Steel Structures - PB University Home Pageondemandweb.pbworld.net/.../lrfd_steel_structures_final_handout.pdf · 1 V 1.0 – Rev. 09.10.07 AASHTO- Load and Resistance Factor Design

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V 1.0 – Rev. 09.10.07

AASHTO- Load and Resistance Factor Design (LRFD)

Steel Structures

Mahir Sen, P.E. and Paul Pilarski, P.E.

Credits

The content for this course has been provided by the following PB employees:

Narrated by Greg Metzger, PB University

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• For your convenience and future reference, you may download a PDF version of this course.

• Click on the ATTACHMENTS link located in the upper right corner of this course window to access the document and save the file to your desktop.

Download Information

Course Navigation

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Throughout the course review questions are provided. These are designed to reinforce what you have learned in this course and help prepare you for the final assessment. The review scores will not count toward your final grade on the assessment.

Reinforcement

• After completing the content within the course you will be asked to take a final test to ensure that you mastered the key training objectives.

• You will need to make a minimum scoreof 80% on the final assessment to receive credit for passing the course.

• Successful completion of the course will earn 0.1 IACET CEU.

• Please refer to your state’s specific continuing education requirements regarding applicability.

Successful Completion

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This class is the fourth class in the Structures TRC curriculum for LRFD Design, developed internally at PB. The curriculum focuses on the following ten areas of major change introduced bythe LRFD Bridge Design Specifications:

• Foundations

• Decks & Deck Systems

• Joints and Bearings

• Abutments, Piers, and Walls

• Railings

• Introduction to LRFD

• Loads and Load Factors

• Concrete Structures

• Steel Structures

• Buried Structures

LRFD Design Curriculum

After successfully completing the course you will be able to identify:

• The three areas where the LRFD code changes past practice of steel structures design

• The basic proportioning of economic steel structures

• Applications of the LRFD design philosophy to basic steel structures

• The checks required in steel girder design

• The resources and alternate examples available to designers

The class will take you approximately one hour to complete.

Objectives

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• Present rules of thumb for economic construction and selection of steel plates

• Provide guidance on analysis procedures

• Walk through the basic implementation of LRFD

• Demonstrate checks required by the LRFD Specifications as applied to a simple span steel girder

• Present and explain an example problem

• Identify resources and alternate examples available to designers

The LRFD Steel Structures section covers a wide variety of applications. This course will give guidance on some basic principles in steel design and demonstrate LRFD by focusing on one of the most basic applications – a simple span composite steel girder bridge. The course will:

Course Description

1. LRFD Changes from Prior Specifications

2. Analysis Options

3. Proportioning Recommendations

4. Checks Required by the LRFD Specifications

5. LRFD Design of Simple Span Steel Girder

6. Resources

Course Outline

Course Outline- LRFD Steel Structures

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Topic 1

LRFD Changes from Prior Specifications

Topic 1

LRFD Changes

• LRFD has removed 25 ft. max. diaphragm spacing requirement

• Live Load deflection check is optional and per owner preference

• LRFD produces lower Live Load stress range

• LRFD requires higher number of load cycles


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LRFD Changes, cont.

• Simplified Fatigue Design Provisions

• LRFD is more reflective of actual conditions

• Unified Specs for Straight and Curved Steel Girders

• Parallel Commentary


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Review

Question Review

LL deflection check is optional

LRFD is more reflective of actual conditions

Shear design has been simplified

Maximum diaphragm spacing of 25 feet has been removed

The correct option to the question on the last slide is: Shear design has been simplified. Unfortunately, shear design has not been simplified. The other items listed have been implemented in LRFD.

Which of the following is not a change implemented in LRFD?

Course Outline


2. Analysis Options




6. Resources


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Topic 2

Analysis Options

Topic 2

• Model using beam + deck as composite section throughout length

• Larger negative & smaller positive moments

• Model using beam + deck reinforcement as composite section in negative moment region

• Larger positive & smaller negative moments

• LRFD 4.5.2.2 & commentary allow top model to be used.

2. Analysis Options

Analysis Options

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• Preliminary Design - Line Girder Analysis• Useful to investigate various span arrangements• Determine optimum girder depth

• Final Design• Line Girder Analysis

• Use for short, square bridges• Use for skew < 20 degrees

• Refined Analysis• Curved or severely skewed bridges• Flared bridge framing• Long or long span bridges

2. Analysis Options

Analysis Options

• Grid Analysis• 2-dimensional model

• Assumptions made regarding diaphragm stiffness

• Live loads distributed based on simple span distribution

• 3-D Finite Element Analysis• Actual cross-frame and deck stiffness

may be modeled

• Live load distribution based on refined stiffness

2. Analysis Options

Analysis Options – Refined Analysis

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1. LRFD Changes from prior Specifications

2. Analysis Options



5. LRFD Design of Simple Span Steel Bridge

6. Resources

Course Outline


Topic 3

Proportioning Recommendations

Topic 3

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• Minimum depths shown in AASHTO are often not the most economical (LRFD Table 2.5.2.6.3-1.)

