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«SUPPLEMENTARY
MATERIAL

IN BACK OF BOOK

 
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IN BACK OF BOOK

 

 
 
rresented for B. S. Degree

By
T. De Moss and Ie F. French
Title

DESIGH OF

WICHIGAN CaNTRAL xralbavaD AWD MOUNT HOPE AVENUE

GRADE SirAnATIOL

LANSING,  lICd.

rev I CEA
THESIS
Grade separations have assumed an importance undreamed of a dee
cade ago. The rapid change in the speed of the vehicular traffic
due to the avent of the automobile, has increased accidents at m
alarming rate.

Many of the accidents can be traced to the lack of grade
separations. The public in its new idea of “Safety Always” has
rigen above looking at the cost of the project and are demanding
safer crossings.

Since by a new state law there can be no more railroad
crossings laid without constructing a grade separation it is just
as necessary to build them over old crossings.

This particular grade separation that we are designing
is located at the intersection of the Michigan Central Railroad
and Mount Hope Aveme at Lansing, Michigan. Mount Hope Avenue
runs East and West and the railroad North and South. Mount Hope
is one of the main arteries leading from Lansing and on a traffic
census there were in eight hours, 251 pedestrians, 183 vehicles
and eight trains. The location on the city map in the holder at
the back is indicated.

We first looked over the ground, taking pictures and
getting the location and general contour of the ground in mind.
At this time we also took traffic census to determine if grade
separation was needed.

There are four possibilities in construction at a grade

‘separation. They are as follows:

102916
Change direction of the railroad; change the direction of highe
way; go over the railroad by means of a bridge; or to tunnel
under. The first two are impractical except in rare instances.
This being the case we decided to use a pass-over due to the
profile of the ground which was advantageous to this design.

The next step was the fimal survey which consisted of
taking cross sections and topography. The topography was check-
ed over three times because our work did not agree with city
maps. We checked every thing and found that the city maps were
in error.

With this data and with the advice and aid of Profess-
or's Vedder and Allen we designed the grade separation which
computations follow ;

The overhead clearance necessary for a steam railroad
is eighteen feet from the rail top to the lowest point of a
bridge or other structure. In order to obtain the greatest ne-
cessary elevation for the design of the new grade of the high-
way it was needful to first design the bridge and get the depth
from crown of roadway to under side of bridge. Althe the raile
road at present has only a single track, the necessity of double
tracking railroads is well recognized and in consequence of this
fact the lateral clearance allowed was that needed for two tracks.
The width required for two tracks is thirty three (33) feet, and
as the abutments were to have no batter, this was taken as the
span of the bridge. As the traffic rate is rather high and also as
a protection against future congestion, the roadway over the

bridge was designed twenty-four (24) feet wide.
 

 
 

The general form of the bridge as designed is as
follows ; The roadway is carried by three floor slabs suprort-
ed by two beame transverse to the highway and by two abute
ments. The two beams are supported by two longitudinal girders
which also bear the weight of the side walks and railings.

The latter are attached by cantilever construction to

the girders. The type of abutment adopted was the cantilever

type@e
Ae

Be

Standard Notation.

Rectangular Beams.

f s tensile unit stress in steel.

6

f = compressive unit stress in concrete.

c
E - modulus of elasticity of steel.

8
E - modulus of elasticity of concrete.

Cc
n = Es

Ec
Mes moment of resistance, or bending moment.
A, = area of steel.
b = breadth of beam.
ad = depth of beam to center of steel.
K = ratio of depth of neutral axis to depth, d.
Z = depth below top to resultant of the compressive strecses.
j = ratio of lever arm of resisting couple to depth, d,
jds de-Zs arm of reeisting couple.
Pp ss steel ratio = As
bd

T »~ Beamse
b 8 width of flange.

ob ' w width of stem.

t

thickness of flange.
Mt

+
Standard Notation - continued.

Shear, Bond and Web Reinforcement.

Vs total shear.

V's total shear producing strees in reinforcement.
v 8 shearing unit stress.

v= bond stress per unit area of bar.

o se circumference or perimeter of bar.

Ev 3s sum of perimeters of all bars.

T es #£=total stress in single reinforcing member.

6 = #£=horizontal spacing of reinforced members.

The notations have been made and approved by a come
mittee consisting of the foremost engineers in the United
States.

All formulae's are taken from Hool and Johr.son a

recognized authority on concrete design.
Design of Bridge.

M.S.H. ( Mich. State Highway ) Specifications.-

Mina continuous slab = wi.

1

Mina simple beam = wi?
8

Minimum slab thickness 5 1/30 of the span.
Loading for sidewalks = 100# / eq. fte
Impact = 25% of the live load.
A 20 ton truck gives a load of 140 # per sc. ft. on the
area actually occupied and as this is a short span 140 #
will be taken ae the live load.
Slab Design.
span = ll's 2 Use # 2000 concrete

live load = 140 # fs = 16000 / eq. in.

