STRC05 Steel Part1 0715

7/25/2019 STRC05 Steel Part1 0715

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Structural Engineering Exam Review Course Steel (Part 1)

© 2015 Professional Publications, Inc. 1

STRC ©2015 Professional Publications, Inc.

Steel Part 1

Structural Engineering Review Course

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Steel Part 1

Lesson Overview

Chapter

4:

Structural

Steel

Design

• Introduction

• Load Combinations

• Design for Flexure

• Design for Shear

• Design of Compression Members

2

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Steel Part 1

Learning Objectives

You

will

learn• the differences between ASD and LRFD design methods

• how to design steel members for flexure, shear, and compression

3

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Steel Part 1

Prerequisite Knowledge

You

should

already

be

familiar

with• fundamentals of mechanics of materials

• structural analysis

• material properties of steel

• fundamentals of steel design

4

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Steel Part 1

Referenced Codes and Standards

•

Steel Construction

Manual (AISC,

2011)• Seismic Design Manual (AISC, 2012)

• Specification for Structural Steel Buildings (AISC 360, 2010)

• Minimum Design Loads for Buildings and Other Structures (ASCE/SEI7, 2010)

• International Building Code (IBC, 2012)

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Steel Part 1

Introduction: Design Principles

allowable

strength

design

(ASD)

method

• required strength due to working loads, not to exceed the allowable strength

• factors of safety, Ω

load and resistance factor design (LRFD) method

• required strength due to factored loads, not to exceed the design strength

• resistance factors,

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Steel Part 1

Poll Question: Load Factors

Which

of

the

following

do

LRFD

load

factors

account

for?(A) variability of anticipated loads

(B) errors in design methods and computations

(C) lack of understanding of material behavior

(D) all of the above

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Steel Part 1

Poll Question: Load Factors

Which

of

the

following

do

LRFD

load

factors

account

for?(A) variability of anticipated loads

(B) errors in design methods and computations

(C) lack of understanding of material behavior

(D) all of the above

The answer is (A).

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Steel Part 1

Load Combinations

D dead loads kips or kips/ft

E earthquake load kips or kips/ft

H load due to lateral pressure kips/ft2

L live loads due to occupancy kips or kips/ft

Lr roof live load kips or kips/ft

Q load effect produced by

service load

kips

R load due to rainwater or ice kips or kips/ft

Rn nominal strength kips

S snow load kips or kips/ft

U required strength to

resist factored loads

kips

W wind load kips or kips/ft

γ load factor ‐

resistance factor ‐

Ω safety factor ‐

nomenclature

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Steel Part 1

Load Combinations: LRFD Required Strength

LRFD Required Strength, ∑γQ

• consists of the most critical combination

of factored loads applied to the member

• defined by the seven combinations given

in IBC Sec. 1605.2.1

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Steel Part 1

Load Combinations: ASD Required Strength

ASD Required Strength, ∑γQ

• consists of the most critical combination

of factored loads applied to the member

• defined by the nine combinations given

in IBC Sec. 1605.3.1

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Steel Part 1

Example: LRFD vs. ASD Required Strength

Example 4.1

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Steel Part 1

• nominal strength identical for ASD and LRFD methods

• nominal strength of member = theoretical ultimate strength, per AISC 360 provisions

• member loaded in tension

P n = F y A g

• compact braced beam in flexure

M n = M p = F y Z x

• column in compression

P n = F cr A g

Nominal Strength

AISC 360 Eq. D2‐1

AISC 360 Eq. F2‐1

AISC 360 Eq. E3‐1

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Steel Part 1

LRFD Design Strength vs. ASD Allowable Strength

LRFD design strength [AISC 360 Sec. B3] ASD allowable strength [AISC 360 Sec. B3]

17

l l ( )

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Steel Part 1

Example: LRFD Design Strength

A pin‐ended column of grade A50 steel

and an unbraced length of 10 ft is

subjected to a LRFD factored axial load of

∑γQ = 555 kips. Determine the lightest

adequate W10 shape.

18

St t l E i i E R i C St l (P t 1)

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Steel Part 1

Example: LRFD Design Strength



subjected to an LRFD factored axial load

of ∑γQ = 555 kips. Determine the lightest

adequate W10 shape.

