Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1 Structural Engineering Review Course
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Lesson Overview Chapter 4: Structural Steel Design • Introduction • Load Combinations • Design for Flexure • Design for Shear • Design of Compression Members
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Structural Engineering Exam Review Course
Steel (Part 1)
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
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Structural Engineering Exam Review Course
Steel (Part 1)
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
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Steel (Part 1)
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)
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)
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)
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)
Steel Part 1
Load Combinations nomenclature D
dead loads
kips or kips/ft
Rn
nominal strength
kips
E
earthquake load
kips or kips/ft
S
snow load
kips or kips/ft
H
load due to lateral pressure
kips/ft2
U
kips
L
live loads due to occupancy
kips or kips/ft
required strength to resist factored loads
W
wind load
kips or kips/ft
Lr
roof live load
kips or kips/ft
γ
load factor
‐
Q
load effect produced by service load
kips
߶
resistance factor
‐
R
load due to rainwater or ice
kips or kips/ft
Ω
safety factor
‐
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Steel (Part 1)
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)
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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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: LRFD vs. ASD Required Strength Example 4.1
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Steel (Part 1)
Steel Part 1
Example: LRFD vs. ASD Required Strength
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Steel (Part 1)
Steel Part 1
Example: LRFD vs. ASD Required Strength
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Steel (Part 1)
Steel Part 1
Example: LRFD vs. ASD Required Strength
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Steel (Part 1)
Steel Part 1
Nominal Strength • nominal strength identical for ASD and LRFD methods • nominal strength of member = theoretical ultimate strength, per AISC 360 provisions • member loaded in tension Pn = FyAg
AISC 360 Eq. D2‐1
• compact braced beam in flexure Mn = Mp = FyZx
AISC 360 Eq. F2‐1
• column in compression Pn = FcrAg
AISC 360 Eq. E3‐1
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Steel (Part 1)
Steel Part 1
LRFD Design Strength vs. ASD Allowable Strength LRFD design strength [AISC 360 Sec. B3]
ASD allowable strength [AISC 360 Sec. B3]
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Steel (Part 1)
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.
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Steel (Part 1)
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 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. Rn c Pn 605 kips Q [satisfactory] STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Example: ASD Allowable Strength A pin‐ended column of grade A50 steel and an unbraced length of 10 ft is subjected to an ASD factored axial load of ∑γQ = 370 kips. Determine the lightest adequate W10 shape.
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Steel (Part 1)
Steel Part 1
Example: ASD Allowable Strength A pin‐ended column of grade A50 steel and an unbraced length of 10 ft is subjected to an ASD factored axial load of ∑γQ = 370 kips. Determine the lightest 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. Rn Pn 403 kips Q [satisfactory]
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
Steel Part 1
Design for Flexure: Plastic Moment of Resistance •
yield reached and residual stresses ignored, applied moment is Figure 4.16 Plastic Moment of Resistance
My = SFy •
applied moment at first yielding when residual stresses are accounted for Mr = 0.7FyS
•
plastic hinge formed, nominal strength is Mn = Mp = ZFy
•
The shape factor is defined as
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Steel (Part 1)
Steel Part 1
Nominal Flexural Strength • plastic moment strength: Mn ≤ Mp • flange local buckling • web local buckling • lateral‐torsional buckling • lateral‐torsional buckling modification factor
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Steel (Part 1)
Steel Part 1
Compact, Noncompact, and Slender Sections • compact section, λ ≤ λpf
Flange local buckling
Mn = Mp
• noncompact section, λpf ≤ λ ≤ λrf 0.7FySx ≤ Mn < Mp
• slender section, λ < λrf
Figure 4.2 Variation of Mn with λ
Mn < 0.7FySx
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Steel (Part 1)
Steel Part 1
Compact, Noncompact, and Slender Sections AISC 360 Table B4.1b Width‐to‐Thickness Ratios: Compression Elements Members Subject to Flexure
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Compact, Noncompact, and Slender Sections AISC 360 Table B4.1b Width‐to‐Thickness Ratios: Compression Elements Members Subject to Flexure
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Steel (Part 1)
Steel Part 1