• Use as wide a girder spacing as practical. Staged re-decking requirements may control girder spacing

• Some states limit fascia girder to 1 ft from gutterline for rolled beams and 2 ft from gutterline for plate girders. This is to accommodate contractor forms

• Limit field section lengths to 120 ft for shipping and erection

• Both flanges should meet NSBA rule of thumb for shipping and handling: bf > L/85


Recommendations

• Use constant width bottom flange for straight girders

• Use constant width top flange between field splices

• Economical to change flange section at field splices and where a minimum of 800 lbs of steel can be saved

• Locate bolted splices as near inflection points as possible

• Set –M region diaphragm spacing based on bottom flange under all loads

• Set +M region diaphragm spacing based on top flange under construction loads on non-composite beam section


Recommendations, cont.

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• Size stiffener widths in even inches – allows the use of bar stock

• Avoid longitudinal stiffeners

• Use 7/16” minimum web thickness to control welding distortions

• Use 3/4” minimum flange thickness


Recommendations, cont.

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Review

Question Review

Use a web thickness greater than 7/16”

Use 3/4” minimum compression flange thickness

Use a constant width bottom flange

Keep flange widths smaller than the shipping length/85

Which of the following is NOT a recommendation for an economical structure?

The option Keep flange widths smaller than the shipping length/85 is incorrect. The flange width should be kept greater than L/85. All of the other options are recommendations for economical steel I-girders.


2. Analysis Options




6. Resources

Course Outline


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Topic 4

Checks Required by the LRFD Specifications

Topic 4


LRFD Checks

A. Service Limit State Checks for• Elastic Deformations• Permanent Deformations

B. Constructability Checks for• Flexure• Shear

C. Fatigue and Fracture State Check for• Fatigue• Fracture• Special Fatigue Requirements for Web

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D. Strength Limit State Check for

• Flexure• Compact or Non-compact Section• Ductility Requirement

• Shear• Check whether Web Stiffeners are required• Transverse Stiffener Design• Flange-to-Web Welds• Bearing Stiffeners• Shear Connectors for Composite Sections


2. Analysis Options



5. LRFD Design Example - Simple Span Steel Girder

6. Resources

LRFD Steel Structures

Course Outline

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Topic 5

LRFD Design Example - Simple Span Steel Girder

Topic 5


Framing Plan: 100’-0” Long Simple Span

Preliminary Design Steps

1. Choose girder spacing and preliminary diaphragm spacing

Stringer (Typ.)

Diaphragm (Typ.)

4 Spa. @ 25’-0”

3 @

8’-4

”= 2

5’-0

”

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Framing Plan: 100’-0” Long Simple Span

Typical Section

2. Choose preliminary sizes for web and flange plates

3. Determine deck thickness requirements for selected configuration

4. Analyze to determine moments and shears

5. Check section proportion limits per LRFD 6.10.2

Refer also to LRFD Chapter 6, Appendix C for basic steps and code references

Preliminary Design Steps


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Trial Section

20”

PLASTIC NEUTRAL AXISOF PLATE GIRDER

ELASTIC NEUTRAL AXISOF PLATE GIRDER

23.3

3”38

.92”

60”

51.3

2”10

.93”

3 / 4”

7/16”

16”

11/ 2”


Check web proportions per LRFD 6.10.2 (Webs without longitudinal stiffeners in (+) flexure):

D / tw ≤ 150 Art. 6.10.2.1.1-1

where D is depth of the web between flanges, in

tw is web thickness, in

60 / (7/16) = 137 < 150 OK

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Flexural Components will be proportioned such that:

0.11.0 ≤≤yt

yc

II

where Iyc = moment of inertia of the compression flange about the vertical axis, in4

Iyt = moment of inertia of the tension flange about the vertical axis, in4

Iyc = (3/4) 163 / 12 = 256 in4

Iyt = (11/2) 203 / 12 = 1000 in4

0.1 < Iyc / Iyt = 256/1000 = 0.26 < 1.0 OK

Compression Flange:bf / 2tf ≤ 12.0 16 / (2 x 0.75) = 10.7 < 12.0 OK

Tension Flange:bf ≥ D/6 16 > 60/6 = 10 OK

tf ≥ 1.1tw 0.75 > 1.1 (7/16) = 0.48 OK

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Review

Question Review

b/(2*tf) ≤ 12.0

bf ≥ D/6

tf > 1.1twD/t ≤ 500

Which of the following is NOT a proportioning limit given in LRFD 6.10.2?

The option D/t ≤ 500 is the correct answer. The actual limit of 300 forms the upper limit on webs with longitudinal stiffeners. For webs without longitudinal stiffeners the limit is 150.


Elastic Section PropertiesLinear Elastic Beam Theory:

•Plane sections remain plane•Stress directly proportional to strain (homogeneous elastic)•No deformation remains upon removal of loading

•Elastic Neutral Axis (NA) at centroid of section (first moment of area)•Bending stiffness and bending moments relate to moment of inertia, I (second moment of area)

Composite Section:•Dissimilar materials (Concrete and Steel)•Non-homogeneous elastic•Section is made composite by means of shear connectors

•Convert to equivalent homogeneous elastic section using modular ratio, n•n = Es / Ec•Transformed Section is equivalent all-steel section•Replace concrete (deck slab) by equivalent amount of steel•Concrete is assumed to carry no tensile stress

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Computation of Elastic Section Properties (Steel girder alone)

Material A d Ad Ad 2 Io I

Top Flange 16 x ¾ 12.0 30.38 365 11,090 11,090

Web 7/16 x 60 26.3 7,880 7,880

Bottom Flange 20 x 1 ½ 30.0 -30.75 -923 28,370 28,370

68.3 -558 47,340

ds = -558/68.3 = 8.17 in. -8.17 x 558 = -4,560

INA = 42,780

Distance from neutral axis of steel section to:

Top of steel = 30 + 0.75 + 8.17 = 38.92 in.

