design of full integral bridge
TRANSCRIPT
VOL. V - PART 2 DATE: 11May2007 SHEET 1 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
GENERAL INFORMATION FILE NO. 20.02-1
GENERAL INFORMATION: This section of the chapter establishes the practices and requirements necessary for the design and detailing of integral abutments. For general requirements and guidelines on the use of integral abutments, see File Nos. 20.01-1 thru -5. Included are sample design calculations and details to assist the designer. The bridge shall be within the limits for length, skew and thermal movement as stated in File No. 20.01-1:
maximum span length (single span) = 180 feet maximum skew = 30º
Use a single row of vertical steel H-piles (Preferably HP10x42) and orient piles for weak axis bending whenever possible. The web of the H-pile shall be perpendicular to the centerline of the beams/girders regardless of the skew. This will facilitate the bending about the weak axis of the pile. Minimum spacing of the piles is 4’-0”. Maximum pile spacing shall not exceed beam/girder spacing. Avoid high abutment walls except for short spans where anticipated movements are small and can be easily tolerated. Limit the total abutment height to a maximum of 17 feet from finished grade. Wingwalls supported by the abutment shall be limited to 6 feet for straight wings (elephant ear length beyond the 6 feet extension of footing to support wingwall shall be considered). Length of U-wings shall be limited to 8 feet. The portion of the wall beyond shall be designed as a free-standing retaining wall. If not, an independent retaining wall system, which does not move, shall be considered. For design/detailing check lists for integral abutments, see File Nos. 20.02-23 thru -24 and File Nos. 20.02-27 thru -28.
VOL. V - PART 2
DESIGN OF FULL INTEGRAL BRIDGE
Given and Assumptions:
γ = 145 pcf Unit weight of soil (select backfill material) (See Manual of S&B Division Vol. V – Part 2, file no. 17.102-2)
Kp = 4 Assumes the use of EPS material behind backwall
WBridge = 43.33 ft Total bridge width
LBridge = 150.0 ft Bridge length
LThermal = 75.0 ft Length of thermal expansion
HBackwall = 6.33 ft Backwall height
TBackwall = 2.5 ft Backwall thickness
SBeam = 9.33 ft Girder spacing
Overhang = 3.0 ft Slab (and integral backwall) overhang
Cover = 3.5 in Cover over reinforcing steel in backwall
CS = 0.0208 Crown rate (cross slope) of bridge deck
Δh = 2CS(SBeam) Change in height of footing
Hftg = 3.0 ft Height of footing
f’c = 4,000 psi Compressive strength of backwall and wing concrete
f’cf = 3,000 psi Compressive strength of footing concrete
fy = 60,000 psi Yield strength of reinforcing steel
θ = 30 deg Bridge skew angle
α = 6.5 x 10-6 per deg F Coefficient of thermal expansion
DAS = 1.5 ft Depth of approach slab at backwall
Tbottomflange = 1 in Thickness of bottom flange
'c
1.5
sb
fw
En33
= = 8 Modular ratio of concrete to steel for backwall and wing
'cf
1.5
scf
fw
En
33= = 9 Modular ratio of concrete to steel for footing
DATE: 11May2007 SHEET 2 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-2
VOL. V - PART 2
Δh = 4.66 in = 0.39 ft
Hftg = 3.0 ft
Ph = γ Kp (HBackwall + Δh) Passive soil pressure at hinge level Ph =4(145 pcf)(6.33 ft + 0.39 ft) = 3.9 ksf
Pf = γ Kp (HBackwall + Hftg) Passive soil pressure at bottom of footing
Pf = 4(145pcf)(6.33 ft + 3.0 ft) = 5.4 ksf
DATE: 11May2007 SHEET 3 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-3
VOL. V - PART 2
Determine Backwall Moments and Shears: 2
Backwallp h(HK 21w )Δ+= γ Earth pressure resultant per foot
(above hinge)
2ft 0.39ft 33lbs)(4)(6. k/1000 1 x pcf 14521w )( += = 13.1 klf
θ cosS
L Beam= Girder spacing along skew
ft 10.7730 cos
ft 9.33L ==o
For simplicity, use the following equations to determine moments, shear, and reaction.
