[ Pobierz całość w formacie PDF ]

T 259
Ffat = = k/in.
= 0.18
2Ao 2(59)(12)
Therefore,
Vsr =
Vfat + Ffat
( )2 ( )2
2 2
=
(0.39) + (0.18) = 0.43
Compute the required shear connector pitch for fatigue for 3 studs per row.
nZr 6.315
Shear stud pitch = = in./row
= 14.7
Vsr 0.43
Although not illustrated here, the number of shear connectors that is provided must also be checked at
the strength limit state according to the provisions of Article 6.11.10 and subsequently Article 6.10.10.4.
D-22
Girder Stress Check Section 8-8 G2 Node 48
Shear Connectors - Maximum Transverse Spacing
Check the maximum transverse spacing, st, between shear connectors on composite box flanges using
Eq 6.11.10-1. This limit insures that local buckling of the flange is prevented when it is subject to
compression. In positive bending regions, the maximum torque occurs at Section 8-8, therefore, the
maximum transverse shear connector spacing is determined at this location.
st Fyf
Eq (6.11.10-1)
d" R1
tf kE
where:
k = 4.0 according to Article 6.11.8.2 for the plate-buckling coefficient for uniform normal
stress
R1 = limiting slenderness ratio for the box flange determined using Eq 6.11.8.2.2-8.
0.57
= Eq (6.11.8.2.2-8)
îø 2 ùø
fv k 2
ïø ëø öø úø
1
2 ëø öø
ïø" + " + 4ìø Fyc ìø ks úø
2
ðø íø øø íø øø ûø
where:
Load Torque (k-ft)
Steel 1.25(72) = 90.0 (Table C3)
Deck 1.25(211) = 263.75
SupImp 1.25(125) = 156.25
FWS 1.5(164) = 246.0
LL + IM 1.75(839) = 1468.25
Total factored torque = 2224.25 k-ft
Section Properties at Section 8-8
Top flange: 16 in. x 1.0 in. (compression flange)
Web: 78 in. x 0.5625 in.
Bottom flange: 83 in. x 0.75 in.
[ 120 + [ 83 - 2(1) ] ] 0.75 1 1
ëø öøëø öø
Ao = ft2
+ 78 + = 55
ìø ìø
2 2 2 144
íø øøíø øø
T 2224.25(12)
fv = = ksi
= 1.69
2Ao tfc 2(55) 144(1.0)
D-23
Girder Stress Check Section 8-8 G2 Node 48
Shear Connectors - Maximum Transverse Spacing (continued)
2
2
fv
ëø öø
1.69
ëø öø
= - 3 1
Eq (6.11.8.2.2-5)
" = - 3 = 0.998
1
ìø ìø
Fyc 50
íø øø
íø øø
Therefore,
0.57
R1 =
= 0.57
îø 2 2ùø
ïø úø
1 1.69 4.0
2 ëø öø ëø öø
ïø0.998 + 0.998 + 4ìø 50 øø ìø 5.34øø úø
2
ðø íø íø ûø
Solve for the maximum transverse shear connector spacing, s .
t
st 50
= 0.57
1.0 4.0(29000)
1.0(0.57)
st = in. > shear connector spacing provided = 14.7 in. from fatigue
= 27.45
50 calculations
4.0(29000)
D-24
Girder Stress Check Section 5-5 G2 Node 36
Strength - Bottom Flange
Check the bottom (box) flange for strength at this section according to the provisions of Article 6.11.8.2
for compression flanges in negative flexure. The section will be checked for the Strength I limit state in
the following computations. Assume one longitudinal compression flange stiffener.
Load Moment
Steel -3,154 k-ft All values are from Table C1
Deck -12,272 k-ft Unfactored results are shown
Total noncomposite -15,426 k-ft
Superimposed DL -1,932 k-ft
FWS -2,541 k-ft
LL + IM -8,127 k-ft
The dynamic load allowance has been applied to the live load according to Article 3.6.2. Multiple
presence factors, specified in Table 3.6.1.1.2-1 were also considered in the live load analysis.
Compute the factored vertical bending stress in the bottom flange due to dead and live load. For loads
applied to the composite section, assume a cracked section, as specified in Article 4.5.2.2. Section
properties are from Table C5. Shear lag need not be considered since the box flange width does not
exceed one-fifth of the span of the bridge (Article C6.11.1.1). The longitudinal vertical bending stress is,
therefore, assumed to be uniform across the flange because shear lag need not be considered and
because it is assumed that the spacing of the internal bracing is such that the longitudinal warping
stress at the strength limit state is limited to 10 percent of the stresses due to major-axis bending
(Article C6.7.4.3).
()
îø³ MDC Cnc ³DC2 MDC2 + ³DW MDW C3n ³LL MLL Cnùø
DC
fbot flg = fbu =
+ +
ïø úø(12) ·
Inc I3n In
ðø ûø
1.25(-15426)(38.81) [ 1.25(-1932) + 1.5(-2541) ]39.76 1.75(-8127)(41.55)
îø ùø(12)(1) = -41.6 ksi
=
+ +
ïø úø
438966 454805 484714
ðø ûø
(C)
Compute the factored St. Venant torsional shear stress, f , in the bottom flange due to the noncomposite
v
loads. Torques are taken from Table C3.
Load Torque
Steel 1.25(-22) = -28 k-ft
Deck 1.25(48) = 60 k-ft
Total Noncomposite Torque = 32 k-ft
D-25
Girder Stress Check Section 5-5 G2 Node 36
Strength - Bottom Flange (continued)
The nominal flexural resistance of the compression flange of a longitudinally stiffened flange is determined
according to Article 6.11.8.2.3.
Compute the enclosed area within the noncomposite box section, A .
o
[ 120 + [ 83 - 2(1) ] ] 1.5 3 1
ëø öøëø öø
Ao = ft2
+ 78 + = 56
ìø ìø
2 2 2 144
íø øøíø øø
T 32(12)
fv = = ksi Eq (6.11.8.2.2-6)
= 0.016
2Ao tfc 2(56)(144)(1.5)
where: T = internal torque from factored loads (k-ft); tf = bottom flange thickness (in.)
Compute the factored torsional shear stress in the bottom flange due to the composite loads. Torques
are taken from Table C3.
Load Torque (-) Torque (+)
SupImp DL 1.25(-149) = -186 k-ft 1.25(193) = 241 k-ft
FWS 1.50(-197) = -296 k-ft 1.50(254) = 381 k-ft
LL + IM 1.75(-863) = -1,510 k-ft 1.75(980) = 1,715 k-ft
Total Comp. Torque = -1,992 k-ft = 2,337 k-ft
Since -1992
, use positive torque.
Compute the enclosed area of the composite box, A .
o
(120 + 81) 1
ëø öø
Ao = ft2
(80.25 + 7.25) = 61.1
ìø
2 144
íø øø
Therefore, the factored torsional shear stress is:
T 2337(12)
fv = = ksi
= 1.06
2Ao tfc 2(61.1)(144)(1.5)
fv tot = ksi
0.016 + 1.06 = 1.08
Check the applied torsional stress against the factored torsional shear resistance of the flange, F .
vr
Fyf
Fvr = Eq (6.11.1.1-1)
0.75Æv
3
50
ëø öø
= ksi > fv tot = 1.08 ksi OK
0.75(1.0) = 21.65
ìø
3
íø øø
D-26
Girder Stress Check Section 5-5 G2 Node 36
Strength - Bottom Flange (continued)
Although the torques on the noncomposite and composite box act in opposite directions, the resulting [ Pobierz całość w formacie PDF ]