Bottom of steel = 30 + 1.50 – 8.17 = 23.33 in.

Section modulus, top of steel

Sst = 42,780 / 38.92

= 1,100 in3

Section modulus, bottom of steel

Ssb = 42,780 / 23.33

= 1,830 in3


Computation of Elastic Section Properties (Composite Section)

243n

rounded 0.86.7ksi 3834ksi 000,29

ksi 3834ksi 4)kcf 150.0(000,33000,33 5.1'5.1

=

⇒===

=××=××=

c

ccc

EEn

fwE Eqn. 5.4.2.4-1

Eqn. 6.10.1.1.1b-1

See Comm. C6.10.1.1.1d

Use 3n = 24 for long-term dead loads such as sidewalks, parapets, future overlay, utilities

Use n = 8 for short-term loads such as live loads


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Computation of Elastic Section Properties (Composite, 3n = 24)


Steel Section 68.3 -558 47,340

Concrete 100 x 8 ½ / 24 35.4 37.00 1,310 48,470 210 48,680

103.7 752 96,020

ds = 752/103.7 = 7.25 in. -7.25 x 752 = -5,450

INA = 90,570


Top of steel = 30.75 - 7.25 = 23.50 in.

Bottom of steel = 31.50 + 7.25 = 38.75 in.

Top of Concrete = 23.50 + 2.00 + 8.50 = 34.00 in.

Top of steel

Sst = 90,570 / 23.50

= 3,850 in3

Bottom of steel

Ssb = 90,570 / 38.75

= 2,340 in3

Section modulus

Top of Concrete

Sc = 90,570 / 34.00

= 2,660 in3


Computation of Elastic Section Properties (Composite, n = 8)


Steel Section 68.3 -558 47,340

Concrete 100 x 8 ½ / 8 106.3 37.00 3,933 145,520 640 146,160

174.6 3,375 193,500

ds = -3,375/174.6 = 19.33 in. -19.33 x 3,375= -65,240

INA = 128,260


Top of steel = 30.75 – 19.33 = 11.42 in.

Bottom of steel = 31.50 + 19.33 = 50.83 in.

Top of Concrete = 11.42 + 2.00 + 8.50 = 21.92 in.

Top of steel

Sst = 128,260 / 11.42

= 11,230 in3

Bottom of steel

Ssb = 128,260 / 50.83

= 2,520 in3

Section modulus

Top of Concrete

Sc = 128,260 / 21.92

= 5,850 in3


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Review

Question Review

Creep in the steel will create higher stresses in the concrete due to permanent loads

Shear connectors must be designed assuming increased stresses at the flange interface

Creep in the concrete will create higher stresses in the steel due to permanent loads

Reduced section properties are required to accurately compute LL deflection

Why must one compute section properties for a modulus ratio of 3n?

The correct option is Creep in the concrete will create higher stresses in the steel due to permanent loads. Concrete creep redistributes stresses to stiffer steel sections; using 3 times the modulus ratio is an approximate way to account for this.

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Computation of Elastic Section Stresses – Unfactored Loads

(a) Service Limit State Stresses, ksi

Top of Steel (Compression) Bottom of Steel (Tension) Top of Concrete (Compr.)

DC = 2,063(12)/1,100 = 22.51 2,063(12)/1,830 = 13.53

DW = 204(12)/3,850 = 0.64 204(12)/2,340 = 1.05 204(12)/(2,660x24) = 0.04

LL + IM = 1,911(12)/11,230 = 2.04 1,911(12)/2,520 = 9.10 1,911(12)/(5,850x8) = 0.49

Total 25.19 23.68 0.53

0.9234.4032.65Total

0.863,344(12)/(5,850x8) =15.923,344(12)/2,520 =3.573,344(12)/11,230 =LL + IM =

0.06306(12)/(2,660x24) =1.57306(12)/2,340 =0.95306(12)/3,850 =DW =

16.912,579(12)/1,830 =28.132,579(12)/1,100 =DC =

Top of Concrete (Compr.)Bottom of Steel (Tension)Top of Steel (Compression)

(b) Strength I Limit State Stresses, ksi


Plastic Section Properties

Plastic Theory:• Plastic capacity is the amount of load required to completely yield a

member• Plastic hinges start to form, rotations take place• Material is at yield strength• Some deformation remains upon removal of loading

Composite Section:• Calculate plastic forces in steel using yield strengths of flanges and web• Calculate plastic forces in concrete based on rectangular block using 0.85f’c• Concrete is assumed to carry no tensile stress

• Plastic Neutral Axis (PNA) at equilibrium condition (no net axial forces in section)

• Using the forces in each component determine whether the PNA is in concrete deck, top flange or web

