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T. S. Kahlonich -

' > 1948‘

 

 

 

 

An Analysis of a Rolled Beam,

Deck Type Vighway Bridge

A Thesis Submitted to

The Faculty of
MTCPTGAN STATE COLLEGE
of
AGRTCULTURE AND APPLIED SCIENCE

by

T. S. Katalenich
hun-
Candidate for the Degree of

Bachelor of Science

December 1948

J - anew:

./

ACKNOWLEDGMENT

The writer wishes to take this opportunity to express
his appreciation to Charles A. Miller, Associate Professor
of Civil Engineering, Michigan State College and H. R. Puffer,
of Michigan State Highway Deptment, for their wholehearted

_ co-oheration and counsel.

.1153

«13
m4
2;:
;3
sea-9

(J

(.0

INTRODUCTION

The bridge selected for analysis is located 1.8 miles
northwest of Reed City, Michigan, on U.S.lO. This struc-
ture is a single span, rolled beam, deck type highway bridge.
Tt was designed for 3-20 loading and width of roadway to be
of sufficient width to accommodate three lanes of traffic,
although a.two lane pavement exists at the present time.

The design of this structure was governed by, "Specifi-
cations for the Design of Fighway Bridges" adopted by Mich-
igan State Highway Department November, 1936. As in the de-
sign of most structures, there are a few instances where ex-
perience and practical application of'construction methods
has permitted a deviation from strict application.of the
specifications. Such instances will be made a note of in the
analysis.

The bridge provides a crossing of Johnson Creek, a
stream which drains a basin¢3f approximately 12 sq. miles.
The basin top soil is composed of a very sandy loam with a

low runoff coefficient.

L.’

.4

I: r. an w‘ L*

“CA.

SYMBOLS and MTATTON
area of tensile reinflarcement
width of rectangular beam
distance from extreme fiber to compressive resultant
effective depth of flexural members
eccentricity
eccentricity measured fnam gravity axis.
compressive stress in extreme fiber
ultimate compressive strength of concrete
stress in tensile reinforcement
moment of inertia
ratio of distance (jd)
ratio of distance between extreme fiber & N.A. to eff. d
%fc 3k (Reinforced Concrets-—Sutherland & Reese, p. 520)
length of span; length of anchorage of reforcement bars
external moment
ratio of modulus of elasticity of steel to concrete
neutral axis
sum of perimeters of bars
external concentrated load

spacing,of reinforcement bars

bond stress

shearing stress

total shear

permissible soil pressure; uniformly distributed load
distance of moment arm of a horizontal force resultant

distance of moment arm of a vertical.!orce resultant

ANALYSIS of CFANT’FL OPFVTNG
Specifications:

Art. 11! 'Tn general, the waterway provided shall be suf-
ficient to insure the discharge of flood waters with-
out undue backwater head and at a velocity which will
not increase the erosive action.of the stream to such
an extent as to endanger the structure, or cause dam-
ages to Upstream property.

Art. 132 Fridges proposed over streams where future
drain projects are a probability shall have footings
at a sufficient depth estimated to prevent instability
or undermining when the drain project is carried out.

Art. 149 Natural obstructions, such as islands, rocks,
trees and brush which retard or deflect the stream in
the vicinity of a bridge shall be removed and such
portions of the channel shall be cleaned out as are
necessary to straighten out the stream at the bridge
and prevent eddies or scour.

Art. 152‘ The clear width of all openings and the clear
vertical distance between the superstructure and the
flood water elevation shall be sufficient for the pas-
sage, without damage to the structure, of ice floes and
of the largest drift or debris which may be expected.
Ordinarily a minimum of’one foot vertical clearance

shall be provided above extreme high water.

A. Computation of the Cross-Section needed:
Area of basin = 12 sq. mi.
Coefficient taken from Fighway Map .25
Talberts Formula: I
a=127 c V A113 ' where;
a=area in sq. ft.
C: Coefficient taken from weather map.
Am=Drainage Basin in square miles.
a==137 (.SS) 13 ='203 sq.ft. needed
B. Area Provided by Channel Opening:

30’ 4" i
#1

TUna/er clearance [I /060. 4.?

J/alae Jon (3 p
K {C fianfie/ Haifa/f7 [IX/fly)

 

 

4M7349"

 

 

 
       

 

 

5'2 ‘ 20'
1. To find it! . Fig. T

52?

x=‘tan. 26 30' (5.17')= .498 (5.17')==8.67'
2. Area.pnovided:
A: (20+30.33)_(3.67)+ 30.33 (7.1r):ggs.o> qu‘t.
Tt is obvhgus that sufficient channel area has been

provided as flood areas and.othsr drainage facilities in

the area have not been discoumted from the 203 sq.ft. needed.

The clear width of 30' 4" provides sufficient width for pas-

sage of largest drift or debris to be expected. Abutment
footing is 4.8' below channel bottom which is sufficient.

oa/

Spec:

A1

SLAB ANALYSTS

Specifications:

Art. 18: Tn no case shall the roadway on a bridge be
made less in width than that provided for traffic on
the bridge approaches and preferably the width of road—
way should be not less than the following:

Two lane highway (pavement) 84 ft.

Three lane highway 33 ft.

Art. 20: Substantial curbs shall be built on.sach side
of the roadway and they shall have a width not less
than 6 in. and a height of not less than 9 in. measured
above the wearing surface at a point adjacent to the
curb.

Art. 22: Approach grades shall preferably not exceed
4 % and intersecting grades shall be joined by a par-
abolic vertical curve.

Art. 24: The roadway shall be crowned to fit the ap-
proach pavement and sloped to provide effective drain-
age of the roadway surface.

Suitable gutters or provisions for drainage shall be
constructed at the ends of the approaches adjacent to
the bridge so as to prevent erosion of approach shoul-
ders and fills by water draining,off the ends of the
structure.

Bridge roadway surfaces of 40 ft. width or less
shall be constructed with transverse parabolic crown of

height not less than given by the following formula:

C = 0.0018? R2 where:
C = Crown of roadway in inches.
R = Width of roadway in feet.

The minimum crown shall be three-fourths (5;) inch.
4

Art. 25: Substantial railings shall be provided along
each side of the bridge for the protection of traffic.
The top of railing shall be not less then 3' -O" above
the top of the curb, and when on a sidewalk, not less
than 3' -O" above the top of the sidewalk. Railings
shall contain no opening of greater width than eight
(8) inches. Ample provision shall be made for inequal-
ity in the ratio of movement of the railing and the sup-
porting superstructure, due to temperature or erection
conditions.

Art. 26: Where no separate wearing surface is provided
on concrete floor slabs, an additional slab thickness
of one-half inch shall be provided. This one-half inch
thickness on the top of the slab shall be disregarded
in computing strength of floor slabs.

Art. 29: The dead load shall consist of the actual weight-
of all materials and construction comprising the com-
pleted‘design.

The following weights of materials may be used in
estimating dead load:

Steel 490 lb./cu. ft.

Concrete, plain or reinforced 150 lb./cu. ft.

Loose sand and earth 100 lb./cu. ft.

