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''#4@6.5''

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

7#10@11.25''

on both sides

5#10@11.25''

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