Slenderness Limitations:

.,

300

inA

Ir

r

rL

rL

d except rods (see user note)

slenderness ratio

radius of gyration


51

D2. Tensile Strength:

The design tensile strength, , of tension members,

shall be the lower value obtained according to the limit

states oftensile yieldingin the gross section and tensile

rupture in the (effective) net section.

D2.(a) For tensile yielding in

the gross section:

gyn AFP 90.0tI

D2.(b) For tensile rupture in

the net section:

eun AFP 75.0tI

ntPI

ntu PP Id


52

D3. Area Determination:

Ag = gross area

An = net area (gross area hole area)

B

B'

A

A'Sec. A-A'

Gross Area (Ag)

Sec. B-B'

Net Area (An)



14/91




53


U=1.0 if all the elements in the section are connectedby bolts or welds. If one or more elements are not

connected, calculate U according Table D3.1

Ae = effective area

U = shear lag factor

UAA ne


54



55



56




15/91




57


U accounts for unequal stress distribution in the

transverse section of the tension member. If the

section has unconnected elements, the effective area

is less than the net area (i.e. U


16/91




61


More examples:

uT

Length of bolted connection:

uT

62

More examples:

Length of welded connections:


uT uT

63

Staggered Fasteners:

In a bolted connection, the reduction in the net area,

An, can be minimized if the bolts are staggered. The

net area is the minimum area calculated, following all

the possible paths where tension fracture can occur.


64

Staggered Fasteners:A

A

B

B

C

C

min

nA

long. spacing between

consecutive holes (pitch)

transv. spacing between

consecutive holes (gage)

number of diagonals (n=2

for the conn. shown)area of one holehA

t

g

snAAA

AAA

AAA

hgCC

hgBB

hgAA

43

2

2

thickness of material at the hole


s

g

n

t


17/91




65

Example:

Calculate the net area of the element shown below.

AC

CAB

B

.in.r

.in.x

.in.g

.in.s

.ind

in.A

.in.t

.in.d

y

b

g

w

7790

6740

52

62

347

3870

0012

43

2

C12x25


"8

52

"212

66

2

2

2

2

2

2

2

161

161

43

526

716

3870524

6252

433905347

52

3870524

6252233904347

66633902347

33903870

in.A

in.

..

.

..A

in.

..

...A

in...A

in..'A

n

FE

BB

AA

h

Example:


AC

CAB

B

"852

"2

12

AC-C

67

D4. Built-up Members:

300d

z

z

yxz

r

L

randrrx

y z

principal axes

For a single angle:


x

y

300

300

d

d

y

x

r

L

r

L

For two angles:

68

2.3 Analysis and Design Examples:

Example 1:

Calculate the tension

design strength of the I-

shape shown in the figure.

W18x35

.7.17

.425.0

.300.0

30.102

ind

int

int

inA

f

w

g

ASTM A992

ksiF

ksiF

u

y

65

50

critical section



18/91



6

Example 1:

425043010

01

750

87

87

161

161

43

..A

.ind

.U

AUA

AF.P

n

'h

ne

eun

I

kips.

..

AF.P gyn

50463

301050900

900

I

Rupture strength:

all elements connected


Yielding strength:

kips.

..P

in.AA

n

ne

5429

81865750

8182

I

70

Example 1:

Design strength:

kips.P

kips.P

kips.PP

n

rupture_n

yielding_nminn

5429

5429

5463

I

I

II

rupture controls


71

Example 2:

AC

CAB

B

Determine if the element is adequate to resist the loads

shown and satisfies the deflection limit. The member

length is 15 ft.

C12x25, A36

(see previous example)


"8

52

"212.in

kipsLkipsD

L 163

105

50

d

'

.in.r

.in.x

in.A

in.A

y

n

g

7790

6740

526

347

2

2

72

Example 2:

Design strength:

Yielding strength:

kips

..

AF.P gyn

330

34750900

900

I

Required strength:

kips

..

L.D.Pu

228

105615021

6121

By inspection the second loadcombination controls


Di d E t t d A P t I


19/91




73

Example 2:

Design strength:

Rupture strength:

kips

..P

in...A

..

.

L

xU

AUA

AF.P

n

e

ne

eun

269

19658750

196526950

95062525

674011

750

2

I


74

Example 2:

the element can resist the applied loads

kipsP

kipsP

kipsPP

n

rupture_n

yielding_nminn

269

269

330

I

I

II

rupture controls

kipsPkipsP nu 269228 d I


75

Example 2:

Slenderness limit:

300231

7790

1215

d

.r

L

y

OK


76

Example 2:

Deflection limit:

.in..in..EA

LP

g

LL

18800890

29000347

1215105

163 d

'

OK

C12x25, A36, is adequate


Diseo de Estructuras de Acero Parte I


20/91




77


Workshop 1

78

79 80



21/91




81

Compression Member Design

Limit states: 1. Flexural (lateral) buckling

2. Flexural-torsional buckling

3. Torsional buckling

4. Local buckling

Lateral buckling Torsional buckling Lat.-tor. buckling

3.1 Introduction:

Web local buckling Flange local buckling

In this course we

will focus on limit

state 1. and define

limit state 4.