2pos 0.08wlM = = 0.08(13.1 klf)(10.77 ft)2 = 121.6 ft-kip Maximum positive moment
2
neg 0.10wlM = = 0.10(13.1 klf)(10.77 ft)2 = 152.0 ft-kip Maximum negative moment
0.6wlVmax = = 0.6(13.1 klf)(10.77 ft) = 84.7 k Maximum shear
1.1wlRmax = = 1.1(13.1 klf)(10.77 ft) = 155.2 k Maximum reaction at girder Check to make sure overhang does not govern.
=⎟⎠⎞
⎜⎝⎛=
2
OH cosOverhang0.5wM
θ
2
30 cosft 3.0klf) 0.5(13.1 ⎟
⎠⎞
⎜⎝⎛
o
MOH = 78.6 ft-kip < Mneg Interior support governs
=⎟⎠⎞
⎜⎝⎛=
θ cosOverhangwVOH ⎟
⎠⎞
⎜⎝⎛
o30 cosft 3.0klf) (13.1 = 45.4 k < Vmax Interior support governs
Design Integral Backwall:
Group IV load combination controls. Group IV allowable overstress is 125%. ** For this example, Group IV loading controls the design. It shall be the responsibility of the designer to verify which load case controls, and design accordingly. fs = 125%(0.4Fy) fs = 30,000 psi Allowable stress of steel
fc = 125%(0.4f’c) fc = 2,000 psi Allowable of stress of concrete
)f125%(0.95v 'cec.allowabl = vc.allowable = 75 psi Allowable shear stress in
concrete
DATE: 11May2007 SHEET 4 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-4
VOL. V - PART 2
Flexure design using negative moment: n = 8 Modular ratio of backwall concrete to steel
h = TBackwall h = 30.0 in Height of section resisting flexure
b = HBackwall – DAS = 76 in – 18 in = 58.0 in Width of section resisting flexure
d1 = h - Cover = 30 in - 3.5 in = 26.5 in Depth to first mat of reinforcing steel
d2 = h - (2 x Cover) = 30 in - 2(3.5 in) = 23.0 in Depth to second mat of reinforcing steel
d’ = 3.5 in Depth to compression steel
Try minimum reinforcing, # 6 bars at ~10” spacing. For this backwall height, there are 6 bars in each tension layer.
As1 = 2.64 in2 As2 = 2.64 in2
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=s2s1
2s21s1
AAdAdA
d
⎟⎟⎠
⎞⎜⎜⎝
⎛++
= 22
22
in 2.64in 2.64in) (23.0in 2.64in) (26.5in 2.64d = 24.75 in Depth to centroid of tension steel
Asc = 2.64 in2 6 bars in compression layer Performing section analysis (including the compression steel) Fs = 15,040 psi < fs.allowable = 30,000 psi OK In first layer of tension steel
Fc = 510 psi < fc.allowable = 2,000 psi OK
Shear Design: V = Vmax V = 84.7 k
( ) psi 59in) in(24.75 58.0
k) lbs/1 k(1000 84.7bd
Vv max === Actual shear stress in concrete
v < vc.allowable, Shear reinforcement not required vc.allowable = 75 psi Use stirrup spacing of 12”
DATE: 11May2007 SHEET 5 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-5
VOL. V - PART 2
Shear Stud Design at Girder Ends: Zr = 8.12 k For 7/8” φ AASHTO Sec. 10.38.5.1.1 Zr = 125%(8.12 k) Horizontal shear capacity per stud, with 25% overstress for Group IV loading Zr = 10.15 k
15.3k 10.15k 155.2
ZRn
r
maxstuds ===
Therefore, use 8, 7/8” φ studs on each side of beam web, for a total of 16 studs. Design Overhang and Wing:
DATE: 11May2007 SHEET 6 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-6
Check thinner wing portion only: Lw = 7.0 ft True length of wing [ ] klf 6.5ft) 0.75(6.33pcf)4 (145
21)H x (0.75K γ
21w 22
Backwallpw ===
MOH = 0.5ww(Lw)2 = 0.5(6.5 klf)(7.0 ft)2 = 159.3 ft-kip VOH = ww x Lw = 6.54 klf(7.0 ft) = 45.8 k
b = HBackwall b = 76.0 in Width of section resisting flexure
tw = 16.0 in Thickness of wing
d = tw - cover = 16.0 in – 3.5 in = 12.5 in Depth to first mat of reinforcing steel
VOL. V - PART 2
DATE: 11May2007 SHEET 7 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-7
ry 7 # 6 bars at 10" Spacing
s
s i > fs.allowable = 30,000 psi Redesign
ased on trial and error using 8 # 8 bars at 9” spacing
As = 6.32 in
Fs = 26,800 psi < fs.allowable = 30,000 psi OK
Check shear in wing:
T
= 3.08 in2 A
= 53,600 psF
B
2
Fc = 1,300 psi < fc.allowable = 2,000 psi OK
psi f0.95v '
ca = va = 60 psi
( ) psi 48in) in)(12.5 (76.0
lbs 45,800v =bdVOH == tirrups required
*Since the As required for this section exceeds that calculated for the integral backwall, additional
Allowable shear stress in concrete
< va No s
reinforcing steel will be required for the wing portion of the integral backwall.
