Design Reinforcement for Steel Members Part 2.pdf

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

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AISC Live Webinar Design of Reinforcement for Steel Members

Part II

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AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

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AISC Live Webinars

Design of Reinforcement for Steel Members Part II

Presented by Bo Dowswell, Ph.D., P.E.

February 13, 2014

This presentation is a continuation of the AISC live webinar, Design of

Reinforcement for Steel Members. Several topics will be discussed which affect

the strength and serviceability of reinforced steel members. Deflection of

beams with pre-load will be discussed briefly, and further information will be

provided on the design of built-up columns. Other topics include: weld design,

including the calculation of weld strength for built-up columns; how intermittent

welds affect the section properties of a built-up member; local buckling,

including the effect of stitch welds on the local buckling of plates. An example

problem will be used to illustrate some of the design principles for a composite

beam with reinforcement at the tension flange.

Course Description

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

• To learn and understand the AISC Specification strength and

serviceability requirements for designing reinforced structural steel

members.

• To gain familiarity with the deflection of beams with pre-load when

safely reinforcing steel members for structural steel buildings.

• To learn and understand material and code requirements for welding

of built-up structural steel beams and columns.

• To learn and understand the design principles for reinforcing composite

beams through the presentation of a design example.

Learning Objectives

DESIGN OF

REINFORCEMENT FOR

STEEL MEMBERS-

PART 2

Bo Dowswell, P.E., Ph.D.

Principal

ARC International, LLC

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

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• Reinforced Beam & Column Design

 Reinforcement types

 Pre-load

 Partial-length reinforcement

• Course video and presentation slides are

available on the AISC Website:

www.aisc.org/content.aspx?id=37256

Part 1 Summary

• Beam Deflection

• Section Properties of Built-Up Members

• Weld Design

• Built-Up Columns: Additional Topics

• Flexural Plate Buckling (Plate Buckling

Between Intermittent Welds)

• Composite Beam Example

Course Description

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

BEAM DEFLECTION

Beam Deflection

Deflection of Beams with Pre-Load

• Structural analysis models are built with

reinforced member properties

• Deflection must be adjusted to account

for pre-load.

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

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Pre-load deflection: deflection before

reinforcement is added.

Beam Deflection

= proportion of load applied before the

member is reinforced

Deflection of reinforced member: only

loads applied after reinforcement is added.

Beam Deflection

= proportion of load applied after the

member is reinforced

AISC Live Webinar Design of Reinforcement for Steel Members

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Total Deflection:  = 

+ 



= deflection of non-reinforced member

using loads applied before the

member is reinforced



= deflection of reinforced member using

loads applied after the member is

reinforced

Beam Deflection

The adjusted deflection can be calculated

using the value from a finite element model



= total deflection from the finite element

model using the reinforced member

properties

Beam Deflection

mr n







  













AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

= moment of inertia of non-reinforced

member

= moment of inertia of reinforced

member

Beam Deflection

Deflection of Beams with Partial-Length

Reinforcement

Finite Element Model

Beam Deflection

AISC Live Webinar Design of Reinforcement for Steel Members

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SECTION PROPERTIES OF

BUILT-UP MEMBERS

AISC Specification Section F4

Equations F4-5 and F4-8 require the

calculation of J



1 0.078

xc o t

CE L

Sh r











AISC Specification

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

Torsion Constant

The torsion constant, J, can be calculated

based on individual element behavior or

integral element behavior

individual behavior integral behavior

Torsion Constant

Individual Element Behavior (1 Plate)

= J

+ J

= torsion constant of the W shape

= torsion constant of the plate

= plate thickness

= plate width

ppp

Jbt

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

Torsion Constant

Individual Element Behavior (2 Plates)

= J

+ J

= torsion constant of Plate 1

= torsion constant of Plate 2

Torsion Constant

Integral Element Behavior (1 Plate)

b = minimum of b

and b

= flange thickness

= flange width





iwp pfpf

JJ Jbtttt 

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

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Torsion Constant

Integral Element Behavior (2 Plates)