Art. 30: For structures with concrete slab floors with-
out separate wearing surface, a minimum allowance of 20
poumds per square foot of roadway shall be made, in ad-
dition to the weight of’any'nonolithically placed con-
crete wearing surface, to provide for future wearing
surface.

Art. 31: The live load.or1roadways shall consist of a
train.of standard motor trucks, in each traffic lane,
as pictured in Figure 2.

The truck trains shall be assumed to occupy traffic
lanes, each having a width of 9 feet corresponding to
the standard truck clearance width. Within the curb to
curb width of the roadway, the traffic lanes shall be
assumed to occupy any position which will produce the
maximum stress, but which will not involve overlapping
of adjacent lanes, nor place the center of the lane
nearer than 4 feet 6 inches to the roadway face of the
curb.

All trucks in the same, or adjacent, traffic lanes

shall be assumed headed in the same direction.

 

 

 

 

   

 

Fig. 2 Roadway Loads

 

Art. 32: The Standard Motor Truck, as specified above,
is a motor truck with dimensions and weight distribu-

tion as shown in Fig. 3.

 

 

 

 

 

 

 

 

 

 

 

 

I I _ - _ g
[— __J
L___J
. . 9 o
73.4“ 5' o 5: .4 I" O" A
Lil!” wmm j

 

 

.au ii: .Fig..3 ‘

Art. 33. When roadway provides for two lanes of traffic
or less, the design shall provide for the maximum load
that can be placed simultaneously in all traffic lanes.

lhen provision is made for three lanes of traffic,
the design shall provide for 90% of the simultaneous
maximum loading of all lanes.

Art. 35. Railings shall be designed to resist a horizon-
tal force of not less than 150 pounds per lineal foot,
applied at the top of the railing, and a vertical force
of not less than 100 pounds per lineal foot. For rail-
ings adjacent to the roadway, the bottom rail shall be
designed for a horizontal force of 500 pounds per lin-

eal foot of rail.

Art. 36. Curbs shall be designed to resist a force of
not less than 500 pounds per lineal foot of curb ap-
plied at the top of the curb.

Art. 37: All live load stresses, shall be increased by an

allowance to provide for dynamic effect. The impact

Art.

a.

b.

allowance shall be determined by multiplying the max-
imum live load stress by the coefficient determined
by the following formula:
I: L+20
637725' where
I = Impact Coefficient
L = Span Length

54 General Assumptions in concrete Structure Design:
Calculations are made with reference to unit Working
stresses and safe loads, as elsewhere specified here—
in, rather than with reference to ultimate strength
and ultimate loads.

A sectiorlplane before bending remains plane after

bending.

The modulus of elasticity of concrete in compressions

is constant within the limits of working stresses;

distribution of compressive stress in flexure is

therefore rectilinear.

The ratio 'n' of the modulus of elasticity of the

steel to that of the concrete shall be taken as fol-

lows (applies also to compression members)
“fiddle-939'

Concrete shall be assumed as offering no tensile re—

sistance.

The bond between concrete and metal reinforcement is

assumed to remain unbroken throughout the range of

Working stresses.

Ml

g. Ynitial stress in the reinforcement due to contrac-
tion or expansion of the concrete is neglected.

h. Nomenclature and formulas for design shall be those
common usage as given in the Standard Specifications
for Concrete and Reinforced Concrete of the Joint
Committee.

Art. 58. The maximum spacing of principal reinforcement
in walls and slabs shall not exceed two times the thick—
ness of the slab or wall, nor more than 24 inches.

Art. 59. For slabs the distance from the surface of the
concrete either top or bottom, to the center of the
nearest bar shall be not less than one and one-half
times the diameter of the bar nor less than one and
one-half inches.

Art. 61. The length of bar added for anchorage may be
either straight or bent. The inside radius of bend
shall not be less than four diameters of the bar. A
standard book as referred to in these specifications
shall consist of a semi-aircular bend, whose inside
radius is four bar diameters, and a final tangent
length of four bar diameters.

Art. 64. Where reinforcement is used to resist tensile
stresses developed by beam action, the bond stress.
shall be taken not less than that computed by the fol-

fo wing fo rmula:

u: V___
7%; d

Art. 66. Horizontal reinforcement for shrinkage and tem-
perature stresses normal to the principal reinforcement
shall be provided where the principal reinforcement ex-
tends in one direction only. Such reinforcement shall
be placed at exposed surfaces and shall provide not
less than one eighth square inch of reinforcement per

foot of width of surface.

Crown Analysis: (Spec. Art. 24.)
Width of roadway 38 Ft.
0 = 0.00187 R3

0 0.00187 (38)3 = 2.7 in.

C 2 3" =-' e (Parabolic Curve)

4

‘0.

(b

n.
9“".

 

‘0.
--___-_§ ..-....._...

..-¢-.- --

Crossecticn of Road Crown
Fig. 4

Computations: (quarter-sections of a right_triangle)

ff': 2 e f'f" 3' e
dd' 2 3 2e: 3e d'd" = 3 3 e _._. 9e
(2;) or (I) To“
cc' = (_1_) Be: e c'c": (1)3 e = _e_
2 ' 8 3 4
bb' = (l) 86 = _e_ b'b" = (l) e = e
4 3 " 4 "1‘6

Crown Dimensions:

bb"=_e_—_e_ a 19 =(2.75_1=1§__~
2 16 16 18 16
co" = e - g 3 3e 3 2.75)_3¢=2 _1___"
4 4 4 16
dd“~ e-9e e:_-_(2.751_5=2 "

1
onto:

|l~
H H
CD 01

Veiw Showing Roadway 'idth and Slab Support:

l'~_6_; W 38' - 0" Rdwy. 1's"

y.

 

23”." Modified Parabolic Crown

‘11" 4 +

 

 

7: Minimum Slab ‘ 10' Curb 1

 

 

 

L‘ 8 - Spaces @ 5' 2' 41' 4" 27 WF 94 %’J

‘Q

 

Fig. 5

Maximum Loading on Slab occurs when rear wheel is over

point midway between stringers:

16k ‘ 16k 16k 16k
.' 3' 6'

 

 

 

 

A.

B.

Effective width: Spec. 41: Wheel concentration shall be
assumed as uniformly distributed on a line perpendicular
to the main reinforcement over an effective width not
more than the value given by the formula:
B: 0.7S+2 where:
B Effective width of slab in feet.
S Span.of‘s1ab, center to center of supports.
8:0.7 (5.17')+2=5.6'
Maximum Dead Load.Moment:
1. Dead Load per foot of slab:
Depth of slab

design depth 7.00"
crown 2.?5
camber .19
wearing surface .59
total depth 10.44H

D.L._—_-10.44 (1) (1) (150): 131.0 lb. per 1n. n.

Extra allowance 20.0
Total D.L. 151.0 lb. per 1n. ft.

2. Maximum D. L. Moment:

Max. 110.: 1 111. (151) (5.17) (12): 4000 lb. in.

12
Mamimum Live Load Moment:

3-20 loading used in analysis. The beam is regarded as
a continuous loaded beam and maximum positive and negative
moments are computed using fo rmulae in Kirkham's "Highway

Bridges" page 74.

