Lecture12 Abutments ( Highway Engineering )

Lecture12 Abutments ( Highway Engineering )

1. Lecture 17

Sub-structure:
Design of the Abutments

2. EXAMPLE 4.1

DESIGN OF A STUB ABUTMENT WITH pet |!
SEISMIC DESIGN CODE 14

PROBLEM: Design a stub abutment to accommodate given

ns from a composite steel superstructure.

LS

A 3 span (29'-63'-29') essential bridge crossing a highway.
1'-0' diameter concrete piles - 40 ft long. Capacity = 30 tons
18 pairs of piles at 6'-8° center-to-center along length of footer.
Concrete strength f,’ = 3,000 psi.
Grade 60 reinforcement 1, = 24,000 psi
) Total reaction from all stringers R = 315 k.

Deck Weight = 21.74 k/ft

® Geographic area has acceleration coefficient: A = 0.19.
Soil tests indicate stiff clay with angle of friction: & = 30°.

DESIGN EXAMPLE
GIVEN PARAMETERS

In this example we will design a stub
abutment to accommodate both struc-
ture and seismic loads. The abutment
geometry in cross section is illustrated
in the accompanying calculation sheet
In plan, the abutment is equipped with
18 pairs of 1 ft diameter, 40 ft long
concrete piles spaced at a distance of
6'-8" center to center. The face of the
backwall is assumed to be smooth

It can be assumed that the abutment
can be displaced horizontally without
significant restraint. Given this condi-
tion, we will utilize the pseudo-static
Mononobe-Okabe method for comput.
ing lateral active soil pressures under
seismic loading. All references to the
AASHTO seismic code contained in
Division I-A of that specification will
be followed by the "I-A" identifier (e.g.,
AASHTO 3.3 I-A)

The loads from 16 stringers which are
transmitted to the abutment are given
as 315 k. The bridge is located in a
geographic area which, when refer-
enced to Figure 3.15, provides an ac-
celeration coefficient of 0.19. Assume
earth with a unit weight of 120 pounds
per cubic foot and concrete with a unit
weight of 150 pounds per cubic foot

3. | DESIGN OF A STUB ABUTMENT WITH
EXAMPLE 41° |SEISMIC DESIGN CODE BET

Determine Type of Seismic Analysis and Other Criteria

2
14

STEP 1:
Hak an “essential bridg by AASHTO 3.3 I-A we have:

= Importance Classification = 1
With an importance classification of "I" by AASHTO 3.4 I-A for:
0.09 < A < 0.19 and IC =1
SPC = Seismic Performance Category = B

The bridge has an unchanging cross section, with similar sup-
ports, and a uniform mass and stiffness, so it is considered to be:

Regular (By AASHTO 4.2 I-A)
For a Regular Bridge, SPC = B, and 2 or more spans, use
Method 1 = Single-Mode Spectral Analysis (By AASHTO 4.2 I-A)
For stiff clay, by AASHTO 3.5 I-A: u

= Site Coefficient = 1.2
Response a Factor (AASHTO 3.6 I-A)
Abutment Stem: R= 2 (Treat as a wall-type pier)

Abutment Footing: R = One-Half R for Abutment Stem
R/2=2/2=1 (AASHTO 4.7.2 1-A)

Soil Profile Type Il:

Combination of Seismic Forces (AASHTO 4.4 I-A)
Case |: — [100% Longitudinal Motion] + 130% Transverse Motion]

Case II: [30% Longitudinal Motion]

Load Grouping (AASHTO 4.7.1 I-A):

1.0(DL + Buoyancy + Stream Flow + Earth Pressure + Earthquake)
(N/A) (N/A)

100% Transverse Motion]

DESIGN EXAMPLE 4.1
STEP 1: DETERMINE SEISMIC

ANALYSIS AND CRITERIA
In Section 3.5.3, Part 2 we discussed the
nature of temporary loads, in general,
and seismic loads in particular. Prior to
beginning design, we need to determine
the seismic criteria which will determine
which type of analysis will be required:
either the more straightforward single-
mode spectral analysis or the more com-
plex multimode spectral analysis.

The type of analysis used is dependent
on the geometry of the structure (i.e., is
the cross section uniform or not) and the
function of the bridge in the transporta-
tion network (i.e., essential or non-es-
sential). Based on our given parameters
for the structure, we determine that the
single-mode spectral analysis method
will be sufficient.

Other seismic criteria which need to be
calculated are the response modification
factors which will be used to determine
seismic design forces for individual
members by dividing elastic forces by
the R factors. Finally, the load combina-
tions and groupings are defined. Since
the bridge crosses a highway, we don’t
need to consider buoyancy and stream
flow pressure loads.