82


Ideal columns: Assumptions

3.2 Flexural Buckling:

1.Elastic:

2. Perfectly straight:Pe

Pe

P

P

0

0

"

"

6

yPyIE

yIEyPM

M

x

o

Eulers solution:

2

2

L

EIPe

S

83


2rI now,

e

Eulers buckling

stress

2

2

L

EIPe

S

2

2

2

22

rL

e

e

e

E

AL

EAr

A

P

SV

S

V

L/r

y

2

2

rL

e

ESV

but steel yields

at some point

The result:


Ideal column behavior

L/r

y

84


Real columns:

T

C

C


1.Residual stresses due to manufacturing process:

actual line ref. line

rt

rc = 10 18 ksi

rt = tensile residual stresses

rc = compressive residual stresses



22/91




85


Real columns:

P

P


1.Residual stresses due to manufacturing process:

In stub columns:

y

y-rc

rc + P/A

Inelastic

Elastic

86


Real columns:


2. Out-of-straightness:

reference line real line

3. Other factors:

- end conditions

- unknown eccentricity

- out of plumbness

87


Design of real columns:


inelastic elastic

empirical curve

ideal curve

e

0.877e

r

L

V

yV

88


Define:

3.3 Local Buckling:

member section

This section has

five elements

elements



23/91




89


Define:

3.3 Local Buckling:

unstiffened elements

(one end connected)

stiffened element

(both ends connected)

90


Define: There are two types of elements, stiffened and unstiffened

3.3 Local Buckling:

1 stiff. elmt. 4 unst.

2 unst. elmts. 3 unst. elmts. 4 stiff. elmts. 1 stiff. elmt. 2 unst.

7 stiff. elmt. 4 unst. 8 stiff. elmt. 4 unst.

91


Local buckling occurs when the cross section of a

compression member has elements so thin that they

buckle locally, with a lower load than the load required

to make the column buckle laterally.

3.3 Local Buckling:

local buckling flex. (overall) buckling

Only those members thathave thin (slender)

elements can buckle

locally. If there are not

slender elements, the

only failure mode is

overall buckling.

92


In an I-section member there are two types of local

buckling: flange local buckling and web local buckling.

3.3 Local Buckling:

Web local buckling (WLB) Flange local buckling (FLB)



24/91



93


3.3 Local Buckling:

Local slenderness ratio

A member can collapse because of local buckling if its section

has slender elements. The element slenderness ratio is

computed by dividing its length (b) by its thickness (t).

t

b

f

f

t

b

t

b

2

wt

h

t

b

Example:

flange slenderness ratio

web slenderness ratio

94


More examples:

3.3 Local Buckling:

flanges:

web:

f

f

tb

tb

wt

h

t

b

flanges:

tb

tb

t

h

t

b

legs:

tb

tb

flange:

stem:

f

f

tb

tb

wt

d

t

b

95


3.3 Local Buckling:

Table B4.1 of the AISCs specification gives the limit rfor sections subject to compression. If the slendernessratio, b/t, is more than r, the element is consideredslender.

rt

bO!

slender element

local buckling may occur

before flexural buckling

If

Note: A section is considered slender if one or more of its

elements are slender.

96


3.3 Local Buckling:



25/91



97


Determine if a W21x93, A992 is a slender section.

Example:

332

534

930

428

580

621

2

.

.

.in.t

.in.b

.in.t

.in.d

w

f

f

th

tb

f

f

w

W21x93 ASTM A992

ksiF

ksiF

u

y

65

50

98


Example:

the flange is non-slender

Flange: From Table B4.1, Case 1:

135

549302

428

2

513

50

29000560560

..

..

.

t

b

t

b

.

.F

E.

f

f

y

r

O

99


Example:

Web: From Table B4.1, Case 5:

935332

332

935

50

29000491491

..

.t

h

t

b

.

.F

E.

w

y

r

O

the web is non-slender

Neither the flanges nor the web are slender elements.

Therefore, the section is a non-slender section. Local

buckling will not control the design of this member.

100


Determine if the built-up section shown below is a

slender section. The material is A572 Gr. 50

Example:

.in.t

.in.b

.in.t

.in.h

.in.d

f

f

w

01

09

50

119

022

Section ASTM A572 Gr. 50

ksiF

ksiF

u

y

65

50



26/91



101


Example:

the flange is non-slender

Flange: From Table B4.1, Case 2:

m

m

41254

54012

9

2

41250

29000650640

760650350

65044

640

50119

.

..t

b

t

b

..

.

...

.k

F

Ek.

f

f

r

..

th

c

y

cr

w

O

O

OK (footnote [a])

102


Example:

Web: From Table B4.1, Case 5:

935238

238500

119

935

50

29000491491

..

..

.

t

h

t

b

.

.F

E.

w

y

r

!

O

the web is slender

Since the web is slender, the section is slender. Local

buckling may occur prior to overall member buckling.

103


3.3 Design Criteria:Chapter E addresses members subject to axial

compression through the centroidal axis.

E1. General Provisions

E2. Slenderness Limitations and Effective Length

E3. Compressive Strength for Flexural Buckling ofMembers without Slender Elements

E4. Compressive Strength for Torsional and Flexural-Torsional Buckling of Members without SlenderElements

E5. SingleAngle Compression Members

E6. Built-up Members

E7. Members with Slender Elements

104


E1. General Provisions:

The design compressive strength, IcPn, is determined asfollows:

The nominal compressive strength, Pn, shall be thelowest value obtained according to the limit states offlexural buckling (E3.), torsional buckling (E4.), andflexural-torsional buckling(E4.).



27/91



105


E1. General Provisions:

(a) For doubly-symmetric and singly-symmetricmembers the limit state of flexural buckling is

applicable.

(b) For singly-symmetric and unsymmetric members,

and certain doubly-symmetric members, such as

cruciform or built-up columns, the limit states of

torsional or flexural-torsional buckling are also

applicable.