I
I
- I I
. -.-
9
a
' - -.
I l .
O
0
J .
.u b - u .m—
f ‘ s a
I
. I
. ' . ,. x r
(
o I
a
.. , - g (V-
.' "
1 ,
O ' A.
l
.' \. o ‘
| I
o r '- . ‘ L ,
'I

’ 1

‘ifl-I‘

 

1.

8.

Max. posit ive mo ment :

M: 1 Pl: 1 (16000) (5.17) (12): 198,000 lb. in.

5 5

Max. negative moment:

ll: 1 P1: 1 (16,000) (5.17) (12)=165,000 1b. in.
6 6 .

Maximum is due to positive moment. Tnasmuch as the
steel in the slab is the same in top and bottom, we

will consider analysis for positive moment only.

 

Max. L.L. moment for 1' width of slab:
Max. Mo.= 198,000 ::35,400 lb. in.
5.6'
Impact:

1: L+20 z 3 .58 20 = .24
- 6L+20 6(32.5‘8)+20

Impact = 35,4000 (0.24) = 8600 lb.in.

5. Total Moment:

Live Load 35,400 lb. in.
Impact j 8,600
Dead Load 4,909

Total Moment 48,000 lb. in.

Steel needed:

d:

7-1.5:5.5 H
b

 

T

 

 

 

 

A. d
__ d _ if
_L“ 11: j m
.'_______ .‘T __

 

‘

 

 

 

 

f
-Fig. 7 f3

1. To find jd :
x: 1200 (5.5) 32.2"
3000

= 2.2 $.73"
3
jd = d—z = 5.5 ... .73 = 4.77"
3. To find Age:

A 48,009 :';56 sq. in. per ft.

.. M _
' r, (‘de ' " 18, 000 (4'.‘7')’
E- Steel provided: (per foot of'concrete)
éf' round bars spaced at 6".
8
A3: .307 sq. in. (2):: .62 sq. Tn.
F. Check stress:

f — (2) 48,000 = 775 lb. per sq. in.

=' 2M
° (b) (3&7de 12 (4.777 (2.2) less than 1200 p.-s.i.

f.= I = 4 000 = 16,500 lbs per age, in.
AB (jd) .53 (4,77) less than 18,000 p.s.i.

 

G. Check Shear:
Formula for maximum shear from Steel Construction Manual

of A180 5th edition.1947 page 376.

v=gw 14. _1__1 P .. 5 (151) (5. 17),, _1_1_ (15,000)
15 8 , 15 5.17

v- - 487. 5 4- 2130. 0: 2517. 5 lbs.
V = = 8618 = 46.8 p. 8.10
b' (33) ‘1'2‘1'4'.7"7) less than 50 13.5.1.

u: 8v

7""? z... 1.8 (2618? 1:53 1.
':° - =lesspthan 150 p. s. i.

it

H. Temperature steel provided.normal to principal rein-
forcement:
f in. circular bars equally spaced at (5,12')
As.= 1.963 - .114 sq. in. per 1n. ft.(sop)

_‘P

1.72

As: 1,963 :: .157 Sq. in. per 1n. ft.(bottom)
1029 '

As required Spec. art. 66;
.125 sq. in. per 1n. ft.

Temperature steel in top of slab is .011 sq. in.=
less than required. Since it is desired to maintain
one bar directly over a stringer, it was necessary to
space the bars as they are. This deficiency is so
emall that it is negligible. Temperature steel in bot-
tom of slab is more than sufficient. Maximum spacing

is 20 in. which is within specificathgns.

rn

STRINGER ANALYSIS
Specifications:

Art. 43. The bending moment carried by each interior beam
or stringer shall be taken not less than that deter-
mined by the following formulae:

M :1 Bending moment for one traffic lane.

14 = Width of Traffic Lane (not to exceed 10')
Spacing of stringers or beams

C : Coefficient based on type of floor.
15': Pending moment on one beam or st ringer.

M': C(11)
N

Value of C is one for reinforced concrete slab.

Tn determining the end shear on longitudinal beams
or stringers, the floor slab or flooring shall be as-
sumed to act as a simply supported beam.

Art. 47. Structural steel:
Axial tenshgn (net section) 18,000 p.s.i.
Tension.on.extreme fibre in flexure 18,000 "
Shear on power-driven rivets and pins 13,500 H
Shear on gross area of webs of beams &

girders ,‘§,‘where the clear distance

between flanges is not more than 60

times the thickness of the web 12,000 "

Art. 79. Depth ratio of rolled beams not more than
1 /25 of the span.
Art. 80. Fblled beams shall be proportioned by the m0-

ments of inertia of their net sections.

Art. 88. The minimum thickness of structural steel
shall be 5/16 in. except for gusset plates, fillers
and railings. Gusset plates shall in no case be less
than 3/8 in. in thickness.

Art. 98. The diameter of rivets in angles carrying cal-
culated stress shall not exceed one-fourth.of the
width of the leg in which they are driven.

The minimum distance between centers of 3/4 in.
rivets shall be not less than 2% in.

The minimum edge distance for 3/4 in. rivets shall
be 1% in.

Art. 9?. A11 connections shall be proportioned to de-
velop not less than the full strength of the members
connected provided that the full strength does not ex-
ceed the maximum.computed stress by more than 50%,
in which case the latter shall govern.

No connection, except fer lacing bars and handrails,
shall contain less than three rivets. Connectkpns
shall be made symmetrical about the axes of the mem-
bers in so far as practicable.

Art. 110. Provision shall be made for expansion and con-
traction, to the extent of l/8 in. for each 10 feet of
span. Expanskpn ends shall be firmly secured against
lifting or lateral movement.

Art. 111. Spans of less than 70 feet may be arranged to

slide upan metal plates with smooth surfaces.

Art. 117. Superstructure steelwork shall be securely
anchored to the substructure. Anchor Bolts shall be
not less than one inch in diameter embedded not less
than ten inches in the masonry and shall be swedged or
threaded to secure a satisfactory grip in the masonry
or otherwise suitably anchored.

Art. 184. Diaphragms shall be provided at the third
points of all T-Beam spans of forty feet or more.
Art. 127 (1). Sole plates shall each be not less than
3/4 in. thick and not less than the thickness of the

flange angles plus 1/8 in. Preferably, they shall
not be longer than 18 in.

A. Maximum Dead Load Moment:
1. Dead Load per la. ft. of stringer
(distance between stringers - 5' 3")
161 lb. (5.17') 780
wt. per ft. of 27 WF‘ 94 Beam __9;4__

Total dead load 874 lbs. per ft. of
“ stringer
2. Ms; W 13: _1_ (s74) (32.67)3g’117,000 ft. lb.
8 s

8. Maximum Live Loading. (H-BO loading)

1. haximum loading¢3n stringer (A) will occur when two
trucks are side by side with the wheel of one truck
directly on the stringer and the respective wheel of
the other truck 3' to the side of the first wheel.

. (shown below)

i: 6' .I. 3' II 6' 1'1 Ll

 

 

r.