• Forces in longitudinal reinforcement may be neglected in compression section (Conservative)


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LRFD Appendix D, Table D6.1-1: Calculation of Y and Mp in Positive Flexure_


Computation of Plastic Section PropertiesNeglecting longitudinal deck reinforcement, Prb and Prt:

Plastic section properties: Variable definitions per LRFD Appendix D6.1

App. D6.1 Case IIICheck if the PNA is in concrete deck

Pt + Pw + Pc ≥ Ps

AbfFy + AwFy + AtfFy ≥ 0.85 f’cbeffts


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Computation of Plastic Section Properties

App. D6.1 Case III

Therefore PNA is in deck slab

Compute Mp by summing moments about PNA

AbfFy + AwFy + AtfFy ≥ 0.85 f’cbeffts

30(36) + 26.3(36) + 12(36) ≥ 0.85(4.0)(100)(8 ½)

1080 + 947 + 432 = 2459 < 2890 kips

2459

2890Y = 8.5” x = 7.23 in. from top of slab, or 8.5 in - 7.23 in = 1.27 in. from bottom of slab_

Mp = Y2 Ps / (2 x ts ) + Pcdc + Pwdw + Ptdt

= (7.23)2(2,890)/(2 x 8.5) + 1,080(61.50 + 2 + 1.27) + 947(30.75 + 2 + 1.27) + 432(0.38 + 2 + 1.27)

= 8,890 + 69,950 + 32,220 + 1,580 = 112,640 in-kips

Mp = 9,387 ft-kips


_




D. Strength Limit State Check for• Flexure

• Compact or Non-compact Section• Ductility Requirement

• Shear• Check whether Web Stiffeners are required• Transverse Stiffener Design• Flange-to-Web Welds



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Service Limit State Checks for:• Elastic Deformations

• Deflection check in LRFD is optional except for orthotropic and lightweight decks

• Owner may require deflection check

• All design lanes are loaded

• Service I live loads plus IM are considered

• All girders deflect equally

• Deflection limit = span length/1000 unless owner specifies differently

• Permanent Deformations• Service II loading combination

• Flanges of composite sections should satisfy the following:

• Top flange, ff ≤ 0.95RhFyf

• Bottom Flange, ff + fl /2 ≤ 0.95RhFyf

Art. 6.10.4.2

Art. 6.10.4.1


Maximum live-load deflection for Service I Load Combination

Greater of

• Deflection due to the design truck loading alone, or

• Deflection due to design lane loading plus 25% of design truck loading

Design Truck Loading Deflection

includes dynamic load allowance, IM

Mid-span deflection may be obtained with acceptable accuracy from

δT+IM = 324PT+IM(L3 – 555L + 4780) / EI for a simply supported stringer

PT+IM = 8 X 0.676 (1 + 0.33) = 7.2 kips

δT+IM = 324(7.2)[1003 – 555(100) + 4780] / [(29,000)(128,260)] = 0.6 in.

Elastic Deformations


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Maximum live-load deflection for Service I Load Combination

Design Lane Loading Deflectiondoes not include dynamic load allowance, IM

mid-span deflection due to uniformly distributed load from

δL = 5ωL4 / 384EI for a simply supported stringer

ωL = 0.64 X 0.676 = 0.43 kip/ft or 0.036 kip/in

δL = 5(0.036)(12004) / [384(29,000)(128,260)] + 25%(0.6) = 0.4 in.

Since δT+IM = 0.6 in > δL = 0.4 in., Truck Design Loading governs.

δT+IM = 0.6 in < L / 1000 = 1.2 in. OK

The trial section satisfies the elastic deflection requirement.










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Flexure for Service II Load CombinationTop Flange

stresses in top flange due to short and long-term loads

[using the stresses already calculated in Table (a)]

ff ≤ 0.95RhFyf

ff = 22.51 + 0.64 + 1.3(2.04) = 25.80 ksi < 0.95(1.0)(36) = 34.2 ksi OK

Bottom Flangestresses in bottom flange due to short and long-term loads

[using the stresses already calculated in Table (a)]

ff + fl /2 ≤ 0.95RhFyf

ff + fl /2 = 13.53 + 1.05 + 1.3(9.10) + 0 = 26.41 ksi < 0.95(1.0)(36) = 34.2 ksi OK

The trial section satisfies the permanent deflection requirement.

Permanent Deformations










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Constructibility• During critical stages of construction, nominal yielding or reliance on post-buckling resistance not permitted for main load-carrying members

• For investigating the constructibility of flexural members, all loads shall be factored

• For the calculation of deflections, the load factors shall be taken as 1.0

Constructibility Checks• Flexure Checks Art. 6.10.3.2

• Shear Check Art. 6.10.3.3


Flexure Checks• For discretely braced flanges in compression

• Flange nominal yielding: fbu + fl ≤ ΦfRhFyc

• Flexural resistance: fbu + 1/3 fl ≤ ΦfFnc

• Web bend buckling: fbu ≤ ΦfFcrw

• For discretely braced flanges in tension

• Flange nominal yielding: fbu + fl ≤ ΦfRhFyt

where fbu = calculated stress due to factored dead loads = 28.13 ksi

fl = lateral bending stress in flange, ksi = 0

Φf = resistance factor for flexure = 1.00

Fnc = nominal flexure resistance of the flange = 36 ksi

Fcrw = nominal bend-buckling resistance for webs, ksi

Rh = hybrid factor = 1.0 for homogenous built-up sections

Fyt = yield strength of tension flange = 36 ksi


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28.13 + 0 < 1.00 x 1.0 x 36 ksi

28.13 < 36 ksi OKThe trial section satisfies the flange nominal yielding requirement.




