Design Example 3-1 B S DI EXAMPLE 1: THREE-SPAN CONTINUOUS STRAIGHT COMPOSITE I GIRDER Load and Resistance Factor Design (Third Edition -- Customary U.S. Units) by and Michael A. Grubb, P.E. Robert E. Schmidt, E.I.T. Bridge Software Development International, Ltd. SITE- Blauvelt Engineers Cranberry Township, PA Pittsburgh, PA B S DI DESIGN PARAMETERS SPECIFICATIONS: LRFD Third Edition (2004) STRUCTURAL STEEL: - ASTM A 709 Grade HPS 70W for flanges in negative-flexure regions - ASTM A 709 Grade 50W for all other girder and cross-frame steel CONCRETE: f' c = 4.0 ksi REINFORCING STEEL: F y = 60 ksi ADTT: 2,000 trucks per day
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Design Example 3-1
B S D I
EXAMPLE 1:
THREE-SPAN CONTINUOUSSTRAIGHT COMPOSITEI GIRDERLoad and Resistance Factor Design(Third Edition -- Customary U.S. Units)
by andMichael A. Grubb, P.E. Robert E. Schmidt, E.I.T.Bridge Software Development International, Ltd. SITE-Blauvelt EngineersCranberry Township, PA Pittsburgh, PA
B S D I
DESIGN PARAMETERS
SPECIFICATIONS: LRFD Third Edition (2004)STRUCTURAL STEEL:
- ASTM A 709 Grade HPS 70Wfor flanges in negative-flexure regions
- ASTM A 709 Grade 50Wfor all other girder and cross-frame steel
CONCRETE: f'c = 4.0 ksiREINFORCING STEEL: Fy = 60 ksiADTT: 2,000 trucks per day
Design Example 3-2
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BRIDGE CROSS-SECTION
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CROSS-FRAMES(Article 6.7.4.1)
“The need for diaphragms or cross-frames shall be investigated for all stages of assumed construction procedures and the final condition. The investigation should include, but not be limited to the following:
w Transfer of lateral wind loads from the bottom of the girder to the deck & from the deck to the bearings,
w Stability of the bottom flange for all loads when it is in compression,
w Stability of the top flange in compression prior to curing of the deck,
w Consideration of any flange lateral bending effects, andw Distribution of vertical dead & live loads applied to the
structure.”
Design Example 3-3
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BRIDGE FRAMING PLAN
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CROSS-SECTION PROPORTIONS
Web Depth
Span-to-Depth Ratios (Table 2.5.2.6.3-1)
0.032L = 0.032(175.0) = 5.6 ft = 67.2 in.Use 69.0 in.
Web Thickness (Article 6.10.2.1.1)
( ) .in46.015069
150Dt .minw ===
Design Example 3-4
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CROSS-SECTION PROPORTIONS (continued)
Flange Width (Article 6.10.2.2)
Flange Thickness (Article 6.10.2.2)
( ) .in5.116/0.696/Db .minf ===
( ) .in1.1485
)12(10085Lb .minfc ===
( ) ( ) .in62.05625.01.1t1.1t wminf ===
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Design Example 3-5
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CROSS-SECTION PROPORTIONS (continued)
Flange Width-to-Thickness (Article 6.10.2.2)
Flange Moments of Inertia (Article 6.10.2.2)
( ) ok0.123.10875.02
18t2
b
f
f <==
( )
( )51.0
1218375.1
12161
I
I3
3
yt
yc == ok1051.01.0 <<
B S D I
DEAD LOADS(Article 3.5.1)
Component Dead Load (DC1)DC1 = component dead load acting on the
Basic LRFD Design Live LoadHL-93 - (Article 3.6.1.2.1)
Design Truck: ⇒or
Design Tandem:Pair of 25.0 KIP axles spaced 4.0 FT apart
superimposed onDesign Lane Load0.64 KLF uniformly distributed load
Design Example 3-7
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LRFD Negative Moment Loading(Article 3.6.1.3.1)
For negative moment between points of permanent-load contraflexure & interior-pier reactions, check an additional load case:n Add a second design truck to the design lane
load, with a minimum headway between the front and rear axles of the two trucks equal to 50 feet.
n Fix the rear-axle spacing of both design trucks at 14 feet, and
n Reduce all loads by 10 percent.
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LRFD Fatigue Load(Article 3.6.1.4.1)
Design Truck only =>
n w/ fixed 30-ft rear-axle spacing
n placed in a single lane
Design Example 3-8
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LOAD for OPTIONAL LIVE-LOAD DEFLECTION EVALUATION
Refer to Article 3.6.1.3.2:
Deflection is taken as the larger of:- That resulting from the design truck by
itself.- That resulting from 25% of the design
truck together with the design lane load.