Lateral Support (Assuming Cb = 1.0) • maximum nominal moment capacity of compact rolled I‐shape, Mn = Mp
Figure 4.3 Variation of Mn with Lb for Cb = 1.0
• nominal flexural strength decreases with increasing Lb • three phases: plastic hinging, inelastic buckling, and elastic buckling
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Lateral Support (Assuming Cb = 1.0) Example 4.5
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Lateral Support (Assuming Cb = 1.0) Example 4.5
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Steel (Part 1)
Steel Part 1
Example: Lateral Support (Assuming Cb = 1.0) Example 4.5
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Steel (Part 1)
Steel Part 1
Inelastic Phase, Lp < Lb ≤ Lr • Mr = 0.7FySx • linear interpolation between Mp and Mr
Figure 4.3 Variation of Mn with Lb for Cb = 1.0
AISC 360 Eq. F2‐2
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Steel (Part 1)
Steel Part 1
Inelastic Phase, Lp < Lb ≤ Lr LRFD method Figure 4.3 Variation of Mn with Lb for Cb = 1.0
ASD method
Values of BF are tabulated in AISC Manual Part 3 STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Example: Inelastic Phase, Lp < Lb ≤ Lr Example 4.6
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Steel (Part 1)
Steel Part 1
Example: Inelastic Phase, Lp < Lb ≤ Lr Example 4.6
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Steel (Part 1)
Steel Part 1
Example: Inelastic Phase, Lp < Lb ≤ Lr Example 4.6
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Steel (Part 1)
Steel Part 1
Elastic Phase, Lb > Lr elastic lateral‐torsional buckling M n Fcr S x Mp
AISC 360 Eq. F2‐3
Figure 4.3 Variation of Mn with Lb for Cb = 1.0
AISC 360 Eq. F2‐4
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Elastic Phase, Lb > Lr Example 4.7
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Elastic Phase, Lb > Lr Example 4.7
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Steel (Part 1)
Steel Part 1
Lateral‐Torsional Buckling Modification Factor, Cb Cb accounts for influence of moment gradient on lateral‐torsional buckling of beam.
Figure 4.5 Typical Values of Cb
AISC 360 Eq. F1‐1 Figure 4.4 Determination of Cb
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Determination of Cb Example 4.8
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Determination of Cb
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Determination of Cb
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Determination of Cb
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Steel (Part 1)
Steel Part 1
Example: Variation of Mn with Lb for Cb > 1.0 Example 4.9
The beam is braced only at the supports.
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Steel (Part 1)
Steel Part 1
Example: Variation of Mn with Lb for Cb > 1.0
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Steel (Part 1)
Steel Part 1
Example: Variation of Mn with Lb for Cb > 1.0
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Variation of Mn with Lb for Cb > 1.0
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Structural Engineering Exam Review Course
Steel (Part 1)
Steel Part 1
Example: Variation of Mn with Lb for Cb > 1.0
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Steel (Part 1)
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 •
Fy > 65 ksi
• axial force exceeds 0.15ϕcFyAg • axial force exceeds 0.15FyAg/Ωc
[LRFD] [ASD]
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Steel (Part 1)
Steel Part 1
Example: Continuous Beams Example 4.10
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Steel (Part 1)
Steel Part 1
Example: Continuous Beams
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Steel (Part 1)
Steel Part 1
Example: Continuous Beams
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Steel (Part 1)
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.
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Steel (Part 1)
Steel Part 1
Example: Biaxial Bending Example 4.11
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Steel (Part 1)
Steel Part 1
Example: Biaxial Bending (LRFD) Example 4.11
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Steel (Part 1)
Steel Part 1
Example: Biaxial Bending (ASD) Example 4.11
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Steel (Part 1)
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
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Steel (Part 1)
Steel Part 1
Example: Shear in Beam Webs Example 4.12
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Steel (Part 1)
Steel Part 1
Example: Shear in Beam Webs Example 4.12
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Steel (Part 1)
Steel Part 1
Example: Shear in Beam Webs Example 4.12
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Steel (Part 1)
Steel Part 1
Block Shear • nominal resistance to block shear AISC 360 Eq. J4‐5
Figure 4.7 Block Shear in a Coped Beam
• reduction coefficient
• See AISC Spec. Commentary Fig. C‐J4.2 for additional illustrations for Ubs values. • resistance factor, • safety factor, STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
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 lh = lv = 1.5 in and s = 3 in. The bolt holes are standard and the bolt diameter is ½ in.