90.0cIFor all limit states,

106


E2. Slenderness Limitations and Effective Length:

r

L

K

Slenderness limit:

effective length factor

laterally unbraced length of the member

governing radius of gyration

md

200r

LKrecommended

107


E2. Slenderness Limitations and Effective Length:

108


E3. Compressive Strength for Flexural Buckling of

Members without Slender Elements:

gcrn AFP

crF flexural buckling stress

yF

E

r

KL71.4d(a) When y

F

F

cr FFe

y

658.0

yF

E

r

KL71.4!(b) When ecr FF 877.0



28/91



109




elastic critical buckling stress

(Eulers buckling stress).

2

2

r

KL

EFe

S

eF

110


Elasticregion

Inelastic

region



2

2

r

KL

EFe

S

ecr FF 877.0

200 r

KL

yF444.0

yF yF

F

cr FFe

y

658.0

yF

E71.4

crF

111



Example 1:

Determine the available strength, , of a W14x99.

A992 steel.ncPI

ncPI Properties:

W14x99

523

349

713

176

129

2

2

.

.

.in.r

.in.r

in.A

w

f

f

th

tb

y

x

g

112



Example 1:

51350

29000560349

2...

t

b

f

f

Overall slenderness:

5.3171.3

1275.90.1

0.19

17.6

1275.90.1

y

yy

x

xx

r

LKr

LK

controls

Element slenderness:

Flange:

93550

29000491523 ...

t

h

w

Web:

The section is non-slender



29/91



113



Example 1:

:ncPI

?! 5.3111350

2900071.4 we are in the

inelastic zone

ksiF

ksiEF

FF

cr

rKL

e

y

F

F

cre

y

5.4650658.0

2885.31

29000

658.0

288

50

2

2

2

2

SS

kips

...

AF.P gcrnc

1217

12954690

90

I

114



Example 1:

:ncPI

Notice that . Table 4-22 lists for

various values.

From Table 4-22, with and :

kipsP

PksiF

nc

nc

crc

1218

1.2985.4185.41

I

II

yrKL

crc FfF ,I crcFIyF

ksiFy 50 5.31rKL

115



Example 1:

:ncPI

116



Example 2:

Determine the available strength, , of a W12x65,

braced at mid-height as shown below. Use A992 steel.nc

PI

ncPI Properties:

W12x65

9.24

92.9

.02.3

.28.5

1.19

2

2

w

f

f

th

t

b

y

x

g

inr

inr

inA

ncPI

'26xxLK '13yyLK



30/91



117



Example 2:

Check local slenderness:

513929

51350

29000560

560

9292

..

..

F

E.

.t

b

t

b

y

r

f

f

O

rt

bOdIf the section is non-slender

Flange:

935924

93550

29000491

491

924

..

..

F

E.

.t

h

t

b

y

r

w

OTable B4.1 Case 3

Case 10

Web:

this section will not failby local buckling

118



Example 2:

Flexural buckling:

751023

121301

159285

122601

..

.

r

LK

..

.

r

LK

y

yy

x

xx

controls

1.59rKLksiFcrc 87.34Iwith from Table 4-22

kipsP

P

nc

nc

666

1.1987.34

I

I

119



Example 3:

Select the lightest W14 section if Pu = 480 kips. The

member length is 12 ft and both ends are pinned. The

member material is A992.

Assume

.45.2

9.17 2

inr

inA

y

g

.92.1

6.15 2

inr

inA

y

g

80rKL

; from Table 4-22 ksiFcrc 2.28I

0.172.28

480

80.180

)1212(0.1

crc

ug

y

F

PA

r

I

with these values, the

possible sections are:

W14x61

W14x53

120



Example 3:

Try the W14x53 section:

0.7592.1

12120.1

r

KL; from Table 4-22 ksiFcrc 8.29I

kipsPkipsP

AFP

unc

gcrcnc

480465

6.158.29

I

II

N.G.

Try the W14x61 section:

; from Table 4-22 ksiFcrc 8.32I

kipsPkipsP

AFP

unc

gcrcnc

480587

9.178.32

!

I

II

OK

8.5845.2

12120.1

r

KL



31/91



121



Example 3:

Check local buckling:

5.1375.7

5.1350

29000

56.0

56.0

75.72

r

y

r

f

f

F

E

t

b

t

b

O

O

Flange:

9.354.30

9.3550

29000

49.1

49.1

4.30

r

y

r

w

F

E

t

h

t

b

O

OTable B4.1 Case 3 Case 10

Web:

this section will not fail bylocal buckling

Use W14x61, A992

122



Example 4:

Select a W12 , A992 if the required strength is 745 kips.

The column details are shown below.

kipsPu 745

'24xxLK '18yyLK

kipsPu 745

123



Example 4:

Assume y-direction controls,

.88.275

1812inry

.07.3

.38.5

6.252

inr

inr

inA

y

x

g

y

yy

r

LK

75

y

yy

r

LK; from Table 4-22 ksiFcrc 8.29IAssume

20.258.29

745inA

g Required ; Required

Try W12x87:

124


5.5307.3

12240.1

x

xx

r

LK


Example 4:

Check assumptions done before:

kipsPkipsP

P

AFP

unc

nc

gcrcnc

745801

6.253.31

!

I

I

II

3.7007.3

12180.1

y

yy

r

LK; from Table 4-22 ksiFcrc 3.31I

3.7007.3

12180.1

y

yy

r

LKOK

OK

>



32/91



125



Example 4:

Check slenderness:

Use W12x87, A992

8.359.18

4.135.72

r

w

r

f

f

t

h

t

b

O

O OK

OK

126



Example 5:

Find ncPI for the section shown below. The material is A36.

ncPI

98

3 xPL

98

3 xPL

12

16

5 xPL

127



Example 5:

Check slenderness:

Web:Flange:

m

614012

61436

29000650640

76065044

640

012

2

9

2

165

12

8

3

..