B S D I
WIND LOADS(Article 3.8)
Eq. (3.8.1.2.1-1)
PB = base wind pressure = 0.050 ksf for beamsVDZ = design wind velocity at elevation ZVB = base wind velocity at 30 ft height = 100 mph
For this example, assume the bridge is 35 ft above low ground & located in open country.
000,10V
PVV
PP2DZ
B
2
B
DZBD =
=
Design Example 3-9
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WIND LOADS (continued)
Eq. (3.8.1.1-1)
Vo = friction velocity = 8.2 mph for open countryV30 = wind velocity at 30 ft above low ground = VB = 100
mph in absence of better informationZ = height of structure above low ground (> 30 ft)Zo = friction length of upstream fetch = 0.23 ft for open
Assume Span 1 of the structure resists the lateral wind force as a propped cantilever with an effective span length of 120′-0″ (i.e. assume top lateral bracing provides an effective line of fixity 20′-0″ from the pier):
Proportion total lateral moment to top & bottom flanges according to relative lateral stiffness of each flange. Then, divide total lateral moment equally to each girder:
Separate calculations indicate that lateral bending stresses in the top (compression) flange may be determined from a first-order analysis (i.e. no amplification is required).
Separate calculations (see example) indicate a compressive force of -19.04 kips in the diagonal to the self-weight of the steel. Therefore, the total compressive force in the bracing diagonal is:
Estimate the maximum lateral deflection of Span 1 of the structure (i.e. the propped cantilever) due to the factored wind load using the total lateral moments of inertia of the top & bottom flanges of all four girders at Section 1-1:
Eq. (6.10.3.2.1-1) – Top flange 0.844Eq. (6.10.3.2.1-2) – Top flange 0.835Eq. (6.10.3.2.1-3) – Web bend buckling 0.697Eq. (6.10.3.2.2-1) – Bottom flange 0.535
Flexure (STRENGTH III – Wind load on noncomposite structure)Eq. (6.10.3.2.1-1) – Top flange 0.261Eq. (6.10.3.2.1-2) – Top flange 0.170Eq. (6.10.3.2.1-3) – Web bend buckling 0.085Eq. (6.10.3.2.2-1) – Bottom flange 0.272
Flexure (STRENGTH IV)Eq. (6.10.3.2.1-1) – Top flange 0.955Eq. (6.10.3.2.1-2) – Top flange 0.977Eq. (6.10.3.2.1-3) – Web bend buckling 0.836Eq. (6.10.3.2.2-1) – Bottom flange 0.602
Shear (96′-0″ from the abutment) (STRENGTH IV) 0.447
In regions of negative flexure, the constructibility checks for flexure generally do not control because the sizes of the flanges in these regions are normally governed by the sum of the factored dead and live load stresses at the strength limit state. Also, the maximum accumulated negative moments during the deck placement in these regions typically do not differ significantly from the calculated DC1negative moments. Deck overhang brackets and wind loads do induce lateral bending into the flanges, which can be considered using the flexural design equations.Web bend-buckling and shear should always be checked in these regions for critical stages of construction (refer to the design example).
Calculate the longitudinal concrete deck tensile stress at the end of Cast 1 (use n = 8):
Therefore, provide one-percent longitudinal reinforcement (No. 6 bars or smaller @ ≤ 12″). Extend to 95.0 feet from the abutment.Tensile force = (0.453)(100.0)(9.0) = 408 kips
ksi432.0ksi453.0)8(518,161
)12)(20.23)(403,1)(5.1(0.1fdeck >=
−=
Design Example 3-41
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Service
Limit State
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SERVICE LIMIT STATEElastic Deformations (Article 6.10.4.1)
SERVICE LIMIT STATEPermanent Deformations (Article 6.10.4.2)
Under the SERVICE II load combination:1.0DC + 1.0DW + 1.3(LL+IM)
Top steel flange of composite sections: yfhf FR95.0f ≤Eq. (6.10.4.2.2-1)
Bottom steel flange of composite sections: yfhf FR95.02ff ≤+ l
Eq. (6.10.4.2.2-2)
Web bend-buckling: crwc Ff ≤ Eq. (6.10.4.2.2-4)
B S D I
SERVICE LIMIT STATEPermanent Deformations (continued)
Check top flange (Section 1-1):
ksi50.47)50)(0.1(95.0FR95.0 yfh ==
yfhf FR95.0f ≤
0.469)(Ratio
okksi47.50ksi22.30
ksi22.301213,805
1.3(3,510)4,863
322)1.0(3351,581
1.0(2,202)1.0ff
=
<−
−=
++
+=
Design Example 3-43
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SERVICE LIMIT STATEPermanent Deformations (continued)
Check bottom flange (Section 1-1):
For composite sections in positive flexure with D/tw ≤ 150, web bend-buckling need not be checked at the service limit state.