  

Fig. 7

Maximum load rear axle:

Ra: 16,000+ 3 (16.000): 16,000 + 6,780: 22,720
5.17 lbs.

Maximum load front axle:

Ra: 4,000+ 3 (4,000) : 4,000 + 1680 = 5,680 lbs.
5.17

Maximum Live Load Moment:
Loading to produce maximum live load moment on.stringer
will occur when rear wheels of the two trucks will ei-
multaneously reach a point the distance (5) passed the

center of the span. (absolute moment)

22.72k 38.4k
[gal= x

$.11: 14' 5.6k
4 .

 

 

 

 

 

 

 

16' 312! _ I 16' 3%"
I.» _
s
a Fig. 8 Rb
1. To find R, and Rb!
an: 22.720 (313.29'14- 5.680 (8—°.2:.9_.'.L= 11,750 lbs.

32.58'

Rb: 28,400 - 11,750 = 16,650 lbs.
2. To find x:

x: 11.750 (38.58') = 13.5:
38,400

3. To find a:
a: 16.29' - 13.5' = 2.79'

4. To find Absolute Maximum Moment: (Formula from

“Structural Theory” Sutherland & Bowman, 3rd. edition

1944, page 116)

lig:§?_fi%$3 -’L§L§14L(%i§h “’RL b
= a 3 - W
M 28 400 E38535 ) 5 3 79 +(_3.é19.)a -0

 

32.58'
M : 196,000 lb. ft.

U
in. _..

0
I
. , g
s
O
I - ‘..n.-fl
‘ e
..
.
.,
O
. ..' _ '_
w .; '
- -..9 .

f \
‘ ..

5. Spec. Art. 43:

M': c (g); ; (155,000 1b.rt.)_—_- 155,000 lb.ft.

N 9 5.17
11' above is much less than the moment computed in
4 above. Therefore I will use the larger moment

in this analysis.

D. Beam size:

1. Total moment on stringer:

 

Dead Load moment 117,000 lb.ft.

Live Load noment 196,000 "

Impact (196,000 x .24) 47.000 '
Total moment 360,000 lb.ft.

8. To find T (section modulus)

5
1.— 14 = 246040001191 = 240 in.3
c" a 18,000 psi

87 NF 94 beam has 1 of 848.8 in.3 Therefore the
C
beam is suitable to carry the maximum loading on

t 1'10 bridge.

3. Depth Ratio: Spec. Art. 79.

42

Hat is :

713.5,). Therefore the beam is satisfactory for depth.

 

r . 1 ‘* __.. 1
3215's ' (12) -:. 27 " 14

The ratio is greater than specifications call for

Check Shear: Spec. Art. 47.
87 73- .49 = 58 (less than 60 times 'web thickness)

17 = 16,650 (1.24),. 874 (32.59) = 34,900 lbs.
" 2

9: 1:: 34 9'00 = 2 650 .s.i. (less than 12 000
A .49 ('2'6".9) ’ p ’

therefore 0 .K.)

.
D e
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A .
4 t. .
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a ... .
. _ . c C V ,
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0 \ e . . e I » .
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\ .IHJ.’ Ila! ‘l!,,fu.l.f|ll!.l-Il. I!
m. _ m...

s W s...
|l|.

 

 

E. Diaphragms: Provided at 1/3 distance of span.
Spec. Art. 184

Connection.angles 4 x 4 x 3/8
Horizontal angles 3 x'3 x 3/8
Diaphragm plate 18 x 3/8

Force resisted by Diaphragms: Spec. Art. 38
Area of substructure : 9(87) (30.33) :: 307 sq.ft.
Area of side elevation: 5(30.33)(1.5) =__8__8_§_ "
Total area 535 sq.ft.
Lateral force, wind on structure-535(30) 8 16,050#
Lateral force, wind on Live Load-150(30.33) = 4,§OO£
Total force 20,650#

Area steel required:

As: 80,650 g 1.15 sq. in.
18,000 ~
Area steel provided: (8 plates)
A = 8 (l8) (3) = 13.5 sq.in.
‘ 8
This is much more steel area than is necessary.
Stress in rivets:

80 650 : 4,700 p.s.i. . (allowed is 13,500)

8:
2 (5) (.442)

F. Bearing Plates:
Plate size: 12" x 4" x 19%"
Hearing on concrete:
fc'z 34L900 lb. :2 149 p.s.i. (this is very low)
12 (19.53
C. The ends of the stringers are not connected but are em-
bedded in concrete a depth of one f0ot, with two steel

bars passing through the web of the stringers. This
gives the necessary stability.

ABUTMENT ANALYSIS
Specifications:

Art: 39. Retaining walls, abutments and structures built
to retain fills shall be designed to resist pressures
determined in accordance with the "Rankine" theory of
pressure distribution in non-cohesive granular mate-
rial, provided that no structure shall be designed for
an equivalent fluid pressure of less than 30 pounds
per square foot.

To puovide for live load an equivalent earth sur—
charge of four feet shall be applied over all surfaces
subject to highway traffic.

Art: 59. In footings and.other principal structural
members where the concrete is in direct contact with
soil, the reinforcement shall have a minimum covering

-of 3 in. of concrete measured to the center of bar.

Art. 63. (b) In cantilever'footings all bars shall

be anchored by means of standard hooks at the outer

ends of the bar.

 

O
I\

a.

In analysis of the Abutment, four types of loadings will

be considered as follows: ‘

Case 1: Abutment before stringers are placed and only
pressure is due to retained soil without allowance
for live loading over fill. ‘

Case 8: Pressure due to retained soil 1.115.! allowance
for live loading over fill and stabilizing moment
due to dead load of bridge.

Case 3: Pressure due to retained soil not allowing for
live loading surcharge and stabilizing moment due to
dead and live loading of bridge.

Case 4: Pressure due to retained soil 11331 allowance
for live. loading surcharge and stabilizing moment due

to dead and live loading of bridge.

Total Height of Fill:

Surcharge live load 4' 0"
Surcharge (18' pavement) 1' 6'
Distance,pavement a. abutment 8' 6 _5_"
Total surcharge F5575— (Call it 8') 8' 0"
Feight retaining 15' 11"
___B_
Total height of fill 83' 11"
8
Weight of reinforced concrete 150 lb. per cu.ft.
Weight of fill 100 lb. per cu.ft.

Angle of repose of fill 30 degrees

1:1»).0413AI .. .01... ’7..l..lnil........1 Ill?
. l.

_e-

Case 1

Abutment befiore stringers are placed and.only pressure

is due to retained 9011 without live load surcharge:

433' . 1254’

 

 

 

m")
' W1 *W E \\
:2 y it 6" : \
N uq———a-.—p-) I \
\

 

 

 

'/.5/7’
5' .
%

 

E.

 

 

 

 

 

9': th= .33 (100) 15.17: 504 1b.