25% Impact = 35 # fc 650 / sq. ine

Dead load : Consists of 12" packed gravel = 120 #
Total load = 140+35+120 = 295 #

Assume that slab is 8” thick.

Then unit weight or "w" = 295 §x150 = 395 #/ ft.
12

From specifications Ms wi?

1

and "b" = 12" ( assume a section one foot wide ) and

"K*® = 107.4 ( from tables on pere 355 Hool and Johnson).

 

bd? = Mu
K
a ae Py j= __ s 28 x12 s 5.9"
IbK 12x12x107.4 12x107.4
Since it is necessary for fire protection to adda
certain amount to the beam ( table on page 300 Hool and Johnson)

14" additional was added and D & 7-15" say 7”

382.5 # / ft.

our neww= 295+ 7 x 150
12

 

oO

ds 1 382s x1l2 = 5.75 and De® 7.0"
12 x 107.4 (which is 0.Ke bee

cause we assumed an 8" thickness and 395 a / ft and both figured
results are under this).

Then the area of steel must be found.
As® pbd #8 .0077 x12 x 5.75 = x531 in.? Or AB x55
Use + bars spaced 5$" longitudinally
Use ~ bars spaced 18” transversely for shrinkage and temperate
cracks.
(The size and spacing of the rods is found on page 357 of Hool and
Johnson).

Testing for shear is necessary to see if slab is strong
enough to take all shear.

ve V - 382 25 x 5 5 < 34.9 # which is 0.Ke
| 12x0 374x515

allowable 40 #
In appendix B Hool and Johnson it says * For slabs with
horizontal bars only and without web reinforcement, 2% of the
compressive strength may be allowed". Since we are using a 2000 #

concrete it would be 02 x 2000 = 40 #
it}
 

 

 

 

It is now necessary to see if the bond is great enough
- between the steel and concrete so that they will not pull apart.

us _V_ = _382.5 x 565 = 96 #
Eo jd A3OX » B74X5 075 too large for straight

bars but can be made safe with hooking the ends.

In Appendix B ( page 846 Hool and Johnson) under the
title BOND . “The bond stress between concrete and plain rein-
forcing bare may be assumed at 4% of the compressive strength of
the concrete". "But by hooking them or deforming 5% of compress~
ive strength may be allowed”.

.04 x 2000 = 80 # eo it is too large (96#) for plain bars but
with hooks on .05 x 2000 = 100 # hooks make it safe.

Design of T = Beams

(2 span length = 11 x 12 = 33"
b § (12 x slab,t, = 84" Use b & 33"

Mot wit - Woo ll x 382.5 = 4208 # / lin. ft.

“8

Ll = 24%

 

d = 3(M = 14208 x 242 x12 5 10.15" D® 11.55"
DK 8 x 107.4 x 33
SAY 12" 2D

Ki = 2ndAs+bt? = 22x15 x 10.25 x 2.75 + 33 x 49 = 4.22"
2n AsBt2Bt 30 x 2.73 + 2x 33x 7

Case I because the neutral axis comes in the flange.
As & pbds .0077 x 33x 1005 = 2.67 in 2
Use 7 =~ 5/8" bars Ass 2.73 in2 ( p. 357 Had).
b * needed ¢ 3(n-1)4 4 | d 23 (7-1 )+4, 5/8 = 1375

W= 4208 (12 =~ 7 ) 13.75 x 150 = 4279 #
144

—

d cM = 14279 x 242 x12 = 10.25
/ 8 x 107-4 x 33 DZ 11.75

This is OeK. because assumed D was 12” .

<
36
<3

= 4279 x 12 ~- 417
bjd 1375 x 0874 x 10.25

 

Stirrups are necessary because shrinkage is liable to part.

Use 5/8" W - Stirrups As = 1.227 Hook them.

—*
éf

3/4 x 1.227 x 16000 x .974 x 10.25 = 5.14
4279 x 12

space stirrups 5 feet apart.

Side Walk Slab.
L.L = 100 #/ ft 2 1 = 6%
Suppose slab t+ = 53" including 2" finish and 1” steel

protection.

 

12
d = -/M = -/A72 x 36x12 = 2.70 04K.
y bK / 12 x 10764 x 8

As S&S pbd s&s .0077 x12 x 2.75 8 .25
Use 3" bars snzced 3° transversley.
Yee +" bars spaced 18" longitudinally.
The railing used is an approved design of the United

States Office of Public roads,
 
Steel necessary in railing to furnish support for

outer edge of sidewalk and its own weight.