Controlled by the smaller of:

Local buckling resistance of compression flange, or

Lateral torsional buckling resistance of compression flange

Local buckling resistance of compression flange, Fnc, is given by

Fnc = RbRhFyc when λf ≤ λpf

Or, otherwise,

Fnc = 1 - 1 - RbRhFyc

Fyr λf - λpf

RhFyc λfr - λpf


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For this example,

λf = compression flange slenderness ratio, bfc / 2tfc

λpf = limiting slenderness ratio for compression flange, 0.38 √ E / Fyc

λf = 16 / (2 x 0.75) = 10.7 < λpf = 0.38 √ 29,000 / 36 = 10.8

Then, Fnc = 1.0 x 1.0 x 36 = 36 ksi




Lateral torsional buckling resistance of compression flange, Fnc, is given by

Fnc = Cb 1 - 1 - RbRhFyc ≤ RbRhFyc

where Lb = unbraced length, in., for this example 300 in.

Lp = limiting unbraced length, in., 1.0rt √ E / Fyc

Lr = π rt √ E / Fyc , in.

rt = effective radius of gyration for lateral torsional buckling, in

bfc

√ 12 (1+ 1/3(Dctw / bfctfc))

Fyr Lb - Lp

RhFyc Lr - Lp

rt =


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Dc = web depth in compression, in

Dc = d - tfc ≥ 0

Using service limit state stresses for elastic range,

Dc = 62.25 – 0.75 = 31.34 in.

then

16

√ 12 (1 + 1/3 [ (31.34 x 7/16 ) / (16 x 0.75) ] )

-fc

fc + ft

rt = = 3.93 in.

- (-25.19)

|-25.19| + 23.68


- (-25.19)

|-25.19| + 23.68



Dc = web depth in compression, in

Dc = d - tfc ≥ 0

Using service limit state stresses for elastic range,

Dc = 62.25 – 0.75 = 31.34 in.

then

16

√ 12 (1 + 1/3 [ (31.34 x 7/16 ) / (16 x 0.75) ] )

- (-25.19)

|-25.19| + 23.68



Lp = 1.0 x 3.93 √ (29,000 / 36) = 111.5 in.

Lr = π rt √ E / Fyc where Fyc is the smaller of 0.7Fyc = 0.7 x 36 = 25.2 ksi

or Fyw = 36 ksi

but not less than 0.5 Fyc = 0.5 x 36 = 18.0 ksi

then Lr = π 3.93 √ 29,000 / 25.2 = 418 in.

Substituting these values and conservatively taking Cb = 1.0, and Fnc = 29.4 ksi

28.13 + 0 < 1.0 x 29.4 ksi OK

The trial section satisfies the flexural resistance requirement.


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• Web bend buckling: fbu ≤ ΦfFcrw

For webs without longitudinal stiffeners

Fcrw = ≤ RhFyc or ≤ Fyw / 0.7, whichever is smaller

where k = 9 / (Dc / D)2 = 9 ( 38.17 / 60) = 22.24

in which Dc is the depth of the web in compression in elastic range

Fcrw = 0.9 x 29,000 x 22.24 / 1372 = 30.9 ksi < 1.00 x 36 ksi = 36 ksi

Fcrw = 36 / 0.7 = 51.4 ksi > 30.9 ksi therefore Fcrw = 30.9 ksi

and, since fbu = 28.13 ksi < Fcrw = 30.9 ksi

The trial section satisfies the web bend buckling requirement.

0.9 Ek

(D / tw)2










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Constructability Checks• Shear Check

• Interior panels of the transversely stiffened webduring the critical stages of construction shall satisfythe following requirement:

Vu ≤ ΦvVcr = ΦfCVp

where Vu = shear due to factored permanentloads and factored constructionloads on the non-composite section,kips

Φv = resistance factor for shear = 1.00

Vcr = shear buckling resistance, kips

C = ratio of shear buckling resistance toshear yield strength

Vp = plastic shear force = 0.58FywDtw, kips


Shear CheckVu ≤ ΦvVcr = ΦfCVp

Vu = 1.25 x 82.5 kip + 1.50 x ½ x 8.33’ x 0.100 ksf x 100’ = 144.8 kips

assuming a construction load of 0.100 ksf

C =

when D / tw ≥ 1.40 √ Ek / Fyw k = 5 +

since 60” / (7/16”) = 137 > 1.40 √ 29,000(5) / 36 = 89 k = 5 + = 5.8

assuming a transverse stiffener spacing (d0) of 150 in. for this example.

C = = 0.39

Then, Vp = 0.58 x 36 x 60 x 7/16 = 548 kips

And Vu = 144.8 < 1.00 x 0.39 x 548 = 213.7 kips OK

The trial section satisfies the shear requirement.

