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TRANSCRIPT
1
Part 1
Introduction To Bridge Design
2
How Do Bridge Engineers Decide
On What Type Of Bridge To Build? Bridge Survey
• flood plain cross sections
• inspection reports
• existing bridge (scour, etc)
• water elevations
• photos
• existing roadway profile
Geotechnical Report
• soil / geological formations
• slopes and grading
• foundation problems
• soil prop.’s - phi angles etc
Factors affecting choice of superstructure
• location, city or rural
• span length
• vertical clearance
• maintainability
• environmental concerns
• transportation to site issues
• cost
Factors affecting choice of substructure
• location and geometry
• subsoil conditions
• height of column
3
Bridge Design Process
Preliminary Design Process
• Bridge Survey
• Geotechnical Report
1. Determine the most
economical type structure and
span arrangement
2. Hydraulic Analysis
3. Preliminary Cost Estimate
4. Foundation Borings
5. Determine Foundation Type
Final Design Process
• Top to Bottom Design (twice)
• Design methods per AASHTO and
MoDOT Bridge Manual
• Analysis via
•computations
•spreadsheets
•computer programs
• Detail plans are produced by technicians
(Micro-Station)
• Plans are checked
• Quantities computed
• Special Provisions written
• Plans are advertised for bidding
• Low Bid Contractor builds the bridge
4
Types of Superstructures
Bridges are often referred to by their superstructure types.
The superstructure system of members carry the roadway over a crossing
and transfer load to a substructure.
Superstructures are categorized by;
• Support type (simply supported or continuous)
• Design type (slab on stringer, slab, arch. Rigid frame, etc)
• Material type (steel, concrete, timber)
5
Slab on Stringer Bridges
• Most common type of bridge in Missouri.
• Consist of a deck, resting on the girders. The deck distributes the
loads transversely to the girders.
• The girders carry the loads longitudinally (down the length of the
bridge) to the supports, (abutments and intermediate bents).
• Concrete
• Deck Girder
• Prestressed I Girder
• Prestressed Double Tee
• Prestressed Box
• Steel
• Plate Girder
• Wide Flange
• Steel Box Girder
6
I I I I ---- GIRDERGIRDERGIRDERGIRDER
BULB TEEBULB TEEBULB TEEBULB TEE
Prestressed Girders
7
Prestressed Concrete I-Girder
8
Prestressed Concrete I-Girder Bridge
9
Prestressed Concrete Panels
10
Prestressed Double Tee Girders
11
Steel Plate Girder / Wide Flange Beam / Box Beam
12
Steel Plate Girder Bridge
13
Slab Bridges
In slab bridges the deck itself is the structural frame or the entire deck is a thin
beam acting entirely as one primary member. These types are used where
depth of structure is a critical factor.
Typical Slab Bridges : Concrete Box Culverts Solid Slabs Voided Slabs
14 Box Culvert
Triple Box Culvert
15
Voided Slab Bridge
16 Solid Slab
Voided Slab Bridge
17
Substructures
The substructure transfers the superstructure loads to the foundations.
End Abutments
• Integral Abutment - girders on beam supported by piles, girders “concreted” into the
diaphragm
• Non-Integral Abutment - diaphragms of steel cross-frames, uses expansion devices
• Semi-Deep Abutment - used when spanning divided highways to help shorten span
• Open C.C. Abutment - beam supported by columns and footings, rarely used
Intermediate bents
• Open Concrete Bent - beams supported by columns and footings (or drilled shafts)
either a concrete diaphragm (Pre-Stressed Girder) or steel diaphragm (Plate Girder)
This is the most common type of Pier MoDOT uses.
• Pile Cap Bent - beams supported by piling (HP or C.I.P.) and are used when the
column height is less than 15 feet and usually in rural areas.
• Hammer Head Bent - single oval or rectangular column and footing.
• Spread footings - are used when rock or soil can support the structure.