 

 

b s 8*
Wel2ax3+8x48x150 = 916
144
de -M> = y96x1k2x12> 5 11.4%
,; OK 8 x 12x 107.4
As = pbdd s ,.0077 x 8x11.4 = .70

Use 3 4" rods As & o75
Design of Cantilever Beam -«- to supvort sidewalk and railing.
Let d s 10,25 *

bd? & M or bs _M Say b = 6 #" w& 6,375 "
K ad 2K 8

5103 # / lin. ft.

=
as

31

 

6.875 x ( 11015 = 3015 ) x 150

N

it 12 = 780,000 in. f/ db.

”

MS "916 x ll <x 6-1/3 + 57.3 x 6.65
- 2

Ce
t8

190,000 s 6.87 *
10.25 x 107.4

As = p bd =: .0077 x 6.87 x 10.25 =) 0543
Use 3 = 3" rods.
yeofuy : = 53 # / in.?

pa x ll + 57.3 x 6
bjd 087 x 0874 x 10.2
Stirrups are necessary.

Using 5/16" ° YU stirrups with hooks Ass ,.1534

S = 3 Asfsjd #& 3 x 21534 x 16000 x .874 x 10.25 = 3.16"
2 Vv 2 10450 7

Space stirrups 33" .
di
 

1/

Girder Design.

M of concentrated loads s ( 916 x 11+ 57.3 x 644279 x 12)

x 16.5 = 186000 ft. / b
5 5

Say Girder is 20" x 50"

Wo2 172 x3+4+20x 50 x150 = 1556 #/ ft.
144

M due to ‘istributed load = wi? = 1556 x 33% = 212000 ft. #

8 8
Total M & 180000 t 212000 = 398000 ft. / b.

d = Ms 7398000 x12 = 47.5" D & 50"
bK 20 x 10764 OK. because

this size is equal to the assumed size.
As = pbd & .0077 x 20 x 47.5 = 7.31

Use 5-14" rods Ass 7.8

The Joint Committee recommends that the lateral spacing
of parallel bars should not be less then 3 diameter C -C and
that the distance from the side of the beam to the center of the
nearest bar should not be less than 2 diameter.
Testing ; b & E (ne-l ytala » (3x4%44)1¢ &S 20°

OK. because this is equal to the assumed size.

ve V e 1556 x 16.5 x 61900 = 105.5 # / in. 2
bjd 20 x e874 x 4705

Stirrups are necessary because the allowable unit shear
is 40#.

Using $” Round hooked W - stirrups Ass 4x .1963 = .786

a
58
ope

Aspfsejd = 3x 786 x 16000 x .874 x 4765 5 895 *
V 2 87500

Space stirrups 8 7”
8
iq
1k

Girder Design - continued.

Point where no more stirrups are needed.

(ley) 2 16.5 (1+ 49 ) = 10.25 ' from end.
v 105 5 |

nm]

From a point 10! = 3” from the end spxuce etirrups 24"

to take care of shrinkage cracks due to temperature changes.
13

Abutment Design.

In this climate the depth to which a footing must go
to be safe from frost action is usually taken as four and one
half ( 4¢ ) feet. This makes the total height of the abutment
18 +44 8 224 feet.

As an expansion joint separates the bridge from one
of the abutments, the abutment must. be designed to resist the
earth pressure and to sustain the weight of the bridge, the
bridge furnishing no bracing for the two abutments.

Nse a factor of limitation "f" of 2 and let "K" ,
the proportion which the length of the toe bears to the
breadth of the base "b” , equal 1/3 . Take the unit weight of
the fill "w" to be 120 # / ft. The angle of friction of the fill
is 30° . With theee assumptions the abutment was designed ace
cording to Hool and Johnson formulae's pages 5838596 .
C$ is a constant depending on the material of the fill and may
be read from Diag. ” page 588.

Lateral unit pressure of earth = Wits CGg Ws .333 x 120
- 40 # / ft.

= ctw: 07d Y 2 x 40 = 2662

b 2 .662 x 22.5 & 14.9 + Say 15 °

2
h

From the right hand side of Dig. 6 page 587 Hool and
Johnson it is found that for f 5 2 and K =1/3 the resultant
will strike the base at a point .50 x 15 or.7.5 ' from the toe ‘
which will make the structure safe from overturning due to earth

pressure and cause an evenly distributed pressure over the base.
 

 

 

 

 

 

/4-

Abutment design - continued.

Stem.
say that the base is 18” thick.
h * is the height of the abutment wall from the top of the base.

h' = hel" s 21 1

 

M of stem = wht2xh' = 40x 212 = 61700 ft. #

 

2 3
d s7M_ = ,61700x12 © 24" D = 26.5"
(DK /12 x 10724

As = pbd =e .0077 x12 2 24 3 2.22

Use 7" rods spaced 4" As = 2.30
8

As the greatest bearing on the abutment is beneath a
girder, the necessary thickness of the abutment wall will eq-

ual the girder shear divided by the allowable unit stress and

by the width of the girder = 11300 = 6,73"
50 x 20

Taper the abutment wall to 18” at top.

