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BRIDGE SUBSTRUCTURE NOTES
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1
Section 3
Substructure
Abutment Pier
Wingwall
2
I) ABUTMENTS a) DEAD LOAD (from super structure)
INTERIOR BEAM
Asphalt ( )( ) 1866.014.0812
2=
Concrete Slab ( )( ) 75.015.0812
25.7=
Steel BM + Plate 22.0= 1.16 k/ft.
Reaction 1.55)16.1(2
95== K
EXTERIOR BEAM
Asphalt ( ) 1633.014.02
83
12
2=!
"
#$%
&+
Conc. Slab ( )( ) 656.015.072̀1
25.7=
Barrier, Curb+Railing ( ) 47.002.015.012
94 =+!
"
#$%
&
Steel Beam 0.22 1.51 k/ft.
Reaction ( ) 73.7151.12
95== K
Total Dead Load ( ) ( ) 66.25373.7121.552 =+= K b) LIVE LOAD (from Super Structure – No Impact)
32k 32k 8k
14' 14'
Truck Loading Reaction = 64.93 K For 2 Lanes 129.86 K Lane Loading 56.4 K / Lane (table)
c) LONGITUDINAL FORCE ( ) 49.605.086.129 == K
3
d) LEFT ABUTMENT Preliminary dimensions were considered in deciding on Talbot’s Constraints Left and Right Abutment had almost same depth, then same procedure applies for both of them. Left Abutment Pier and Abutment should be at least 6 ft. below stream bed. Using Coulomb’s Earth Pressure
!"
#$%
&'=2
45tan2
22 ()H
Pa
o30=! 120=! lb/ft2
( ) 2222030tan
2
120HHPa =!=
If the width of Abutment = 30’
4
5
1' 3'
PLAN
3'
30'
3'
3' 1' 3' 3'
Neglect710 lb/ft2
(120)(20.75)/3=830 lb/ft
3' 4' 3'
3'
DL+LL
6.5k
Overturning Moment: Loading Magnitude M. Arm Moment (k-ft)
Back Soil ( )
( )302
2183.0! K 6.917’ -1808.45
Long. Force -6.5 K 17.0’ -110.5 DL 253.66 K 4.5’ 1141.5 LL 129.9 K 4.5’ 584.6 Wt. Of Footing 162 K 5’ 810 Wt. Of Abutment 269 K 5.094’ 1371.3
( ) 5.114475.31 !!+! 1988.5 k-ft. Overturning Moment Including LL = 1988.5 k-ft.
6
Factor of Safety against Overturning:
F-S for D.L ( )( )
5.173.15.11045.1808
5.11413.1371810>=
+
++= O.K
F-S for D.L+L.L ( )( )
0.210.25.11045.1808
6.5848.3328>=
+
+= O.K Slightly Larger
Stresses at Tip of Footing:
( )( )( ) ( ) ( )( )( ) ( )( )5.01307.253115.03075.3145.692.530
2
75.17710.0++!+"#
$%&
'=M
= 1384.4 k-ft. ( ) 8151622691307.253 =+++=P K
M
P
A B
5' 5'
Stresses:
( )( )
043.0308.2264.2
12
1036
54.1384
3610
815
3!=!=
""
#
$
%%
&
'(
!=A) k/ft2 (tension)
( )( )
725.24615.0264.2
12
1036
4.1384
3610
815
3=+=
!!
"
#
$$
%
&'
+=B( k/ft2 (comp.)
Since Tension is observed under Footing Increase size of footing to 12 ft. instead of 10 ft.
7
Load Magnitude Arm Moment (k-ft)
Soil ( )
( )302
2183.0! 6.917’ -1808.5
Long. Force -6.5 17’ -110.5 DL 253.7 5.5’ 1395.4 LL 130 5.5’ 715 Wt. Of Footing 194.4 6’ 1166.4 Wt. Of Abut. 252.0 6’ 1512.0 Wt. Of Abut. 16.9 7.5’ 126.8 2996.6 k-ft. Overturning Moment = 2996.60 k-ft.
F-S for D.L ( )
5.12.25.1105.1808
8.12615124.11664.1395>=
+
+++= O.K
F-S for L.L 256.25.1105.1808
6.4915>=
+= O.K
8
Stresses:
( ) 8477.2534.1942529.16 ++++=P K
( )( )
359.0602.1961.1
12
1236
64.1384
3612
847
3=!=
"
!=A# k/ft2 (comp.)
( )( )
562.3602.1961.1
12
1236
64.1384
3612
847
3=+=
!
+=B" k/ft2 (comp.) O.K
II. DESIGN OF PIER a) Dead Load Reaction Interior beams = 2(55.1) = 110.2 K Exterior beams = 2(71.73) = 143.5 K Total D.L = 2(110.2+143.5) = 507.4 K b) Live Load Reaction
Load / wheel ( ) 05.3395
149516416 =!