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Steel (Part 1)
Steel Part 1
Example: Block Shear The hole diameter for a ½ in diameter bolt is defined in AISC 360 Sec. B4.3b and Table J3.3 as 1 1 1 d h d B in in in 8 2 8 0.625 in
tw 0.305 in Anv tw lv 2 s 2.5d h tw 1.5 in 2 3.0 in 2.5 0.625 in 5.9375 in 2 tw
Ant tw lh 0.5d h tw 1.5 in 0.5 0.625 in 1.1875 in 2 tw
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Steel (Part 1)
Steel Part 1
Example: Block Shear The tensile stress is uniform, and the reduction coefficient is U bs 1.0 kips U bs Fu Ant 58 2 1.1875 in 2 tw in 68.875 kips tw
kips 0.6 Fu Anv 0.6 58 5.9375 in 2 tw 2 in 206.625 kips tw
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Steel (Part 1)
Steel Part 1
Example: Block Shear Agv tw lv 2 s tw 1.5 in 2 3 in 7.5 in 2 tw
kips 0.6 Fy Agv 0.6 36 7.5 in 2 tw 2 in 162 kips tw 0.6 Fu Anv [governs]
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Steel (Part 1)
Steel Part 1
Example: Block Shear LRFD Method
ASD Method
Shear yielding governs, and the design strength for block shear is given by AISC 360 Eq. J4‐5 as
Shear yielding governs, and the allowable strength for block shear is given by AISC 360 Eq. J4‐5 as
Rn 0.6 Fy Agv U bs Fu Ant
Rn 0.6 Fy Agv U bs Fu Ant Ω Ω kips kips 68.875 0.305 in 162 in in 2 35.21 kips
kips kips 0.75 0.305 in 162 68.875 in in 52.81 kips
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Steel (Part 1)
Steel Part 1
Web Local Yielding • bearing plate used to distribute concentrated loads to prevent local web yielding
Figure 4.8 Web Local Yielding
• capacity if load applied < d from end of beam AISC 360 Eq. J10‐3
• capacity if load applied > d from end of beam AISC 360 Eq. J10‐2
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Steel (Part 1)
Steel Part 1
Web Local Yielding • Design strength is given by ϕRn with ϕ = 1.0.
Figure 4.8 Web Local Yielding
• AISC Manual Table 9‐4 tabulates
• Allowable strength is given by Rn/Ω with Ω = 1.5. • AISC Manual Table 9‐4 tabulates
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Steel (Part 1)
Steel Part 1
Example: Web Local Yielding Example 4.14
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Steel (Part 1)
Steel Part 1
Example: Web Local Yielding Example 4.14
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
Steel Part 1
Example: Web Local Yielding Example 4.14
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
Steel Part 1
Web Crippling • nominal strength for load applied > d/2 from end of beam AISC 360 Eq. J10‐4
• nominal strength for load applied < d/2 from end of beam and for lb/d ≤ 0.2 AISC 360 Eq. J10‐5a
• nominal strength for load applied < d/2 from end of beam and for lb/d > 0.2 AISC 360 Eq. J10‐5b
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Steel (Part 1)
Steel Part 1
Web Crippling • Design strength is given by ϕrRn, with ϕr= 0.75. AISC 360 Eq. J10‐5a AISC 360 Eq. J10‐5b
• Allowable strength is given by Rn/Ωr, with Ωr = 2.00. AISC 360 Eq. J10‐5a AISC 360 Eq. J10‐5b
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Steel (Part 1)
Steel Part 1
Example: Web Crippling Example 4.15
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Steel (Part 1)
Steel Part 1
Example: Web Crippling Example 4.15
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
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?