..

.

..k

F

Ek.

.

t

b

r

twh

c

y

cr

f

f

O

O

m

342438

34236

29000491

491

43812

16

5

..

..

F

E.

.

t

h

y

r

w

OTable B4.1 Case 4 Case 10

OK

OK

128



Example 5:

Section properties:

Need

.in..

.

A

Ir

in.

I

in.A

y

y

y

g

082510

645

645

1212

19

12

12

5101292

4

3

1653

83

2

165

83

:, yg rA



33/91



129



Example 5:

Flexural buckling:

kips.

..

AFP gcrcnc

0230

510921

II

5.8608.2

12150.1

y

yy

r

LK; from Table 4-22 ksiFcrc 9.21I

130


Workshop 2

131 132



34/91



133

Beams - Laterally Supported

4.1 Overview:Define: Laterally supported beam

Pu

Lateral supports Pu

Pu

The beam can deflect vertically, butrotation and horizontal displacement

are prevented by lateral supports

Deflection

Rotation

and

H. disp.

y'

134


4.1 Overview:Define: Laterally unsupported beam

The beam deflects vertically, has ahorizontal displacement, and rotates

Pu x'

y'

I

135


4.1 Overview:

Laterally supported beam design study

Topic Section

- Bending strength 4.2, 4.3, and 4.4

- Shear strength 4.5

- Deflection limits 4.6

- Web crippling strength 4.7

- Web yielding strength 4.7

- Beam bearing plate strength 4.8

136


4.1 Overview:


Shear strengthuV

uM Bending strength Deflection limits

'nvu VV Id

nbu MM Id max'd

uq

arrugamiento



35/91



137


4.1 Overview:


Web crippling Web yielding Beam bearing plate strength

uq

nu RR d Inu RR d I nu RR d I

yielding here

uR uR uR

138


4.2 Bending Theory:Assumptions:

a) Plane sections remain plane

b) The member is prismatic

c) The material is elastic perfectly plastic

yV

yH H

139


4.2 Bending Theory:

Strain

Stress

yHH yHH yHH ! yHH !!

yVV

yV

yV yV yV

140


CL

4.2 Bending Theory:

Plastic moment

Moment at 1st yield

maxM

pM

yM

'

EI

wLCL

384

54

'

CL'

yM

pM

w

Fluye toda laseccin



36/91



141


4.2 Bending Theory:

maxVReview of Elastic Beam Theory:

StressStrain

e.n.a.

e.n.a.= elastic neutral axis

I

yM

I

yM

M ane

maxmax

... 0

6

V

V

142


4.2 Bending Theory:

maxy

IS

Define:

2d

x

x

IS

S = elastic section module

For H and C shapes:

2fb

y

y

IS

For T shapes:

Review of Elastic Beam Theory:

143

Ac

AtTy

Cy


4.3 Calculation of Plastic Moment Capacity:

yV

yVp.n.a. = plastic neutral axis

pMcy

ty

p.n.a

0

AAA

AA

TC

F

tc

tycy

yy

VV

one half of the area is above

and below the p.n.a144



iix yAZ

x

xyp

iiyp

ttccyp

tytcycp

tycyp

tycyp

anp

Z

ZM

yAM

yAyAM

yAyAM

yTyCM

yTyCM

M

V

V

V

VV

0...

plastic section modulus

with respect to p.n.a.



37/91



145



x

x

xy

xy

y

p

S

Z

S

Z

M

Mf

f

V

V

xyr

xyp

xyy

SF.M

ZM

SM

70

V

V

shape factor

Definitions:

plastic section modulus

elastic section modulus

Define:

x

x

r

p

y

Z

S

M

M

M

plastic moment

yield moment

first yield moment

146



f shape factorDefine:

The shape factor is a measure of the remaining

bending strength beyond the yield moment.

CL

yM

pM

maxM

'

reserve strength

147

ftkips.

SM

in.S

in.

I

IS

xyy

x

x

d

xx

19612

26536

265587

58712

17

4

1

2

176

2

1

12

5062

3

21

217

43

165

23

2

V



f

Example 1:

Find ,and for strong axis bending. The

member material is A36.py MM ,

PL1

2 x 6

PL12 x 6

PL 516 x 17

148

151

196

225

225

36

1754

17

2

17

16

5

4

1

2

176

2

12

12175

3

.f

M

Mf

ftkips

ZM

in.

yAZ

y

p

.xyp

iix

V


p.n.a.


Example 1:

with respect to p.n.a.

CL

15%

yMpM

maxM

tiene quever conlos esfuerzosresiduales entre 10y 18 ksi



38/91



149

kipsftM

SM

inI

S

inI

y

yyy

b

y

y

y

f

0.18

12

0.636

00.63

0.18

0.1812

17

12

65.02

3

2

4

3

1653

V



f

Example 2:

Compute ,and for the previous section

with respect to y-axis.py

MM ,

PL12 x 6

PL12 x 6

PL 516 x 17

x

y

150

53.1

0.186.27

6.27

36

21.92

17

2

235.02

1221.9

2

3252

f

MMf

ftkipsM

ZM

Z

yAZ

y

p

p

yyp

y

iiy

V



Example 2:

with respect to y-p.n.a.

yM

pM

maxM

CL

53%

'

151

70

70

.