VOL. V - PART 2
DATE: 11May2007 SHEET 8 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-8
nalyze Pile Group:
A
ft 50.03ft 43.33W
L Bridge === 30 cos cosftg oθ
ftg = TBackwall Wftg = 2.5 ft Width of footing
Depth of footing
.0 k Live load reaction per truck/per
R = Lftg x Hftg x Wftg x 150 pcf Reaction of footing
Rftg = 50.03 ft(3.0 ft)(2.5 ft)(150 pcf x 1 k/1000 lbs) = 56.3 k
Dneat = 6.33 ft – (8.5 in x 1 ft/12 in) = 5.62 ft Depth of neatwork
Rneat = Lftg x Wftg x Dneat x 150 pcf Reaction of abutment neatwork
Rneat = 50.03 ft(2.5 ft)(5.62 ft)(150 pcf x 1 k/1000 lbs) = 105.4 k
Rss = 604 k Superstructure dead load
ss + Rftg + Rneat Total dead load reaction for
RDL = 604 k+ 56.3 k + 105.4 k = 766 k
gth of footing Len
W Hftg = 3.0 ft VLL = 64 lane ftg
RDL = R footing
VOL. V - PART 2
DATE: 11May2007 SHEET 9 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-9
σp = 9000 psi Maximum pile stress
p = 12.4 in Pile area
APpile = Ap σn Pile capacity
APpile = 111.6 k
2 A
C C
3.6ft 12
ft 43.33ft 12
WN Bridge
L === NL = 3 Number of lanes
LL = 0.75(NL VLL ) = 0.75(3)(64.0 k) = 144.0 k Total live load
R = R + RDL = 144 k + 766 k = 910 k Total load on piles
R ∑ LL
8.2k) (111.6
k 910Cap
RNpile
Piles ==Σ
= Therefore, use 9 piles
etermine Lateral Loading on Piles:
f = 0.5(Ph + P )Dftg = 0.5(3.9 ksf + 5.4 ksf)(3.0 ft) = 14.0 klf
= Rf + Rbw = 14.0 klf + 11.6 klf = 25.6 klf Total resultant along integral
p =
D R f q abutment
k 739.530 cos
)30 ft)(tan klf(43.33 25.6 costanθqWBridge ==
o
o
θF
p is the theoretical force required to restrain all lateral movement. Lateral restraint should be
Fconsidered when transverse displacement can interfere with performance, or with adjacent structures. The designer may wish to employ software such as COM624 or L-Pile to determine the actual forces in the pile. If the force must be restrained (due to proximity to adjacent structures, utilities, retaining/MSE walls, etc. or any facility which may be damaged by lateral displacements), the designer shall provide sufficient restraint in the structure (either through the piles or other supplementary means) to mobilize the lateral force.
