Elastic Stresses in Unshored Composite Section The elastic stresses at any location shall be the sum of

stresses caused by appropriate loads applied separately Steel beam

Permanent loads applied before the slab has hardened, are carried by the steel section.

Short-term composite section Transient loads (such as live loads) are assumed to be carried

by short-term composite action. The short-term modular ratio, n, should be used.

Long-term composite section. Permanent loads applied after the slab has been hardened are

carried by the long-term composite section. The long-term modular ratio, 3n, should be used.

Elastic Stresses (6.10.1.1)

Original section Transformed section

b b/n

yt

yb

t t

tr

bsb I

Myf

tr

tct nI

Myf

The procedure shown in this picture is only valid if the neutral axis is not in the concrete. Use iterations otherwise.

Elastic Stresses (6.10.1.1)

Effective Width (Interior) According to AASHTO-LRFD 4.6.2.6.1, the effective width

for interior girders is to be taken as the smallest of:

One quarter of the effective span length (span length in simply supported beams and distance between permanent load inflection points in continuous beams).

Average center-to-center spacing.

Twelve times the slab thickness plus the top flange width.

Hybrid Sections 6.10.3, 6.10.1.10

The web yield strength must be: 1.20 fyf ≥ fyw≥ 0.70 fyf and fyw≥ 36 ksi

The hybrid girder reduction factor = Rh

Where, =2 Dn tw / Afn

Dn = larger of distance from elastic NA to inside flange face

Afn = flange area on the side of NA corresponding to Dn

fn = yield stress corresponding to Afn

212

)3(12 3

hR

Additional sections

6.10.1.4 – Variable web depth members

6.10.1.5 – Stiffness

6.10.1.6 – Flange stresses and bending moments

6.10.1.7 – Minimum negative flexure concrete deck rft.

6.10.1.8 – Net section fracture

Web Bend-Buckling Resistance (6.10.1.9)

For webs without longitudinal stiffeners, the nominal bend buckling resistance shall be taken as:

When the section is composite and in positive flexure Rb=1.0

When the section has one or more longitudinal stiffeners, and D/tw≤ 0.95 (E k /Fyc)0.5 then Rb = 1.0

When 2Dc/tw ≤ 5.7 (E / Fyc)0.5 then Rb = 1.0

2

2

0.9

9,

/

,

crw

w

c

c

E kF

D

t

where k bend buckling coefficientD D

where D depth of web in compression in elastic range

Web Bend-Buckling Reduction (6.10.1.10)

If the previous conditions are not met then:

21 1.0

1200 300

, 5.7

2

wc cb rw

wc w

rwyc

c wwc

fc fc

a DR

a t

Ewhere

F

D tand a

b t

Calculating the depth Dc and Dcp (App. D6.3)

For composite sections in positive flexure, the depth of the web in compression in the elastic range Dc, shall be the depth over which the algebraic sum of the stresses in the steel, the long-term composite and short term composite section is compressive

In lieu, you can use

f

n

IMLL

n

WSDC

steel

DC

IMLLWSDCDCc t

cf

c

ff

c

fffff

D

3

21

21

0

, sec

cc fc

c t

c t

fD d t

f f

where d depth of steel tion

f and f are the compression and tension flange stresses

Calculating the depth Dc and Dcp (App. D6.3)

For composite sections in positive flexure, the depth of the web in compression at the plastic moment Dcp shall be taken as follows for the case of PNA in the web:

1

85.0

2

'

wyw

ryrsccyctytcp AF

AFAfAFAFDD

6.10 I-shaped Steel Girder Design

Proportioning the section (6.10.2) Webs without longitudinal stiffeners must be limited to

D/tw ≤ 150 Webs with longitudinal stiffeners must be limited to

D/tw≤ 300 Compression and tension flanges must be proportioned

such that: / 6

12.02

1.1

0.1 10

f

f

f

f w

yc

yt

b D

b

t

t t

I

I

Slender

Noncompact

Compact

Moment

Curvature

Mp

My

Section Behavior

6.10 I-Shaped Steel Girder Design

Strength limit state 6.10.6 Composite sections in positive flexure (6.10.6.2.2) Classified as compact section if:

Flange yield stress (Fyf ) ≤ 70 ksi

where, Dcp is the depth of the web in compression at the plastic moment

Classified as non-compact section if requirement not met Compact section designed using Section 6.10.7.1 Non-compact section designed using Section 6.10.7.2

23.76cp

w yc

D E

t F

6.10.7 Flexural Resistance Composite Sections in Positive Flexure

Compact sections At the strength limit state, the section must satisfy

If Dp≤ 0.1 Dt , then Mn = Mp

Otherwise, Mn = Mp(1.07 – 0.7 Dp/Dt)

Where, Dp = distance from top of deck to the N.A. of the composite section at the plastic moment.