Solution

From AISC Manual Table 4‐1, for an

unbraced length of 10 ft, a W10 x 54

column provides the design axial

strength.

605 kips [satisfactory]n c n R P Q 19


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Steel Part 1

Example: ASD Allowable Strength



subjected to an ASD factored axial load of


adequate W10 shape.

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Steel Part 1

Example: ASD Allowable Strength



subjected to an ASD factored axial load of


adequate W10 shape.

Solution

From AISC Manual Table 4‐1, for an

effective height of 10 ft, a W10 x 54

column provides the axial strength to

support the stated load of 370 kips.

403 kips [satisfactory]n n R P

Q 21

Reproduced from Steel Construction Manual , Fourteenth ed., 2012.

American Institute of Steel Construction, Inc., Chicago, IL.


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Steel Part 1

Design for Flexure: Plastic Moment of Resistance

• yield reached and residual stresses ignored,

applied moment is

M y = SF y

• applied moment at first yielding when

residual stresses are accounted for

M r = 0.7 F yS

• plastic hinge formed, nominal strength is

M n = M p = ZF y

• The shape factor is defined as

Figure 4.16 Plastic Moment of Resistance

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Steel Part 1

Nominal Flexural Strength

• plastic moment strength: M n ≤ M p

• flange local buckling

• web local buckling

• lateral‐torsional buckling

• lateral‐torsional buckling modification factor

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g g ( )



Steel Part 1

Compact, Noncompact, and Slender Sections

• compact section, λ ≤ λ pf

M n = M p

• noncompact section, λ pf ≤ λ ≤ λrf

0.7 F yS x ≤ M n < M p

• slender section, λ < λrf

M n < 0.7 F yS x

Flange local buckling

Figure 4.2 Variation of Mn with λ

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Steel Part 1


AISC 360 Table B4.1b Width‐to‐Thickness Ratios: Compression Elements Members Subject to Flexure


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Steel Part 1


26

AISC 360 Table B4.1b Width‐to‐Thickness Ratios: Compression Elements Members Subject to Flexure


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Steel Part 1

Lateral Support (Assuming C b = 1.0)

• maximum nominal moment capacity of

compact rolled I‐shape, M n = M p

• nominal flexural strength decreases with

increasing Lb

• three phases: plastic hinging, inelastic

buckling, and elastic buckling

27

Figure 4.3 Variation of Mn with Lb for C b = 1.0


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Steel Part 1

Example: Lateral Support (Assuming C b = 1.0)

Example 4.5

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Steel Part 1


Example 4.5

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Steel Part 1


Example 4.5

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Steel Part 1

• Mr = 0.7F y S x

• linear interpolation between M p and Mr

Inelastic Phase, L p < Lb ≤Lr

AISC 360 Eq. F2‐2


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Steel Part 1

LRFD method

ASD method

Values of BF are tabulated in AISC Manual

Part 3

Inelastic Phase, L p < Lb ≤Lr


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Steel Part 1

Example: Inelastic Phase, L p < Lb ≤Lr Example 4.6

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Steel Part 1


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Steel Part 1

Elastic Phase, Lb > Lr elastic lateral‐torsional buckling

Figure 4.3 Variation of Mn with Lb for C b = 1.0 AISC 360 Eq. F2‐3

AISC 360 Eq. F2‐4

36

n cr x

p

F S


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Steel Part 1

Example: Elastic Phase, Lb > Lr Example 4.7

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Steel Part 1

Example: Elastic Phase, Lb > Lr Example 4.7

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Steel Part 1

C b accounts for influence of moment gradient

on lateral‐torsional buckling of beam.

Lateral‐Torsional Buckling Modification Factor, C b

Figure 4.4 Determination of C b

Figure 4.5 Typical Values of C b

AISC 360 Eq. F1‐1

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Steel Part 1

Example: Determination of C bExample 4.8

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Steel Part 1

Example: Determination of C b

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Steel Part 1

Example: Variation of Mn with Lb for C b > 1.0

Example 4.9

44

The beam is braced only at the supports.