P = is; (15.17) = 3820 1b.
8 :1, agm. moment
1'1 = 150 (2.33) (12.75') .-.- 4460 lbs. 5.5 = 24,800 lb.ft.
'2 :. 150 (2.50) (10.5) = 3940 5.85 = 20,660 "
v3 .-.- 100 (3.84) (12.75) = 4299 8.58 = 42,000 '
Total wt. 13300 lbs. Total Mo. 87,460 1b.ft.

To find 'x':

x 3 87,460 = 6.57'
13,300

To find "e"

e': 5.96 (3820):. 1.46'
13,300

0 = 5.25 - 5.11 = .14'

Earth Pressure:

6.57 -- 1.46 = 5.11'

1170#/eq.ft.

P; 1.3.30.2 _[11- ‘ 14 =1270"(1+“..08)"-_113704/3q.r't.-9~
10.5' ‘, I 0.5 "

Overturning Factor:

Ratio : 874460 1b.f3- : 4.4 (very safe)
3,820(5.06)

Sliding Factor:

Ratio: 13,399,,(_._59_) : 1. 74 (Just on th borderline of
3,880 being safe

Tt is obvious that the resultant intersects the base within
the middle third since ”e" is only .14'.
Stem Moment: 5

p :: mm = .33 (12.75)(100) = 421 lbs.

M=(481) (12.75) (12.75): 11,400 lb. ft.
'2 3

Toe Moment:

p: 6.17 (355) + 1090 : 1899 lbs. (earth pressure at
10.5 edge of stem)

 

Moment upward due to earth pressure:
2199 (4. 33)(4. ___3_23) : 12, 050 ft. lb.

.1... (4. 33) (8H4. 33)-‘- 896 a
2 3
Total Moment = 18,946 ft. lb.

Moment downward due to concrete in toe:

M: 150 (4.33)(2.5) (4.33): 3,470 ft. lb.
2

Toe Moment = 18,946 - 3,470= 9,476 ft. lb.
Heel Moment:

9: 384 (355): 130 lb. /sq. ft. (earth spreesure at edge
100 5 cm.)

Upward 14: 1090(3.84)(3.84;+130(3.84)(3.84)2= 8369 ft.lb .
2 2 '3". -

Downward M: 3.84(2.5)(150)+ 4900 x 3.84 = 12,180 ft.lb.
”8"

Feel Moment 21' 18,180 - 8,369 = 3,811 ft. lb.

Case 8
Pressure due to retained soil with allowance for live

load surcharge and stabilizing moment due to dead load of

 

 

 

 

 

 

 

 

bridge. , __.
rtbl *5— -E\\ P’
. F aL--‘EH\
. ‘\
Q ‘0 1:51 We: \
"5 ‘11 X i. t t \
N 1.6 ‘——>0 | \
\ i I p \
1'1"“. 4 .
t i l 1)" WP)“.

 

‘ z
1 e
To find horizontal force:
P': th= .33(100)(8) = 864 lb.

I": own: .33(100)23.25= 768 lb.
264 (15.2mm = 30,700 ft.lb.
1‘2 ' ..

(mfliass) (1.51382) = Winn.

Total Moment 3 50,850 ft.1b.
P: 364 (15.25)+ (768 - §64H15.35) : 7870 lb.
8

Y: 59,850 11.19.: .4:
O b. 6

Dead Load of deck per foot of Abutment:

D.L._-: 874 1b. (33.87') _,
8 5.17' "" 2860 lbs.

fl]:

.l‘l'fllm Jul. I?!»
'r .

To find Resultant Vertical Force:
D.L. = 2860 lbs. 5.46' '= 15,600 lb.ft.

W1 150 (2.33)(12.75) = 4450 " 5'5 ' = 34,800 '

W :3940 " 5.25' = 20,660 "

2

'4 20.75 (3.83)(100) =7950 . 8.58' =-68,250 .

Total weight 19210 lbs: Total M 129,310 1b.ft.

x: 129,310 lb.f‘t._ :
19,210 lbs. - 6°73

e': 6.4‘ (7870) = . . . - ._ ..
19,210 2.62 . 6.72 2.62 _ 4.10

 

6‘: 5.25' - 4.10' = 1.15'

Earth Pressure: Pressure greater at toe.

P = _.._._,._l9 310 (1 + 66) " 3020 lb
_ e»! -' I eper Sq. t.
10.5 620 1b.per Esq-gt-

Overturning Factor:

Ratio: 129,310 1b.ft. _._._ 3.6 ( this is satisfactory).
~787O 1b.(6.4)

Sliding Factor:

Ratio: WblbLSO) ;-_- 1,3 ( this is unsatisfactory)

 

Check if Resultant intersects base within middle third:

l91§..:'3.5'~ 9 £291:

3 1'7'(greater than 1.15' therefore

within middle third)
Stem Moment: (tending to overturn)

_*= CV’h: 20.75(100)(.33) = 685 1b..

 

264 (12.75) = 3370 15. 12.75' = 21,500 1b.ft.
2

@85;264)(13.75)= 2680 lb. 12.75! =11,400 1b. ft.
3

Total 171:. 3' 6050 lb. Tot sl Mo.=32,900 lb. ft.

Stem Moment:(tending to stablize )‘

2860 (1.125') = 3210 lb.ft.
4460 (1.165') =- 5309 lb.ft.

Total Mm: 8410 lb.ft.
Total Stem Moment:
M= 38,900 lb.ft. 5-84'1‘0' lb.ft.= 84,490 lb.ft.
Total Toe Moment:

p— 6. 17 (2400) + 680: 2040 lb. per sq. ft. (earth pressure
10. 5 . , - at edge. of stem)

Moment due to Earth Pressure:
arm oment

 

2040 (4.33) = 8780 lbs. ““4333 =18 ,5’0‘0“?t'. lb.
2
(2400 .. 2040)(4.33) = 795 lbs. 2 4.33 = 2,260 *
3 3

Total weight: 9575 lbs: Total Mo ‘-‘- 81,160 ft.lb.
Moment due to toe concrete:

150 (2.5) (4.33) '= 1615 lbs. 4.33 .4. 2,470 ft.lb.
- 2

Total Moment: 81,160 - 3,470 = 17,690 ft.lb.
Total Shear := 9,575 - 1615 _ 7,960 lbs.

Total Peel Moment:

= mlfiéflfll+ 620: 15-00 lbs. per sq. ft. (earth
.5 pressure at edge of stem)

Farth Pressure Moment upward:
_ 620 (3.83) = 2370 lb. 3 9.3
(880)(3.83)
8

4,550 ft.lb.

 

1685 lb. 5.3;: = 2,150 ft.lb.
6, 700 ft.lb.

3
Total wt? 4055 lb. 2 Total M0.

Earth and Concrete Moment Downward:

7950+ 1435 = 9,385 lbs. 3.83 a 18,000 ft.lb.
-—--—-2 .
Total Moment = 18,000 - 6,7000 .-.-. 11,300 ft. lbs.

 

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Case 3
Pressure to retained 6011 without allowance for live load

surcharge. Stabilizing moment due to dead and live loading

 

of deck.
A

 

 

.42.237'
/:i.?di'

 

 

 

 

J
_A

 

 

 

F”: .33 (4)(100) = 132 lb.
9": .33 (l9.25)(100) = 636 lb.