SF.M xyr

CL

slender

non-compact

compact


4.4 Flexural Design of Beams:

4.4.1 Basic Beam Behavior and Response Curve:

pM

rM

maxM

'

xyp ZFM

accounts forresidual stresses

local buckling

yM

yM yield moment

rM moment at first yielding

152



4.4.1 Basic Beam Behavior and Response Curve:

p

M

rM

nM

pO

Types of elements :

a) compact

b) non-compact

c) slender

General response curve

compact

element

non-compact

elementslender

element

straight line

hyperbola

rO tbO

tipos de seccionespara vigas

esbeltez



39/91



153


Define:


4.4.2 Strong Axis Bending of H-Shapes:

r

rp

p

OO

OOO

OO

!

dd

If

If

If

compact element

non-compact element

slender element

Section Elements

f

w

O

O web slenderness ratio

flange slenderness ratio

154




Table B4.1 indicates

the limits p and r to

determine the type of

element within a

section

155




nbu MM I

Design criterion:

90.0bIuM Required flexural strength (from moment diagram)

n

M Nominal flexural strength (Chapter F specification)

, in a laterally supported beam, is function of the type

of elements (compact, non-compact, slender) within the

section

nM

156




, compact flange and compact web:nM

Compact web:

y

pw

w

wFE

th 76.3d OO

Compact flange:

y

pf

f

f

fF

E

t

b38.0

2d OO

xypn ZFMM

Table B4.1 Case 15(rolled and built-up section)

Table B4.1 cases 10 and 11(rolled and built-up section)

AISCS (F2-1)

Un elementorigidizado y 4sin rigidizar



40/91



157




, non-compact flange and compact web:nM

Non-compact flange: rffpf OOO d

Table B4.1 Case 10(rolled section)

yL

th

c

L

crf

y

rf

F.F

.k.

F

Ek.

F

E.

w

70

7604

350

950

01

dd

O

O

Table B4.1 Case 11

(built-up section)

Table B4.1 footnote (a)

Table B4.1 footnote (b)

158




, non-compact flange and compact web:nM

Notice that:

pfrf

pff

xyppn SF.MMMOO

OO70

xyr SF.M 70

AISCS (F3-1)

159




, slender flange and compact web:nM

rff OO !

AISCS (F3-2)2

9.0

f

xcn

SEkM

O

160


pM

rM

nM

pfO

Summary for beams with compact webs:

compact

flange

non-compact

flangeslender

flange

rfOf

f

t

b

f 2

2O



xy ZF

pfrf

pff

xypp SFMMOO

OO75.0

2

9.0

f

xcSEk

O



41/91



161




Example 3:

Compute for a W21x68. The member material is A992.nbMI

W21x68

3

3

2

160

1406.43

04.6

inZ

inS

x

x

t

h

tb

f

f

162




Example 3:

Check slenderness:

web:

?

6.906.43

6.90

50

2900076.376.3

6.43

y

p

w

w

F

E

t

h

O

O

flange:

?

15.904.6

15.9

50

2900038.038.0

04.62

y

p

f

f

f

F

E

t

b

O

O

ftkips.ZF.M xynb

600

12

160509090IDesign strength:

compact web compact flange

163




Example 4:

Compute for a W14x90. The member material is A992.nbMI

W14x90

3

3

2

157

143

9.25

2.10

inZ

inS

x

x

th

tb

f

f

164




Example 4:

Check slenderness:

web:

?

6.909.25

6.9050

2900076.376.3

9.25

y

p

w

w

F

E

th

O

O

flange:

?

1.242.1015.9

1.2450

290000.10.1

15.950

2900038.038.0

2.102

y

r

y

p

f

ff

F

E

F

E

t

b

O

O

O

compact web

non-compact flange



42/91



165




Example 4:

Design Strength:

ftkips..

...M

ftkips.SF.M

ftkipsZFM

SF.MM.M

nb

xyr

xyp

pfrf

pff

xyppnb

574159024

15921041765465490

41712

143507070

65412

15750

75090

I

OO

OOI

166




Example 5:

Compute for a section shown below. The member

material is A572 Gr. 50.nbMI

Properties:

3

3

3233

1135.4925.0225.9105.02

1035.9

977

97725.9105.012

5.0102

12

1825.0

inyAZ

inc

IS

inI

iix

xx

x

PL12 x 10

PL12 x 10

PL14 x 18

167




Example 5:

Check slenderness:

web:

?

6.9072

6.9050

2900076.376.3

7225.0

18

y

p

w

w

F

E

th

O

O

compact web

168




Example 5:

Check slenderness:

760471044

15950

29000380380

10502

10

2

25018

..k

..F

E.

.t

b

.th

c

y

p

f

f

f

w

O

O

non-compact

flange

flange:

?

81810159

8185070

290004710950

950

..

..

..

F

Ek.

L

crO



43/91



169

ftkips..

..M

ftkips.SF.M

ftkipsZFM

MMM.M

nb

xyr

xyp

pfrf

pff

rppnb

410159818

1591030047147190

30012

103507070

47112

11350

90

I

OO

OOI




Example 5:

Design Strength:

170



4.4.3 Weak Axis Bending:

Web local buckling is not a limit state

Flange local buckling is the same as strong axis bending

e.n.a=p.n.a.

yV yV

171



4.4.4 Analysis and Design Examples:

Select the lightest H-shape if Mu = 305 kips-ft, A992.

uxynbMZFM !? 9.0I

33.81509.0

12305

9.0in

F

MZ

y

urequired

Example 6:

Lets assume that the section is compact

From Manual Table 3-2, two possible options are:

W16x45

W10x68

33.82 inZx 3

3.85 inZx

The W16x45 (45plf) is the best option

since it is lighter than the W10x68

(68plf). Need to check compactness.