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Steel Part 1

Moment Redistribution in Continuous Beams

• redistribution of bending moment [AISC 360 Sec. B3.7]

• negative moments at supports reduced by 10%, positive moments increased by 10%

of average adjacent support moments

• not permitted if

• F y > 65 ksi

• axial force exceeds 0.15ϕc F y A g [LRFD]

• axial force exceeds 0.15 F y A g /Ωc [ASD]

49


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Steel Part 1

Example: Continuous Beams

Example 4.10

50


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Steel Part 1


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Steel Part 1

Biaxial Bending

A beam subjected to bending moments about both the x and y‐axes may be designed by

using the following interaction expressions.

For beams without axial loads, AISC 360 Eq. H1‐1b reduces to the equations above.

53


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Steel Part 1

Example: Biaxial Bending

Example 4.11

54


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Steel Part 1

Example: Biaxial Bending (LRFD)

Example 4.11

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Steel Part 1

Example: Biaxial Bending (ASD)

Example 4.11

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Steel Part 1

Shear in Beam Webs

• shear stress uniformly distributed over area

of the web

• nominal shear strength governed by

yielding of the web, provided that

• nominal shear strength

• design shear strength

• allowable shear strength

AISC 360 Eq. G2‐1

57


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Steel Part 1

Example: Shear in Beam Webs

Example 4.12

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Steel Part 1


Example 4.12

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Steel Part 1


Example 4.12

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Steel Part 1

Block Shear

• nominal resistance to block shear

• reduction coefficient

• See AISC Spec. Commentary Fig. C‐J4.2

for additional illustrations for Ubs values.

• resistance factor,

• safety factor,

Figure 4.7 Block Shear in a Coped Beam AISC

360

Eq.

J4

‐5

61


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Steel Part 1

Example: Block Shear

Determine the resistance to block shear of the

coped

W16

x

40

grade

A36

beam

shown.

The

relevant dimensions are l h = l v = 1.5 in and

s = 3 i n. The bolt holes are standard and the bolt

diameter is ½ in.

62


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Steel Part 1


The hole diameter for a ½ in diameter bolt is

defined

in

AISC

360

Sec.

B4.3b

and

Table J3.3

as1 1 1

in in in8 2 8

0.625 in

h Bd d

2

0.5 1.5 in 0.5 0.625 in

1.1875 in

nt w h h w

w

A t l d t

t

0.305 inwt

2

2 2.5 1.5 in 2 3.0 in 2.5 0.625 in

5.9375 in

nv w v h w

w

A t l s d t

t

63


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Steel Part 1


The tensile stress is uniform, and the reduction

coefficient

is

1.0bsU

2

2

kips58 1.1875 in

in

68.875 kips

bs u nt w

w

U F A t

t

2

2

kips0.6 0.6 58 5.9375 in

in

206.625 kips

u nv w

w

F A t

t

64


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Steel Part 1


2

2

1.5 in 2 3 in

7.5 in

gv w v

w

w

A t l s

t

t

2

2

kips0.6 0.6 36 7.5 in

in

162 kips

0.6 [governs]

gv w

w

u nv

F A t

t

F A

65


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Steel Part 1


LRFD Method

Shear yielding governs, and the design

strength for block shear is given by AISC 360

Eq. J4‐5 as

ASD Method

Shear yielding governs, and the allowable

strength for block shear is given by AISC 360

Eq. J4‐5 as

0.6

kips kips

0.75 0.305 in 162 68.875in in

52.81kips

n y gv bs u nt R F A U F A

0.6

Ω Ω

kips kips0.305 in 162 68.875in in

2

35.21 kips

y gv bs u nt n F A U F A R

66


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Steel Part 1

Web Local Yielding

• bearing plate used to distribute

concentrated loads to prevent local web yielding

• capacity if load applied < d from end

of beam

• capacity if load applied > d from end

of beam

AISC 360 Eq. J10 ‐3

AISC 360 Eq. J10 ‐2

Figure

4.8 Web Local Yielding

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Steel Part 1

Web Local Yielding

• Design strength is given by ϕ Rn with

ϕ = 1.0.

• AISC Manual Table 9‐4 tabulates

• Allowable strength is given by Rn/Ω

with Ω = 1.5.