• Pile footings - rectangular c.c. supported by HP or Cast in Place piles
• Drilled Shafts - holes drilled into bedrock filled with concrete
18
Integral End Abutment
19
Semi-Deep End Abutment
20
Prestressed I-girder intermediate bent
21
Steel girders with open intermediate
bent diaphragms
22
Footing
Pile Cap Column Footing
23
Column Footing
24
Preliminary Design
• Bridge location
• Hydraulic design to determine required
bridge length and profile grade
• Bridge type selection
25
Stream Gage Data
26
Flood-Frequency Rating Curve
0
40000
80000
120000
160000
0 20 40 60 80 100
Return period (years)
Discharge (cfs)
27
Q = discharge (cfs or m3/s)
kc = constant (1.0 for English units or
0.00278 for metric units)
C = Runoff Coefficient
I = Rainfall Intensity (in/hr or mm/hr)
A = Drainage Area (acres or hectares)
Rational Method
AICkQ c ⋅⋅⋅=
28
Drainage Area Delineation
29
n1 n2 n3
Left
Overbank
Right
Overbank
Channel
Stream Valley Cross-sections
30
Manning’s Equation
03
2486.1SRA
nQ ⋅⋅⋅=
n = Roughness Coefficient
A = Area
R = Hydraulic Radius = A / P
P = Wetted Perimeter
S = Hydraulic Gradient (channel slope)
31
n1 n2 n3
Left
Overbank
Right
Overbank
Channel
Stream Valley Cross-sections
32
Energy Equation
Elevation
1 2
Datum Elevation
Pressure
Pressure
Velocity
Velocity
Headloss EGL
HGL
z1
z2
y1
y2
V12/2g
V22/2g
hl
lhg
Vyz
g
Vyz +++=++
22
2
222
2
111
33
Constriction of Valley by Bridge
Opening Length
Bridge Deck/Roadway
34
Encroachment by Roadway Fill
Flood elevation
before encroachment
on floodplain
Fill Fill
Bridge Opening
Encroachment
Backwater
Encroachment
35
Backwater
Normal Water
Surface
Water Surface through Structure
Affect of Bridge on Flood
Elevations
Design High Water
Surface (DHW)
36
Part 2
Slab Design
37
Geometry & Loads
16k 16k
Deck Weight = Width x Thickness x Unit Weight
1 ft x (8.5in x12 in/ft) x 150 lb/cf = 106 lb/ft
38
39
40
Design Moment
• MDL1 = wS2/10 = 0.106 x 82 / 10 = 0.678
• MDL2 = wS2/10 = 0.035 x 82 / 10 = 0.224
• MLL = 0.8(S+2)P/32 = 0.8(8+2)(16)/32 = 4
• MImp = 30% x MLL = 1.2
• Mu = 1.3[0.678+0.224+1.67(4+1.2)] = 12.4
Design For 12.4 k-ft/ft
41
Statics, Moment, Shear, Stress?
42
Reinforced Concrete Design
• Basic Equations For Moment Utilize Whitney
Stress Block Concept
Design Moment = Capacity
12.4 k-ft/ft = φ As fy(d-a/2) φ = 0.90
Compression = Tension
0.85f’cba = As fy
Two Simultaneous Equations, Two Unknowns (a & As)
d
c
Comp.
Tens.
c = a / ββββ1
43
Reinforced Concrete Design
• (0.85)(4ksi)(12in)(a)=(As)(60ksi) a=1.47As
• 12.4k-ft=(0.9)(As)(60ksi)(6in-1.47As/2)/(12in/ft)
• 12.4=27As-3.31As2
• ax2+bx+c=0 a=3.31, b=-27, c=12.4, x=As
• As = [-b - (b2 - 4ac)1/2]/2a
• As = [-27 - ((-27)2-(4)(3.31)(12.4))1/2]/[(2)(3.31)]
• As = 0.49 in2/ft
• 5/8” rebar at 7.5 in centers
d
c
Comp.
Tens.
c = a / ββββ1
44
Part 3
Steel Beam Design
45
Simple Span Beam – 50 ft span
46
Dead Load = Beam Weight + Deck Weight
47
Live Load = HS20 Truck x Distribution Factor
Distribution Factor = S/5.5
48
Design Moment = 2358 kip-ft
49
Design Shear = 214 kips
50
Steel Girder Design • Design Moment = 2358 k-ft
• Design Shear = 214 kips
• Limit Bending Stress
Due To Moment
• Limit Shear Stress
Due to Shear
51
52
Girder Design
• Moment Of Inertia (I)
– 1/12bh3+Ad2
– Parallel Axis Theorem
• Section Modulus = S = I/c
• Stress = Moment/Section Modulus (M/S)
• For Strength Design – Limit Stress to Fy
• Find Shape With S > M/Fy • S > (2358k-ft)(12in/ft)/50ksi = 566 in3
• A W36x170 Provides 580 in3
53
Part 4
Intermediate Bent Design
54
Load Cases
• Permanent Loads:
– DD = Downdrag
– DC = Dead Load
Component
– DW = Dead Load
Wearing Surface
– EH = Horizontal Earth
– ES = Earth Surcharge
– EV = Vertical Earth
– EL = Locked In Forces
• Transient Loads:
– SE = Settlement
– BR = Braking
– CE = Centrifugal Force
– CT = Vehicular
Collision
– CV = Vessel Collision
– EQ = Earthquake
– IC = Ice Load
– FR = Friction
55
Load Cases (Cont.)