172

Beams - Laterally Supported4.4 Flexural Design of Beams:


Example 6:

primera fluencia

Plastico



44/91



173

?

159236

159

50

29000380380

2362

..

..

F

E.

.t

b

y

p

f

f

f

O

O




Example 6:

Use W16x45

Check slenderness:

web:

?

690141

690

50

29000763763

141

..

..

F

E.

.t

h

y

p

w

w

O

O

flange:

compact web compact flange

174




Example 7:

Design a H-shape for the beam shown below. Use A992.

self-weightkipsPkipsP

LD30,10

4PL

2wL

uM

-

175




Example 7:

Factored load and moment:

max

.4 1.4 10 141.2 1.6 1.2 10 1.6 30 60

Du

D L

P kipsPP P kips

Estimate the beam weight = 60 plf plfwu 72602.1

ftkips

.Mu

3044300

8

200720

4

2060 2

176




Example 7:

Assume the section is compact:

Estimate the beam section:

3

1815090

12304

90 in..F.

M

Z y

u

d'req

.25.0

.5.0

.8

int

int

inb

w

f

f

PL12 x 8

PL14 x 2y

PL12 x 8



45/91



177




Example 7:

Determine the minimum web height:

.in..d

.in.y.in.y

y..y..

yA.Z

y

iix

01750282

00897

2502505082181

681

2

o

m

t

mt

.in.in

Ld

171024

1220

24Rule of thumb

OK

PL 12 x 8

PL14 x 2y

PL 1 x 8

178




Example 7:

Check slenderness:

m

690064

690

50

29000

763763

64250

117

..

.

.F

E

.

.t

h

yp

w

w

O

O

m

1598

159

50

29000

380380

08502

8

2

..

.

.F

E

.

..t

b

yp

f

f

f

O

O

compact, OK

web:

compact, OK

flange:

179




Example 7:

Check beam weight assumption:

m

plfplf.

Aw

inA..A

g

g

g

60840

12

12490

12250165082

2

2

G

OK

180




Example 7:

Use PLx8 for the flanges and

PLx16 for the web, A992

PL12 x 8

PL14 x 16

PL12 x 8

(must check shear and deflection)

Para una primeraaproximacin se puededespreciar el peso propio,ya definido se comprueba



46/91



181




Example 8:

Design the filler beams for the bay shown in the figure. Use

A572 Gr. 60.Live load = 100 psf

Slab + deck = 107 psf

Mech. = 5 psf

Ceiling = 3 psf

Miscellaneous = 2 psf

Dead load = 117 psf

182




Example 8:

Factored load and moment:

ftkipsLw

M

plfLD

plfDw

plfw

plfw

uu

u

L

D

m

1538

20064.3

8

306410006.112202.16.12.1

170812204.14.1

100010010

12205011710

22

max

Assume beam weight = 50 plf:the beam self-weight

is 5% of the dead load

183




Example 8:

Assume the section is compact:


3

'0.34

609.0

12153

9.0

in

F

MZ

y

udreq

.25.0

.375.0

.6

int

int

inb

w

f

f

PL 3 8 x 6

PL 14 x 2y

PL 3 8 x 6

184




Example 8:


.25.12

.75.5.62.5

25.0375.0620.34

0.34

22375.0

ind

inyiny

yy

yAZ

y

iix

o

mt

mt

.in..inft

Ld

ft

.in

25121020

2rule of thumb

OK

PL 3 8 x 6

PL 14 x 2 y

PL 3 8 x 6



47/91



185




Example 8:

Check slenderness:

m

690046

690

50

29000763763

46250

2755

..

.

.F

E.

.

.

t

h

yp

w

w

O

O

m

1508

159

50

29000

380380

0837502

6

2

..

.

.F

E

.

..t

b

yp

f

f

f

O

O

compact, OK

web:

compact, OK

flange:

186




Example 8:

Check beam weight assumption:

m

plfplfw

Aw

inA

A

g

g

g

501.25

12

38.7

490

38.7

25.075.52375.062

2

2

G

OK

187




Example 8:

Use PL3/8x6 for the flanges and

PLx11 for the web, A572 Gr.60

PL 38 x 6

PL 14 x 111

2

PL 38 x 6

(must check shear and deflection)

188


Workshop 3



48/91



189


V

4.5 Shear Strength:Limit states:

maxW

It

VQW

Design approximation:

a) Shear Yielding

b)Shear Buckling

Shear stress:

ww td

V

A

V

maxW

Shear yield stress:

y

y

y VV

W 577.03 Von Misses

theory of failureyy

F6.0W for design

190


4.5 Shear Strength:4.5.1 Design Criteria:

Chapter G addresses webs of singly- or doubly-

symmetric members subject to shear in the plane

of the web, single angles and HSS sections, and

shear in the weak direction of singly or doubly

symmetric shapes.

nvu VV Id 90.0vI

191


4.5 Shear Strength:

4.5.1 Design Criteria:

AISCS section G2.1:

vwyn CAFV 6.0

for shear yielding:

for shear buckling:

AISCS (G2-1)

0.1v

C

0.1v

C

192

inelasticbuckling

elasticbuckling

G2-4

G2-5

yielding

G2-3

1.0


4.5 Shear Strength:


AISCS section G2.1:

is a function ofw

webt

hO

vC

wth

vC

y

v

F

Ek.101

y

v

F

Ek.371



49/91



193



AISCS section G2.1:

w

y

v

v

y

v

wy

v

v

y

v

w

th

F

Ek

C

F

Ek

t

h

F

Ek

C

F

Ek

t

h

10.1

37.110.1

0.1

10.1

d

dFor

For

AISCS (G2-3)

AISCS (G2-4)

194



AISCS section G2.1:

yth

vv

wy

v

F

kEC

t

h

F

Ek

w

2

51.1

37.1

For

AISCS (G2-5)

stiffenersFor sections without

stiffeners, 5v

k

195


4.5 Shear Strength:


AISCS section G2.1:

:24.2yw F

E

t

hd

0.1vI

Special case:

For hot-rolled sections with

vwyn CAFV 6.0 0.1vC

Note: Shear will only control for large loads and short

spans.