yfhf FR95.02f
f ≤+ l
)50.77(Ratiookksi47.500ksi36.80
ksi36.80122,706
1.3(3,510)2,483
322)1.0(3351,973
1.0(2,202)1.0ff
=<+
=
+++=
B S D I
SERVICE LIMIT STATEPermanent Deformations (continued)
Check Section 2-2 (interior pier):n Article 6.10.4.2.1 -- for members with shear connectors
provided throughout their entire length that also satisfy the provisions of Article 6.10.1.7 (i.e. one percent longitudinal reinforcement is provided in the deck wherever the tensile stress in the deck due to the factored construction loads or the SERVICE II load combination exceeds the modulus of rupture), flexural stresses caused by SERVICE II loads applied to the composite section may be computed using the short-term or long-term composite section, as appropriate, assuming the concrete deck is effective for both positive and negative flexure.
Design Example 3-44
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SERVICE LIMIT STATEPermanent Deformations (continued)
Check Section 2-2 (interior pier):n Flange major-axis bending stresses at Section 2-2 and
at the first flange transition located 15′-0″ from the interior pier are checked under the SERVICE II load combination and do not control. Stresses acting on the composite section are computed assuming the concrete is effective for negative flexure, as permitted in Article 6.10.4.2.1.
n Web bend-buckling must be checked for composite sections in negative flexure under the SERVICE II load combination:
Eq. (6.10.4.2.2-4)crwc Ff ≤
B S D I
WEB BEND-BUCKLING RESISTANCE (Article 6.10.1.9)
Eq. (6.10.1.9.1-1)
where: Eq. (6.10.1.9.1-2)
According to Article D6.3.1 (Appendix D), for composite sections in negative flexure at the service limit state where the concrete is considered effective in tension for computing flexural stresses on the composite section, as permitted in Article 6.10.4.2.1, Dc is to be computed as:
Eq. (D6.3.1-1)
)7.0F,FRmin(
tD
Ek9.0F ywych2
w
crw ≤
=
0tdff
fD fc
tc
cc ≥−
+
−=
( )2c DD9k =
Design Example 3-45
B S D I
WEB BEND-BUCKLING RESISTANCE (Article 6.10.1.9)
Check the bottom-flange transition (controls):
ok Ratio = (0.979)
ksi88.3812463,2
)709,2(3.1246,2
)358373(0.1789,1
)656,2(0.10.1ff −=
−
+−+−
+−
=
ok 0.in25.430.10.7151.2388.38
)88.38(D c >=−
+−
−−=
( )9.22
0.6925.439
k 2 ==
ksi72.39
5625.00.69
)9.22)(000,29(9.0F 2crw =
= ksi72.3988.38 <−
B S D I
SERVICE LIMIT STATEPerformance Ratios
POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1)Service Limit State
In the LRFD Specification, 75% of the stress range due to the fatigue load is considered to be representative of the effective stress range; that is, the stress range due to an HS-15 truck weighing 54.0 kips with a fixed rear-axle spacing of 30′-0″ (FATIGUE load combination -- load factor = 0.75)The maximum stress range is assumed to be twice the effective stress range (i.e. stress range due to a 108-kip truck)
Design Example 3-50
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NOMINAL FATIGUE RESISTANCE(Article 6.6.1.2.5)
In the LRFD Specifications, the nominal fatigue resistance is specified as follows:
Eq. (6.6.1.2.5-1)
Eq. (6.6.1.2.5-2)
A = detail category constant (Table 6.6.1.2.5-1)n = cycles per truck passage (Table 6.6.1.2.5-2)
where ADTT = number of trucks per day in one direction averaged over the design life (assumed to be 2,000 for this example)
For a 3-lane bridge: p = 0.80 (Table 3.6.1.4.2-1)
∴ (ADTT)SL = 0.80(2,000) = 1,600 trucks/day
Design Example 3-54
B S D I
n Article 6.6.1.2.1 -- for flexural members with shear connectors provided throughout their entire length, and with concrete deck reinforcement satisfying the provisions of Article 6.10.1.7 (i.e. one percent longitudinal reinforcement is provided in the deck wherever the tensile stress in the deck due to the factored construction loads or the SERVICE II load combination exceeds the modulus of rupture), the live-load stress range may be computed using the short-term composite section assuming the concrete deck is effective for both positive and negative flexure.