• AISC Manual Table 9‐4 tabulates

Figure

4.8 Web Local

Yielding

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Steel Part 1

Example: Web Local Yielding

Example 4.14

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Steel Part 1


Example 4.14

70




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Steel Part 1


Example 4.14

71




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Steel Part 1

Web Crippling

• nominal strength for load applied > d /2 from end of beam

• nominal strength for load applied < d /2 from end of beam and for l b/d ≤ 0.2

• nominal strength for load applied < d /2 from end of beam and for l b/d > 0.2

AISC 360 Eq. J10 ‐4

AISC 360 Eq. J10 ‐5a

AISC 360 Eq. J10 ‐5b

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Steel Part 1

Web Crippling

• Design strength is given by ϕr Rn, with ϕr = 0.75.

• Allowable strength is given by Rn/Ωr , with Ωr = 2.00.





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Steel Part 1

Example: Web Crippling

Example 4.15

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Steel Part 1

Example: Web Crippling

Example 4.15

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Steel Part 1

Poll Question: Effective Length

Assuming that columns (a), (b), and (c)

differ

only

in

the

conditions

of

restraints,

which column can carry the highest axial

compressive load without buckling? The

lowest?

(a) (b) (c)

76


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Steel Part 1

Poll Question: Effective Length

Assuming that columns (a), (b), and (c)

differ

only

in

the

conditions

of

restraints,

which column can carry the highest axial

compressive load without buckling? The

lowest?

• highest – (c)

• lowest – (a)

(a) (b) (c)

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Steel Part 1

Effective Length

• The effective length factor, K , is used to

account

for

the

influence

of

restraint

conditions at each end of a column.

• The available strength of an axially

loaded column depends on the

slenderness ratio, KL/r . [AISC 360

Sec. E2]

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Steel Part 1

Alignment Charts

• For compression members forming part

of a frame with rigid joints, AISC 360

Commentary App. 7.2 presents alignment

charts for determining the effective

length.

• To use alignment charts, calculate

stiffness ratio at the two ends of the

column.

Figure 4.10 Alignment Charts for Effective Length Factors

Adapted from American Institute of Steel Construction, Specifications for Structural

Steel Buildings, Commentary Fig. C‐A‐7.1 and Fig. C‐A‐7.2.

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Steel Part 1

Example: Effective Length

Example 4.16

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Steel Part 1

Example: Effective Length

Example 4.16

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Steel Part 1

Axially Loaded Members

• design strength in compression

• allowable strength in compression

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Steel Part 1


short column

• Inelastic buckling governs if KL/r ≤ 4.71( E / F y)0.5 or F y/ F e ≤ 2.25.

• critical stress

• λ is defined by

• elastic critical buckling stress

AISC 360 Eq. E3‐2

AISC 360 Eq. E3‐4

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Steel Part 1


long column

• Elastic buckling governs if KL/r ≤ 4.71( E / F y)0.5 or F y/ F e ≤ 2.25.

• critical stress

• Once the governing slenderness ratio of a column is established, the available critical

stress may be obtained from AISC Manual Table 4‐22.

AISC 360 Eq. E3‐3

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Steel Part 1

Example: Buckling around Minor Axis

Example 4.17

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Steel Part 1

Example: Buckling around Minor Axis

Example 4.17

86




l

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Steel Part 1

Poll Question:

Is the following statement true or false?

A compression member always buckles around the minor axis.

(A) true

(B) false

87


S l P 1

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Steel Part 1

Poll Question:

Is the following statement true or false?

A compression member always buckles around the minor axis.

(A) true

(B) false

The answer is (B) false.

88


St l P t 1

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Steel Part 1

Equivalent Effective Length

• Calculate slenderness ratios ( KL/r ) x and ( KL/r ) y. The larger ratio will control the

design.

• Divide the effective length about the x‐axis by the ratio r x/r y, to obtain an equivalent

effective length about the y‐axis, ( KL) x/(r x/r y).

• Use AISC Manual Table 4‐1 to obtain the available design strength using the

equivalent length value, ( KL) x/(r x/r y).

89


Steel Part 1

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Steel Part 1

Example: Buckling around Major Axis

Example 4.18

90

The unbraced length along the minor axis is 6 ft.


Steel Part 1

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Example 4.18

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Example 4.18

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Steel Part 1

Other Steel Compression Members

AISC Manual Table 4‐22 is useful for non‐

standard compression members, such as

built‐up sections and laced compression

members.

AISC Manual Table 4‐22 tabulates λc F cr and

F cr / Ωc against KL/r for steel with yield

stresses of 35 kips/in2, 36 kips/in2, 42

kips/in2, 46 kips/in2, and 50 kips/in2,

respectively.