= minimum of b

and b

= minimum of b

and b

= thickness of Plate 1

= thickness of Plate 2







12111

22 2

iwp p pfpf

pf p f

JJ J J bttt t

bt t t t

   



Effective torsion constant for members with

intermittent welds, J

eff

< J

eff

< J

eff

= effective torsion constant

= torsion constant for independent

behavior

= torsion constant for integral behavior

Torsion Constant

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

• Generally, commercial software

packages provide J

• J

can be used as a conservative

estimate of J

eff

, but this can be extremely

conservative

• See Chang and Johnston (1952) for a

more accurate solution

Torsion Constant

WELD DESIGN

AISC Live Webinar Design of Reinforcement for Steel Members

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Intermittent Welds

• Lower cost

• Less weld shrinkage distortion

• Greater corrosion potential

Intermittent Welds

AISC Specification Section J2.2b

• L ≥ 4w

• L ≥ 1.5 in.

L = segment length

w = weld size

AISC Live Webinar Design of Reinforcement for Steel Members

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Maximum weld spacing for intermittent welds

in AISC Specification Sections D.4 and J3.5

Intermittent Welds

1. Ensure close fit-up

over entire faying

surface

2. Prevent corrosion

between connected

elements

AISC Specification Sections D.4 and J3.5

(a) P

max

= 24t ≤ 12 in.

(b) Unpainted Weathering Steel:

max

= 14t ≤ 7 in.

max

= weld spacing

t = thickness of the thinner part

Intermittent Welds

AISC Live Webinar Design of Reinforcement for Steel Members

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Members Welded Under Load

Elevated temperatures near the weld cause

a reduction in material properties of the

base material

Welding to Loaded Members

General Recommendations

• Use small diameter electrodes and low

welding current

• Use stringer beads only (not weave

beads)

• Specify intermittent welding in short

lengths

Welding to Loaded Members

AISC Live Webinar Design of Reinforcement for Steel Members

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General Recommendations (continued)

• Where appropriate, specify a maximum

interpass temperature of 300 °F

• Allow time for welds to cool between

passes

• Use temperature crayons or other

suitable means to monitor the

temperature of the base metal near the

weld

Welding to Loaded Members

600 °F

800 °F

1,600 °F

1,100 °F

Concept of Inactive Areas

Standard Weld Temperature Field

AISC Live Webinar Design of Reinforcement for Steel Members

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Welding to Loaded Members

600 °F

800 °F

1,600 °F

1,100 °F

Section Normal to Arc Travel

Standard Weld Temperature Field

Example Temperature Curve for Low Heat

Input Weld

Welding to Loaded Members

Section Through

Weld Normal to

Arc Travel

AISC Live Webinar Design of Reinforcement for Steel Members

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Example Yield Strength Curve for Low

Heat Input Weld

Welding to Loaded Members





Section Through

Weld Normal to

Arc Travel

Inactive Area Normal to Arc Travel

Conservative Values:

=1 in. for SMAW

=2 in. for FCAW

See Huenersen et al. (1990)

for a more rigorous solution

Welding to Loaded Members

AISC Live Webinar Design of Reinforcement for Steel Members

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Inactive Area Parallel to Arc Travel

Conservative Estimate:

Full flange width plus

the portion of the web

defined by w

See Huenersen et al. (1990)

for a more rigorous solution

Welding to Loaded Members

BUILT-UP COLUMNS:

ADDITIONAL TOPICS

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How can we design the

reinforcement welds for

columns?

What are the loads?

Weld Design for Columns

The load in the cover plate

must be transferred to the non-

reinforced section at the

theoretical cutoff point.