5. AASHTO SPECIFICATION
DIVISION I-A
3.3 IMPORTANCE CLASSIFICATION
3.4 SEISMIC PERFORMANCE
CATEGORY

Essential bridges with an acceleration
coëficient greater han 0.29

an importance classification (IC)

Alo

An esse

mined to be critical to

and securiy/delense

KA, 3.3). Based on the IC and the acc
eration coefficient at the bridge site, a
seismic performance catagory (SPC) can
be determined from the table below
(AASHTO, FA, 34)

——
AASHTO SPECIFICATION
DIVISION I-A

4.2 CHOOSING THE APPROPRIATE
SEISMIC ANALYSIS METHOD

Once a seismic performance calegory
(SPC) has been assigned, the type 0
analysis requited is identified based on
the SPC and whether the bridges regular
or irregular. Regular bridges are those
with an unchanging bridge cross sec-
tion, similar supports, and a uniform
mass and stitiness. Bridges which do no
salisty these iter are irregular. Thetwo
methods are applicable to multispan
bricges only (AASHTO, I-A, 42)

Method 1 = Single-Mode Spectel Analysis
Method 2 = Mullimoce Spectral Analysis

AASHTO SPECIFICATION
DIVISION I-
3.6 RESPONSE MODIFICATION —
FACTORS (R FACTOR)
For bridges with a SPC = B, C, or D, the
seismic design forces for individual mem-
culated by dividing the els

Conc. Pile Bent (Vertical Piles)
Conc. Pile Bent (1+ Baïer Piles)
Single Columns

Steel Pile Bent (1+ Batter Piles)
Multi

Conn

2
~ Bridges with 2 or More Spans Only

Superstructure to Abutment 08
Exp. Joints within Span of Super. 0.8
Columns, Piers, or Pile Bents to
Cao Beam or Superstructure 1.0
Colums or Piers to Foundations

AASHTO SPECIFICATION
DIVISION I-A

3.5 SITE COEFFICIENT Bu
As stated at th ginning of this sec-
tion, the type of soil present at the
bridge site plays an important role in
the forces an earthquake exerts on a
structure. The site coefficient is deter-
‘mined by selecting one of three soil
profile tyoes that best fits the condi-
tions atthe site. (AASHTO, I-A, 3.5.1).

SOIL PROFILE TYPE I: S=1.0
IF rock of any type is present, this profile
‘ypeaoplies. Shalelike orcrystallinetypes
wilh a shear wave velocity greater than
2,500 fs (762 m/s). Stitt soil on top of
rock witha depth less than 200 ft (61 m)
Consisting of stable deposits of sands,
graves, or stil clays

SOIL PROFILE TYPE I: S=1.2
For stif clay or deep cohesionless soll
Conditions along with sites where the
soll depth on top of rock is greater than
200 t (61 m) consisting of stable depos
gravels, or tif clays. I the

oil proper-

the three types, this

1.2) is used for all

SOIL PROFILE TYPE
Soft to medium-stff clays

S=15
sands

6. A counter-fort retaining wall

W.= WEIGHT OF M

7. DESIGN OF A STUB ABUTMENT WITH
SEISMIC DESIGN CODE

STEP 2: Compute Seismic Active Earth Pressure

Using the Mononobe-Okabe Equation (AASHTO C6.3.2 I-A):

1 2
Po (1)

Where the seismic active pressure coefficient is defined as
mn u c0s?(9-0-B)
AE w-cosd- cos?B- cos(5 +B +0)
where:
© = Angle of Friction of Soil = 30° (Given)
3 = Angle of Friction Between Soil and Abutment = 0 (Smooth)
i = Backfill Slope Angle = O (Level Backfill)
B = Slope of Soil Face = 0 (Vertical Rear Face of Stem)

EXAMPLE 4.1

13

and where
= Horizontal Acceleration Coefficient
e= ara 1.5A = (1.5)(0.19) = 0.285
, = Vertical Acceleration Coefficient
) 0.3k, < k, < 0.5k,
(0.3)(0.285) < k, < (0.5)(0.285)

un 0.0855 < k, < 0.1425 so use:
6 =17.57° = 18° k, = 0.10

/
= ATAN

-0.1

Check horizontal acceleration coefficient (AASHTO C6.3.2 I-A):
K, < (1 - k)tan(@ - i) so: 0.285 < (1 - 0.10)(tan(30 - 0))< 0.52 u

+B+@)-cos| aT

(Step 2: Continued on Next Sheet)

DESIGN EXAMPLE 4.1
STEP 2: COMPUTE SEISMIC
ACTIVE EARTH PRESSURE

The AASHTO seismic code specifies that,
for abutments, the pseudo-static Monon-
obe-Okabe analysis method is to be used
(AASHTO 6.3.2 I-A). This method uti-
lizes a modified Coulomb equation to
compute the active earth pressure acting
on the abutment.