VOL. V - PART 2
DATE: 11May2007 SHEET 10 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-10
Design Footing for Bending in the Vertical Plane:
Point loads are girder loads
Design as continuous beam with a combination of both point loads and distributed loads:
oh
= 1.37 ft Cantilever length for distributed L
loads
klf 3.2ft 50.03
k 105.4k 56.3L
RRw
ftg
neatftg =+
=+
= Distributed loads
Sp = 5.0 Pile spacing
For Distributed Loads:
( ) kip-ft 3.02
ft) klf(1.37 3.22
LwM22
ohoh === Overhang moments
intP = 0.08w(Sp )2 =0.08(3.2 klf)(5.0 ft)2 = 6.4 ft-kip Positive moments
intN = 0.1w(Sp )2 = 0.1(3.2 klf)(5.0 ft)2 = 8.0 ft-kip Negative moments
egative moment controls
ftg p
M M N
= 0.6wS = 0.6(3.2 klf)(5.0 ft) = 9.6 k V
VOL. V - PART 2
DATE: 11May2007 SHEET 11 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-11
or Point/Girder Loads:
rom superstructure analysis: P3 = P2
F FP1 = 185.5 k P2 = 241.8 k
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡= 2
p
2p
1girder1 S in 17.5-Sin 17.5
0.1PM
( )
⎥⎦
⎤⎢⎣
⎡= 2
2
girder1 ft) (5.0 n) 1ft/12i x in (17.5-ft 5.0in) ft/12 1 x in (17.5k) 0.1(185.5 M = 13.6 ft-kip
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡= 2
p
2p
2girder2 S in 8.75-Sin 8.75
0.1PM
( )
⎥⎦
⎤⎢⎣
⎡= 2
2
girder2 ft) (5.0 in) ft/12 1 x in (8.75-ft 5.0in) ft/12 1 x in (8.75k) 0.1(241.8 M = 12.9 ft-kip
girder = 13.6 ft-kip Controlling girder moment
M = MintN + Mgirder = 8.0 ft-kip + 13.6 ft-kip = 21.6 ft-kip
esign Integral Abutment Footing (Vertical Plane):
= H h = 36.0 in Height of section resisting flexure
= Wftg b = 30.0 in Width of section resisting flexure
= h – cover d = 32.5 in Depth to first mat of reinforcing
= 9
ry 4 # 6 bars As = 1.76 in2
s = 3,680 psi < fs.allowable = 30,000 psi OK
heck Shear in Footing:
M ∑ D h ftg b
d steel n T F
Fc = 81 psi < fc.allowable = 1,500 psi OK C psi f0.95v '
ca = va = 52 psi
Allowable shear stress in concrete
( ) ==bdV
v ftg [ ]in) in(32.5 30.0
k) lbs/1 (1000 k 9.6 v = 10 psi a ps required
< v No stirru
VOL. V - PART 2 DATE: 11May2007 SHEET 12 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-12
esign footing for bending in the horizontal plane:
D
f = 0.5(Ph + P )Hf = 0.5(3.9 ksf + 5.4 ksf)(3.0 ft) = 14.0 klf Resultant on footing in
p = 5.0 ft Loh = 1.37 ft
R f horizontal plane S
( )2
LRM
2ohf
oh = = ( )=
2 ft 1.37klf 14.0 2
13.1 ft-kip Overhang moments
intP = 0.08R (Sp )2 = 0.08(14.0 klf )(5.0 ft)2 = 28.0 ft-kip Positive moments
intN = 0.01R (Sp )2 = 0.1(14.0 klf)(5.0 ft)2 = 35.0 ft-kip Negative moments
egative moment controls
ftg = 0.6R Sp = 0.6(14.0 klf )(5.0 ft) = 42.0 k Shear in footing
M f M f N
V f
VOL. V - PART 2
DATE: 11May2007 SHEET 13 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-13
esign Integral Abutment Footing (Horizontal Plane):
= 2.5 ft Height of section resisting flexure
= 3.0 ft Width of section resisting flexure