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Steel Part 1

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Steel Part 1

Example: Built‐Up Sections

A laced column consisting of four 6 6

½ angles of A572 grade 50 steel is shown.

The column is 30 ft high with fixed ends

and may be considered a single integral

member. Determine the maximum design

axial load.

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Steel Part 1

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Steel Part 1


A laced column consisting of four 6 6

½ angles of A572 grade 50 steel is shown.

The column is 30 ft high with fixed ends

and may be considered a single integral

member. Determine the maximum design

axial load.

95

The relevant properties of a 6 6 ½ angle

are

The relevant properties of a laced column are

2

2

5.77 in

19.9 in

1.67 in

A

I

y

2

2

4 4 5.77 in23.08 in

A A

2

24 2

4

42

4 19.9 in 23.08 in 15 in 1.67 in

4181in

d I I A y

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Steel Part 1

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Steel Part 1


LRFD Method

From AISC

Manual Table

4‐22,

the

design

stress is

The design axial strength is

244.1 kips / inc cr F

2

2

kips44.1 23.08 in

in

1018 kips

c n c cr P F A

97




Steel Part 1

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Stee a t


ASD Method

From AISC

Manual Table

4‐22,

the

allowable

stress is

The allowable axial strength is

229.3 kips / incr F

2

2

kips29.3 23.08 in

in

676 kips

cr n

c c

F A P

98




Steel Part 1

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Composite Columns

• Concrete‐filled hollow structural sections and concrete‐encased rolled steel sections

are reinforced with longitudinal and lateral reinforcing bars designed by using

AISC 360 Sec. I2.

• Design axial strength values for typical column sizes are tabulated in AISC Manual

Part 4.

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Example: Composite Columns

Example 4.20

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Example 4.20

101




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Example 4.20

102




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Second‐Order Effects

• Secondary moments and axial forces

caused by P‐delta effects must be added to

the primary moments and axial forces.

• P‐δ effect

• amplified moment due to

eccentricity, member effect

• moment magnification factor, B1

• P‐Δ effect

• amplified moment due to drift, frame

effect

• moment magnification factor, B2

Figure 4.11 P‐delta Effects

103


Steel Part 1

d d ff

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Second‐Order Effects

B1‐B2 procedure

•

final forces obtained as the summation of the two analyses, sway

and non‐sway

Figure 4.12 Determination of Secondary Effects

AISC 360 Eq. A‐8 ‐1

AISC 360 Eq. A‐8 ‐2

104


Steel Part 1

M ifi i F B

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Magnification Factor B1

• From ACI 360 Eq. A‐8‐3,

• α = force level adjustment factor

= 1.0 for LRFD load combinations

= 1.6 for ASD load combinations

• for member not loaded transversely

• P e1 = Euler buckling strength in the plane

of bending

= π2 EI /( K 1 L)2

• P r = required second‐order axial strength

105


Steel Part 1

M ifi ti F t B

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Magnification Factor B2

• B2 multiplier for each story and each

direction of lateral translation

α = force level adjustment factor

= 1.0 for LRFD load combinations

= 1.6 for ASD load combinations

• elastic critical buckling strength for the

story in the direction of translation

• H and ΔH may be based on any lateral

loading that provides a representative

value of story lateral stiffness, H/ΔH.

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e story M

H

HL P R

ACI 360 Eq. A‐8 ‐7 ACI 360 Eq. A‐8 ‐6


Steel Part 1

A l i M th d f S d Eff t

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Analysis Methods for Secondary Effects

• effective length method [AISC 360 App. 7.2]

•

second‐order elastic analysis [AISC 360 App. 8]

• direct analysis method [AISC 360 Sec. C2 and C3]

• first‐order elastic analysis [AISC 360 App. 7.3]

• simplified method [ AISC Manual Part 2]

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Steel Part 1

Effective Length Method

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Effective Length Method

• This method is restricted to structures with a sidesway amplification factor of

• Notional lateral loads that are applied at each story are given by

• The notional loads are applied solely in gravity‐only load combinations.

• Use the appropriate effective length factor, K .