1.57 Ek

(D / tw)2 Fyw 5

(d0 / D)2

5

(150 / 60)2

1.57 29,000 ksi (5.8)

1372 36 ksi


37

Review

Question Review

Post-buckling resistance is included in the construction stage resistance

Flexural resistance may be controlled by lateral-torsional buckling resistance

Flexural resistance may be controlled by local buckling

Web bend-buckling must be checked at the constructability level

Which is NOT a true statement regarding the LRFD constructability check?

The option Post-buckling resistance is included in the construction stage resistance is the incorrect item. LRFD does not allow web post-buckling strength to be utilized during the construction stages.

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Fatigue and Fracture Limit State• Fatigue Art. 6.10.5.1

• Fracture Art. 6.10.5.2

• Special Fatigue Requirements for Web Art. 6.10.5.3


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Fatigue Art. 6.10.5.1

• Details shall be investigated for fatigue as specified in Art. 6.6.1

Table 6.6.1.2.5-3 Constant Amplitude Fatigue Limit (CAFL)


Fatigue - Higher fatigue stress range by computation

( ) ( ) 213

1

thn FNAF ∆≥⎟⎠⎞

⎜⎝⎛=∆

where A is a detail category constant as taken from Table 1 and (∆F)th is the CAFL

6.6.1.2.5-1

N = 365 (75) n (ADTT)SL 6.6.1.2.5-2

Table 6.6.1.2.5-1Detail Category Constant, A


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Fracture


• Quality measure

• Charpy V-notch test requirements

• Fracture-critical members require

• More frequent inspections

• Higher quality control

• Increased toughness









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Special Fatigue Requirements for Web Art. 6.10.5.1

Interior panels of webs with or without longitudinal stiffeners shall satisfy the following requirement:

Vu ≤ Vcr

where Vu = shear in the web at the section under consideration due to un-factored permanent loads plus factored fatigue load, kips

Vcr = shear-buckling resistance determined from Vu = Vcr = CVp, kips

Vp is the plastic shear force, Vp = 548 kips and

C is the ratio of shear-buckling resistance to shear yield strength, C = 0.39 as determined earlier.

Then Vu = 0.39 x 548 = 213.7 kips


Special Fatigue Requirements for Web Art. 6.10.5.1At the first interior panel, 12.5 ft from bearing, un-factored permanent load shears are

VDC = 82.5k – 12.5’ x 1.65k/’ = 61.9 kips

VDW = 8.2k – 12.5’ x 0.163k/’ = 6.2 kips

Similarly,

VFatigue = = 28.8 kips

AASHTO requires that, for the purposes of this check, the fatigue load be taken

as twice that calculated from the fatigue load combination. Then

Vu = 61.9k + 6.2k + (0.75 x 1.15 x 28.8k) 2 = 117.8 kips

Vu = 117.8 kips < Vcr = 213.7 kips OK

The web will not be subjected to significant elastic flexing.

Then Vu = 0.39 x 548 = 213.7 kips

The trial section satisfies the special fatigue requirement the web.

72k (87.5’ – 30’ + 11.78’)

100’

0.693

1.20


42









Strength Limit State Check for• Flexure

• Compact or Non-compact Section

• Ductility Requirement

• Shear• Check whether Web Stiffeners are required

• Transverse Stiffener Design

• Flange-to-Web Welds

• Bearing Stiffeners

• Shear Connectors for Composite Sections


43

Flexure- Compact or Non-compact Section

Compact Section: A section in satisfying specific steel grade, web slenderness and ductility requirements that is capable of developing a nominal resistance exceeding the moment at first yield, but not to exceed the plastic moment.

Noncompact Section: A section in flexure for which the nominal resistance is not permitted to exceed the moment at first yield.

Composite section compactness requirements: Art. 6.10.6.2.2

•Yield strength of flanges, Fy < 70 ksi, Fy = 36 ksi < 70 ksi OK

•Web proportion, D / tw < 150 D / tw = 137 < 150 OK

•Web slenderness limit, 2Dcp / tw ≤ 3.76 √ E / Fyc

where Dcp is the web depth in compression.

Previously it was determined that the PNA was in the deck slab. As a result, Dcp = 0

So, web slenderness limit is also satisfied.

The section is compact provided that the following flexure requirement is met.


Flexure

Flexural Resistance Art. 6.10.7

At the strength limit state, Mu + 1/3fl Sxt ≤ Φf Mn

where Φf = resistance factor for flexure = 1.00

Mn = nominal flexural resistance of the section, k-ft

Mu = factored bending moment about the major axis, k-ft

fl = lateral bending stress in flange, ksi

Sxt = elastic section modulus taken as Myt / Fyt, in3

Myt = yield moment with respect to the tension flange, k-ft

Fyt = yield strength of tension flange, ksi

In this example, Mu = 6,229 k-ft as calculated in “Loads and Load Factors” module.


44

Flexure

Flexural Resistance at the strength limit state, Mu + 1/3fl Sxt ≤ Φf Mn

fl is lateral bending stress in flange due to lateral bending moment, Ml

eccentric loadings such as

- cantilever brackets of the deck forms

- loading from deck concrete pouring sequence, or

- wind loading on exterior girders between the brace points

may be calculated from Ml = Fl Lb2 / 12 or Pl Lb / 8, k-ft

In this example, it will be assumed that Ml = 0 and fl = 0.