Welds must develop the cover

plate’s portion of the required

axial load in the column

Weld Design for Columns

AISC Live Webinar Design of Reinforcement for Steel Members

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The load in each plate, based on equal axial

stress over the cross sectional area, is

= area of the reinforcing plate

= total area of the reinforced section



Weld Design for Columns

Shear Flow

= area of the reinforcing plate

= strong-axis moment of inertia of the

built-up member

= required shear force

v = shear force per unit length



QAy



Weld Design for Columns

AISC Live Webinar Design of Reinforcement for Steel Members

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Anchor Force



Weld Design for Columns

How can we determine V

and M

Apply a uniform notional load to develop a

bending moment equal to that caused by

the initial out-of-straightness. For δ

L/1,000

= required axial load

125



AISC Live Webinar Design of Reinforcement for Steel Members

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Weld Design for Columns

Apply w

in the proper direction

The first-order moment,

, is calculated at the

theoretical cutoff point.

If the column is subjected

to axial load only, the

maximum shear, V

, is at

the theoretical cutoff

point.

Weld Design for Columns

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

AISC Manual Table 3-23,

Case 1.



lx

Vw x









Weld Design for Columns

The required moment, M

, can be determined

with a second order analysis or calculated using

multiplier, B

from AISC Specification

Appendix 8, Section 8.2.

1 α





1rc c

BM

Weld Design for Columns

AISC Live Webinar Design of Reinforcement for Steel Members

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= elastic lateral buckling load of the

column

= required axial strength

= 1.0

 = 1.00

for LRFD

= 1.60 for ASD

Weld Design for Columns



The total load in the plate at the theoretical

cutoff point is the anchor force for the

moment plus the plate’s portion of the axial

load.

Weld Design for Columns



rpA





AISC Live Webinar Design of Reinforcement for Steel Members

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must be developed

over distance,

The shear force per unit

length,

v, must me

developed between the

theoretical cutoff points

Weld Design for Columns

FLEXURAL PLATE

BUCKLING

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Column Local Buckling





140

.

Column Flange Plates

Specification Table B4.1a, Case 7

Beam Local Buckling





140

.

112

.

Beam Compression Flange Plates

Specification Table B4.1b, Case 18

AISC Live Webinar Design of Reinforcement for Steel Members

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Flexural Plate Buckling

Flexural Buckling Between Intermittent

Welds

Local Buckling Flexural Buckling

AISC Specification Section E6.2

Non-Staggered:

t = plate thickness

075 12in.

max

.

Flexural Plate Buckling

AISC Live Webinar Design of Reinforcement for Steel Members

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The buckled shape of the plate changes

when staggered welds are used

Flexural Plate Buckling

AISC Specification Section E6.2

Staggered:

50% increase in

max

when staggered

1 12 18 in.

max

.

Flexural Plate Buckling

AISC Live Webinar Design of Reinforcement for Steel Members

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Norris et al. (1951) found critical C/b ratios,

where local buckling will be the controlling

limit state If

C/b < (C/b)

Flexural Plate Buckling

C = clear distance

between fasteners

b = plate width

For Stiffened Elements,

For Unstiffened Elements,

Flexural Plate Buckling

050

















AISC Live Webinar Design of Reinforcement for Steel Members

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140

 .

05 05 070

rr r

Cbt t

.. .

Beam Compression Flange Plates

050









For Stiffened Elements

Specification Table

B4.1b, Case 18

Flexural Plate Buckling

112

 .

05 05 056

pp p

Cbt t

.. .

Beam Compression Flange Plates

Specification Table

B4.1b, Case 18

Flexural Plate Buckling

AISC Live Webinar Design of Reinforcement for Steel Members

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







For Unstiffened

Elements

Ct.

Select 

and 

from AISC

Specification Table B4.1a or

B4.1b.

Flexural Plate Buckling

COMPOSITE BEAM

EXAMPLE

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An existing composite beam must carry an

additional live load of 40 psf. This example

shows calculations for the required

reinforcement using LRFD.