= h – cover d = 26.5 in Depth to first mat of reinforcing
ry 5 # 6 bars (Minimum reinforcement for integral abutment)
s = 2.2 in
s = 5310 psi < fs.allowable = 30,000 psi OK
:
D h b
d steel T
2A F
Fc = 133 psi < fc.allowable = 1,500 psi OK Check Shear in Footing
psi f0.95v 'ca = v = 52 psi Allowable shear stress in concrete
a
( )bdVftg
( )in) in(26.5 36.0k) lbs/1 (1000 k 42.0v = = = 44 psi < va No stirrups required
etermine Number of Dowels Required:
D
φ = 0.875 in Diameter of dowels
y = 36,000 psi Yield strength of dowels
shear = 0.4Fy Allowable shear in dowels
shear = 14.4 ksi
F F
F
VOL. V - PART 2
4 Adowel
2φπ= Adowel = 0.60 in2 Area of dowel
APdowel = Adowel Fshear = 0.60 in2 (14.4 ksi) = 8.7 k Capacity of one dowel
f = 14.0 klf Passive reaction of soil against
olve for Pile Reaction:
p Effective pile length
perature in one direction
LT = 0.75(α120°LThermal)
T = 0.75(6.5x10 g. F)120°F[75.0 ft(12 in/ 1ft)] = 0.53 in
C
R footing S
= 20.0 ft L
= 71.7 in4 Moment of inertia of piles Ip
ΔLT = Change in length due to tem Δ
-6 ΔL per de
( )( ) ⎥
⎥⎦
⎤
⎢⎢⎣
⎡= 5.1
3ER
Δ3
p
pTp L
I L Pile reaction at top of pile
( )
( ) ⎥⎦
⎤⎢⎣
⎡= 3
4
p ft in/1 12 x ft 20.0 in 71.7in 0.53ksi) 3(29,000R 5.1 = 0.4 k
olve for Dowel Spacing:
S
> Min spacing of 4φ in 7.4 ft 0.62
ft 5.0k 0.4klf 14.0
k 8.7==
⎟⎠⎞
⎜⎝⎛ +
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
p
pf
doweldowel
SR
R
CAPS - OK
DATE: 11May2007 SHEET 14 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-14
VOL. V - PART 2
DATE: 11May2007 SHEET 15 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-15
TEMPORARY SUPPORT BOLT DETAIL
heck Load on Temporary Support Bolts:
Rsteel = 9.0 k Reaction of steel beam, diaphragms, etc.
CT = 4.0 k Reaction of construction tolerances and other
Rbeam = Rsteel + RCT Rbeam = 13.0 k Total reaction
H = 13.66 in Unsupported height of temporary support
F = 55 ksi
C
R
loads
bolt
Bolt
y
φ bolt = 0.875 in Temporary support bolt diameter
=⎟⎠⎞⎛
4boltφ
⎜⎝=
42 I bolt
π4
4in 0.875 ⎞⎛πin 0.03
42 =
⎟⎠
⎜⎝ Temporary support bolt moment of inertia
==4
2
boltboltA
φπ ( ) 20.60in
4in 0.875
=⎟⎟⎠
⎞⎜⎜⎝
⎛ 2
π Temporary support bolt area
== bolt
boltIr
boltAin 0.22
in 0.6 2
4
=⎟⎟⎠
⎞⎜⎝
Radius of gyration of temporary support bolt
= 1.0 k factor for bolt acting as column
Lr Factor =
in 0.03⎜⎛
k
= ⎟⎠⎞
⎜⎝⎛
in 0.22in 13.661.0
⎟⎟⎠
⎞⎜⎜⎝
⎛
bolt
bolt
rH
kK = 62.1
Hbolt = 9 in + Δh
VOL. V - PART 2
Axial compression in temporary
bolt
beama 2A
R=
)in 2(0.6k) lbs/1 (1000 k 13.0
2=σ = 10,800 psi
support bolt
==y
2
c FE2
Cπ
102ksi 55
ksi) (29,0002 2
=π
Since KLr Factor < Cc, then allowable compression in temporary support bolt:
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−=
EFFactor KLrF
Comp y2
yallow 44
1122 π.
psi 25,000psi) 0(29,000,00psi) (55,000(62.1)psi 55,000 2
=⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−= 44
1122 π.