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Steel Part 1

Example: Effective Length Method

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Example: Effective Length Method

Example 4.21

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Steel Part 1

Example: Effective Length Method (LRFD)

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Example 4.21

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Steel Part 1


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Steel Part 1

Example: Effective Length Method (ASD)

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Steel Part 1


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Steel Part 1


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Steel Part 1

Direct Analysis Method

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Direct Analysis Method

• must be used when

• structure analyzed using reduced flexural and axial stiffnesses

• stiffness reduction coefficient

• additive notional loads applied to

each

story,

,

when

• available strength of members

determined using an effective length

factor of K = 1.0

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Steel Part 1

Example: Direct Analysis Method

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Example: Direct Analysis Method

Example 4.22

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Steel Part 1

Example: Direct Analysis Method (LRFD)

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p y ( )

Example 4.22

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Steel Part 1


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p y ( )

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Steel Part 1


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p y ( )

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Steel Part 1

Example: Direct Analysis Method (ASD)

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Example 4.22

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Steel Part 1Example: Direct Analysis Method (ASD)

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STRC ©2015 Professional Publications, Inc. 122


Steel Part 1Example: Direct Analysis Method (ASD)

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Structural Engineering

Exam

Review

Course Steel

(Part

1)

Steel Part 1First‐Order Elastic Analysis

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• restricted to structures with

• additive notional loads applied at each

story

• limit of required axial compressive

strength

• available strength of members

determined using an effective length

factor of K = 1.0, with some exceptions to columns in moment frames [see

AISC Table 2‐2]

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Exam

Review

Course Steel

(Part

1)

Steel Part 1Example: First‐Order Elastic Analysis

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Example 4.23

125


Exam

Review

Course Steel

(Part

1)

Steel Part 1Example: First‐Order Elastic Analysis (LRFD)

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Example 4.23

126


Exam

Review

Course Steel

(Part

1)


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Structural

Engineering

Exam

Review

Course Steel

(Part

1)


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Structural

Engineering

Exam

Review

Course Steel

(Part

1)

Steel Part 1Example: First‐Order Elastic Analysis (ASD)

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Example 4.23

129

Structural

Engineering

Exam

Review

Course Steel

(Part

1)


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Structural

Engineering

Exam

Review

Course Steel

(Part

1)


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Steel Part 1Simplified Method

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simplified method

• quick, conservative way to determine

member size (final designs should use

more rigorous second‐order analyses)

• use only when

• second‐order analysis not required

• nominal stiffness of members used in

analysis with no reduction for inelastic

softening effects

• ratio

Table 4.1 Amplification Factor B2 for Use with the

Simplified Method

Copyright © American Institute of Steel Construction. Reproduced with

permission. All rights reserved.

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Steel Part 1Example: Simplified Method

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Example 4.24

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Steel Part 1Example: Simplified Method (LRFD)

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Example 4.24

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Steel Part 1Example: Simplified Method (LRFD)

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Simplified Method



135


Steel Part 1Example: Simplified Method (ASD)

E l 4 24

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Example 4.24

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Steel Part 1Example: Simplified Method (ASD)

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Simplified Method



137


Steel Part 1Combined Compression and Flexure

• For P /P ≥ 0 2

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For P r / P c ≥ 0.2,

• For P r / P c < 0.2,

• Values of p, b x, and b y are tabulated in AISC Manual Table 6‐1 for W shapes with a

yield stress of 50 kips/in2 and assuming a bending coefficient of C b=1.0.

AISC 360 Eq. H1‐1a

AISC 360 Eq. H1‐1b

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Steel Part 1Example: Combined Compression and Flexure

Example 4 25

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Example 4.25

139



Example 4.25

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Example 4.25

140



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Steel Part 1Column Base Plates

required base plate thickness is given by Figure 4.14 Column Base Plate

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the largest of Figure 4.14 Column Base Plate

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Steel Part

1

Example: Column Base Plates

Example 4.26

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Steel Part

1


Example 4.26

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Steel Part

1


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Steel Part

1

Learning Objectives

You have learned

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• the differences between ASD and LRFD design methods

• how to design steel members for flexure, shear, and compression

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Steel Part

1

Lesson Overview

Chapter 4: Structural Steel Design

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• Introduction

• Load Combinations

• Design for Flexure

• Design for Shear

• Design of Compression Members

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STRC05 Steel Part1 0715

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Transcript of STRC05 Steel Part1 0715