This method makes use of the accel-
eration coefficient (based on the geo-
graphic location of the bridge), which
is decomposed into horizontal and
vertical components. For free stand-
ing abutments, the horizontal accel-
eration coefficient k, is taken as half of
the acceleration coefficient A (i.e., A /
2). However, if batter piles are present
as exist in our case, then the value for
k, is taken as 1.5A. The range for the
vertical acceleration coefficient given
is a generally accepted limit.

To compute the active earth pressure
we also need to identity certain physi-
cal parameters concerning the soil
present and the abutment physical
characteristics. The horizontal accel-
eration coefficient is checked to en-
sure that it falls under the AASHTO
specified limit.

8. DESIGN OF A STUB ABUTMENT WITH
| SEISMIC DESIGN CODE

STEP 2: Compute Seismic Active Earth Pressure (Continued)
nn ft {sin(30- 0): sin(30- 18-0) |
1 cos(0+0+18)-cos(0—0)
Recall that the seismic active pressure coefficient is defined as
me cost(9-0-B)
AE w- cos@- cos’B - cos(5 +B +0)

se cos” (30-180)
~ (1.7705). cos(18)- cos?(0)- cos(0+0+18)

EXAMPLE 4.1 oer | 4

= 1.7705

= 0.5974

Also recall that the Mononobe-Okabe Equation is defined as:

1
Pero? H*.(1—k,)-K

So for the whole wall:

PL + (120 1b /1t3)C11 ft)” (1—0.10)(0.5974) — 3,903.4 Ib / ft

STEP 3: Compute Static Active Earth Pressure

AE

The static active earth pressure coefficient is defined as:
cos*(~B)
A y-cos*B-cos(8 +B)
where:

Y c0s(0)- costo)

EEE Le {sin(30)- sin(30) |

bé loas) cos(i—p) | —

y=2.25

(Step 3: Continued on Next Sheet)

DESIGN EXAMPLE 4.1
STEP EISMIC ACTIVE EARTH
PRESSURE (CONTINUED) _
The seismic active earth pressure re
quires the computation of one more
constant, after which the active earth
pressure coefficient K,, and the result-
ant P,¿ value can be computed

We will need to calculate two values,
one for the stem and the other for the
whole wall. These pressures will be
used to describe the dynamic pressure
acting on the abutment. The siatic pres-
sure acting on the abutment will be
computed in the next step using the
traditional Coulomb expression (i.e.
=k,=0)

The dynamic pressure will be taken to
act at a distance of 0.6H above the top
of footing and the static a distance of
H/3 above the top of footing.

STEP 3: COMPUTE STATIC ACTIVE

EARTH PRESSURE
As stated above, for the static active
earth pressure, € and k, = 0. So, we
need to recompute the active earth
Pressure coefficient for the static con-
dition (designated by K,). Firstwe must
recompute the value of y for the static
condition.

9. DESIGN OF A STUB ABUTMENT WITH

SEISMIC DESIGN CODE L

STEP 3: Compute Static Active Earth Pressure (Continued)

Recall that the static active pressure coefficient is defined as:
cos’(p-B)

A wy-cos*p- cos(6 +B)

u c0s?(30- 0)
(2.25)- cos*(0)- cos(0 + 0)

EXAMPLE 4.1

= 0.3333

DESIGN EXAMPLE 4.1
STEP 3: STATIC ACTIVE EARTH
PRESSURE (CONTINUED)

We now compute the static active earth
pressure coefficient using @ and k, = 0.
Substituting in, we determine the ‘final
static active earth pressure acting over
the whole wall. The reader should keep
in mind that these values are per unit
foot of wall length.

So the static active earth pressure is defined as:

1 2
Pager HK,

So for the whole wall:

= (120 tb / #t?)(11 10) (0.3333) = 2,420.0 Ib / ft

STEP 4: Compute Equivalent Pressure
Determine a single, equivalent pressure based on:

Static pressure acting at H / 3
Seismic pressure acting at 0.6H

«(rs -P,)-0.50H]

2420 lb / ft ue +[(8903 lb /ft- 2420 Ib / ft). 0.60.11 ft]

(2420 tb / tt) (3)

Use an equivalent pressure of: P,. =F,’ P, = (2.103)P,

STEP 4: DETERMINE EQUIVALENT
ACTIVE PRESSURE

The static component of the soil force
acts at the traditional H/3 distance from
the bottom of the abutment. The addi-
tional dynamic effect is taken to act at
the height of 0.60H from the bottom of
the abutment.