Since σ mpAllow Bolt diameter is sufficient
Check Bending in Shortest Temporary Support Bolt:
Hbolt = 9.0 in
a < co
3
4
bolt
boltbolt in 0.07
2in 0.875
in 0.03
2
IS =⎟⎠⎞
⎜⎝⎛
=⎟⎠⎞
⎜⎝⎛
=φ
Section modulus of bolt
s = 0.6 Coefficient of static friction
ind P-Load associated with ΔT induced at top of bolt and associated moment:
p = 9 Number of piles
b = 5 Number of beams
LT = Change in length due to temperature
LT = 0.75(α120°LThermal) = 0.75(6.5x10 per deg. F)120°F(75.0 ftx12 in/1ft) = 0.53 in
µbetween the beam flange and the steel bearing plates
F n n Δ
-6 Δ
DATE: 11May2007 SHEET 16 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-16
2RF s
beamfμ
k 3.92
0.6k 13.0 =⎟⎠⎞
⎜⎝⎛==
Friction force of superstructure loading per bolt
VOL. V - PART 2
DATE: 11May2007 SHEET 17 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-17
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟
⎟⎠
⎞⎜⎜⎝
⎛= TEΔL
P
pp
bolt2p
boltb
3bolt
pp
3p
b
I2nHL
I3nH
I3nL
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
)in 2(9)(71.7in) (9.0ft)] in/1 ft(12 [20.0
in 3(5)(0.03in) (9.0
in 3(9)(71.7ft) in/1 ft(12 (20.0
ksi) in(29,000 0.53P
4
2
4
3
4
3b
))
b = 1.7 k Resultant load at top of bolt
b = Pb Hbolt =1.7 k[9.0 in(1 ft/12 in)] = 1.3 ft-kip Moment induced into bolt
P M
( )bolt b
bb S2n
M=σ ( )3in 2(5)(0.07
k) lbs/1 ft)(1000 in/1 kip(12-ft 1.3= Bending stress
= 22, 300 psi
∑σ = σa + σb = 10,800 psi + 22,300 psi = 33,100 psi Sum of axial compression
wable = 125%(0.55Fy) σallowable = 37,800 psi Allowable bolt stress
∑σ < σallowable OK
va = 0.4Fy va = 22 ksi
bσ
and bending stresses σallo
vb
bolt
b
2AP
= ksi 1.4)in 2(0.60
k 1.72 == < Allowable shear OK
heck Shear in Temporary Support Bolts: C
VOL. V - PART 2
DATE: 11May2007 SHEET 18 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-18
esign Plate A:
D
bf = 12.0 in
de = 1.5 in Clearance between the plate
ends and temporary support bolt centerline
in 12.130 cos
in 1.5 - in (12.0==
)30 cos d-(b e f
e)
=L
initial plate size: (use minimum values)
p = 5 in Plate width
= 0.5 in Minimum plate thickness
A = tp = 2.5 in Area of plate
Assume an w tp
p wp
12
3pp
p
twI = 4
3
in 0.05in) in(0.5 5.0==
12 Moment of inertia of plate
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
2tI
Sp
pp
34
in 0.20
2in 0.50in 0.05
=⎟⎠⎞
⎜⎝⎛
Section modulus of plate
Beam/girder flange width
Span between temporary support bolts
VOL. V - PART 2
DATE: 11May2007 SHEET 19 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-19
heck Plate Stresses: C
plf 13,000ft 1.0
k) lbs/1 (1,000 k 13.0b
Rw
f
beam === Loading along length of Plate A
k 6.62
in) ft/12 1 x in klf(12.1 13.02
wLv ep ===
Shear force at temporary support bolt
psi 2,640in 2.5
k) lbs/1 k(1,000 6.6Av
p
p ===τ
τ < 12,000 psi – OK
kip-ft 1.78
in) ft/12 1 x klf(12.1in 13.08
wLM22
ep === Bending moment in plate
psi 102,0000.20in
ft) k)(12in/1 lbs/1 kip(1,000-ft 1.7SM
3p
pb ===σ
bσ > 27
Redesign:
= 5.0 in Revised plate width
tp = 1.0 in Revised plate thickness
p= 5 in2 Area of plate
,000 psi-Increase plate size wp
A
12
3pp
p
twI = 4
3
in 0.4in) in(1.0 5.0==
12 Moment of inertia of plate
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
2tI
Sp
pp
34
in 0.8
2in 1.0in 0.4
=⎟⎠⎞
⎜⎝⎛
Section modulus of plate
kip-ft 1.78
in) ft/12 1 x in klf(12.1 13.08
wLM22
ep === Bending moment in plate
psi 23,900in 0.8
ft) in/1 k)(12 lbs/1 kip(1,000-ft 1.7SM
3p
pb ===σ
bσ < 27,000 psi - OK
Check Plate Stresses:
VOL. V - PART 2 DATE: 11May2007 SHEET 20 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
SAMPLE DESIGN CALCULATIONS FILE NO. 20.02-20
ote: Plate B will function as a plate washer and no structural design is necessary.