The safe bearing capacity of the soil is 8000 # / ft.°

The allowable pune hing shear is 120 # / in. 2

The concentrated load of the girders becomes evenly
distributed over the base as the stress is transmitted down
through the wall.

Distributed weight of girder and bridge =

51509 + 4279 = 7779 #/ ft.
4

Weight of wall per lin. ft = 9 x 26.5 x 21.5 x 150 # 69500 #
12
*
Abutment Design ~- continued.

Amount of punching shear s

 

 

 

1S

60500 + 7779 = 24 x 1 x 8000

12

= 52279.

Thickness of base needed for punching shear =

The unit pressure on the base due to weight

wall = 10 x 1 x 22.5 x 120 = 1900 #/ ft.

15
The bearing capachty of the soil,

vious figures, will take care of all the load due to

no moment need be figured for this load.

M to be designed for at the toe =

d 23M = 722500 x12 = 14.5"
ee rae 107.
A&B = p bd s .0077 x 14.5 x12 s 1.34
Use 3/4" rods spaced 5" Ass

The M at the heel.

as can be

wl? = 1800 x 25
2 2

22
2x12x120

= 18"

of fill and

seen from pre-

the bridge so

= 22500 ft. #

1.35

Overturning M of stem 3 61700 ft. lb.

Resisting M of toe s 39200 ft. lb.

M taken by heel & 39200 ft. lb.

d s,/M 8 200 x 12 @w 19. 1"
bK j12 x 107¢4
As S pbd = .0077 x19. x12 = 1.76
Use 7” rods spaced 5” As 8 1.84

8
Ni
 

76

Design of Grade.

After the bridge had been designed and the greatest
elevation necessary for the centerline of the road found, the
new grade of the highway was designed. The grade was designed
on the profile sheet (which may be found in the pocket at the
back ) . In choosing the maximum grade to be used several ave
thorities were consulted and the concensus of opinion was that
a grade of 444 » while slightly more costly, was the maximum
that could be used and still have a roadbed give its proper
wear and service. In other words the longer life of the road
more than makes up for the extra cost of the fill. Bearing
this in mind the grade was designed with smooth curves at the
breaking points and with a slope of 44% .

The cross section adopted (shown in typical cross-
section in pocket at the back) was largely copied from the
standard cross section used by the Michigan State Highway Dept,
but due to the large amount of traffic at this point and ale
lowing for future expansion of the city the M.S.H. cross section
was slightly enlarged upon including the depth of road metal. In
order to avoid any unnecessary fill it was decided to run the
sidewalk along what would ordinarily be the shoulder of the road,
separating the sidewalk from the roadway by a standard 6” x 20"
curb projecting 6" above the road surface. This curb protects the
sidewalk from the water draining off of the roadway and also
sives protection to the pedestrians. The water collected from the
roadway is drained through curb catch basins into sewers at Donora

street and Bailey street at the east and west ends respectively,
17

Design of Grade - continued.

It was decided that as there was a sidewalk on the
south side of the road only, there would be no sidewalk pla-
ced at present on the north side of the street. A curb was
placed on the north side however and enough fill put in place
to support a sidewalk if that becomes necessary.

The cross sections taken in the final survey were
plotted for every fifty ( 50 ) foot station and over these the
new cross sectiéns were plotted at the proper grade elevation.
The area between the new and old cross sections was measured
by means of a planimeter and the amount of fill and excavation
found by the "average end areas” method.

Tne estimate of cost was accomplished by figuring the
amount of fill and the distance it must be hauled, by computing
the cubie yardage of concrete and the number of pounds of steel,
and mutiplying these values by the unit costs of each. These

unit costs were taken from the current Engineering News Record.
1

Cost Estimate.

 

By the "average end areas" the amount of fill
was found to be 12,542.8 cueyd. The amount of gravel
necessary is 222.1 cu.eyd.- of 75.0 pebble and 208 cueyde
of 60%. gravel. The amount of concrete to be used in the

bridge is 227.9 cueyd. and the weight of steel is 40584.7#-

12542.8 eueyd. Fill at 3.73 ----------- 39157200

e22.1 "  " 75,0 Grevel at 31.80 ------- 400.00
208.0 " " 60:0 " at G1.50 ------- 312.00

£2729 " " Concrete including
form work at ~l5~200 —_— $418.50
40584.7F Steel in place at 5.05 --------- 2029.24

 

Total $15316.77
a Lr Pocket. - 3 Su vel.

Tracing ~ Crossec trons

44

- General Plan of Sridge

44

— rd a? 44 Wore

re
_

 
 

 
9438

iN

3 031