"
#$%
& '++= K
For Max. Eccentricity of Live Load, Consider One Lane Loading only.
9
Select a Hammer-Head Pier
3'
4'
13'
4'
4' 8' 4'
24'=(595.7-583.7)+6'+2'+4'
ELEVATION
A A
D.L---
L.L---
143.5k 110k110k 143.5k
33k 33k
2' 8' 8' 8' 2'
10
3'
3'
3'
4' 4'8'
PLAN Stability in the Transverse Direction
L.L
D.L
1
2
3
5 4
6
7
Tip
11
Load Type Force M Arm Moment (k-ft) D.L 143.5 K 20 2870 D.L 110.0 12 1320 D.L 110.0 4 440 D.L -143.5 4 -574 L.L 33 18 594 L.L 33 12 396 Wt. of Pier (footing) 86.4 (1) 8 691.2 Wt. of Pier 61.2 (2) 8 489.6 Wt. of Pier 35.1 (3) 13 456.3 Wt. of Pier 9 (5) 15.3 138 Wt. of Pier -5.4 (7) 2 -10.8 Wt. of Pier 5.76 (4) 1.8 10.4 Wt. of Pier -8.1 (6) 3 -24.3
6796.4 k-ft.
Overturning Moment = 6796.4 k-ft.
F-S due to D.L ( )
5.153.93.248.10574
3965944.6796>=
++
!!= O.K
F-S due to L.L + D.L 22.111.609
4.6796>== O.K
Stresses: Full L.L M=0 & P=784 K Partial L.L M=528 k-ft. & P= 751 K Partial L.L:
( )( )
590.6
12
169
8528
916
751
3=
!
+=A" ksf (comp.)
( )( )
840.3
12
169
8528
916
751
3=
!
"=B# ksf (comp.)
Full L.L:
( )44.5
916
784=== BA !! ksf (comp.)
12
13
III. DESIGN OF WING WALLS:
10'
3'
4.75'
8'
Wing
Wall
3.75'girder
4'
4'
Abutment
Wing Wall Abutment
Bridge
8'
14
Bending @ X-X:
( ) ( ) 79.200.33
8
2
75.4875.733
3
14383
3
1
1000
120=!"#
$
%&'
()*
+,-
. "+++""")
*
+,-
.=M k-ft.
8'
3'
4.75'
X
X
15
16
DESIGN OF PIER DEAD LOAD REACTION: Interior beams = 2(55.64) = 11.28 K Exterior beams = 2(75.04) = 150.08 K Total (D.L) reaction = 2(111.28+150.8) = 522.72 K LIVE LOAD REACTION:
Load / wheel ( ) 0.3395
149516416 =!
"
#$%
& '++= K
For max. Eccentricity of Live Load, consider only one lane loading.
17
18
Overturning Moment: Load Type Force M Arm Moment (k-ft) @ Toe DL 150.1 K 20 3002 DL 111.3 12 1335.6 DL 111.3 4 445.2 DL -150.1 4 -600.4 LL 33.1 4 132.4 LL -33.1 2 -66.2 Wt. of Footing 86.4 (1) 8 691.2 64.8 (2) 8 518.4 35.1 (3) 13 456.3 9 (5) 15.3 138 -5.4 (7) 2 -10.8 5.76 (4) 1.8 10.4 -8.1 (3) 3 -24.3 6027.8
19
If L.L is placed on right side; before, when LL on left 6796.4 k-ft ∴Overturning Moment = 6027.8 k-ft
F-S against D.L ( )
5.13.87.701
2.5833
3.248.102.664.600
2.28313002>==
+++
+= O.K
20
F-S against L.L+D.L 25.87.701
4.1322.5833>=
+= O.K
Notice: That for previous case when L.L was considered on left side provided higher safety factors (F-SDL = 9.5 + F-SLL+DL = 11.2) ∴case is more critical (i.e. LL on right side of pier) Stresses:
M
P
A B a) Partial L.L ( ) 4.4634101.33 =+=M k-ft & 56.803=P K
( )
( )79.6
12
169
84.463
916
6.803
3=
!
+=A" ksi &
4.421.158.5 =!=B" ksi O.K b) Full Live Load M = 0 P = 870 K
( )
0.6916
870=== BA !! ksi (comp) O.K
21
DESIGN OF REINFORCEMENT IN THE PIER:
3'
4'
2' 2' 6'
2' 8'
150.1k
111.3k
33.1k 33.1k
L.L
D.L
Exteririor
girder
Interiorgirder
1 2
3
6'
a) Shear f’c = 3000 psi , Grade 60 Consider section directly underneath exterior girder Section 1-1: At 2’
( ) 1.1993.115.032
8.023321.150 =!