hickness of the EPS layer:
File No. 20.06-6:
L thermalαΔt) = 75.0ft(12in/1ft)(6.5x10-6 per deg. F )120°F = 0.7in Total range of move-
t Backwall L
.7 in)] = 12.3 in
t
NOTE: DESIGN FOR PRESTRESSED CONCRETE BEAMS IS SIMILAR
N T
hickness of EPS layer as per T
t = 120°F Δ
= (LΔ ment at abutment due to temperature
PS = 10(0.01H + 0.67Δ ) E
PSE t = 10[(0.01)(76.0 in)+ (0.67)(0
herefore, use EPS = 13 in. T
VOL. V - PART 2
CHECK LIST FOR FULL INTEGRAL – STEEL BEAMS / GIRDERS
1 Show plan and elevation views of integral abutment and footing plan at a preferred scale of
DATE: 11May2007 SHEET 23 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
CHECK LIST – STEEL BEAMS / GIRDERS FILE NO. 20.02-23
3/8” = 1’-0”. Regardless of skew, the elevation view should be projected down from the plan view. Sections taken through the integral abutment shall be shown at 3/4” = 1’-0” min.
2 Label the location centerline/baseline as shown on the title sheet.
3 “End of slab” shall be used as the reference line for layout of integral abutments. Any
stations and elevations required shall be referenced to this line.
4 Label the skew of abutment in relation to the end of the slab and the skew of the piles in relation to the centerline of integral abutment.
5 The minimum width of integral abutment shall be 2’-6”. If this width is not sufficient, a 3’-0”
width may be used.
6 All ST series and SV series bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of these bars shall be 12”. ST0602 bars between the backwall and the approach slab (where applicable) are not required outside of the exterior beam / girder.
7 Dowels shall be a minimum of 7/8” ø and shall be spaced at a maximum of 12” center-to-
center along the centerline of integral abutment. Locate first dowel 6” beyond the edge of bottom flange and dimension from the centerline of beam/girder. Dowels shall be embedded 12” into footing.
8 The approach slab seat (7”) shall be provided on all integral abutments regardless of whether
or not the bridge will have an approach slab.
9 Show elevation at bottom of beam/girder along centerline of integral abutment. Show elevation at bottom of footing.
10 For details of when to slope the abutment footing or to detail the abutment footing level, see
File No. 20.06-8.
11 The wing shall project a minimum of 12” below the top of footing. For alternate wingwall details, see File No. 20.06-7.
12 AF04 bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of
these bars shall be 12”.
13 The footing shall be designed with a minimum height of 3’-0” and a maximum height of 4’-0”.
14 A 4” clearance between the end of the beam/girder and the end of slab shall be provided. The flanges may be clipped to provide this clearance. In the case of a minimum width abutment and a skew approaching the maximum of 30° the entire end of the beam/girder may need to be clipped to provide this clearance and to allow the anchor bolts to be located on the centerline of the integral abutment.
VOL. V - PART 2
CHECK LIST FOR FULL INTEGRAL – STEEL BEAMS / GIRDERS (Continued)
15 A series of 1
DATE: 11May2007 SHEET 24 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
CHECK LIST – STEEL BEAMS / GIRDERS FILE NO. 20.02-24
1/2” ø holes shall be provided for the SH series bars to pass through. These holes shall be located a minimum of 3” and a maximum of 6” from the flanges and shall be spaced at a maximum spacing of 12”. See File Nos. 20.05-1 thru -4.
16 A distance of 9” minimum to 2’-0” maximum shall be provided between bottom of flange and
top of footing.
17 The fill in front of the integral abutment shall be set at a minimum 9” distance below the top of footing with the edge of berm set at a minimum 3’-0” distance from the face of abutment.
18 Temporary support bolts shall be a minimum 7/8” diameter and shall be embedded a
minimum of 1’-0” into footing.
19 Plate A shall be a minimum of 1/2” thick and 5” wide.
20 The minimum width of plate B is 3” and the minimum length is 5”. The plate shall be 1/2” thick in all cases.
21 For determining the thickness of EPS material, see File No. 20.06-6.
22 For instructions on completing the title block, see File No. 03.03.
23 For instructions on completing the project block, see File No. 03.02.
24 For instructions on developing the CADD sheet number, see File Nos. 01.01-7 and 01.14-4.
25 ST0501, ST0602 and SV0402 shall be galvanized. All other backwall reinforcing steel shall
be epoxy-coated.