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(c)
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Steel (Part 1)
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)
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(c)
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Steel (Part 1)
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)
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
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Steel (Part 1)
Steel Part 1
Example: Effective Length Example 4.16
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Steel (Part 1)
Steel Part 1
Example: Effective Length Example 4.16
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Steel (Part 1)
Steel Part 1
Axially Loaded Members • design strength in compression
• allowable strength in compression
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Steel (Part 1)
Steel Part 1
Axially Loaded Members short column • Inelastic buckling governs if KL/r ≤ 4.71(E/Fy)0.5 or Fy/Fe ≤ 2.25. • critical stress AISC 360 Eq. E3‐2
• λ is defined by
• elastic critical buckling stress AISC 360 Eq. E3‐4
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Steel (Part 1)
Steel Part 1
Axially Loaded Members long column • Elastic buckling governs if KL/r ≤ 4.71(E/Fy)0.5 or Fy/Fe ≤ 2.25. • critical stress AISC 360 Eq. E3‐3
• Once the governing slenderness ratio of a column is established, the available critical stress may be obtained from AISC Manual Table 4‐22.
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Steel (Part 1)
Steel Part 1
Example: Buckling around Minor Axis Example 4.17
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Steel (Part 1)
Steel Part 1
Example: Buckling around Minor Axis Example 4.17
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
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
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Steel (Part 1)
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.
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Steel (Part 1)
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 rx/ry, to obtain an equivalent effective length about the y‐axis, (KL)x/(rx/ry). • Use AISC Manual Table 4‐1 to obtain the available design strength using the equivalent length value, (KL)x/(rx/ry).
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Steel (Part 1)
Steel Part 1
Example: Buckling around Major Axis Example 4.18
The unbraced length along the minor axis is 6 ft.
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Steel (Part 1)
Steel Part 1
Example: Buckling around Major Axis Example 4.18
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Steel (Part 1)
Steel Part 1
Example: Buckling around Major Axis Example 4.18
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Steel (Part 1)
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 λcFcr and Fcr/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)
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)
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.
The relevant properties of a 6 ൈ 6 ൈ ½ angle are A 5.77 in 2 I 19.9 in 2 y 1.67 in
The relevant properties of a laced column are
A 4 A 4 5.77 in 2
23.08 in 2
d I 4 I A 2 y
2
4 19.9 in 4 23.08 in 2 15 in 1.67 in
2
4181 in 4 STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Example: Built‐Up Sections A laced column consisting of four 6 ൈ 6 ൈ The radius of gyration of the laced column is ½ angles of A572 grade 50 steel is shown. I 4336 in 4 r 13.71 in The column is 30 ft high with fixed ends A 23.08 in 2 and may be considered a single integral member. Determine the maximum design The slenderness ratio of the laced column is axial load. in
0.65 in 30 ft 12 KL ft 2 r 13.71 in 17.07 200 [satisfactory]
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Steel (Part 1)
Steel Part 1
Example: Built‐Up Sections LRFD Method From AISC Manual Table 4‐22, the design stress is
c Fcr 44.1 kips / in 2 The design axial strength is
c Pn c Fcr A kips 44.1 2 23.08 in 2 in 1018 kips Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL. STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Example: Built‐Up Sections ASD Method From AISC Manual Table 4‐22, the allowable stress is Fcr 29.3 kips / in 2 The allowable axial strength is Pn Fcr A c c kips 23.08 in 2 29.3 2 in 676 kips Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL. STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
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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Steel (Part 1)
Steel Part 1
Example: Composite Columns Example 4.20
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Steel (Part 1)
Steel Part 1
Example: Composite Columns Example 4.20
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
Steel Part 1
Example: Composite Columns Example 4.20
Reproduced from Steel Construction Manual, Fourteenth ed., 2012. American Institute of Steel Construction, Inc., Chicago, IL.
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Steel (Part 1)
Steel Part 1
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
Figure 4.11 P‐delta Effects
P‐Δ effect •
amplified moment due to drift, frame effect
•
moment magnification factor, B2 STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Second‐Order Effects B1‐B2 procedure
Figure 4.12 Determination of Secondary Effects
• final forces obtained as the summation of the two analyses, sway and non‐sway AISC 360 Eq. A‐8‐1 AISC 360 Eq. A‐8‐2
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Steel (Part 1)
Steel Part 1
Magnification Factor B1 • From ACI 360 Eq. A‐8‐3,
• Pe1 = Euler buckling strength in the plane of bending = π2EI/(K1L)2
• α = force level adjustment factor
• Pr = required second‐order axial strength
= 1.0 for LRFD load combinations = 1.6 for ASD load combinations • for member not loaded transversely
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Steel (Part 1)
Steel Part 1
Magnification Factor B2 • B2 multiplier for each story and each direction of lateral translation ACI 360 Eq. A‐8‐6
α = force level adjustment factor
• elastic critical buckling strength for the story in the direction of translation Pe story RM
HL H
ACI 360 Eq. A‐8‐7
• H and ΔH may be based on any lateral loading that provides a representative value of story lateral stiffness, H/ΔH.