"
#$%
&'(
)*+
,-
-+--+=uV K
b = 3 ft. h = 3.8 ft. d = 3.4 ft.
( ) ( )( ) 137124.31231000
3000285.0 =!!=" cV K
cu VV !>
621371.199 =!="!=" cus VVV K and shear reinforcement is needed. Using
No. 4 bars (closed stirrups)
22
( )
( )( )( )4.13
62000
124.3600004.0=
!=
"#
"=
cu
yv
VV
dfAs in.
∴Use # 4 @ 13 in.
max ( )( )
''13''3.133650
10604.0
50
3
>===w
yv
b
fAs O.K (0-1)
At Section 2-2:
( ) ( ) 27615.032
6.143341.1503.117.21.33 =!
"
#$%
&''(
)**+
,'(
)*+
,-
-+--++uV K
b = 3 ft. , h = 4.6 ft. d = 4.3 ft.
( ) ( )( ) 0.173123.41231000
3000285.0 =!!=" cV K
∴Need Shear reinforcement Since 103=!"=#<! cusuc VVVVV K
Using #4 stirrup (closed) (0.44 in2)
( )( )( )
''22.10103
123.4600004.085.0=
!=
"#
"=
cu
yv
VV
dfAs
∴ Use #4 @ 10 in. (10 in.< smax=13 in.) O.K (1-2) Section 3-3:
( ) ( ) 72.51217.22.663.13.11115.032
10433101.150 =+!
"
#$%
&+'
(
)*+
,-
-+--+=uV K
b = 3 ft. h = 7 ft. d = 6.7 ft.
( ) ( )( ) uc VV <=!!=" 5.269127.61231000
3000285.0
2.243=sV K
( )( )( )''74.6
2.243
127.660004.085.0=
!="s
Use #4 @ 6 in. (2-3) b) Flexure Consider bending moment at section 3
( )( )( ) ( ) ( ) 234061.3317.23
103
2
104515.031081.1503.1 =!+"
#
$%&
'()
*+,
-!++!=uM k-ft.
Using 3 in. cover & #10 bars d = 84-3-0.63 = 80.3 in.
23
( )( )( )
( )( )!"
#$%
&
'('='
3.801233
606.013.80609.0234012
ss
AA
048.61015.423
=+!"!
ss AA
As = 6.66 in2 Using #10 bars
No. of bars 24.527.1
66.6== bars
∴Use #10 bars @ 6 in. c) Design of Stem (Column)
Check slenderness ( )( )
225.883.0
172.1<==
!
kl
∴It is not slender
( ) ( ) ( )( )( ) 59515.083173.11121.1502 =+++=DLP K
( ) 2.661.332 ==LLP K
( ) ( ) 4.4631.33101.334 =+=LLM k-ft.
( ) ( ) 2.9172.6617.25953.1 =+=uP K
( ) 10064.46317.2 ==uM k-ft.
e = 1.1 in. (M = P.e)
24
Assume 1% steel ( )( ) 6.34100
0.11238
2 =!!=sA in2
Using #10 bars No. of bars 2827.1
6.34!=
( ) 103712381000
30001.0'1.02.917
2 =!!"#
$%&
'=<= ccu AfP K
Arrangement of Bars: 5 #10 on each 3 ft. side 7 #10 on each 8 ft. side
8'
3'As=24#10
( )( )( )
0019.0127.7123
27.15=
!!=
bd
As
( )( )( ) 240513
600019.06.017.76027.157.0
'6.017.0 =!
"
#$%
&'(
)*+
,--.=
!!"
#
$$%
&
''(
)**+
,.=
c
ysysu
f
f
bd
AdfAM
10062004 >=uM O.K
(Mu was based only on As = 5 #10 at bottom only, ignoring compression steel and steel on both sides) As ties use #4 at b = 36 in. (min. dimension of column)
25
DESIGN OF FOOTING (PIER):
Bending: Moment at face of pier
( )( ) 8.3523
89457.029414.4 =+!!!=M k-ft.
( ) 2.5298.3525.1 ==uM K (mostly DL)
d = 4-0.4 = 3.6 ft.
!"
#$%
&'=
'6.019.0
c
ysysu
bdf
fAdfAM
26
4.6350=uM k-in.
8.23329.0 =df y
!"
#$%
&
'''(=)
3126.39
606.01772.2
ss
AA
0105.2772.223=!+"
"ss AA
then As = 2.75 in 2 Use #10 bars at 18 in. in both direction (long. and tran.)