We will calculate an equivalent active
pressure by taking the ratio shown in
the accompanying calculation sheet.
This ratio is known as a thrust factor.
This thrust factor is multiplied by the
static active earth pressure to provide
a single, equivalent load acting on the
abutment.

As we will see later, the active pressure
will be taken to act over a unit length
defined by the center to center dis-
tance between piles.

10. DESIGN OF A STUB ABUTMENT WITH
SEISMIC DESIGN CODE

STEP 5: Compute Abutment Loads
For all dimensions refer to figure on Calculation Sheet 1

per ©

EXAMPLE 4.1 |

DL ley NOTE:

E Do For all loads W, and DL there is

Y, y a corresponding load KW, and

—” [EW k DL acting upward (k, > 0) or
downward (k, < 0)

pile pair basis. Recall that the
distance between piles along
the length of footing was given
as 6'-8" = 6.667 ft.

Use the following unit weights:

CONCRETE = 150 Ibyft?
SOIL = 120 Ib/ft?

W, -
= AlIW, loads are based on a per
y

Compute all W, loads from abutment and soil:

W = (Height)(Width)(Pile Distance)(Weight)

w, (4.0 ft)(1.667 ft)(6.667 #X0.150 k / ft?) = 6. 7k
W, = (4.0 ft)(3.333 ft)(6.667 ft)(0.150 k / ft°) = 13.3 k
W, = (3.0 ft)(6.0 ft)(6.667 ft)(0.150 k / ft?) = 18.0 k
W, = (8.0 ft)(0.333 ft)(6.667 1t)(0.120 k / ft°) = 2.1 k

pi) = =
Recall that the dead load from the superstructure was given as:
DL = 315 k/ 18 Pairs of Piles = 17.5 k / Pair

14

DESIGN EXAMPLE 4.1.
STEP 5: COMPUTE ABUTMENT
LOADS

We must now compute the loads due
to the abutment and soil at the rear face
over the footing along with the super-
structure dead loads acting on the abut
ment. For this example we will assume
the effects of passive pressure to be
negligible

The loading diagram shows the loads
induced by the weight of each section
of the abutment (W,) as well as the
horizontal component of the seismic
loads (k,). The reader should keep in
mind, however, that a vertical compo-
nent of seismic loads, k,, isalso present
but not illustrated in the diagram. These
loads will either act upward or down-
ward, depending on the sign of k,

The loads are computed based on a
per pile pair basis. We were given a
distance of 6-8" between piles at the
beginning of this example. Also given
were the unit weights of concrete and
soil to be used.

The last load to be computed in this
step is that of the superstructure dead
loads, which are distributed over the
18 pairs of piles.

11. DESIGN OF A STUB ABUTMENT WITH per | Z
SEISMIC DESIGN CODE |

STEP 6: Compute Active Earth Pressure for Stem and Wall
O

EXAMPLE 4.1

STEM)

(

8-0"

0" (WHOLE WALL)

H

/
/

Using Rankine Equation for:

Level Backfill
Vertical Rear Face
No Friction Between Backfill and Backwall

414

Recall from Step 3: K, = 0.3333

(0.120 k/ #t°)- (8.0 ft)? - (0.833) = 1.28 K/ ft

>, = (1.28 k / ft) - (6.667 ft Between Piles) = 8.533 k / Pair

1
= (0.120 k/ ft°)- (11.0 ft)? - (0.3333) = 2.42 k / ft
2

A
P, = (2.42 k / ft). (6.667 ft Between Piles) = 16.133 k / Pair

_ DESIGN EXAMPLE 4.1
: COMPUTE ACTIVE EARTH
PRESSURES STEM/WALL

We will compute two separate active
earth pressures, one acting on the stem
alone and the other acting on the entire
wall. The resultant forces will act at a
distance of one-third the height of the
member under consideration (see ac-
companying figure).

To compute the active earth pressures,
we will utilize the Rankine equation,
which reduces to the expression shown
in the calculation sheet because of the
geometry of the abutment and rela-
tionship between the backwall and
backfill with regard to friction.

The active pressure coefficient was
computed earlier in Step 3. The pres-
sures for both the stem and entire wall
are computed over a unit length which
is defined by the distance between
piles. This results in two active earth
pressure values which are given in
units of kips per pair of piles.

These active pressures will be multi-
plied by the thrust factor F,” = 2.103
computed earlier in Step 4 to produce
an equivalent dynamic and seismic
pressure, Pg.