= 1.0 for LRFD load combinations = 1.6 for ASD load combinations
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Steel (Part 1)
Steel Part 1
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)
Steel Part 1
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)
Steel Part 1
Example: Effective Length Method Example 4.21
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (LRFD) Example 4.21
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (LRFD)
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (LRFD)
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (ASD)
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (ASD)
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Steel (Part 1)
Steel Part 1
Example: Effective Length Method (ASD)
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Steel (Part 1)
Steel Part 1
Direct Analysis Method • must be used when • structure analyzed using reduced flexural and axial stiffnesses
• additive notional loads applied to each story, , when • available strength of members determined using an effective length factor of K = 1.0
• stiffness reduction coefficient
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method Example 4.22
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (LRFD) Example 4.22
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (LRFD)
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (LRFD)
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (ASD) Example 4.22
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (ASD)
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Steel (Part 1)
Steel Part 1
Example: Direct Analysis Method (ASD)
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Steel (Part 1)
Steel Part 1
First‐Order Elastic Analysis • limit of required axial compressive strength
• restricted to structures with
• additive notional loads applied at each story • 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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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis Example 4.23
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (LRFD) Example 4.23
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (LRFD)
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (LRFD)
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (ASD) Example 4.23
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (ASD)
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Steel (Part 1)
Steel Part 1
Example: First‐Order Elastic Analysis (ASD)
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Steel (Part 1)
Steel Part 1
Simplified Method simplified method • quick, conservative way to determine member size (final designs should use more rigorous second‐order analyses)
Table 4.1 Amplification Factor B2 for Use with the Simplified Method
• use only when • second‐order analysis not required • nominal stiffness of members used in analysis with no reduction for inelastic softening effects • ratio
Copyright © American Institute of Steel Construction. Reproduced with permission. All rights reserved.
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Steel (Part 1)
Steel Part 1
Example: Simplified Method Example 4.24
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Steel (Part 1)
Steel Part 1
Example: Simplified Method (LRFD) Example 4.24
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Steel (Part 1)
Steel Part 1
Example: Simplified Method (LRFD) 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 1)
Steel Part 1
Example: Simplified Method (ASD) Example 4.24
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Steel (Part 1)
Steel Part 1
Example: Simplified Method (ASD) 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 1)
Steel Part 1
Combined Compression and Flexure • For Pr/Pc ≥ 0.2, AISC 360 Eq. H1‐1a
• For Pr/Pc < 0.2, AISC 360 Eq. H1‐1b
• Values of p, bx, and by 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 Cb=1.0. STRC ©2015 Professional Publications, Inc.
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Steel (Part 1)
Steel Part 1
Example: Combined Compression and Flexure Example 4.25
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Steel (Part 1)
Steel Part 1
Example: Combined Compression and Flexure Example 4.25
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Steel (Part 1)
Steel Part 1
Example: Combined Compression and Flexure
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Steel (Part 1)
Steel Part 1
Example: Combined Compression and Flexure
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Steel (Part 1)
Steel Part 1
Example: Combined Compression and Flexure
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Steel (Part 1)
Steel Part 1
Column Base Plates required base plate thickness is given by the largest of
Figure 4.14 Column Base Plate
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Steel (Part 1)
Steel Part 1
Example: Column Base Plates Example 4.26
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Steel (Part 1)
Steel Part 1
Example: Column Base Plates Example 4.26
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Steel (Part 1)
Steel Part 1
Example: Column Base Plates
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Steel (Part 1)
Steel Part 1
Learning Objectives You have learned • 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)
Steel Part 1
Lesson Overview Chapter 4: Structural Steel Design • Introduction • Load Combinations • Design for Flexure • Design for Shear • Design of Compression Members
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