composite steel girder design
DESCRIPTION
nnnTRANSCRIPT
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”