12. | EXAMPLE 4.1

DESIGN OF A STUB ABUTMENT WITH
SEISMIC DESIGN CODE

8
DA

| STEP 7: Compute Abutment Stiffness

Compute Deflection Considering Effects of Shear with P

CE
SEI

here:

= 3.33 ft
= 115.17 ft

KNOWN GEOMETRY:

The following are known geo-
metric parameters regarding
the abutment and bridge:
Backwall Width = 115'-2'
Span 1 = 29"

Span 2 = 63

Span 3 = 29°-0'

Modulus of Elasticity for con-
crete is given by:
E. = 33w!5 JE
= 33(150 pef)'* 3, i
= 3,300,000 psi
475,200 ksf

G= 0.4E = 190,080 ksf

he 12h 12h, 12h

3E1 * 0.4EA ~ 3Ebd° * 0.4Ebd

| The equation above can be rewritten as:

(n)

Pg) 4-0.2°+3:012)

E-b (475,200)(115.17)

DESIGN EXAMPLE 4.1 _

STEP 7: COMPUTE ABUTMENT
STIFFNESS

In order to compute the effects of lon-
gitudinal earthquake motion on the
abutment, we first need to determine
the stiffness of the abutment. To calcu-
late the stiffness we will apply a unit
load of P = 1 on the abutment assum-
ing it acts as a cantilevered beam fixed
at the footing. We will compute the
deflection due to this unit load and, by
definition, the stiffness will be the in-
verse of this value.

We are given a total width of backwall
of 115'-2" and an end span length of
29'-0" (which will be used later). Be-
cause the height to depth ratio is not
great, we need to take into account the
effects of shear when computing the
deflection. To simplify the calculation,
the general deflection equation is re-
written using a parameter of h/d

The value of “h" is taken as 4.00 ft and
the top portion of the abutment ig-
nored as a result of the beam action at
the bench. Therefore this part of the
abutment height is not included when
computing the stiffness for the abut-
ment transverse to earthquake motion
(i.e., longitudinal motion).

13. DESIGN OF A STUB ABUTMENT WITH
EXAMPLE 41 | sEigMic DESIGN CODE
| STEP 8: Compute Earthquake Load on Abutment
2-0" 63-0"

Weight of Deck
21.74 kt

P.

pa

Esa Ez

Compute static displacement v, with P, = 1:

PL ((29.0ft) _
k 5,206,315 k / ft

Compute single-mode factors ax, B, y (AASHTO Division I-A 5.3):

=5.57x 10% ft

a= Jiv.ooax = v,L = (5.57 x 10% ft)(29.0 ft) = 1.6153 x 10% ft?

B = fimcov, Godx = cow = (1.6153 x 10° #2)(21.74 k / ft)
3.51176 x 10% ft-k

Y= Jjucov oo? dx = Bv, = (3.51176 x 10° ft- k)(5.57 x 10° ft)

= 1.956 x 10° ft? -k

(Step 8: Continued on Next Sheet)

DESIGN EXAMPLE 4.1
STEP 8: EARTHQUAKE LOAD ON
— ABUTMENT O 3/3
Recall that, in our force diagram shown
earlier in Step 5, a horizontal force of
V, was applied at the stringer support
point. This is the load caused by longi-
tudinal earthquake motion. Since we al-
ready have determined our seismic cri-
teria (Step 1) and abutment stiffness
(Step 7), we are now ready to compute
this force.

The first step is to calculate the static
displacement v, based on a unit load of
P, = 1. Recall that the length of the end
span was earlier given to be 29 ft and the
unit weight of deck as 21.74 k/ft.

The diagram in the accompanying cal-
culation sheet illustrates the bridge un-
der longitudinal earthquake loading with
the static displacement depicted at the
pier locations.

Once the static displacement is known,
we can compute the three single-mode
factors which were defined earlier in Sec-
tion 3.5.3, Part 2 (Equations 3.1 through
3.3). The integrals are solved from x
equals O to L, where L is the length of the
end span, v, is the static displacement,
and w is the unit weight of deck.

14. z
B= fwnomswmdx

en

where Z = length of bridge

With these factors known, the fundamental period of the bridge

can be computed with the following:

cf
V Pogo

where p,-1
2 = acceleration of gravity length/time>

where A = Acceleration Coefficient
S = Site Coefficient

Dp (x) = LE wind, (x)

y

15. ASSUMED TRANSVERSE LOADING
(PLAN)

Vs 00

paras |

ASSUMED LONGITUDINAL LOADING
(ELEVATION)

Assumed Loading for Single-Mode Spectral Analysis.