• Transient Loads:
– LL = Live Load
– IM = Dynamic Load
– LS = Live Load
Surcharge
– PL = Pedestrian Load
– WL = Wind On Live
Load
– WS = Wind On
Structure
• Transient Loads:
– TG = Temperature
Gradient
– TU = Uniform
Temperature
– CR = Creep
– SH = Shrinkage
– WA = Water Load
56
Load Combinations
Load Combination
Limit State
DC
DD
DW
EH
EV
ES
EL
LL
IM
CE
BR
PL
LS WA WS WL FR
TU
CR
SH TG SE
Use One of These at a Time
EQ IC CT CV
STRENGTH I
(unless noted) γγγγ p
1.75 1.00 -- -- 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
STRENGTH II γγγγ p 1.35 1.00 -- -- 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
STRENGTH III γγγγ p -- 1.00 1.40 -- 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
STRENGTH IV γγγγ p -- 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --
STRENGTH V γγγγ p 1.35 1.00 0.40 1.0 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
EXTREME EVENT I γγγγ p γγγγEQ 1.00 -- -- 1.00 -- -- -- 1.00 -- -- --
EXTREME EVENT II γγγγ p 0.50 1.00 -- -- 1.00 -- -- -- -- 1.00 1.00 1.00
SERVICE I 1.00 1.00 1.00 0.30 1.0 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
SERVICE II 1.00 1.30 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --
SERVICE III 1.00 0.80 1.00 -- -- 1.00 0.50/1.20 γγγγTG γγγγ SE -- -- -- --
SERVIE IV 1.00 -- 1.00 0.70 -- 1.00 0.50/1.20 -- 1.0 -- -- -- --
FATIGUE – LL, IM &
CE ONLY -- 0.75 -- -- -- -- -- -- -- -- -- -- --
57
Water (WA) – Strength
M = (Pbh)(½h)
= ½ Pbh2
½ h
Resultant
P C
on
tra
ctio
n S
co
ur
10
0 y
ea
r P
ier
Sco
ur
10
0 y
ea
r
Q100
b
M
58
Water (WA) - Extreme Event
(Cont.)
( )(b)1000
0.7VForce2
=
Co
ntr
actio
n S
co
ur
50
0 y
ea
r P
ier
Sco
ur
50
0 y
ea
r
Q500
b
B
A ( )(B)
10000.5VForce
2
=
A = ½ Of Water Depth ≤ 10’
B = ½ Sum Of Adjacent Span Length ≤ 45’
Drift Mat
Pressure = CDV2/1000
CD=0.7
CD=0.5
59
Wind on Structure (WS)
P(WS)Vert.
W
¼W
P(WS)Trans. H ½
H
P(WS)Long.
PSub.
PVert. = (20psf)(W)(L)
PTrans. = (50psf)(H)(L)
PLong. = (12psf)(H)(LT)(%)
PSub. = (40psf)(b)
L = Tributary Length
LT = Total Bridge Length
% = Long. Distribution %
b = Column Or Cap Width
60
Wind on Live Load (WL) PTrans. = (100plf)(L)
PLong. = (40plf)(LT)(%)
L = Tributary Length
LT = Total Bridge Length
% = Long. Distribution %
P(WL)Trans.
P(WL)Long. 6’
61
Int. Bent Analysis
62
Cap Beam - Strength Limit State
• Basic Equations For Moment Utilize Whitney
Stress Block Concept
– φ Mn = φ As fy(d-a/2)
– φ = 0.90
de
c
Comp.
Tens.
c = a / ββββ1
63
Cap Beam – Service Limit State • Crack Control
–
– dc = Concrete Cover To Center Of Closest Bar
– fs = Service Tensile Stress In Reinforcement
– h = Overall Section Thickness
– γγγγe = 1.00 For Class 1 Exposure (Crack Width = 0.017”)
= 0.75 For Class 2 Exposure (Crack Width = 0.013”)
)d0.7(h
d1
c
cs
−−−−++++====ββββ2dc
700s
ss
e −−−−≥≥≥≥fββββ
γγγγ
64
Cap Beam Service Limit State
• Crack Control Is Based On A Physical Model
x
h
dc
fc1
fc2
fs/n
l l Crack Spacing
Primary Tension Reinforcement
fc1
fc2
fs/n
fc1
fc2
fs/n
l = =16.03”
s s
( )2 2 c 2
sd2 +
dc
65
Simplified Shear Design
• LRFD
– φ Vn = φ (Vc + Vs + Vp)(kips) φ = 0.90
–
– αααα Set At 90°
– Set: ββββ=2.0, θ θ θ θ =45°
–
– Results In:
vvcc d b ' 0.0316V fββββ====s
)sincot(cotdAV
vyv
s
ααααααααθθθθ ++++====
f
Lbs To Convert To 1000By Multiply V c
vvcc d b ' 2.00V f====s
dAV
vyv
s
f====
0.0
66
Simplifed Shear Design
Section A-A
5 -
#6
’s
(Ea
ch
Fa
ce
)
6 - #9’s
6 - #9’s
#5’s @ 12” or 6” A
A
-400
-200
0
200
400
67
Column Design
Column 42” Diameter
-1000
3500 P (kip)
(P max)
(P min)
1800
M (k-ft)
Controlling Point
Axial Load – Moment Interaction Diagram
18-#9 Bars