Dt = total depth of composite section

For continuous spans, Mn = 1.3 My. This limit allows for better design with respect to moment redistributions.

1

3 nu l xt fM f S M

6.10.7 Flexural Resistance Composite Sections in Positive Flexure

Non-Compact sections (6.10.7.2) At the strength limit state:

The compression flange must satisfy fbu ≤ f Fnc

The tension flange must satisfy fbu + fl/3 ≤ f Fnt

Nominal flexural resistance Fnc = Rb Rh Fyc

Nominal flexural resistance Fnt= Rh Fyt

Where, Rb

= web bend buckling reduction factor

Rh = hybrid section reduction factor

Ductility requirement. Compact and non-compact sections shall satisfy Dp ≤ 0.42 Dt

This requirement intends to protect the concrete deck from premature crushing. The Dp/Dt ratio is lowered to 0.42 to ensure significant yielding of the bottom flange when the crushing strain is reached at the top of deck.

6.10.7 Flexural Resistance Composite Sections in Positive Flexure

6.10 I-Shaped Steel Girder Design

Composite Sections in Negative Flexure and Non-composite Sections (6.10.6.2.2)

Sections with Fyf ≤ 70 ksi Web satisfies the non-compact slenderness limit

Where, Dc = depth of web in compression in elastic range. Designed using provisions for compact or non-compact web

section specified in App. A. Can be designed conservatively using Section 6.8

If you use 6.8, moment capacity limited to My

If use App. A., get greater moment capacity than My

25.7c

w yc

D E

t F

6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section

Discretely braced flanges in compression

Discretely braced flanges in tension

Continuously braced flanges: fbu≤ f Rh Fyf

Compression flange flexural resistance = Fnc shall be taken as the smaller of the local buckling resistance and the lateral torsional buckling resistance.

Tension flange flexural resistance = Fnt = Rh Fyt

1

3 ncbu l ff f F

1

3 ntbu l ff f F

Flange Local buckling or Lateral Torsional Buckling Resistance

Fn or Mn

Lb

Inelastic Buckling (non-compact)

Elastic Buckling(Slender)

Lp

Fmax or Mmax

Inelastic Buckling

(Compact)

pf

Lr

rf f

Fyr or Mr

6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section

Fnc Compression flange flexural resistance – local buckling

0.38 0.562

,

, 1 1

0.7

fcf pf rf

fc yc yr

f pf nc b h yc

yr f pff rf nc b h yc

h yc rf pf

yr yc

b E E

t F F

When F R R F

FWhen F R R F

R F

F F

Fnc Compression flange flexural resistanceLateral torsional buckling

2 2

1 1

2 2

1.0

,

, 1 1

,

,

1.75 1.05 0.3 2.3

b p t rf tyc yr

b p nc b h yc

yr b pb r nc b b h yc b h yc

h yc r p

b r nc cr b h yc

b

E EL L r r

F F

When L L F R R F

F L LWhen L L F C R R F R R F

R F L L

When L L F F R R F

Where

f fC

f f

2

2

12 13

b bcr

b

t

fct

c w

fc fc

C R EF

L

r

br

D t

b t

Lateral Torsional Buckling

Lateral support

Lb

Unstiffened Web Buckling in Shear

yF

E46.2 D/tw

Web plastification in shear

Inelastic web buckling

Elastic web buckling

yF

E07.3

wywpn tDFVV .58.0

21.48n w ywV t EF

D

EtV wn

355.4

6.10.9 Shear Resistance – Unstiffened webs

At the strength limit state, the webs must satisfy:

Vu ≤ v Vn

Nominal resistance of unstiffened webs:

Vn = Vcr = C Vp

where, Vp = 0.58 Fyw D tw

C = ratio of the shear buckling resistance to shear yield strength

k = 5 for unstiffened webs

2

, 1.12 ; 1.0

1.12, 1.12 1.40 ;

1.57, 1.40 ;

w yw

yw w yw yw

w

w yw yw

w

D E kIf then C

t F

E k D E k E kIf then C

DF t F Ft

D E k E kIf then C

t F FD

t

Tension Field Action

d0

D

n cr TFAV V V

Beam Action

Tension Field Action

6.10.9 Shear resistance – Stiffened Webs

Members with stiffened webs have interior and end panels. The interior panels must be such that

Without longitudinal stiffeners and with a transverse stiffener spacing (do) < 3D

With one or more longitudinal stiffeners and transverse stiffener spacing (do) < 1.5 D

The transverse stiffener distance for end panels with or without longitudinal stiffeners must be do < 1.5 D