For the existing beam calculations, see

Example I.1 in the AISC Design Examples,

Version 14.1.

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Given:

The live load increased from 100 psf to 140

psf.

The existing live load will be removed

before the beam is reinforced.

The existing beam is A992.

Composite Beam Example

For the existing beam:

The available flexural strength,



= 769 kip-ft

The lower bound moment of inertia,

= 2,520 in.

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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The concrete is normal weight with a

specified compressive strength,

 = 4 ksi.

There are 46 ¾-in.-diameter headed stud

anchors symmetrically placed about the

midspan.

The existing dead load is 0.900 kip/ft

Composite Beam Example

Solution:

Existing Beam Material Properties

A992

= 50 ksi

= 65 ksi

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Applied Loads

Estimated weight of reinforcement:

20 lbs/ft = 0.02 kip/ft

Dead load:

= 0.900 kip/ft + 0.02 kip/ft = 0.920 kip/ft

Composite Beam Example

Live load:

= (10.0 ft)(140 psf)(0.001 kip/lb)

= 1.40 kips/ft

Required load:

= (1.2)(0.920 kip/ft) + (1.6)(1.40 kip/ft)

= 3.34 kips/ft

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Required Flexural Strength

AISC Manual Table 3-23, Case 1.

Composite Beam Example



3.34 kip/ft 45.0 ft

845 kip-ft

M 



Preliminary Analysis Conclusions

For the existing beam,



= 769 kip-ft

769 kip-ft < 845 kip-ft

(Utilization Ratio = 1.10)

Live load deflection

> L/360 (18% Over)

The beam must be reinforced

Composite Beam Example

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Reinforcement Selection Issues

• Flexural reinforcement is most efficient

when located farthest from the neutral

axis

• Partial-length reinforcement is more

economical than full-length

reinforcement

Composite Beam Example

Horizontal Overhead

Composite Beam Example

Reinforcement Selection Issues (continued)

• Productivity for horizontal welding is

about 4 times that of overhead welding

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Reinforcement Selection Issues (continued)

• An adequate shelf dimension must be

provided for fillet welds

• For this example, the bottom of the beam

must remain clear of obstructions

Composite Beam Example

Two L 2 ½  2 ½  3/8 will be added to the

top of the bottom flange. The material will

be A36 (

= 36 ksi, F

= 58 ksi).

Composite Beam Example

L 2 ½  2 ½  3/8 :

= 1.73 in.

= 0.972 in.

y = 0.758 in.

AISC Live Webinar Design of Reinforcement for Steel Members

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The total weight of the reinforcement is

(2)(5.90 lb/ft) = 11.8 lb/ft = 0.0118 kip/ft

The revised load is

= 1.2(0.912 kip/ft) + 1.6(1.40 kip/ft)

= 3.33 kips/ft

The revised moment, M

= 844 kip-ft.

Composite Beam Example

Available Flexural Strength

AISC Specification Section I3.2a: Plastic

stress distribution shall be used when

The stabilizing effect of the reinforcing

angles will be neglected.

3.76



Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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From AISC Manual Table 1-1, h/t

for a

W21

50 is 49.4.

49.4 < 90.6; therefore, the plastic stress

distribution will be used.

29,000 ksi

3.76 3.76 90.6

50 ksi



Composite Beam Example

Concrete Strength

= (120 in.)(4.5 in.)

= 540 in.

C = 0.85f

A

= (0.85)(4 ksi)(540 in.

)

= 1,840 kips

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Steel Strength

T = A

= (50 ksi)(14.7 in.

) + (2) (36 ksi)(1.73

in.

)

= 860 kips

Composite Beam Example

Steel Anchor Strength

AISC Manual Table 3-21

Assume all anchors are placed in the weak

position

1 anchor per rib:

= 17.2 kips/anchor

2 anchors per rib: Q

= 14.6 kips/anchor

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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Q

= (2)(14.6 kips/anchor)

+ (21)(17.2 kips/anchor)

= 390 kips

The compression load in the concrete is

C = MIN(1,840, 860, 390) = 390 kips

Shear transfer for composite action is

controlled by the anchor strength.