Flexure

Flexural Resistance at the strength limit state, Mu + 1/3fl Sxt ≤ Φf Mn

The nominal flexural resistance, Mn, of the composite section in positive flexural region canbe taken as

Mn = Mp when Dp ≤ 0.1 Dt

where Dp = the distance from top of concrete deck to PNA, in

and Dt = the total depth of the composite section

Dp = Y = 7.23 in from previous PNA calculation

Dt = 8.50 + 2 + 0.75 + 60 + 1.50 = 72.75 in

Thus Dp = 7.23 < 0.1 Dt = 0.1 (72.75) = 7.28 OK

Therefore Mn = Mp = 9,387 k-ft as determined before, and

since Mu + 1/3fl Sxt = 6,229 + 0 < Φf Mn = 1.00 (9,387) OK

The section satisfies the strength limit state for flexure.


45

Review

Question Review

Lateral flange bending stresses due to staggered diaphragms or skewed supports

Lateral flange bending stresses due to overhang form brackets

Lateral flange bending stresses due to wind loading between diaphragms

All of the above

The strength limit state includes a check for flexural resistance: Mu + 1/3 fl Sxt. The term fl is to account for what?

The correct option is All of the above. All statements listed about the term in the flexural resistance check are correct.

46









Flexure

Ductility Requirement Art. 6.10.7

Intended to protect the concrete deck from premature crushing

Dp ≤ 0.42 Dt Art. 6.10.7.3

where Dp = 7.23 in

and Dt = 72.75 in. as calculated before

In this design, Dp = 7.23 < 0.42 Dt = 0.42 (72.75) = 30.56 in. OK

The section satisfies the ductility requirement for flexure.


47









Shear

Shear Resistance Art. 6.10.9

At the strength limit state, straight and curved web panels must satisfy

Vu ≤ Φv Vn

where Vu = factored shear in the web at the section under consideration, kips

Φv = resistance factor for shear = 1.00 Art. 6.5.4.2

Vn = nominal shear resistance of the web, kips

unstiffened webs Art. 6.10.9.2

stiffened webs Art. 6.10.9.3

In this example, Vu = 293 kips at the bearing location as calculated in the“Loads and Load Factors” module.


48

Shear

Shear Resistance Art. 6.10.9The nominal shear resistance of an unstiffened web,

Vn = Vcr = CVp

In which Vp = 0.58FywDtw, kips, and

C is determined from Eqs. 6.10.9.3.2-4, 6.10.9.3.2-5 or 6.10.9.3.2-6 as applicable

with k taken equal to 5.0

In this example, since D / tw = 137 > 1.40 √ Ek / Fyw = 89, C will be obtained from

C = = = 0.337

Thus, Vn = CVp = 0.337 x 0.58 x 36 x 60 x (7/16) = 184.7 kips < Vu = 293 kips.

Therefore, intermediate stiffeners are required for the range where Vu > 184.7 kips.

1.57 Ek

(D / tw)2 Fyw

1.57 29,000 x 5.0

(137)2 36


Shear

General interior panel requirements for stiffened webs

2 D tw

( bfctfc + bfttft )

2 (60) (7/16)

16 (¾) + 20 (1 ½)

≤ 2.5 6.10.9.3.2-1

= 1.25 ≤ 2.5 OK


49









Shear

Transverse Stiffener Design Art. 6.10.11Transverse stiffener spacing for end panels with or without longitudinal stiffeners < 1.5 D

- May consist of plates or angles- May be welded or bolted to the web- May be on one side or both sides- The width should satisfy both of the following conditions:

bt ≥ 2 + d / 30 16 tp ≥ bt ≥ bf / 4

where

d = total depth of steel section, intp = thickness of projecting element, inbt = full width of widest compression

flange in a field section, in


50

Shear

Transverse Stiffener Design Art. 6.10.11bt ≥ 2 + 62.25 in / 30 = 4.08 in

16 tp ≥ bt ≥ 16 in / 4 = 4.0 in

Try a pair of stiffeners plates of 6 x ½ in for the end panels.

bt = 6 in > 4.08 in OK

16 x ½ in = 8 in > bt = 6 in > 4.0 OK


Shear

Transverse Stiffener Design Art. 6.10.11The moment of inertia of the transverse stiffener must satisfy:

It ≥ b tw3 J 6.10.11.1.3-1

Where

It = moment of inertia of the transverse stiffener taken about the mid-thickness of the web for stiffener pairs, in4

b = the smaller of d0 and D, in

d0 = the smaller of the adjacent web panel widths, in

J = stiffener bending rigidity parameter

tw = stiffener thickness


51

Shear

Transverse Stiffener Design Art. 6.10.11

The maximum stiffener spacing in end panels is limited to 1.5D.

Maximum spacing in any other panel is limited to 3D.

3.0 D = 3.0 x 60 = 180 in

1.5 D = 1.5 x 60 = 90 in < 150 in.

Try end panel length of 75 in.