196


4.5 Shear Strength:

4.5.2 Design Examples:

Example 9:

Determine if W18x71, A992, is adequate for shear.

kipsPu 170

kips5.42kips5.127

5.127

5.42

)(kipsVu

W18x71

4.32

.495.0

.5.18

w

w

t

h

int

ind

Neglect the beam weight



50/91



197


4.5 Shear Strength:4.5.2 Design Examples:

Example 9:

0.10.1

9.534.32

9.5350

29000

24.224.2

4.32

6.0

5.127

?m

vv

y

w

vwynv

u

C

F

E

t

h

CAFV

kipsV

I

II

m!

kips.kips.

kips.

.....

.tdF..V wynv

51277274

7274

014950518506001

016001

OK

special case

198


4.5 Shear Strength:4.5.2 Design Examples:

Example 10:

Determine in the following shape. The material is

A572 Gr. 50.nvVI

PL 3 4 x 6

PL 3 4 x 6

PL14 x 16

199


4.5 Shear Strength:

4.5.2 Design Examples:

Example 10:

8.7350

29000537.137.1

2.5950

2900051.11.1

5

6416

6.09.0

41

m

y

v

y

v

v

w

vwynv

F

Ek

F

Ek

kt

h

CAFVI

kips

.....V

..F

Ek.

C

..

nv

th

y

v

v

w

109

9250517250506090

925064

259

11

87364259

?

I

unstiffened webs

inelastic shear

buckling

200


4.6 Serviceability Considerations:

Chapter L addresses serviceability performance

design requirements.

L1. General ProvisionsL2. Camber L3. DeflectionsL4. DriftL5. VibrationL6. Wind-Induced MotionL7. Expansion and ContractionL8. Connection Slip



51/91



201


4.6 Serviceability Considerations:Deflections:

Deflections in structural members and structural

systems under appropriate service load combinations

should not impair the serviceabilityof the structure.

Empirical Rules:

2

240360L

d.ii

span,

span.i LL

t

dd '' (floors) (roofs)

d = depth of the beam

L = length of the beam

202


4.6 Serviceability Considerations:Vibrations:

203



Example 11:

a)Check of filler beams.L'

W16x26

4301 inIx

m

'

.667.0.412.0

.667.0360

1220

360

.412.030129000384

1728200.15

384

5

0.110100

100

44

inin

inL

inEI

Lw

klfftplfw

psfL

LL

L

OK

204



Example 11:

b) Compute due to concrete, deck, and beam.C'

.45.000.1

10.1412.0

10.12610107

26

107

in

klfw

plf

psf

c

c

'

For a beam to be cambered,

slab + deck

beam

camber must be .43 in



52/91



205


4.7 Web Yielding and Web Crippling:

Reactions

Load bearing on flange

failurefailure

*Concentrated Loads

206



yieldinghere

design

ywyn

nu

kk

FtNkFAR

RR

d

5.2

0.1

1

I

I

Web Yielding (J10.2):

AISCS (J10-3)

Exterior:

1

d

207



design

wyn

nu

kk

tFNkR

RR

d

5

0.1II

Web Yielding (J10.2):

AISCS (J10-2)

Interior:

t

208



75.0

d

I

I nu RR

Web Crippling (J10.3):

localbuckling



53/91



209



w

fy

f

wwn

t

tEF

t

t

d

NtR

5.1

2318.0


t

Interior loads:

210



w

fy

f

wwn

w

fy

f

wwn

t

tEF

t

t

d

NtR

dN

t

tEF

t

t

d

NtR

dN

!

d

5.1

2

5.1

2

2.04

14.0

:2.0

314.0

:2.0


Exterior loads:Two cases:

211



nu RR ! IIf , install bearing stiffeners:

2dt

212



nRIDetermine and if bearing stiffeners are required.

kipsPu 150

Example 12:

W16x26, A992

.747.0

.250.0

.345.0

.7.15

ink

int

int

ind

design

w

f



54/91



213



d!

Web yielding:

Example 12:

from the end of the beam:

m!

kipskipsR

R

tFNkR

n

n

wyn

150153

250.0505.8747.050.1

5

I

I

I

OK

No stiffeners required

214



2

d!