26 Approach slabs shall be modified to provide a 10” wide seat.
VOL. V - PART 2
CHECK LIST FOR FULL INTEGRAL – PRESTRESSED CONCRETE BEAMS
1 Show plan and elevation views of integral abutment and footing plan at a preferred scale of
DATE: 11May2007 SHEET 27 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
CHECK LIST – PRESTRESSED CONCRETE BEAMS FILE NO. 20.02.27
3/8” = 1’-0”. Regardless of skew, the elevation view should be projected down from the plan view. Sections taken through the integral abutment shall be shown at 3/4” = 1’-0”.
2 Label the location centerline/baseline as shown on the title sheet.
3 “End of slab” shall be used as the reference line for layout of integral abutments. Any
stations and elevations required shall be referenced to this line.
4 Label the skew of abutment in relation to the end of the slab and the skew of the piles in relation to the centerline of integral abutment.
5 The minimum width of integral abutment shall be 2’-6”. If this width is not sufficient, a 3’-0”
width may be used.
6 All ST series and SV series bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of these bars shall be 12”. ST0602 bars between the backwall and the approach slab (where applicable) are not required outside of the exterior beam / girder.
7 Dowels shall be a minimum of 7/8” ø and shall be spaced at a maximum of 12” center-to-
center along the centerline of integral abutment. Locate first dowel 6” beyond the edge of bottom flange and dimension from the centerline of beam/girder. Dowels shall be embedded 12” into footing.
8 The approach slab seat (7”) shall be provided on all integral abutments regardless of whether
or not the bridge will have an approach slab.
9 Show elevation at bottom of beam/girder along face of integral abutment. Show elevation at bottom of footing.
10 For details of when to slope the abutment footing or to detail the abutment footing level, see
File No. 20.06-8.
11 The wing shall project a minimum of 12” below the top of footing. For alternate wingwall details, see File No. 20.06-7.
12 AF04 bars shall be aligned parallel to the beam/girder centerline. The maximum spacing of
these bars shall be 12”.
13 The footing shall be designed with a minimum height of 3’-0” and a maximum height of 4’-0”.
14 A 6” clearance between the end of the beam/girder and the end of slab shall be provided. This distance is to be shown on the prestressed beam standard.
15 Modify the prestressed beam standard to show the location of 11/2” ø holes in beam web.
The four holes shown are for a PCBT-53. Increase (or decrease) the number of holes by one for every beam depth increment above (or below) the 53” deep section shown. The additional holes shall be placed in 8” increments below the bottom hole shown. Depending on slab thickness, bolster thickness and strand pattern, the designer may need to adjust the hole locations so that adequate cover below the approach slab seat is maintained. Holes must lie within the web.
VOL. V - PART 2
CHECK LIST FOR FULL INTEGRAL – PRESTRESSED CONCRETE BEAMS
(Continued)
16 A distance of 9” minimum to 2’-0” maximum shall be provided between bottom of flange and top of footing.
DATE: 11May2007 SHEET 28 of 28
INTEGRAL / JOINTLESS BRIDGES FULL INTEGRAL ABUTMENTS
CHECK LIST – PRESTRESSED CONCRETE BEAMS FILE NO. 20.02.28
17 The fill in front of the integral abutment shall be set at a minimum 9” distance below the top of
footing with the edge of berm set at a minimum 3’-0” distance from the face of abutment.
18 Top BC0501 shall extend 3” beyond end of beam. Bottom BC0501 and BC0602 shall extend 10”.
19 For determining the thickness of EPS material, see File No. 20.06-6.
20 The insert plate shall be eliminated from the prestressed beam standard.
21 For instructions on completing the title block, see File No. 03.03.
22 For instructions on completing the project block, see File No. 03.02.
23 For instructions on developing the CADD sheet number, see File Nos. 01.01-7 and 01.14-4.
24 ST0501, ST0602 and SV0402 shall be galvanized. All other backwall reinforcing steel shall
be epoxy-coated.
25 Approach slabs shall be modified to provide a 10” wide seat.