27
DESIGN OF PILES: Two cases (1) P = 804 K ; M = 463.4 k-ft. (2) P = 870 K ; M = 0 2' 3' 3' 3' 3' 2'
2'
5'
2'
Use HP 4210! A = 12.4 in2 / pile; Pall = 12.4(9) = 112 K If only vertical load is applied
# of piles 18.7112
804==
Use 10 piles; min spacing = 3 ft. > 2.5 (AASHTO) O.K Edge distance = 24-6 = 18 in > 9 in (AASHTO) O.K
( ) 5.625.2102 ==xI
( ) ( ) 180346422 =+=yI
4.8010
804==
n
P
0=!x
x
I
cM
( )5.15
180
64.463==
y
y
I
cM
Max load / pile, pile no. 1 = 80.4+15.5 = 96 < 112 O.K
28
REINFORCEMENT IN THE WING WALL: Choose a 1 ft. wide wing wall M = 20.21 k-ft. (see page 6/5) ∴Mu = 1.3(20.21) = 26.3 k-ft. (small) Use min reinforcement
( )( ) 16.212125.7002.002. =!== btAs in2
For #4 bars
No. of bars 8.102.0
16.2==
Then use 11 #4 bars @ 8 in. In the horizontal direction where as in the vertical direction use #4 bars @12 in. (cover = 2 in.)
#4@8''
#4@12''
8'
3'
4.5'
Abutment
29
DESIGN OF REINFORCEMENT IN ABUTMENTS: a) Back Seat Retaining Wall Consider a 1 ft. strip (depth)
( )352.0
3
75.3
2
75.315.0./ =!=ftM k-ft / ft.
83.012
21 =!=d ft
( ) 46.0352.03.1 ==uM k-ft / ft
As< As min (small moment) ∴As min = 0.002(12)(12) = 0.28 in2 Use #4 at 8 in in both directions
1'
3.75'
(120)(3.75)/3=0.15
b) Stem
Check slenderness ( )( )
222143.0
148.1<==
!
kl (for this case 8.1!k )
No moment magnification Total over 30 ft width.
1.5298.152524.261 =++=DLP K
130=LLP K
30
( ) ( )( ) ( ) 10585.07.2532.530142
14.071.0=+!"
#
$%&
' +=DLM k-ft.
156=LLM k-ft.
14'
Design / ft Stripe
32330
13017.2
30
1.5293.1 =!
"
#$%
&+!"
#$%
&=uP K
1.5730
15617.2
30
10583.1 =!
"
#$%
&+!"
#$%
&=uM k-ft.
Use min steel ( )( ) 3.1128.3120025.00025.0min =!="= sA# in2
Try #10 @12 in C – T = P PfAbaf ysc =!'85.0
4.3'85.0
=+
=bf
fAPa
c
ys in.
!"
#$%
&'=!
"
#$%
&'=
2'85.0
2
adbaf
adcM cn
( )( )( ) 45672
4.3128.34.312385.0 =!
"
#$%
&'(=nM k-in.
= 380 k-ft > 57.10 k-ft. O.K Then use: #10 @ 12 in all over wall vertically #10 @18 in all over wall horizontally
31
DESIGN OF FOOTING
4.6973.1681.529 =+=DLP K
130=LLP K
1189=uP K
15.1232=DLM k-ft
156=LLM k-ft
1940=uM k-ft
23.4=+=I
Mc
A
P
B! ksf
19.0=!=I
Mc
A
P
A" ksf
Flexure:
( ) ( ) ( )( ) ( ) 5.2425.3
6
55.323.4
2
15.355.3 22
=!
+=uM k-ft / ft.
!"
#$%
&'
'
6.019.0
c
ysysu
bdf
fAdfAM
As = 0.19 in2 < As min Use min reinforcement #10 bars @ 18 in. in both directions Same reinforcement applies for both abutment since they were chosen identical.
32
6#10@6''
#4
3'
variable h
4#10@8''
3''
SECTION B-B
#4@13''#4@10''#[email protected]''
3'
4'
14'
4'
#10@18''
#4@36''
#4@8''
28'
#6@6''
ELEVATION
2'
5'
2'
2' 3' 3' 3' 3' 2'
8'3'
PLAN
HP 10x42 Pile
#10@18''
both directions
Allow for 1%
slope in Roadway
direction
on both sides
on both sides
Stirrup to tie bars
2#4, L=46''
SECTION A-AAs=24#10
cover=3''(typical)
33
30'
ELEVATION
#4@8''
#10@18''
#4@8''
#10@12''
SECTION
#10@18''
#10@12''
#10@18''
3.5' 3.5'
3'
14'
3.75'
1' 3'
for expansion joint
#10@12''
2' 30' 2'
#10@18''
#10@18''
3.5'
4'
3.5'
PLAN
ABUTMENTS(TYPICAL)
34
8'
3'
7.5'
#4@8''
#4@12''
WING WALL