Web crippling:

Example 12:

from the end of the beam:

m

kipskipsR

R

t

tEF

t

t

d

NtR

n

n

w

fy

f

wwn

150106

25.0

345.05029000

345.0

25.0

7.15

5.83125.08.075.0

318.0

5.1

2

5.1

2

I

I

II

N.G., use bearing stiffeners

or change the beam

215



Limit states:

4.7.1 Beam Bearing Plate Design (AISCM, p14-3):

expandinggrout

uV

.4 inN .1 inbB f |t

Web yielding

Web crippling

Plate bending

Concrete crushing

N

ptB

216


on full area

d

d

BNA

psipsif

A

A

AAAfP

AfP

PR

c

cp

cp

pu

1

'

1

2

1

21'

1'

150002500

2

85.0

85.0

60.0II


Concrete Crushing (J8):

on less thanthe full area

for calculation

purposesbut

plan area of base plate

CL

CLA1 A2

A1 A2

11

A1 = A2


Requisitos

aplastamientodel concreto

Es el factor de

resistencia alaplastamientodel concreto



55/91



217


pt

designk

uR


Plate Bending:In the design model only a 1 str ip is

used and is analyzed as a cantileverbeam.

for a rectangle :

pP"1

1A

RP up

m

y

udreq

p

u

F

MZ

mPM

9.0

2

1

'

2

b

d

o

o4

1

4

2

'

2p

dreq

tZ

bdZ solve for

pt

assuming yielding


218

Design the beam and the bearing plate for the member

shown below. Use A36 for the beam and for the bearing

plates. The beam is supported on a 12 in. wide concrete

wall.


ksi.'f

L

ftL

.ind

w

w

c

L

ftkips

D

ftkip

L

03

360

6

10

7

9

d

d

'

4.8 Summary Example:Example 13:

DL ww ,

219


ftkipLw

M

kipsV

LD

D

w

uu

u

ftkips

ftkips

u

6.1028

68.22

8

4.682

68.22

8.2296.172.16.12.1

8.974.14.1

22

max

uu MV

Required strength:

Need to calculate and


220


Bending strength:


3

'0.38

369.0

126.182

9.0in

F

MZ

y

udreq

.25.0

.375.0

.6

int

int

inb

w

f

f

PL3

8 x 5

PL14 x 2y

PL 38 x 5




56/91



221



.75.14

.00.7.83.6

25.0375.0520.38

0.38

228

3

ind

inyiny

yy

yAZ

y

iix

o

m!

t

mt

.in.in.

.inft

d

Ld

.in

ft

.in

37514

32

6

2rule of thumb

OK

PL 38 x 5

PL14 x 2y

PL 38 x 5


222


Check local slenderness:

m

107056

107

36

29000763763

56250

207

.

.F

E.

.

.

t

h

y

p

w

w

O

O

m

810676

810

36

29000380380

67637502

05

2

..

.

.F

E.

..

.

t

b

y

p

f

f

f

O

O

compact, OK

web:

compact, OK

flange:


223

m

kipskips

kipsR

kipsR

R

CAFR

u

nv

nv

vwynv

7.714.68

4.68

7.71

0.125.075.14366.090.0

6.090.0

I

I


Shear Strength:

0.18.6956

8.6936

29000510.1

10.1

5

5625.0

14

?

m

v

y

v

v

w

C

F

Ek

k

t

h

OK

unstiffened web


224


m

.in..in.

.in.L

.in..EILw

klf.w

inI

LL

L

x

2000360

200360

126

360

036025129000384

172860953845

09

251

44

3

'

OK

Check deflection:




57/91



225


Use PL3/8x5 for the flanges and

PLx14 for the web, A36.

PL38 x 5

PL14 x 14

PL38 x 5


226

14


Bearing plate design:

uV

Try PL 7x8 A36:


227


Check web yielding:

m

kips.kips.

kips.R

kips.R

....RtFNk..R

.in.k

u

n

n

wydeignn

design

383468

468

383

250368625052015201

625041

83

I

II

OK


Assume in. welds

228


Check web crippling:

?! 2.027.075.14

4

d

NEnd, Equation J10-5(b):

m

-

-

kips.kips.R

.

.

.

..

....R

t

tEF

t

t.

d

Nt..R

n

.

n

w

fy

.

f

w

wn

468225

250

37503629000

3750

25020

7514

84125040750

204

140750

51

2

51

2

I

I

I

N.G., need to use stiffeners

Stiffener design will be covered in section 10




58/91



229


Check concrete crushing:

m!

kips.kipsP

....P

.A

A

inA

inA

psif

A

AAf..P

p

p

c'

c'p

468131

531560385060

253156

132

13227212

5687

3000

85060

1

2

2

2

2

1

1

21

I

I

I

beam

OK


230


Bearing plate thickness:

.in..Z

t

tbin.

.

.Z

.inkips..

.M

ksi..

A

RP

.in..m

d'req

req,p

p

d'req

u

up

62501

15604

1

4

41560

3690

065

0652

8821221

22156

468

882250

2

3

2

1

27

designk

pt

Use PL 5/8x7x0-8,A36


231



Flanges PL 38 x 5

Web PL 14 x 14A36

PL 58" x 7" x 0'-8" A36

Final design:

232


N

n

n

4.9 Column Base Plate Design (AISCM p14-4):

fb85.0

m

Limit states:

m d95.0

B

uP

1

A

PP up

- Concrete crushing

- Plate bending



59/91



233


pt

y

up

p

y

ur

rynbu

p

p

u

F

Mt

t

F

MZ

ZFMM

nP

mP

M

9.0

4

4

1

9.0

9.0

21

21

2

2

2

max

d

I

pt

2

85.0

6.0

1

2

1

21'

d

d

A

A

A

AAfP

PP

cp

pu II

Concrete crushing (J8): Plate bending:


234

PL15"x24" A36

Pier 24"x48"

W18x119


Check if the column plate shown above is adequate if

.Determine .

Example 14:

uP

kipsPu 1300

psifc 5000'

ptW18x119

.3.11

.0.19

inb

ind

f


235


Concrete crushing:

m!

kipskips

....P

.A

A

inA

inA

A

AAf..P

p

c'p

13001359

4813600585060

2481792

360

7923324

3602415

85060

1

2

2

2

2

1

1

21

I

I

OK

Example 14:


236

Beams - Laterally Sup

part i - design of steel members

Documents