To find "P": horimnta’l force due to earth pressure.

 

132 (15.25) = 2,015 lb. 15.25 = 15,400 lb.ft.
. 2
(636 g 132) (15.25): 3,540 lb. 5 = 19,500 '
Total P : 5,855 lb. Total Mo. = 34,900 lb.ft.

y: 34,900 lb.ft.= 5.96'
5,855 lb.

To find Resultant of vertical forces:
D.L.+L.L. of Deck:E6,650+ §74(33.58]+ 5.17: ezoo#/ft.
8

. 9.3g ment
D.L. L.L. of Deck : 6200 lb. 5.46' .-.- 33,900 lb.ft.
'1 . :. 4460 " 5.5 = 24,800 "
W3 100(3.83)(16.75) = 6480 " 8.58 = 55,000 “
'2 : 534g, " 5.25 :. 80,660 "

Total W. = 31080 lbs. Total Mo.=134,360 lb.ft.

134 360 lb. ft .-

X: _. 6.4'
31,020 159.
e'.-..- 5.96 5 855 = 1.66' : 6.4 - 1.66 = 4.74'

21,020
0 :- 5085 " 4.74 = .51.

Earth Pressure: earth pressure is greater at toe

P: 21,020 (1: .29) = 2580 lb.per sq.ft.
10.5 1430 lb. per sq.ft.

overt urning Facto r :

Ratio : 34 36 lb.ft. = 3.8 (satisfaCtOI‘Y)
34,900 lb.ft.

Sliding Factor:

Bat 10 = 21 .giggépw = 1 . 8 (Sat isfacto ry)
: ° ‘

It is obvious that the Resultant intersects the base within
the middle third as 'e' is very snall.
Stem Moment:

p: .33 (100)(16.'75) = 558 lb.

Overturn Moment: (Due to earth pressure)

132 (18.75): 1,685 lb. 23415 = 10,200 ft.lb.
2
420 {12.75) = 2,580 lb. 12,75 ,___ 11,400 "
. 3 4 3
Total “54,365 lb. Total Mo.= 22,100 ft.lb.

Stablizing meent:(Due to loaded deck and wt. of stem)
6800 (1.125) = 7,000 ft.lb.
4460 (1.165): 5,200 "
Total MOO=12,2OO ft.lb.
Stem Moment :7 22,100 - 12,200 = 49,900 ft.lb.
Toe Moment:

p: §.17 {1160! 4. 1420: 2100 lb.per sq.ft. (earth pres—
10.5 .
sure next to stem edge)

4n!

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I...

 

Moment Upward:
2100 (4.33) 9,100 lb. 4.33 :- 19,700 ft.lb.
489 ' (4.33) 1,030 " (453322

8 Total wt.= 10,130 lb. Totgl Mo.

 

8,980
88,680 ft.lb.

Total To e Moment :
l = 22,680 - 3,470 = 19,810 ft.lb.
Heel Moment:

P: 3.83(1150) + 1420 = 1843 1b./sq.ft. (pressure next
10.5 to stem edge)

Moment Upward:

1420 (3.83) = 5440 lb. 3.83 : 10,400 ft.lb.
2 ,
423 (3.83) z 81] " 3.83 = 1 035 .

Total wt.= 6851 lb. Total Ho.= 11,435 ft.lb.

Moment Downward:

M = 6480 4 1435 = 7855 1b.. 3.83 = 15,030 ft.lb.
Total Moment 8 15,030 - 11,435 3' 3,595 ft.lb.

45

 

Case 4
Pressure due to retained soil with live load surcharge

and stabilizing moment due to dead and live load on deck.

 

A in,~ LLMLF—"R

 

 

 

 

 

 

 

 

 

1 '___I \ P'
F T“)
it1 u) ‘“l “21 \‘
n. b. X. t. t l \x
33 '3 '3 i P\‘
‘0; { -"~re:;W——\\
t «it ’ ' “’2 _‘b P“;

 

 

 

Resultant Forizontal Force= 7870 lb. : y = 6.4'

To find Resultant Vertical Force:

arm mment
n. 3 L. load on deck—-6200 5.46' 33,900 ft.lb.
‘1 = 4460 5.5 24,800 '
1'3 =3940 5.25 20,660 '
w4 = 7950 8.58 68,850 '
Total weight = 22550! .Total Mba=l47,610 ft. 1b.
1: 147,610 ft.lb.: 6.55.1
22,550 lb.
°'= §42411§191-= 2.23! 2 6.55‘ - 2.23'== 4.32'
22,550

= 5.85 - 4.38 :: .93'
Earth Pressure: (greater at the toe)

P: __834__550_ ('1 t .55) : 3380 lb. per sq.ft.
10.5 965 lb. per sq.ft.

0v ert urning Facto r:

Ratio;- 157510 1.15-lb: = 2.9 (SatisfactorY)
- 7870 (5,4'

. . ill-Illllllrllllt) but)...
.. s

Sliding Factor:

Ratio 2 88.5253 1‘5. 1.50.) = 1.4 (this is unsatisfactory)
8 O b.

 

It is obvious that the Resultant intersects the base within
the middle third as 'e' is very small.
Stem Moment:

p = 685 1b.

overturn Moment: 32,900 ft.lb.

Btablizing Moment: 7,000-+ 5,200? = 12,200 ft.lb.

getal Stem Moment: 38,900 —12,200:: 20,700 ft.lb.

Toe Moment:

9: 83:3 56-17 + 965 = 2345 1b.per sq.ft.

Upward Moment:

 

 

2345 (4.33) = 10,150 lb. 4é33 : 31,8001t,1b,
9.7.5. (4.33) = 2,110 lb. (4.33)3 : 6190 .
2 3

total wt.= 18,260 lb. Tota1.M.=27,900 ft.lb.
Total Toe Moment: 87,900" 3,470 = 24,430 ft.lb.

Feel Moment:

P: 3355 3~83 + 965 = 1823 lb.per sq.ft.(earth pressure)
10.5

Upward,Moment:

 

965 (3.83) = 3700 1b. 3.83 = 7,070 ft.lb.
858 (3.83) = 1645 1b. 3.83 = 8,100 ft.lb.
8 . , 3

Total “35345 1b. Total Mo": 9,170 ft.lb.

'Total Moment = 18,000 - 9,170 :: 8,830 ft. lb.

Total Shear : 9,385 - 5,345 .. 4,040 lbs.

. Ell II.l gill")... .i
— .-

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Results of the Invertigations of the Hour Cases:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TTEM CASE 1 CASE 2 CASE 3 CASE 4
Eccentricity .14 1.15 .51 .93
Earth Toe 1,370 3,020 2,580 3,320
Pressure
1b./sqft. Heel 1,170 620 1420 965
Safety Overturning 4.4 2.6 3.8 2.9
Factor Sliding 1.7 1.8 1.8 1.4
Toe 9,476 17,690 19,210 84,430
Moments
Beel 3,811 11,300 3,595 8,830
ftélb. 1
Stem 11,400 24,490 9,900 20,700
Toe 4,891 7,960 8,505 10,635
Shear
Reel 1,898 5,330 1,604 4,040
lbs. ,
Stem 8,690 6,050 4,365 6,050
Conclusions:
1. vMaximum Stem moment occurs at Case 8 loading.
2. Maximum Toe moment occurs at Case 4 loading:
3. Maximum Heel moment occurs at Case 2 loading.
4. Greatest Excentricity occurs at Case 2 loading, which is
within the middle third.
5. Lowest Overturning Factor , 8.6, is satisfactory.
6. Lowest Sliding Factor, 1.2, is unsatisfactory.