J = - 2.0 ≥ 0.5 (6.10.11.1.3-3)

J = - 2.0 = -0.4 < 0.5 Therefore, use J = 0.5

2.5

(d0 / D)2

2.5

(75 / 60)2


Shear

Using It ≥ b tw3 J (6.10.11.1.3-1)

It ≥ 75” (7/16”)3 (0.5) = 3.1 in4

Actual stiffener properties are:

It = Ay2 + I0 = 2 [ 6” x ½” x (3” + 7/32”)2 + ½” (63) / 12] = 80.2 in4 > 3.1 in4 OK_


52

Review

Question Review

Transverse stiffeners increase the nominal shear resistance of an unstiffened web

Transverse stiffeners may consist of plates or angles

Transverse stiffeners may be welded or bolted to the web

Transverse stiffeners may be on one side or both sides of the web

All of the above

Which of the following is correct for transverse stiffeners?

The correct option is All of the above. All statements listed about transverse stiffeners are correct.

53









Shear

Flange-to-web Welds Art. 6.13.3

Fillet welds

• The horizontal shear between web and the flange must be resisted by the weld.

• Flange-to-web fillet-welded connections may be designed without regard to the tensile or compressive stress in those elements parallel to the axis of the welds.

• Design for the lesser of either the factored resistance of the connected material or the factored resistance of the weld metal.


54

Shear

The factored resistance of the weld metal, Rr, is given as

Rr = 0.6 Φe2 Fexx

where Φe2 = resistance factor for shear in throat of weld metal = 0.80

Fexx = classification strength of weld metal = 70.0 ksi (for E70XX weld metal)

Thus, Rr = 0.6 x 0.80 x 70.0 = 33.6 ksi

Check factored base metal resistance in tension:

Fr = 0.95 x 36 ksi = 34.2 ksi Therefore Rf is still controlled by 33.6 ksi


Shear

The shear between the web and flange, v = VQ / I

The factored maximum shear, V = 293.2 kips as determined before

The static moment of the flange area, Q = Af y = 1.50 x 20 x (23.33 – ½ x 1.50)

Q = 677 in3

The gross moment of inertia of the steel section, I = 42,780 in4

Thus, v = 293.3 x 677 / 42,780 = 4.64 kips/in


55

Shear

Flange-to-web Welds Art. 6.13.3

The weld size required to resist the shear may be found from

v = = = 0.10 in

The minimum-size fillet weld required for a base metal thicker than 3/4 inch is 5/16 in

Therefore, use two 5/16-in continuous welds

to connect the web to the bottom flange.

v

2 lines of weld x 0.707 x Rr

4.64

2 x 0.707 x 33.6


56

Review

True or False?

Flange to web welds are designed without considering the longitudinal stresses in the flange or web.

Question Review

True

False

IQVv *

= Weld demand, v, has units of kips/in

The correct option is True. Welds are designed using the maximum factored vertical shear force, not by considering longitudinal stresses in the flange or web. The shear flow formula is V Q over I, where Q is the statical moment of inertia and I is the gross moment of inertia of the steel section.









57


For a complete design

• Bearing Stiffeners

• Shear Connectors for Composite Sections

• Diaphragms and Cross Frames

• Lateral Bracings

• Splices

• Connections


2. Analysis Options




6. Resources

LRFD Steel Structures

Course Outline

58

Resources

Additional Resources

• FHWA Steel Girder Design Example

• http://www.fhwa.dot.gov/bridge/lrfd/us_ds3.htm

• AISC publishing of Highway Structures Design Handbook, Volume II, Chapter 1

• “Four LRFD Design Examples of Steel Highway Bridges”

• A Fatigue Primer for Structural Engineers

• Local agencies such as MnDOT LRFD Manual

• Example of 2-span steel girder bridge

• The three areas where the LRFD code changes past practice of steel structures design

• The basic proportioning of economic steel structures

• Applications of the LRFD design philosophy to basic steel structures

• The checks required in steel girder design

• The resources and alternate examples available to designers

Objective Review

We have now covered all of the objectives of the class. You should now be able to identify the following:

59

Mahir Sen, P.E.

Paul Pilarski, P.E., S.E.

Credits

The content for this class has been provided by the following PB employees:

Reference:

Structural Steel Designer’s Handbook

Brockenbrough - Merritt, Fourth Edition, McGraw Hill, 2006

If you have any questions about the content of this course please contact Mahir Sen or email [email protected].

Instructions

• The assessment consists of 10 multiple choice questions.

• You will need to achieve a minimum score of 80% to receive credit for passing the course.

• If you score below 80%, please go back and review the content of this course, and then retake the assessment to achieve a passing score.

You are now ready to begin the final assessment.

When ready, click the Right arrow below to advance to the assessment.

Final Assessment

60

Thank you for completing this course. If you received a passing score on the assessment, simply close this window to exit the course. Your score will be recorded on your transcript.

If you did not achieve a passing score, please review the content of this course and then retake the assessment to achieve a passing score.

If you have any questions or problems with the functionality of this course, please email us at [email protected] and specify the name of the course and the issue you are experiencing so that we may assist you.

Conclusion

Documents

Steel Structures - PB University Home Pageondemandweb.pbworld.net/.../lrfd_steel_structures_final_handout.pdf · 1 V 1.0 – Rev. 09.10.07 AASHTO- Load and Resistance Factor Design