Composite Beam Example

AISC Specification Section I3.1a

The effective width of the concrete slab is

the minimum of

(1) One eighth of the beam span

= (45 ft.)(2 sides)/8 = 11.3 ft

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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(2) One-half the distance to the centerline

of the adjacent beam

= (10 ft.)(2 sides)/2 = 10.0 ft

(3) Distance to the Edge of the slab

Not Applicable

Composite Beam Example

The effective width of slab is

= b

= 10.0 ft = 120 in.

Composite Beam Example

= 10.0 ft

AISC Live Webinar Design of Reinforcement for Steel Members

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Plastic Analysis of Cross Section

Composite Beam Example

Compression Block Depth

  

0.85

390 kips

0.85 4 ksi 120 in.

0.956 in.







Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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= 390 kips

= (50 ksi)(0.535 in.)(6.53 in.) = 210 kips

= (2 angles)(36 ksi)(1.73 in.

) = 125 kips

= (50 ksi)(0.380 in.)(x  0.535 in.)

= (50 ksi)(0.380 in.)(20.8 in.  0.535 in.  x)

Composite Beam Example

The plastic neutral axis (PNA) is located by

force equilibrium

C = T

+ C

= T

+ T

Solve for x = 3.44 in.

Also: C

= 55.3 kips

= 320 kips

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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= 3.44 in. + 7.50 in.  0.956 in./2 = 10.5 in.

= 3.44 in.  0.535 in./2 = 3.17 in.

= (3.44 in.  0.535 in.)/2 = 1.45 in.

= 20.8 in.  3.44 in.  0.535 in./2 = 17.1 in.

Composite Beam Example

= 20.8 in.  3.44 in.  0.535 in.  2.50

+ 0.758 in.

= 15.1 in.

= (20.8 in.  3.44 in.  0.535 in.)/2 = 8.41 in.

100

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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= (390 kips)(10.5 in.)

+ (210 kips)(3.17 in.)

+ (55.3 kips)(1.45 in.)

+ (320 kips)(8.41 in.)

+ (125 kips)(15.1 in.)

+ (210 kips)(17.1 in.)

= 13,010 kip-in.

= 1,080 kip-ft.

101

Composite Beam Example



= (0.90)(1,080 kip-ft)

= 972 kip-ft

= 844 kip-ft. < 972 kip-ft o.k.

102

Composite Beam Example

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Required Shear Strength

AISC Manual Table 3-23, Case 1.

103



3.33 kips/ft 45.0 ft

74.9 kips

V 



Composite Beam Example

Available Shear Strength

AISC Specification Section I4.2

The available shear strength is based on

the properties of the steel section alone

AISC Manual Table 3-2



= 237 kips > 74.9 o.k.

104

Composite Beam Example

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Location of the Theoretical Cutoff Point

Only partial-length reinforcement will be

provided. The location of the theoretical

cutoff point must be determined so the

flexural strength of the non-reinforced

segments will be adequate.

105

Composite Beam Example

AISC Manual Table 3-23, Case 1.

For the existing beam,



= 769 kip-ft.

l= 45 ft

= 3.33 kips/ft

106

Composite Beam Example

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Solve for x = 15.5 ft

Use a preliminary distance from the end of

the beam to the theoretical cutoff point of

14 ft.

107



3.33 kips/ft

769 kip-ft 45.0 ft

Mlx



Composite Beam Example

Live Load Deflection

Live load deflection will be limited to:

1. For 100% of the design live load:



= L/360 = (45 ft)(12 in./ft)/360 = 1.50

in.

2. For 50% of the design live load:



= 1.0 in.