 

Resistance to Sliding:

 

 

 

 

 

 

 

i

 

Lowest Sliding Factor occurs with case 2 loading:

To secure against sliding there is provided steel sheet
piling driven to a depth of 6' 6' and a secure bond be-
tween footing and piling (fig. above) exists by means
of an angle iron welded to piling and a length of re-
inforcement bar hooked through angle and embedded in
the footing.

Sliding Factor = 19,310 (,50) + 3020(2.5) = 2.2
7870

This factor is satisfactory.

STEM ANALYSIS

Forces acting on Stem:

I

 

[/25 ’
-)I

 

‘-

.4"
1' ’65.; '.

if.

21:

 

 

To find '3 (1":

f0 = 1800 psi.

f :

8 3000 p810

n .: 10

/Zol

 

 

 

 

 

S.L. (Stem Load) 4460 lb.

D.L. (Dead Load of 2860 lb.

structure)

P (Earth Pressure) 5050 lb.

 

kd: 800 (85) :10 in.

3000
3

jd : 25 - 3.33 : 21.67 in.

d!
15¢

 

 

II

cfi5

0/
.id

 

 

Fig. 10

 

Stem Analysis: (moment Case "8")
n = 24,490 ft.lb.
v .-.-. 6,050 lb.

Depth required:

d=‘/_g__'= 4 49 ‘z 11" : stem is 28" )
K b Vgi'e‘c's'fli 2 (

Area of steel required:

 

A9: M 24 490 (12)

a )3 ,5 . ): .75 sq. in.

Area of steel provided:

,‘3! circular bars, spaced at 8' and 1'4' alternately.
‘4

A8: .442 (.121): .663 sq. in.
8
Check f8:

r = M :..- 24 490 12 = 20 500 lb./sq.in.
3 r1357 562731.67) ’

8

The steel above seems to be overstressed; however
Mr. Puffer of Michigan State Fighway explained to me
that it is their policy of allOWIng a 30% over design
stress because they feel that the bridge will probably
never be fully stressed. Also,they feel that the rigid-
ity of the structure resists the force caused by the
loading sufficiently to warrant this policy. The problem
above is beyond the scope of this paper, therefore I will
only mention that the stress above is well within the 30%
allowed. This same condition occurs in the steel placed

in the toe.

q. glirllflln'v I..." I hi: 's‘fl‘e
‘. .

Check fc:

f = M = 2 490 BB =. 886 lb. s .in.
0 35§1b5(}a) 21.57(12)(IU) / q

(1800 psi.allowab1e)

Check shear:

 

V :: ‘ V = 6050 _—-_-, 83.8 psi.(60 psi. allow-
5 (33) 12 (21.67)
able)
Check bond:
)1 = 8‘7 : 8 (6050) : 104 psi. (150 psi. a1-
, : {.3 _...’ .. 77 35
a 4 lowable)

Anchorage is sufficient as a standard hook is provided
which is hooked over another bar and embedded 27'.

Stem analysis in:
3

M = 13,780 - 8410 :: 5,370 ft.lb.
Thickness of stem is constant throughout.

Check f3:

:8 = n z 5370 (12) = 9,000 psi. (18,000 psi
As (id, ' .331 (21.67)
‘ allowable)

The amount of steel used above is just half of the

steel provided in bottom.of stem. Every other bar is out

off 1'8" above the 11:11 point taken above, therefore

sufficient steel is present.

1| with... l.

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I'll '6. 3" is!“ ’1‘
slut.

. --.

Forces acting on base of Abutment:

A. Forces acting on Heel (Case 2):
q

 

_l, ”’6‘" ,9
44 3/0"

"—"'.
m ‘*

 

 

2’60:

 

630
J“ -e-

 

 

.3Ch?¢7

B. Forces acting on Toe (Case 4)

/a’6 u

41 s»-

4 ‘4’] 3701’

9 ‘ DZ"
£155.... 1
J /

c. To find 'jd": kd = 27(12001 = 10.8 in.
3000
jd: 27 - 10,8 = 23.4 in.

 

 

2'6"
3‘

 

 

.96 5

 

 

3 so .t‘
\/.920

.4 _di
k Raw. Z.

t ,t
1% a

 

 

 

 

 

 

 

 

 

1. 51.91 .0!) r. 70' v .r
.

. . riflflue;

Heel analysis: (moment Case '8")
H: 11,300 ft. lb.
v = 5,330 lb.

Depth required:

(1: V11,300 (12$ :1 7.4- (30* is provided)

208 (12)

 

Area of steel required:

AB = 11,300 (12) z .322 sq.in.
8,000 23.4

Area of steel provided:

3' circular bars, spaced at 1'4”

AB -_- .44 (12) z .33 sq.in.

Check f8!

fs = 11 300 18 .-.- 17.500 psi. (18,000 psi. allow-
.33 23.4

able)

Check fc:
r = 131,300 (12) = 45 psi. (veersr low)

’5

V 23.4 (18)(10.8)
Check bond:

u = ;-_ 43 psi. (150 psi. allowable)
7 (93(17)25

Check shear:

v : 15753.77 :- 19 psi. (very low)

Anchorage:

L ,_— 17 500 (3,): 22" required (33" provided)
4 150 4

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Toe analysis: (moment Case '4')
M : 84,430 ft.lb.
V = 10,635 lb.

Depth required:

d: Y24,430 (18) 3 10.8" (30" provided)
808 (18

Area of steel required:

A3 = 24,430 (12) _.,- .695 sq.in.

, u 0
Area of steel provided:

if circular bars, spaced at 8‘ and 1'4“ alternately
4 .

Asf: 1.438.112). = .664 sq. in.

Check f8:
f8 =7 24,430 512; =» 18,900 psi. (18,00 psi. design
.1 . stress allowable)
The overstress here is alnost negligible.

Check fc:

fo 8 (84,430).;§ :' 193 psi. (very low)
, 83.4 (18)10.8

Check bond:

u z 8 10 6351.8, :: 137 psi. (150 allowable)
7977025

.Check shear:

iv = 19,635 = 37.8 (very low)
12 (23.4)

Anchorage:

L : 18 900 (3) = 24" (34* provided)
4 150) 4

WT NGWALL ANALY ST S

The wingwall serves only to retain.soil with a 86 degree

surcharge.

 

 

 

 

 

 

 

 

 

 

 

Angle of repose taken as 30 degrees.