The next step is to calculate the dead weight value w(x) from the superstructure and part of the sub-
structure. It can also include some live load if the bridge is in a heavily traveled urban area. From these

two values, v, and w(x), we can find the fundamental period T of the bridge and the seismic force p,(x).

16. [DESIGN OF A STUB ABUTMENT WITH
SEISMIC DESIGN CODE

STEP 8: Compute Earthquake Load on Abutment (Continued)
Compute period of oscillation: ;

EXAMPLE 4.1 per | 12

al 1.966 x 10% fé -k |
ya k)(32.2 ft / sec? )(1.6153 x 10% ft?)

[x
i |
er, Pogo.

= 1.218472 x 10? sec

Compute elastic seismic response coefficient:

A = Acceleration Coefficient = 0.19 (Given)
S = Site Coefficient = 1.2 (Step 1)

1.208 (1.2)(0.19)(1.2)
Ss” 1% (1.218472x 10? sec)”
AASHTO Division I-A 5.2.1 states that C, need not exceed 2.54:
2.5A = (2.5)(0.19) = 0.475 < 5.167 u Use: C, = 0.475

= 5.167

Compute equivalent static earthquake loading:

p(X) =

BC,
7 WO

(3.51176 x 10% ft-k)(0.475) E
= Tat py (21-74 k/ ft)(5.57 x 10° ft)
(1.956 x 10° ft -k)
= 10.3265 k / ft
Compute force acting on abutment:
P¿00-L _ (10.3265 k / ft)(29 ft)
R, u 2.0

= 149.73 k

14

EARTHQUAKE LOAD ON

___ABUTMENT (CONTINUED)
Next, as defined earlier in Equation
3.4, we calculate the period of oscilla-
tion again using a unit load of 1.0. This
is used in turn to calculate the seismic
response coefficient. To compute this
value we need the acceleration factor
which was given based on the bridge
location and the site coefficient deter-
mined earlier in Step 1

The seismic response coefficient, how-
ever, is not to exceed 2.5A, where A is
the acceleration factor (if a Soil Profile
Type Ill was present where A > 0.30,
the not to exceed value would be 2.0A).

This is the last value needed to calcu-
late the equivalent static earthquake
loading acting on the bridge. This is a
unit load acting over the entire length
of the end span.

To compute the actual force acting on
the abutment we multiply this unit load-
ing by the length of the span. This
value is reduced by the response modi-
fication factor, R, determined earlier in
Step 1. This value in specified by
AASHTO Division I-A 3.6 with a table
of values presented in Section 3.5.3

17. DESIGN EXAMPLE 4.1 _
mn STEP 9: COMPUTE SHEARS AND
Aa MOMENTS

DESIGN OF A STUB ABUTMENT WITH

EXAMPLE 4.1 | seismic DESIGN CODE

DET

STEP 9: Compute Shears and Moments We are now ready to calculate the
— —————— shears and moments acting on the
NOTE: abutment. In this calculation sheet, we

For all loads W, and DL there is
a corresponding load KW, and
k DL acting upward (k, > 0) or
downward (k, < 0).

ia present a free body diagram of the

loads acting on the stem only.

o
W, —— — The abutment stem will be designed
Ww based on a "strip" of abutment defined

by the distance between two pairs of
piles (6'-8")

k

Prior to actually calculating the shears
and moments, however, we must com-
pute the equivalent active earth pres-
sure acting on the abutment. Recall
En that we computed a thrust factor to
Previously Determined Values combine the static and dynamic effects
DL= 175k k= 0.285 into a single load acting on the abut-

67k k =0.100 ment (Step 4). This factor is multiplied

133k | bythe active earth pressure under static
conditions only, computed earlier in
Step 6.

8.533 k/Pair (Step 6)
(2.103)P, = (2.103)(8.533 k/Pair) (Step 4)
17 NPA E, ture due to longitudinal earthquake
rstructure loads acting on a pair of piles: = motion, which was determined in the
149.73 k o (Step 8) previous step, must be reduced to the
Vy = (149.73 k) / 18 Pairs of piles value acting over a strip of abutment

8.3 k/Pair o ol with a pair of piles.