The nominal shear resistance of end panel is

Vn = C (0.58 Fyw D tw) For this case – k is obtained using equation shown on next

page and do = distance to stiffener

Shear Resistance of Interior Panels of Stiffened Webs

2

2

2sec : 2.5

0.87 (1 )0.58

1

,

55

, 0.58

w

fc fc ft ft

n yw w

o

o

o

n

D tIf the tion is proportioned such that

b t b t

CV F D t C

d

D

where d transverse stiffener spacing

k shear buckling coefficientd

D

If not thenV

2

0.87 (1 )

1

yw w

o o

CF D t C

d d

D D

Transverse Stiffener Spacing

Interior panelEnd panel

Ddo 3 Ddo 5.1

D

1.5od D

Types of Stiffeners

D

1.5od D1.5od D

Bearing Stiffener

Transverse Intermediate Stiffener

Longitudinal Stiffener

6.10.11 Design of Stiffeners

Transverse Intermediate Stiffeners Consist of plates of angles bolted or welded to either one or

both sides of the web Transverse stiffeners may be used as connection plates for

diaphragms or cross-frames When they are not used as connection plates, then they shall

tight fit the compression flange, but need not be in bearing with tension flange

When they are used as connection plates, they should be welded or bolted to both top and bottom flanges

The distance between the end of the web-to-stiffener weld and the near edge of the adjacent web-to-flange weld shall not be less than 4 tw or more than 6 tw.

Transverse Intermediate Stiffeners

Less than 4 tw or more than 6tw

Single Plate

Double Plate

Angle

6.10.11 Design of Stiffeners

Projecting width of transverse stiffeners must satisfy:

bt ≥ 2.0 + d/30

and bf/4 ≤ bt ≤ 16 tp

The transverse stiffener’s moment of inertia must satisfy:

It ≥ do tw3 J

where, J = required ratio of the rigidity of one transverse stiffener to that of the web plate = 2.5 (D/do)2 – 2.0 ≥ 2.5

It = stiffener m.o.i. about edge in contact with web for

single stiffeners and about mid thickness for pairs.

Transverse stiffeners in web panels with longitudinal stiffeners must also satisfy: 3.0

tt l

l o

b DI I

b d

6.10.11 Design of Stiffeners

2

2

0.15 (1 ) 18

,

0.31

, 1.0

1.8 sin

2.4 sin

ywus w

w v n crs

crs

crs ys

t

p

FVDA B C t

t V F

where F elastic local buckling stress

EF F

b

t

and B for stiffener pairs

B for gle angle stiffener

B for gle plate stiffener

The stiffener strength must be greater than that required for TFA to develop. Therefore, the area requirement is:

If this equation gives As negative, it means that the web alone is strong enough to develop the TFA forces. The stiffener must be proportions for m.o.i. and width alone

6.10.11 Design of Stiffeners

Bearing Stiffeners must be placed on the web of built-up sections at all bearing locations. Either bearing stiffeners will be provided or the web will be checked for the limit states of: Web yielding – Art. D6.5.2 Web crippling – Art. D6.5.3

Bearing stiffeners will consist of one or more plates or angles welded or bolted to both sides of the web. The stiffeners will extend the full depth of the web and as closely as practical to the outer edges of the flanges.

The stiffeners shall be either mille to bear against the flange or attached by full penetration welds.

6.10.11 Design of Stiffeners

To prevent local buckling before yielding, the following should be satisfied.

The factored bearing resistance for the fitted ends of bearing stiffeners shall be taken as:

The axial resistance shall be determined per column provisions. The effective column length is 0.75D It is not D because of the restraint offered by the top and

bottom flanges.

yspt F

Etb 48.0

1.4sb pn ysnR A F

6.10.11 Design of Stiffeners

Interior panelEnd panel

Ddo 3 Ddo 5.1

D

bt

tp

9tw 9tw

9tw

General Considerations

Shear studs are needed to transfer the horizontal shear that is developed between the concrete slab and steel beam.

AASHTO-LRFD requires that full transfer (i.e. full composite action) must be achieved.

Shear studs are placed throughout both simple and continuous spans.

Two limit states must be considered: fatigue and shear. Fatigue is discussed later.

Strength of Shear Studs

uscccscn FAEfAQ '5.0

Cross-sectional are of the stud in square inches

Minimum tensile strength of the stud (usually 60 ksi)

nscr QQ

0.85

Placement

A sufficient number of shear studs should be placed between a point of zero moment and adjacent points of maximum moment.

It is permissible to evenly distribute the shear studs along the length they are needed in (between point of inflection and point of maximum moment), since the studs have the necessary ductility to accommodate the redistribution that will take place.

Miscellaneous Rules

Minimum length = 4 x stud diameter Minimum longitudinal spacing = 4 x stud diameter Minimum transverse spacing = 4 x stud diameter Maximum longitudinal spacing = 8 x slab thickness Minimum lateral cover = 1". Minimum vertical cover = 2”. Minimum penetration into deck = 2”