0%: .946 .946 - (L9 — .755 a ,3 .4
. .14 T09 " e755)

Soil pressure against footing:

 

Resultant Vertical Force:

71 = (12.75)(1.5)(150) := 2870 afm' = ¥%%5%%'ft.lb,
‘wg : (12.75)(,%3)(150) := 793 6.1 = 4,840 ’ 7
w; = (12.75)(,§_3_)(100) = 530 6.33 = 3,370 "
w4 = (18.75)(3?83)(100)==4880 8.58 = 42,000 '
I5 = (10.5)(2.5)(150) ::3840 5.25 = 20,700 .
We 2 (4.67)(§§§)(100) == 537 8.94“ 51 4,800 7

Total wt213450 Total M580,310 ft.lb.

x = M: 5.96'
13,450

Resultant Horizontal Force:
p= WCh 2' 100 (.4) (17.55) = 703 lb.

P =: 703 (17.55) = 6,170 lb.
8

 

flfljfis 11)” .

9h = 6,170 (.9) = 5,550 lb.
2., = 5,170 (.44): 2,720 lb.
To find 7'
v: (10.5 - 5.96) tan. 26 = 2.22'
y'= 5.85' - 2.22' = 3.63'
To find e:

e'-- 11.31141 3 3.63 (5550) = 1.25'
R+Pv 9

5.96' - 5.25' z: .71' (eccentricity of Rh)

e = 1.25'- .71‘ =' .54' (Distance R intersects base
from base center.)
.54' less than 3.5' ;(therefore.R intersects base
8

within middle third)

Earth Pressure on Base:

P = 16,170 (1: .309) = 2020 1b./sq.ft.Eat toe;
10.5 1065 lb./sq.ft. at hee

Moment against overturning:

F _.._. 16,170 (5.961 -.-_- 4.8 (good safety factor)
5,550 (3.63)

Moment against sliding:
F ___._ £170 (.50) : 1.46 (This is a little low)
Sliding gagggr is a little low but there is sufficient
fill over toe to provide passive resistance sufficient

to keep wall from sliding.

 

41 a--- H 1....

Stem analysis:
p: 100 (.4)(15.05) .-. 6021b.

Px 602(15.05) = 4,503 lb.
2

15h: 4,503(tan 26") = 4070 lb.
Overturn moment: 4,070 (5') = 80,350 ft.lb.
StabiliZing moment:

14.: 18.75(1.5)(150H1é52+"§§(12.75)(150)(1.98)
M = 3,670 ft.lb.

Total Moment: 80, 350 - 3, 670‘: 16, 680 ft.lb.

d: V16 680 12) z 10" (30" provided)
208 (127'

As: 16 680 _—_ .518 sq. in. (required)
18 ,000(21'. 67‘)’

Steel Area provided:

 

,3' circular bars, spaced at 8"

4
A8: .442 (12) = .663 sq. in.(provided)
8
Check f8:

1’45—" .19 580 (13) : 13, 900 psi. (18, 000 psi.

.663 (21.67)
allowable)
Check to:

fc -..: 8 (16,680) : 119 psi. (very low)
0.8 18 81.67

Check shear and bond:

v = 4,070 = 15.6 psi. (60 psi. allowable)
11: s 70 8 = 52.7 psi. (150psi. allowable)

7T

1. ’T‘Ld .,
.

INF

Outvfbu.
II.

y.-.\.

Check steel 4' above base:
p: 11.05 (100) (.4) = 448 1b.

P: 442 (11.05): 2,450 lb.
2

Overturning Moment:
M = 8,450 (11.05)~ : 9,030 ft.lb.
Total Moment: 3
M = 9,030 - 2150 = 64,880 ft.lb. (allowing for
stabilizing moment)
To find “(1":
d: M?- 6" (narrowest portion of stem
808 (18
, . A9”;therefore depth is 0.K)
Steel provided: half of the steel in bottom of stem
is cut off 1" 8" above ,1_h, therefore we will use one-
half the steel providedsin bottom of stem in deter-
ming the steel stress.
Check fs’
f3 ; 6,880 (12) = 15,000 psi.(18,000
.331 .87 19.1

psi. allowed)

Sufficient steel is present in the stem.

Heel Analysis:

P= (2020 - 1065) (3.83)+ 1065 -

 

._ 1415 lb/sq.ft. (pres-
10.5
‘ sure next to stem)

Moment upward:

1065 (3.83)(3.83) :: 7,840 lb.ft.

2
350 (3.83)(3.83) = 633 w
8 3

Total Moment = 8,473 lb.ft.

Moment downward:

M: (48804-1440) (1.92) = 11,360

Total Moment:

lb.ft.

u = 11,360 - 8,473: 2,887 lb.ft.

Total Shear:

V = 6320 - 4575 = 1,745 lb.
, Steel Provided:

,Q'circular bars, spaced at 8' 0'.

4
A8: .2218 (12) .-.: .22 sq.in. (provided)
Check f3:

f8 = 18(8887) :: 675 psi. (very low)
.88 83.4

Obvhous that‘shear is satisfactory.
Check bond:

u:= 745 6
87

68.8 psi. (150 psi. allowed)

q; bl-.’lg wi'wull'b'L-kllj

.
' .

Toe Analysis:
P 2. 955 (6.17) + 1065 = 1625 1b./sq.ft. (earth presure
10.5 7 edge of stem)

Moment upward:
1625 (4.33) = 7040 (4,33) = 15,100 ft.lb.
5.92 (4433) = 1.812 28:313.) = 3,490 '
igTotal wt. = 8250 Totil Mo$=18,590 ft.lb.
Total Moment:

u. = 18,590 - 3,470 = 15,120 ft. lb.
Total Shear:

v = 8,250 451,615 = 6635 lb.
Depth is obviously satisfactory.

Area Steel Required:

As = “ 15,120 €12; .-_- .43 sq.in.

Area Steel Provided:

3F circular bars, spaced at 1' O”.
4

A3 = .448 sq.in.
Check f3: and f0 2

rs -_- 15 120 12 = 17,500 psi. (18,000 allowable)
'."‘""442 (‘2'13".’4)

r0 = 2 (15,120) (12): 240 psi. (very low)
23.4 (12) 10.8

Check v: and u:

v -.-. 635 = 23.5 psi. (very 10")
2 23,4)

u 3 8 565635) 8 :. 85.6 psi. (150 allowable)
7 9 CTT 85

Anchorage is sufficient (same as abutment)

“IIIII’IIA I lit-'1')? :w'r'..yi..v
. .0: ,

CONCLUSTON

The analysis of this structure has shown that it is ad-
equately designed and meets all specifications with the ex-
ception of the overstreesed steel in the stem of the shut-
ment while undergoing maximum bending moment.

The policy of allowing a 30% overstress, if justified,
places the stress of the stem steel well within the limit.

Thus the structure can be expected to serve safely and.
adequately all the traffic that may be passing over it and
at the same time provide free passage of all water that may

be expected to flow in Johnson Creek.

 

 

. V .. . T .. s 3). r‘t’g‘lfll'i/LV |

.El. ft)“; having: ’5 '50:! . .1“;

l.(li‘llllliillllilWHIWill)“ a

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