(Step 9: Continued on Next Sheet) Se ae ae

Finally, the load from the superstruc-

18. EXAMPLE 4.1

SEISMIC DESIGN CODE BET
STEP 9: Compute Shears and Moments (Continued)
Y v=0: (DL +W, + W,)(1-k)

= (17.5 k+ 6.7 k+ 13.3 K)(1 - 0.10)
Yv=0. DL +W, +W,)(1+k)
=-(17.5 k+ 6.7 k+ 13.3 K)(1 + 0.10)
Y H=0: KDL+KW, + KW, +V, + Pe

= (0.285)(17.5k) + (0.285)(6.7k) + (0.285)(13.3k) + 8.3k + 17.9k
36.89 k

Ik, > 0]

-33.76 k
[k, < 0]

41.26 k

Axial Force:

Mao

a = Dist. Between Piles ”

Shear Force:
Ve __ meV = 5.53 Wt
jetween Piles

Em = 0:

4.83V, + 0.67DL(14k,) + 2.67P,g +
(4.83)(8.3k) 40.09 ft-k
(0.67)(17.5k)(1+0.1) = 12.90 ft-k
(2.67)(17.9k) 47.79 ft-k
(6.0)(6.7k)(0.285) 11.46 ft-k
(2.0)(13.3k) (0.285) 7.58 ft-k
(-0.83)(6.7k)(1+0.1 -6.12 ft-k
TOTAL 13.70 ft-k

[k,<0] [k, < 0]
1K, + 2W;K, - 0.83W,(1+K,)
Controlling Moment

_ Worst Case Moment
~ Dist. Between Piles
_ 113.70 ft-k

6.667 ft
M = 17.05 ft-k/t

DESIGN OF A STUB ABUTMENT WITH 12

14

— DESIGN EXAMPLE 4.1
STEP 9: COMPUTE SHEARS AND

MOMENTS (CONTINUED)
Since the direction of earthquake mo-
tion can vary, the values of k, and k
will also vary (i.e., from positive to
negative). First we compute shear oc-
curring in both the vertical and hori-
zontal planes.

These values act over a distance of
6.667 ft (i.e., distance between piles)
so we must divide the shears (and
moments) by the amount to get a value
per foot of abutment.

From the computation of vertical shear
it can be seen that there is no uplift.
The axial force per foot of abutment is
computed using the maximum value.

The calculation for moment shown is
the worst case (i.e., k, < O and k, <0)
When performing a seismic design,
the engineer should check all combi-
nations of sign for k, and k, to deter-
mine which produces the worst case.
For the sake of brevity, we have shown
only the worst case situation. Like the
shear values above, the moment is
divided by the distance between piles
to get a moment per foot of abutment.

19. DESIGN OF A STUB ABUTMENT WITH |
SEISMIC DESIGN CODE

STEP 10: Design Reinforcement for Stem

| EXAMPLE 4.1

Concrete and rcing steel parameters:

3,000 psi (Given) E, _ 29,000 ksi

0.40" E 3,300 ksi ~
(0.40)(3,000 psi)

1,200 psi
24,000 psi (Given)

24,000 psi
1,200 psi

9
60,000 psi (Given) =— =0.31
ner 9+20
(3'-4”) - (2" Cover) 0.31
- 1/2(#6 Bar) =1 =" = 0.897
37.625 in 3

Check for slenderness:

By AASHTO 8.16.5.2.5 effects of
slenderness may be ignored if the
slenderness ratio is less than 22.

unsupported length
4.0 ft

radius of gyration
(0.30)(3.333) = 1.0 | k-2, (20140)

r (1.0)

8<22 w

effective length factor
2.0

Therefore member is not slender.

Design required main reinforcement: —

a = M (17.08 tt-k/ ft of wall)(12 in / ft)
tid (24 ksi)(0.897)(37.625 in)

#6 Bars @ 18 inch Spacing

= 0.253 in? / ft of wall 7 (A, = 0.29 in?)

(Step 10: Continued on Next Sheet)

DESIGN EXAMPLE 4.1

STEP 10: DESIGN REINFORCEMENT

FORSTEM
Prior to beginning the actual design of
reinforcement, we must first define all
of the required concrete and reinforc-
ing steel parameters. In Step 5 of De-
sign Example 3.1 we conducted simi-
lar calculations. For references to the
specific AASHTO and ACI specifica-
tions, the reader is referred to this
step. The modulus of elasticity of con-
crete was computed earlier in Step 7.

We must check the stem to see if the
effects of slenderness must be consid-
ered (AASHTO 8.15.4). The unsupported
length of the stem is 4.0 ft. The effective
length factor can be determined by refer-
encing Figure 4.13. The computed slen-
derness ratio value must be less than 22
in order to ignore the effects of slender-
ness. Since our value of 8 falls well
below this limit, the member is not con-
sidered to be slender.