108

Composite Beam Example

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109

Composite Beam Example



equiv s tr s

IIII



 

Deflection will be calculated using I

eff

AISC Specification Commentary Section I3.2

eff

= 0.75I

equiv

AISC Specification Commentary Equation C-

I3-4

Transformed Moment of Inertia

 = E

= (29,000 ksi)/(3,490 ksi)

= 8.31

= (10 ft)(12 in./ft)/8.31

= 14.4 in.

= 5,500 in.

110

Composite Beam Example

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The moment of inertia of the structural

steel section, including the reinforcing

angles,

= 1,170 in.

111

Composite Beam Example

 Q

= 390 kips

T = A

= 860 kips

C = 0.85f

A

= 1,840 kips

= compression force in slab for fully

composite beam

= minimum of T and C

= 860 kips

112

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

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113

Composite Beam Example



444

in. 5 500 i

390 kips

1 170 1 1n. in.70

860 kip

4 090 n.

equiv s tr s

IIII



 



AISC Specification Commentary Equation

C-I3-4

AISC Specification Commentary Section

I3.2

eff

= 0.75I

equiv

= (0.75)(4,090 in.

)

=3,070 in.

114

Composite Beam Example

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For 100% live load

Using a three-member stepped beam finite

element model with:

= 1.40 kips/ft

= 18.3 ft

= 13.3 ft

E = 29,000 ksi

115

Composite Beam Example

= 3,070 in.

= 2,520 in.



= 1.58 in. > 1.50 n.g.

116

Composite Beam Example

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American Institute of Steel Construction

Options to Limit Deflection

•Calculate

I using a more rigorous

approach

• Increase

by changing the angle size

• Increase the length of reinforcement,

117

Composite Beam Example

Using the three-member stepped beam

finite element model with:

= 39 ft

= 3 ft



= 1.48 in. < 1.50 o.k.

118

Composite Beam Example

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American Institute of Steel Construction

For 50% live load

Using the three-member stepped beam

finite element model with:

= 0.70 kips/ft

= 39 ft

= 3 ft



= 0.73 in. < 1.0 o.k.

119

Composite Beam Example

Welds at the End of the Reinforcement

120

Composite Beam Example

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The anchor force is

However, this equation is valid only in the

elastic range, where

≤ M

= F



121

Composite Beam Example

Due to the approximate nature of the

moment of inertia calculations, the weld

will be designed to develop the available

strength of the angles.

122











0 9 36 ksi 1 73 in. 56 0 kips

P ...

Composite Beam Example

AISC Live Webinar Design of Reinforcement for Steel Members

Part II

American Institute of Steel Construction

Use two 3/16 in. fillet welds extending 8 in.

beyond the theoretical cutoff point. For

concentrically loaded weld groups, AISC

Manual Equation 8-2a is applicable.



= 1.39Dl

= (1.39)(3)[(2 welds)(8 in.)]

= 66.7 kips > 56.0 kips

o.k.

123

Composite Beam Example

The beam web bust be thick enough to

transmit the load from the angles from the

beam. AISC Manual Equation 9-3 is

applicable.

0.286 in. < 0.380 in. o.k.

124









619 3

619

0 286 in.

65 ksi

min

 

Composite Beam Example

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Final Weld Design

125

Composite Beam Example

Questions?

Bo Dowswell, P.E., Ph.D.

Principal

ARC International, LLC

[email protected]

126

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References

Chang, F.K. and Johnston, B.G. (1952), “Torsion of Plate Girders,”

Transactions, American Society of Civil Engineers, April.

Huenersen, G., Haensch, H. and Augustyn, J. (1990), “Repair Welding

Under Load,” Welding in the World, Vol. 28, No. 9/10, pp. 174-182.

Norris, C.H., Polychrome, D.A. and Capozzoli, L.J. (1951), “Buckling

of Intermittently Supported Rectangular Plates,” Welding Research

Supplement, American Welding Society, November, pp. 546-s through

556-s.

127

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