Check the bottom-flange connection-plate weld at 72´-0” from the abutment in Span 1 -- use the n-composite stiffness to compute the stress range as permitted in Art. 6.6.1.2.1:
( ) ( )( )( ) ( )( )
ksi 96.5518,161
31.581249675.0518,161
31.5812337,175.0?f?
=
−+=
( ) ( )TH31
n F21
NA
F ∆≥
=∆ Eq. (6.6.1.2.5-1)
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Design Example 3-56
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(ADTT)SL = 1,600 > 745 trucks per day (Category C′)
∴ ( ) ( )THn F21
F ∆=∆
For a Category C′ detail, (∆F)TH = 12.0 ksi (Table 6.6.1.2.5-3). Therefore:
Fatigue and Fracture Limit StateBase metal at connection plate weld to bottom flange 0.993 (72′-0″ from the abutment)Stud shear connector weld to top flange 0.208
(100′-0″ from the abutment)Special fatigue requirement for webs 0.618
(7′-3″ from the abutment)
INTERIOR-PIER SECTION (Section 2-2)
Fatigue and Fracture Limit StateBase metal at connection plate weld to top flange 0.113 (20′-0″ to the left of the interior pier)Special fatigue requirement for webs (shear at interior pier) 0.636
Determine if the section is a slender-web section (Article 6.10.6.2.3):
Eq. (6.10.6.2.3-1)
Note: Use Dc of the steel section + long. reinforcement (D6.3.1)
Therefore, the section is a slender-web section.=> Go to Article 6.10.8
ycw
c
FE7.5
tD2 ≤
0.11670000,29
7.5 =
0.1160.1305625.0
)55.36(2>=
B S D I
STRENGTH LIMIT STATESection 2-2 (continued)
Use the provisions of Article 6.10.8.For discretely braced compression flanges:
Eq. (6.10.8.1.1-1)
For continuously braced flanges:
Eq. (6.10.8.1.3-1)
ncfbu Ff31
f φ≤+ l
yfhfbu FRf φ≤
Design Example 3-69
B S D I
Calculate the maximum flange major-axis flexural stresses at Section 2-2 due to the factored loads under the STRENGTH I load combination:
ksi64.5512310,3
)040,4(75.1208,3
)664(5.1208,3
)690(25.1149,3
)840,4(25.10.1f −=
−+−+−+−=
Top flange:
Bottom flange:
ksi57.5412703,3
)040,4(75.1194,3
)664(5.1194,3
)690(25.1942,2
)840,4(25.10.1f =
−+−+−+−=
STRENGTH LIMIT STATESection 2-2 (continued)
B S D I
Calculate the hybrid factor, Rh (Article 6.10.1.10.1):
Eq. (6.10.1.10.1-1)
Eq. (6.10.1.10.1-2)
( )β+
ρ−ρβ+=
212312
R3
h
fn
wn
AtD2
=β
0.1f
F
n
yw ≤=ρ
STRENGTH LIMIT STATESection 2-2 (continued)
Design Example 3-70
B S D I
Calculate =>
Dn = larger of the distances from the elastic N.A. of the cross-section to the inside face of either flange
At Section 2-2: Dn = Dc = 36.55 in. (steel + reinf.)
Afn = sum of the flange area & the area of any cover plates on the side of the neutral axis corresponding to Dn. For the top flange, can include reinforcement.
fn = for sections where yielding occurs first in the flange, a cover plate or the longitudinal reinforcement on the side of the N.A. corresponding to Dn, the largest of the specified minimum yield strengths of each component included in the calculation of Afn.
At Section 2-2: fn = Fyc = 70.0 ksi
Otherwise, fn is equal to the largest elastic stress in the component on the side of the N.A. corresponding to Dn at first yield on the opposite side of the N.A.