Atthis point, wecan specify the amount
of main reinforcing steel required. Us-
ing the controlling moment, deter-
mined in the previous step, we come
up with #6 bars spacing at 18 inches

20. DESIGN OF A STUB ABUTMENT WITH 14
EXAMPLE 41 | SEiSMIC DESIGN CODE DER L

STEP 10: Design Reinforcement for Stem (Continued)
Compute design moment strength (AASHTO 8.16.3. 2)

14

OM, = 4 AS

AL, (0.29 in? (60,000 tb /in?)
a sult

so: = = = 0.5686 in
(0.85)(3,000 Ib / in? )(12 in)

a
0.85f/b

[ 0.5686 in ]
2

M,=6M, 0.4 (0.29 in)(60,000 1 in] 37.625 ¡nm

= 584,755 in- Ib = 48.73 1t-k>17.05 ft-k w
Design for shear-friction (AASHTO 8.1.5.4)

Vv
A, = Required Shear -Frition Reinforcement = =
qu

5.53 Wft (Step 9)

24 ksi (Given)
0.602 (AASHTO 8.15.5.4.3(c)) = (0.60)(1.0) = 0.60
5.53 k/ tt

——— = 0.385 in? / ft of wall
(24 ksi)(0.60)
#6 Bars @ 12 inch Spacing
(A, = 0.44 in?)
Temperature steel:

+45 Bars @ 18 inch Spacing
(A, = 0.21 in?)

_ DESIGN EXAMPLE 4.1
DESIGN REINFORCEMENT
FOR STEM (CONTINUED)

Now we will check the strength of the
section to see if it can accommodate
our controlling moment. The strength
is computed in accordance with
AASHTO 8.16.3.2. The strength re-
duction factor y used is 0.90 for flex-
ure (AASHTO 8.16.1.2.2). We see that
our section is well within range.

Next shear-friction must be accounted
for and the temperature steel speci-
fied. With regard to the former,
AASHTO 8.15.5.4 specifies the re-
quired shear-friction reinforcement to
be used. The coefficient of friction pa is
based on concrete placed against hard-
ened concrete not intentionally rough-
ened (design is based on the stem
being placed on top of an existing
footing). The value of A = 1.0 was used
for normal weight concrete.

This concludes the abutment design
example. To be sure, there are many
steps still to be completed for both the
design and detailing of the abutment.
As with all examples in this text, how:
ever, the purpose is to provide the
reader with an understanding of the
fundamental principles rather than a
complete design.

22. Wheel load

Bridge slab Approach slab
Normal pavement

Backfill

Abutment

=
~
=
=
~
=
=
x
-
~
~
~
~

23. D

#4@ 9" front face
#4@ 18" back face

À #4 @ 2-0"

f 6" drains
Y @ 10'-0"

À #6 @ 6"
£

Vert. bars, back face

Vert. bars,

2"x6" key

Provide 3" clear to
all bars except 1} "
clear to bars in
front face of stem

#6 @ 6"

25. The almost finished abutment for a high-speed rail overpass (AVE, St, Sadurniu, Barcelona. The AVE is the
high-speed rail connection between Paris-Barcelona).

26. Abutments damaged by seismic forces

27. San Fernando Earthquake, on Februa . Settlement of the abutment fi

28. Wing-wall damage at abutment

Interchange 210 — Roxford
Street Undercrossing at Foothill
Boulevard.

Single-span, variable depth,
prestressed box girder, slightly
skewed and supported on pile
diaphragm abutments. The

structure was supported entirely
on fill. All eight wingwall-to-
diaphragm connections were
badly damaged.

29. Settlement of backfill

Excessive displacement earth pressure
at seating from
foundation detormation

Substructure failure

from excessive

structural loading

or due to loss of

soil support (0.9.

liquefaction)

Pile failure at depth

from abrupt variation

of soil displacements
(e.g. discontinuity in
soil stiffnesses, land-
slides or liquefaction)

30. Failure in Shear of
Superstructure and End
Diaphragm at Soffit.

Balboa Boulevard Over-crossing —
South Abutment

Seven-span, reinforced concrete box
ler with Class II piles and spread
footings. Abutments were diaphragm

type.

The south abutment was in original
ground and soil supported, while the
north abutment was upon fill and
supported on concrete piles.

31. Abutment Fill Settlement

32. Eureka Earthquake —
November 8, 1980.

Epicenter— 30 miles
northwest of
Eureka, CA

Magnitude — 6.6 to 7.1 on the
Richter Scale

Damage — Collapse of
Fields Landing
Overhead
located on
Highway 101

33. Aerial View of Fields Landing showi ing Collaps ans. Failure caused by earthquake factors which
ribute at the abutme:

34. Abutment 1.

All of the pedestals were
severely shattered, and most
of the steel bearings were
ripped from their
anchorages.

39. Failure of Reinforcing Concrete Pier Wall