0.1f
F
n
yw ≤=ρ 714.00.700.50
==ρ
STRENGTH LIMIT STATESection 2-2 (continued)
Design Example 3-71
B S D I
Calculate Rh:
( )β+
ρ−ρβ+=
212312
R3
h
[ ]984.0
)028.1(212)714.0()714.0(3028.112
R3
h =+
−+=
STRENGTH LIMIT STATESection 2-2 (continued)
B S D I
Calculate web load-shedding factor, Rb(Article 6.10.1.10.2):
Eq. (6.10.1.10.2-3)
0.116FE
7.50.130tD2
ycrw
w
c ==λ>=
0.1tD2
a3001200a
1R rww
c
wc
wcb ≤
λ−
+
−=
fcfc
wcwc tb
tD2a = 028.1
)2(20)5625.0)(55.36(2 ==
990.0)0.1160.130()028.1(3001200
028.11Rb =−
+
−=
STRENGTH LIMIT STATESection 2-2 (continued)
Design Example 3-72
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LOCAL BUCKLING RESISTANCEBottom Flange - Section 2-2 (Article 6.10.8.2.2)
Determine the slenderness ratio of the bottom flange:
From Article 6.10.8.2.3:“For unbraced lengths containing a transition to a smaller section at a distance less than or equal to 20 percent of the unbraced length from the brace point with the smaller moment, the lateral torsional buckling resistance may be determined assuming the transition to the smaller section does not exist.”
Assume Lb = 17′-0″( )
%20%120.17
0.150.17<=
−
Design Example 3-74
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Determine the limiting unbraced length, Lp:
where:Eq. (6.10.8.2.3-4)
Therefore:
yctp F
Er0.1L =
+
=
fcfc
wc
fct
tbtD
31
112
br
.in33.5
)2(20)5625.0(55.36
31
112
20rt =
+
=
ft04.970000,29
12)33.5(0.1
Lp ==
LATERAL TORSIONAL BUCKLING RESISTANCE (continued)
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Determine the limiting unbraced length, Lr:
Eq. (6.10.8.2.3-5)
where:
(≥ 0.5Fyc = 35 ksi ok)
Therefore:
yrtr F
ErL π=
ywycyr FF7.0F ≤=
ksi50ksi0.49)70(7.0Fyr <==
ft95.330.49
000,2912
)33.5(Lr =π=
LATERAL TORSIONAL BUCKLING RESISTANCE (continued)
Design Example 3-75
B S D I
Determine the moment gradient modifier, Cb:
For unbraced cantilevers & members where fmid/f2 > 1.0 or f2 = 0:
Cb = 1.0 Eq. (6.10.8.2.3-6)
Otherwise:Eq. (6.10.8.2.3-7)
Eq. (6.10.8.2.3-10)
3.2ff
3.0ff
05.175.1C2
2
1
2
1b ≤
+
−=
o2mid1 fff2f ≥−=
LATERAL TORSIONAL BUCKLING RESISTANCE (continued)
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f2fmidf0f1
fmid/ f2 = 0.875
f1/ f2 = 0.75
Cb = 1.13Lb
f2fm i df
0f1 = f0
f1 /f2 = 0.375
Cb = 1.40
f2fmid
f0
fmid > f 2
Cb = 1
Examples :
fmid
f2 = 0
f0 < 0Cb = 1
f2f
mid
f0 = 0
f1
fmid /f2 = 0.75
f1/f
2 = 0.5
Cb = 1.3
o2mid1 fff2f ≥−=
When fmid < (fo+f2)/2,f1 = fo
Design Example 3-76
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Since the moment diagram in this region is concave in shape, f1 = fo (therefore, do not need to compute fmid). For STRENGTH I:
Note: fo is positive for compression. f2 is always positive.
ksi34.3112310,3
)615,2(75.1208,3
)321(5.1208,3
)334(25.1149,3
)390,2(25.10.1ff o1 =
−
+−
+−
+−
==
3.225.164.5534.31
3.064.5534.31
05.175.1C2
b <=
+
−=
Bottom flange:
LATERAL TORSIONAL BUCKLING RESISTANCE (continued)
B S D I
Since Lp = 9.04 ft < Lb = 17.0 ft < Lr = 33.95 ft:
Assume Lb = 20′-0″From Article C6.10.8.2.3:“For unbraced lengths containing a transition to a smaller section at a distance greater than 20 percent of the unbraced length from the brace point with the smaller moment, the lateral torsional buckling resistance should be taken as the smallest resistance, Fnc, within the unbracedlength under consideration. This resistance is to be compared to the largest value of the compressive stress, fbu, throughout the unbraced length calculated using the actual properties of the section. The moment gradient modifier, Cb, should be taken equal to 1.0 in this case...”
( )%20%25
0.200.150.20
>=−
LATERAL TORSIONAL BUCKLING RESISTANCE (continued)
Design Example 3-78
B S D I
B S D I
For the smaller section at the flange transition in the first unbraced length adjacent to the interior pier: