MASSPRODUCING SIMPLY SUPTORTED

PRESTRESSED CONCRETE BEAMS

By

Gun Aydin Caldiran

A THESIS

Submitted to the School of Graduate Studies of Michigan
State College of Agriculture and Applied Science
in partial fulfillment of the requirements

for the degree of
MASTER OF SCIENCE

Department of Civil Engineering

1953

 

INTRODUCTION . . . . .

1. Introductory Sta

2. Procedure . .

3. Acknowledgment

NOTATIONS . . . . . .
CONSTANTS . . . .

DETERE'INING BEAM DEPTH
FOR MOMENT . .

DETERMINING BEAM DEPTH
DESIGN........
CABLE POSITIONS . . .
CHECKING FOR SHEAR . .
BLOCK DIMENSIONS . . .
THREADED CABLE SOCKETS

SUMMARY.......

FOR

FOR

te

TABLE

ment

0 o 0

OF CONTENTS

VARIOUS

SHEAR

FLANGE AND WEB THICKNESSES

APPENDIX A: DETERMINING BEAM DEPTH FOR VARIOUS FLANGE AND WEB
THICKNESSES FOR MOMENT . . . . . .

APPENDIX B: DETERMINING BEAM DEPTH FOR SHEAR .

APPENDIX 0: DESIGN . . . . . .

APPENDIX D: CABLE POSITIONS . .

APPENDIX E: CHECKING FOR SHEAR

APPENDIX F: BLOCK DINENSIONS .

BIBLIOGRAPHY..........

300082

 

PME

ON «t: WHHH

15
19
27
32
39
1+3

95
101 ,
120
1M1

197

 

 

INTRODUCTION

1. Introductory Statement

This study describes a graphical procedure for designing blocks
for post tensioned, simply supported prestressed concrete beams. The
blocks designed are I-shaped and can be used for beams with span lengths
from 20 feet to 125 feet. The widths, flange and web thicknesses of the
blocks are expressed in terms of the depth of the blocks. For different
thicknesses of web and flanges the required depths for moment are cal-
culated. After checking for shear the block sizes are determined.

Design tables are derived for post tensioned, simply supported pre-
stressed concrete beams from 20 feet to 125 feet for H20-Sl6-HH loading
of the "American Association of State Highway Officials". The use of
these tables reduces to a minimum the enormous amount of labor necessary

in the design of such beams.
2. Procedure

In this study the beams are placed adjacent to each other as it is

shown in the following figure:

 

L 1 i J

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

The moments are first determined at various points of the beams.
Since there is no tie between the beams it is assumed that the wheels
on one side of the truck are directly on the beam. The difference be-
tween the absolute maximum moment and the moment at the center of the
beam is very small. Therefore, with the assumed conservative loads,
the moment at the center is considered as the maximum moment.

After moments are determined curves are plotted for the depth of
the beams as well as cross sectional areas for various flange and web
thicknesses and various flange widths.

By inspecting the above curves the number of blocks to be used,
from span length 20 feet to 125 feet inclusive, and the flange widths
are decided.

Next, the required depths for shear with various flange and web
thicknesses for the beam are determined.

It is found that for a certain value of web and flange thickness
the required depths for both bending moment and shear are the same.

As a final value for the depths of the beams these values are selected.

The next step in this study is to determine the cable sizes and
positions. To make it practical one cable size and position are used
for a certain kind of block.

Then shear is checked.

After making a summary of the properties of the blocks, the blocks
for the bending points of the cables are designed to give required dis-
tance between the center of gravity of the beam and the center of the

cable.

 

 

 

Finally the and blocks are determined. Here only the practical

aspects of the design are considered.

3. Acknowledgment

The author wishes to express his indebtedness to Prof. W. A.
Bradley for his valuable advice and suggestions and also to acknowb

ledge his thanks to Dr. R. H. J. Pian who read the study.

 

 

 

 

 

 

NOTATIONS

The symbols in this study, defined below conform essentially to

letter symbols for "The Principles and Practice of Prestressed Concrete"

by P. I. Abeles.

020__]

.1.

 

 

 

L 810

e = Distance between the center of gravity of the beam and the
center of the cable.

010 = Distance between the center of gravity of the beam and the
bottom fibers of the beam.

Distance between the center of gravity of the beam and the

top fibers of the beam.

f' = 28 day compressive strength of concrete.

c
Q = Constant

fct = Compressive stress at transfer.

fcw = Compressive stress at working loads.

ftt = Tensile stress at transfer.

 

 

 

 

 

f“, = Tensile stress at working loads.

fsh = Shear stress.
I = Impact load.

t = Web and flange thicknesses.

D Depth of beam.

B = Width of beam.

An = Cross-sectional area of beam.
Io

Moment of inertia of the cross-section of the beam about its

neutral axis.
= Moment of the dead load of the beam.
“LL = Moment of live load.
M. = Moment of working load.
Pt = Total tension in cables.
r = Radius of gyration of the cross-section.
V = Vertical resultant force.
H = Moment of half of the section about the neutral axis of the

entire cross-section.

sle = Section modulus of cross sectional area related to bottom
fibre.
$20 = Section modulus of cross sectional area related to top fibre.
L = Span length.
c‘= 12 Io
EST-
5= °10

W

I’U’ctfi'tw
_—'-'-*6’-——

 

 

 

 

CONSTANTS

In this study “The American Association of State Highway Officials"
specifications (l9h9) are followed.

Loading = s20-516-uu
6000 1b/1n2

’2
u

H = 0.85

2
tot = fc' = 1/3 f3 = 2000 lb/in
1,, = ft' = 0.03 f; = 180 1b/1n2

2
rah = 0.03 :3 = 180 lb/in

= .131..
I L+125 (not over 30% for moment)

For shear adjacent to the support use I = 100%

Since the cross sectional areas of the beams are symmetrical,
the center of gravity of the cross section is on the center line of the
cross sectional area. Therefore:

e10 = e20

Consequently 51° = $29

 

Since 51, = IO and 52¢ = .41..
6lo °20

 

 

 

 

 

 

DETERMINING BEAM DEPTH FOR VARIOUS FLANGE AND WEB

THICKNESSES FOR MOMENT

20 foot span

Moment, using equivalent uniformly distributed load and one con-
centrated load:
Movable Concentrated Load = 18000 pounds for Moment
= 26000 pounds for Shear.
Uniformly Distributed Load = 6H0 pounds per linear foot of Load
Lane.
Impact Load = {562%6——4 = 3H.5§ Use 30%.
18000 x 1.3 = 23N00 pounds
6h0 x 1.3 = 832 pounds.

Influence Line for moment at 1/8 point of span.

 

 

 

For l/h point of span:

3-75

 

5' . 15I

 

For 3/8 point of span:

 

 

 

7.5' 12.5'

 

For 1/2 point of span:

 

10’ 10'

 

 

 

 

 

Maximum Moments: For Lane Load:

For 1/8 Point:

E§:5 I 2-13.+ l1;§_§_2;l2% 532 + 23u00 x 2.19 = 18250 + 21250

 

2
39500 ft-lbs.

For I/M Point:

Elia—«3.5+ ' 2 x 1 ) 532 + 23100 x 3.75 = 31200 + 87700

118900 ft-lbs.
For 3/8 Point:

EE;69 2 7.§_+ n.63 ; 12.5%832 + 23u00 x “.69

3900 + 110000

1h9000 ft-lbs.
For 1/2 Point:
50 x 832 + 23h00 x 5 = M1600 + 117000 = 158600 ft-lbs.
Maximum Moments: For Truck Load:
“000 x 1.3 = 5200 lb.
16000 x 1.3 = 20800 lb.
For 1/8 Point:
2.19 x 20800 + 0.u38 x 20800 = 50625 ft-lb.
For l/M Point:
3.75 x 20800 = 78000 ft-lb.
For 3/8 Point:
0.69 x 20800 = 97750 ft-lb.
.For 1/2 Point:
5 x 20800 = 10h000 ft-lb.

So use truck load.

 

 

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OD lacuna :7?va _ zun Exo—

kua<m 1143.: 73.1.35 3.. 3's. 3!

 

 

 

 

 

ll

1. Assume A Section

t1:: ‘___71

 

‘1: 4-
L._1

Where: t = YD and B = 0.111)

A0 = (0.1m) (2m) + (mun—2m)
= 0.511132 + m2 - 21292 = 1.52m)2 _ 2121)2 = D2(1.8Y - 212)
I0 gig—DE" WM: 112:. (3.112 - 10.8Y2 + 15.213 — M")

a: 121 = 3.1+: - 10.8Y2 -+ 15.2?3 - BY“
3

a: °10 = 1
1372

Q, = 71 fat + ft. = 0.35 I 2000 + 180 = 313.33
‘7‘" “'77—’—

2
"w ' Y1 Mdb = QED é:
Assume M' = Qfinaéi
1011000 x 12 = 313.33 x 0.1.133 £3.11! - 10.13%2 + 15.213 - syn;

 

 

11
1011000 1: 12 =
313.33 x 3.1+ (Y - 3.175!Z+ 11.11713 - 2.3551113
n3 = 1170

 

(Y - 3.175112 + 11.11710 - 2.35519")

 

12

 

 

 

 

Y D Ab
0.05 30.1 77.3
0.1 25.2 101.5
0.15 23.3 122.2
0.2 22.3 139.2

_0.25 21.8 15u.5
0.3 21.6 168
0.h 21.u 179.
0.5 21.3 181

 

2. Assume:
Where: t = YD and B = 0.5D
A0 2.192 + YD(D - 2YD) = 2D2(Y - 22)
14

I6 = (123333" £3.50 " YD)? " 221M; 2‘52-(111! - 1212 + 161:3 - gr")
1 1

 

H

121 u? - 12?2 + 1613 - 8r“

qziafl= 0.5
2

3
10141000 x 12 = 313.33 x 0.5133 EWMQ

D3=-—-9-9§——_..~V‘
Y - 322 + hz3 - 2?

 

 

 

Y D Ab
0.05 23.6 77.7
0.1 23.3 10h
0.15 21.9 122.2
0.2 20.9 139.9
0.25 20.h 156.1
0.3 20.1 170
0.1 20.0 192
0.5 20.0 200

 

3. Assume:

Where: t = YD and B= 9.60.

n
= 0.61)“ _ L0.6D — YD)(D - 2m)3 = 2__ __ 2 1 3 _ 11
IO 12 .....12 .. ... 12 (h.6Y 13.2! 1+ 16.8Y 8Y )

o. ._. 1311 = 3.6! .. 13.21? + 16.213 .. 3}“

303 0.6

 

 

13

, , 2 3 M
. _ , + .8 -
103000 x 12'= 313-33 X 0-6D3 %H bY 1323.6 16 Y 8Y-%

D3 = _‘ 86605 _ 1._
r - 2.86512 + 3.6513’- 1.7ui“

A0 = 1.211)2 + (rn)(n - 210) 202(1.11 - 12)

 

“—— fir

 

1 D Ab
0.05 27.1 77.1
0.1 22.5 111.1
0.15 20.8 123.3
0.2 19.8 1h1.2
0.25 19.2 156.5
0.3 19.0 173.2
0.u 18.8 198.0
0.5 18.7 210.0

 

With these values curves can be plotted "D" versus ”Y” and "A0"
versus ”Y".

For the curves of "D“ and "Ab" see pages 93 and 9k in Appendix.

For the similar calculations for the rest of the spans see
Appendix.

After calculating and plotting the variations of the dimension "D"
and the cross sectional area, for different flange widths, span lengths
and flange and web thicknesses it is possible to select block sections.
In selecting the block sizes practical aspects as well as economy are
considered.

For resisting moment the dimension "D” and for shear the area or
the term I'Y" are the governing elements. Since it would not be very
economical to use one block size for span lengths up to 125 feet, after
thorough study of the curves for beam depths for moment it was decided
to use six different block sizes. ‘The following table shows the six

groups of span lengths from 20 feet up to 125 feet using the selected

block sizes:

 

 

Block No. Span Length
feet __

#1 20-30
#2 30.140
#3 “0-55
#14 55-75
#5 75-100
#6

100-125

 

 

11$

 

6 .

 

IIII. I‘ll-Ill

 

 

DETERMINING BEAM DEPTH FOR SHEAR

As it is seen from the curves on page 93 for values of "Y“ greater
than 0.3 the dimension “D" does not vary greatly and for values of ”Y”
smaller than 0.1 the dimension "D" varies greatly. In selecting the
block sizes, therefore, values for "Y” will be kept between 0.1 and 0.3.

For practical reasons only three flange widths are considered,
namely O.MD, 0.5D, and 0.6D.

Again from the curves on page 9% we can see that the flange
width O.MD gives the smallest value for the cross-sectional area but,
the biggest value for the dimension ”D" for a certain value of ”Y“.

The dimension "D” will vary inversely with the cross sectional area for
different values of ”Y".

When the above mentioned curves are studied carefully, it can be
seen that the curves for cross sectional areas, for flange width of
0.5D are closer to the same curves for flange width of O.hD than to
those for flange width of 0.6D for a certain value of "Y". But the
curves for dimension "D" for flange width of 0.5D are closer to the
same curves for flange width of 0.6D than to those for flange width
of 0.4D for the same value of "Y”.

Therefore, in selecting the block sizes to keep the flange width
0.5D will be both economical and practical.

As far as the bending moment is concerned the following minimum
values for dimension 'D" and cross sectional area are obtained for the

given values of 'Y'.

 

;!;$ *'

 

 

 

 

Block No. Span Length D Area Y
feet in. in2
#1 20. 30 21+.l+ 250 0 . 30
#2 30-110 26. 5 312 0. 26
#3 l+0435 34-9 3116 0.17
#11 55—75 10,5 1+2" 0-15
#5 75-100 M6. 3 ‘+96 0. 13
#6 100—125 119.5 598 0.111

 

16

The reason for variation in values of "Y” will be explained later.

We know the minimum value for dimension ”D” for the bending mement,

we should check to see what will be the minimum value of dimension “D“

for shear for a certain value of “Y".

At present it will be assumed that cables will be bent once, at

the 1/8 points of the spans, Therefore, we can safely assume that the

maximum shear will be at the 1/8 points of the beams.

checked again later, however. The maximum shear, for a certain size

of block will be in its longest span.

where H = (YD)(D - 2m)ED_;s_3_ni)) + ((3 (YD)E%- gig

I 2

1.
n
o FED/2 - (1/2 - 1+1 + 121

t YD

fsh = 180 p510

- 1613 + 815]

Shear will be

Substituting these values and solving for ”D" we have:

D2 = (3 - 6Y + hYa) v .
MSOY (1 - 31 + 1w? - 213)

 

by using the values for shear for the longest span lengths of the blocks

at their 1/8 points and using values from 0.10 to 0.30 for ”Y' we can

Po

 

 

17

determine some values for "D”. In order to be economical, we have to

determine such a value for "D" that it will be the same value of "D"

for moment for the same value of "Y” for a certain span length:

2

when: r = 0.1 D = 0.0789 v
Y = 0.11 0: = 0.0635 m:

Y a 0.12 0 = 0.0583 v
1 = 0.13 02 = 0.0545 vmx'
1 = 0.14 02 = 0.0509 17"“-
1 = 0.15 02 = 0.0’480 v ’
Y = 0.16 D2 = 0.01155 vm'
1 = 0.17 02 = 0.01430 vmx'
1 = 0.18 02 = 0.0010 vm“
Y = 0.19 02 = 0.0386 fun"
1 = 0.20 05 = 0.0375 vm“
Y = 0.21 D = 0.0361 vm“
1 = 0.22. 13‘; = 0.0338 11“"
1 = 0.23 02 = 0.0325 17”"
r = 0.214 02 = 0.0318 vm‘
r = 0.25 02 = 0.0312 vmx‘
1 = 0.26 02 = 0.0305 17“"
r = 0.27 02 = 0.0292 vm’“
1 = 0.28 D = 0.0283 17”“
r = 0.29 D2 = 0.02714 v“"
1 = 0.30 02 = 0.0267 fl:

For block #1, 30 foot span at 1/8 point

 

of the beam:

 

 

 

21150 e In. 21150
1
.875 \
.1109
\
ngsi‘ 26.25 _J

 

If '1‘

Influence Line

( 0 ).
16000 (1 + EVE—T2753 - 21150

 

18

Maximum shear = 21150(O.875 + O.UO9) = 27150 lb.
In order to allow for dead load

say Vfiax. = 30000 lb.

 

”—7

Y D A

 

(L
0.10 48.7 426
0.15 38.0 367
’0.19 34.0 356
0.22 31.8 348
0.25 30.6 351
0.30 28.3 336

 

where Ab = 2D2(Y - Y2)
For above calculations for the rest of the spans see Appendix.
In the following table the required dimensions due to shear and

moment and used dimensions are given:

 

 

 

 

BloCk Span Min. Ab Required ”D” Used

No. Length ”Y“ in2 Shear Moment D Y t

feet in. in. in. in.

(1) (2) (3) (4) (5) (6) (7) (8) (9)
#1 20- 0 0.30 336 28.3 24.4 30 0.30 9

#2 30. 0.27 380 31.0 28.5 32 0.2655 8 1/2
#3 40-55 0.16 428 39.9 35.2 40 0.1625 6 1/2
#4 55-75 0.14 441 42.8 41.0 49 0.1429 7

#5 75-100 0.13 496 45.0 46.3 60 0.1333 8

#6 100-125 0.14 598 44.0 49.5 72 0.139 10

 

The reason for choosing the values for "Y" in column (3) is
that the required "D" for shear and moment, as shown in columns (5) and
(6), are comparatively close to each other. For practical purposes the
value of ”Y“ is kept between 0.1 and 0.3 as mentioned before. In
colmns (7), (8) and (9) the actual values of "D”, ”Y" and "t = YD“ are
given for each block.

These blocks will be checked for these values.

 

 

 

DESIGN

The following formulas are used in the design:

1) 53.19.- Edh.=é: f
A S ct
0 1e

2) 5.211“ +£13.11 2f“
A0 32e

3) “Ptklo - i 5ft.

A0 31 e

1+) M‘tkéf
Ao

cw
Sae

where klo = 1 + °so e10

O

and 1:20

2 _ 10
° A.

O

1'

 

1 - 32%2229

The above formulas can be written as follows:

1+°so°0
1)..!~....= —1:g-L
Pt

Ao(fct+m)

 

 

( $1, )

°so e20 _ 1

2’ -%:-= ”a

t (M b 7
A0 (is... ftt )
11(1 + eso e )
1 ( o )
A6(hs'é'fiL’" ftw )

 

20

(e e )
1 _ V‘('§%g"an ‘ 1)
”’E‘T—fi‘

0(§E;" fcw )

For each equation given above a curve can be plotted for every
span length. These curves will be straight lines when the horizontal
axis is called "ego" and the vertical axis is called "%:r. These
curves will surround an area and from this area we can choose the cable
size or the cable strength.

The curve #1 provides the bottom edge of the area and the curve
#3 provides the top edge of the area. Curves #2 and #9 provide the
sides of the area.

When we plot these four curves for each, the shortest and the long-
est span lengths for a certain size of block, the area surrounded by
the four curves for the longer span is located below the area surrounded
by the four curves for the shorter span length, since the vertical axis
represents "%;P. From these two areas a common "eso" can be selected.
In order to facilitate following the procedure of the design of the
beams, the known factors will be listed for block #1:

f; = 6000 psi

re, and f0" = 1/3 ff: = 2000 psi

ftt and ftw = 0 psi (assumed)
Y\ — 0.85
D = 30 in.

B = 0.5D = 15 in.
‘omin. = 354 inz.

910 = 920 = 15 in.
YD = t = 9 in.

 

the

 

 

21

Actual Ab = 2 x 15 x 9 + 12 x 9 = 270 + 108 =.378 in2.

I =3Qx30x30x15_]_.:2x12x12x6
° 12 12

33750 - 863 = 32887 in“.

_ - I _.'288 .. 3
s1e _ 82e - _2_- 3 1 7 — 2190 in .
010 5
To be able to plot the four curves the points of intersections of
these curves with the axes can be found as follows:

For 20 foot span length:

 

 

l) Crossing the 1.__ axis = :1: = ( 2L 3
P (M ) 2.3.5.911).
t A _gg, 3 8 + 2000
o<$10 + rot ) 7 ( 2190 )
= .00000126

crossing the 6 axis = - i= - .81..= - 5.8

30

 

 

 

 

°10 15
2) Crossing the l__axis = _ l = _ __(v 1 )
P (M ) 2 000
t .82 378 - 0
A6(sle ' ftt ) ( 219° )
= " 00000246
crossing the eSO axis = £fl-= 5.8.
e
10
3) Crossing the L axis g T n 3 = 0.81
P M (1484000 )
t .JL. -——————-+
‘°<s1, * fur) 378< 2190 0)
= .00000332
2
crossing the e$0 axis = _ £9_ ._.. -. 5.8
610

 

 

 

 

\_

 

22

 

 

1+) Crosing the 1_ axis _ VI = _ 0.12%
P ’ ' —(M ) (148 00 )
1' t W ——
Ao<sle — f0") 378( 2190 - 2000)
= .000001705
crossing the 6 axis = 1‘3 = 5.8.
8° 610

And for the 30 foot span length:

 

 

 

 

1) Crossing the L axis = l = ( 1 3
P (“db ) 5M
t ‘°(s—'+ fct) 378( 2190 + 2000)
= .000001175
crossing the eso axis __. _§____ _ 5.8.
E’10
2) Crossing the L axis = _ 1 = _ ( 1 )
P (n ) 531000
t (11)
*o(§1:- ftt ) 378( 2190 - 0 )
= - .0000109
crossing the ego axis = i = 5.8.
e
l
3) Crossing the L axis = ( V\ ) = ( 0-635 )
11,, 210 000
‘o(§'1:* ftw) 373% 2190 * ° )
= .00000182
crossing the 0 axis = __ rc = - 5.8.
so 3—
10
1+) Crossing the l__ axis = _ VT Y1 j=_ ( 2.85 _)
P $9.992.
. 8.8;;- .., 37.. - 2...,
3 .0000028‘4

crossing the e c axis = i = 5.8.
810

 

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2M

As it is shown on page 23 the four curves for both 20 foot span and

30 foot span are plotted.

The two areas have a common part, meaning

that one cable strength could be used for all the span lengths from

20 feet to 30 feet inclusive.

From the table given on page 26 we can select standard cable sizes.

These cables are manufactured by the John A. Roebling's Sons Company

of Trenton, New Jersey.

Use eso = 4 inches

Pt

M
A09“ + .db.)

(868 Page 23)

... Equation 1 for 20 foot span:

 

 

( §1§)
8so 010
1 + 7"“
o
2 000
= 378E2000 + 2190 = 138000 = u71000 lbs.
1 4- %li 1.69

Equation 3 for 30 foot span:

Pt

(14 >
MET? f“)

Y e o )
r1<1 '* £13,510
(2106000 ) '
£81 2190 + O) - “53000 = 325000 lbs.

0.85 ((1 + @371 - 1.Ei

 

Use N - l 1]“ inch cables with Pt = 392000 lb.

325000 < 392000 < M71000 0k.

990 = N inches.

a:

 

 

 

’l "41’

 

 

If these blocks meet the specifications at their longest and short-

33

est designated span lengths with a certain value of "es " then with the

0

I.
same value of eso

fications.

' the intermediate span lengths must meet the speci-

For the same calculations for the other spans see Appendix.

In the following table the selected blocks and their properties

are given:

 

 

 

 

.7:
t Q
”I 4_
[gs—J
Block Span D B t 10 Cable Diem. Pt
No. Length in. in. in. in2 and No. 1b.
ft.
1 20-30 30 15 9 378 u - 1 1/u' 392000
2 3o—uo 32 16 8 1/2 #00 u - 1 3/8” M70000
3 40.55 no 20 6 1/2 H36 u - 1 7/16" M76000
k 55-75 #9 2n 1/2 7 588 b - 1 5/16' 657000
5 75-100 60 3o 8 832 6 - 1 1/2' 858000
6 100-125 72 36 10 12uo 8 - 1 5.8'' 1250000

 

 

 

 

ROEBLING PRESTRESSED CONCRETE STRAND

 

26

 

 

Minimum Design Tensioning
Diameter Area Ultimate Load Load

Strength
0.6 in. 0.215 162 #6000 lb 22500 lb 27000 lb
0.835 in. 0.h09 163 86000 16 R2500 16 51000 lb
0.885 in. 0.u60 in? 97000 16 #8000 16 57000 16
1.0 in. 0.577 1n2 122000 16 60500 16 72000 16
1 1/16 in. 0.663 in? 138000 16 69500 16 8&800 16
1 1/8 in. 0.751 in? 156000 16 79000 16 9 00 16
1 3/16 in. 0.8u3 in 172000 16 88500 16 105000 16
1 1/6 in. 0.931 in? 192000 16 98000 16 116000 16
1 5/16 in. 1.0M in2 212000 16 109500 16 1 0000 16
1 3/8 in. 1.12 in? 232000 16 117500 16 1 000 16
1 7/16 in. 1.2M 162 252000 16 130500 16 155000 16
1 1/2 in. 1.36 in2 276000 16 1u3000 16 170000 16
1 9/16 in. 1.u8 in? 300000 16 155500 16 185000 16
1 5/8 in. 1.60 in? 32u000 16 168000 16 200000 16

 

 

 

 

.4 I :1"

(I)

 

CABLE POSITIONS

After determining the cable sizes the positions of the cables in
the beams should be determined. The dimension “080" is known for each
block, but the position at the ends of the beams is important. The and
positions of the cables are controlled by the shear. The shear will be
checked after the cable positions are determined.

In order to simplify the procedure of assembling the beams the
bending of the cables will be at the same point of the span length for
the same size of block. e.g. For 20 foot beam if the cable is bent
at the l/u point of the span length, it should be bent at the 1/4
points of the span lengths up to 30 feet, since we use the same size of
blocks from 20 foot to 30 foot span lengths inclusive.

In order to determine the positions of the cables we have to de-
termine the top and bottom fiber stresses of the beams at transfer and
working conditions at 1/2, 3/8, l/M and 1/8 points of the beams as
follows:

Block #1

Twenty foot span, top and bottom fibers at the 1/2 point of the

span length:

Mbd _ 233000

_ _ = 2
$18 2190 1038 lb/in

= 106.5 16/1112

 

it = 392000
110 378

11L_ 11182000 _ 2 n? _
510 -———2190 678 lb/in "—110

0.85 1 392000

2
= 882 lb in
378 /

.,.
’1
-.

 

 

 

 

A: 1/1

[,3

‘t 1/5’ ‘

37 ..m

 

28

I

 

 

r 6 e 392000 x u x 15
Hit—1Q: ——72-g8—7——— = 716 113/1112
P e e 0.85 x 392000 x h x 15
n t 3° 1° = = 609 16/1n2

Io 32887

At 3/8 point of the span length:

 

 

 

 

 

 

 

 

 

 

 

M 221000 P 392000
db - __ 2 t . 2
_._ .. _ O, b _ = =
51° 2190 10 6 1 /1n 0 378 1038 lb/in
M, 1396000 2 r\P 0.85 x 392000 2
51° 2190 37 / n 10 378
P s s 392000 x h x 15 .
t so 10 = = 2
10 32887 716 lb/in
“Pt °so e10 0.85 I 392000 x it x 15 - — 2
I0 = 32887 — 609 lb/in
At 1/h point of the span length:
P 392000 2 M 176500 2
J. = = ...db = = .
A0 378 1038 lb/in 81° 2190 80 8 lb/in
r\Pt 0.85 x 392000 2 Mw _ 1111500 _ 2
A0 —T-882 1b/in S—1:-—?1§6—- 508 lb/in
P e s 392000 J: 1+ x 15 .
...—mt 8° = —————————= 716 16/162
Io 32887 -
P . u
Y1 t eso e1Q = 0 85 3: 392000 J: x 15 = 609 lb/ina
10 32887
At l/S point of the span:
F 392000 M 72500
i _ _ 2 .82 _ = 2
AD - 378 — 1038 lb/in $19 - 2190 33 lb/in

 

 

 

 

“1 the

s

 

29

Y\Pt 0.85 x 392000 ”w 728500
.1;:— ""‘T§fl?""' 882 lb/in Sle 2190 331 1b/1n
P e 8 392000 x u x 15

1.3.2.351- .—. 716 115/1112

I ’ 32887

nPt e80 e10 0.85 1: 392000 J: 1+ x 15

Io 32887

 

 

= 609 16/1n2

Thirty foot span, top and bottom fibers at the 1/2 point of the

 

 

 

 

 

 

 

span:
Pt 392 M 531000
—- = = 2 fl = = ' 2
o 378 1038 lb/in $10 2190 2M2 lb/in
P 0.8 2000 2601000
H t = 5 x 39 = 882 1b/in2 ML: _..._—_= 1190 lb/in2
AD 378 51° 2190
P e 9 392000 x u x 15 2
1: so ]Q = =
16 32887 716 lb/in
r}? e e 0.85 x‘392000 x u x 15
t so ]Q = . = 2
10 32887 609 lb/in
At the 3/8 point of the span:
P 392000 Mdb #99000
i: = 2 — = —_—.-= 2
10 378 1038 lb/in 51° 2190 227 lb/in
nPt 0.85 1 392000 _ 2 M' _ 269M000 _ 2
—Ao—-— T—— 882 lb/in :-— -2-1-9-O-—- 1225 lb/in
P e ‘ 2000 1|» l
2.22.112: iii—:2: 716 16/1n2
10 32887
YIPt 8so 810 = 0-85 x 392000 x u x 15 = 009 1b/in2

I 32887

 

 

 

it til

inclusi
length

special ‘
'111 be

F0

 

30

At the 1]“ pnint of the span:

 

 

 

 

 

 

Pt _ 392000 _ 2 Mdb _ 399000 _ 2
I;'- 378 - 1038 lb/in 5;; 2190 182 lb/in
1.1.31 = W: 882 1b/1n2 L4!— = 3163302: 1080 113/1112
378 51° 2190
P e s 392000 x 1+- x 15
t 30 10 = .... = 2
16 32887 716 lb/in
nPt e3o 610 0.85 3: 392000 J: M x 15 2
= m = b 1
10 32887 009 l / 11
At the 1/8 point of the span:

51.: 392000 = 1038 lb/in2 Ens = 232000 = 106 16/1n2
A0 378 $1, 2190

NP 0.85 1: 392000 2 M 1162000
———t=——-——-———-—=8821b1n —'—=——-—-—=6 1111112
10 378 l 516 2190 53 /

P e e 392000 J: )4 x 15

t so 10 _ = 2
-——-i;—-——---—-si§§?7--——- 716 1b/in

VIP e e 0.85 x 392000 J: 1+ x 15

 

Io 35537

When block 41 is used for span lengths from 20 feet to 30 feet

inclusive, the cable will be bent once at the 1/8 points of the span

length as it is shown on page 31. At the bending points of the cables,

special blocks will be used as support to the cables. These blocks

will be designed after shear is checked for all the span lengths.

For the same calculations for other spans see Appendix.

 

 

 

 

 

I)

 

CHECKING FOR SHEAR

In a prestressed concrete beam maximum shear is composed of

three elementary shears:

1)

3)

1) Shear due to dead load.

2) Shear due to live load.

3) Shear due to the vertical component of the tension in the

cable.

Shear due to the dead load is maximum at the ends of the beam and
is equal to zero at the middle of the beam.

Shear due to the live load is also maximum at the ends of the beam
but it is not equal to zero at the middle of the beam due to the
fact the live load is movable.

Shear due to the vertical component of the tension in the cable is
in the opposite direction to the directions of the shears due to
dead load and live load and occurs only where the cable is not
horizontal. Because of this difference in direction the maximum
shear does not necessarily occur at the ends of the beam as in

ordinary reinforced concrete beams.

I For a certain size of block the shear will be checked only for the

sliortest and the longest span lengths for which that particular block

1 8 ‘used. If the shortest and the longest span lengths are safe, natuxa

ally the span lengths in between will also be safe.

 

 

 

Dead

Dead

 

Live

 

 

33

Block #1
20 foot span length
Ab = 378 in2 (see page 21)
Dead load shear at the 1/2 point of the beam:
V = 0

Dead load shear at the support:

_ 8 x 10 x 1 0 =
v _ 1751?"; 3910 lb.

Live load shear at the 1]? point of the beam:

21600

 

10' L - 10'

Influence Line

( 0 = - .
16000(1 + 2,75%- 16000 x 1.315 _ 211200

v = 0.5 x 21600 = 10800 lb.

 

 

 

 

3h

Live load shear at the 3/8 point of the beam:

21600

 

L\ J

7.5' 12.5'

 

Influence Line
V = 0.625 x 21600 = 13500 1b.
Live load shear at the 1/4 point of the beam:

21600 21500
nu

 

 

.05.\\_

 

 

Influence Line

V = 21600( 0.75 + 0.05) = 17250 1b.

.~
a

. 1‘30!

 

 

 

 

 

35

Live load shear at the 1/8 point of the beam:

 

 

 

 

21600 21600
A lu'
V
.875
.1 “X
125
2.5' 17.5'

 

 

 

Influence Line
v = 21600 (0.875 + 0.175) = 22700 lb.

7 Live load shear at the support:

320 21600
luv

 

 

 

20'

.1

 

' Influence Line
V = 32000 + 21600 x 0.3 = 32000 + 6500 = 38500 1b.

 

 

 

1'4-

 

 

36

The vertical component of the tension in the cables between the

support and the 1/8 point of the beam:

 

Jill

 

 

 

 

 

 

 

 

Taxi 9 = 1% = 7,5 9: 82° 21+. 1910

Cos e = 0.13216
Cos e x Pt = 392000 x 0.13216 = 51900 lb.

For checking shear we use the following formula:

1‘

 

sh § I
where fsh = Allowable shear stress = 180 lb/ine.
Vmax = Maximum shear.
H = Moment of the half section of the beam on the
neutral axis. .
Io = Moment of inertia of entire section.
t = Thickness of web.
Vm; 26250 lb (see page 37)
s=9x6x3+9x15x10.5=162+1u20=1532 in}.
I0 = 32887 inn (see page 21)

t=91n (see pagela)

 

,

 

 

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1 1
11. I e 1

 

 

 

 

 

 

 

 

 

 

1 1 I02. ...-HI. 0. IO.

 

.( F 2 Z.IO(3

 

 

 

 

38

fan = gig—Egg = 1111 111/1112 0.1:.

For the same calculations for the other spans see Appendix.

 

 

 

 

 

 

len

mak

 

BLOCK DIMENSIONS

Intermediate Block

Until this point the block sizes are determined and checked. The
lengths of these blocks are not governed by anything. Therefore, to

make it practical we will use a length of 1.5 feet.

Block #1

 

 

 

 

 

‘2
M6 —r‘uZT 'T-us
..oi

 

 

 

 

 

 

For the other blocks see Appendix.

 

Blocks for the Bending Points of the Gables

The cables for span lengths from 20 feet to 30 feet with block #1
are bent once at the 1/8 point of the span. The rest of the span lengths,
that is, from 30 feet to 125 feet using five different kinds of blocks
have cables bent twice. Therefore, we have to have eleven different
kinds of special blocks to be placed at the bending points of the cables:

For block #1 from 20 feet to 30 feet:

 

”61

   

I! 9" I L»!
C; Round to Prevent
Chipping

 

T

 

TA

+—1/8 point of beam length

Block #7
The lengths of these blocks will be made one foot three inches
only, to provide possibility to build beams at different lengths within
20 feet and 30 feet.
For the blocks for the bending points of the cables of the other

span lengths see Appendix.

 

 

 

END BLOCKS

16 1/2" 13 1 "

 

2 1/4"

 

 

 

 

 

Z;

l
7&3

 

 

 

For the other End Blocks see Appendix.

 

Block #18

Ml

To be used with Block #1

 

uz

In designing the end blocks and the blocks to be used at the bend-
ing points of the cables the following items had to be considered.

The design of end blocks is still in experimental stage. That is,
there is no particular procedure to design these blocks. The end blocks
are determined from experience. This conclusion is arrived after read-
ing articles about this matter and a talk with Dr. P. Abeles of England.

The design of the blocks to be used at the bending points of the
cables is very simple. These blocks should have the required dimensions
to give the required “ego” value for the beams these blocks are designed
for.

On the following page the details and the dimensions of the
threaded sockets for bridge strands by John A. Roebling's Sons Co. are

given.

 

 

 

1+3

THREADED SOCKETS FOR BRIDGE STRANDS

£3.
\J

!
III ”in m i
I

 

 

1".

SOCKETS HAY BE OBTAINED WITH
EXTERNAL THREAD, INTERNAL THREAD
OR EXTERNAL AND INTERNAL THREADS

 

 

 

 

1+1!-

 

 

Somaoh .82 5020.5. .ueoapacnon omgnm ..oo meow uhfigeom .4 snobs

:\m H :\H w :\m n- m :\m m w\m m m\H m w\~ H m mH\m :\n mH mH\H oH .mH\HH H
mH\HH H w :\m m- m w\m m :\H m :\m m wH\MH H :\m N mH\m w\H HH mH\HH m .w\m H
mH\HH H w\H H w\m mnw\~ N N\H m w\H m w\m m \m H w\m N mH\m mH\MH NH N\H m gmH\m H
mH\HH H w\m a w\m H.:\m N :\H m w\~ : m\m m mH HH H N\H N mH\m mH\m NH m .N\H H
oH\m H w\H a :\H m-m\m N m m\m : w\H m w\m H N\H N mH\m NH :\H w .mH\N H
mH\m H w\~ w m.N\H N :\H : w\m : w\H : mH\m H N\H N wH\m mH\HH HH N\H w =m\m H
mH\m H mH\HH o :\m N.N\H N mH\m : mH\m : mH\HH : N\H H :\H N mH\m :\H HH :\H w .mH\m H
N\H H w\m o w\m N.:\H N w\H : :\m m w\m : ©H\~ H mH\H N mH\m :\m OH :\H w .:\H H
N\H H ”\H m w\m Num\H N m\H : :\m m :\H : m\m H mH\m N mH\m HH :\H w .oH\m H
m\m H w\~ m w\m N-w\H N w\~ m N\H m : mH\m H :\m H mH\m :\m OH :\H w =m\H H
w\m H wH\HH m :\H N- N :\m m w\m m w\H m :\H H :\m H QH\m :\m OH :\H w .mH\H H
mH\m H mH\m m w\H H.:\m H w\m m m N\H m QH\H H w\~ H mH\m wxa oH :\H w =H

nz<mem

H a e m a a a n o m < .<Ha

 

e mgem mwnHmm mom wafixoom amnamme

 

 

 

 

SUMMARY

The following tables give the dimensions of the blocks as a summary:

 

“W‘VV —j-‘~“--“‘m”1W-—‘ — —

INTERMEDIATE BLOCKS

 

 

 

 

 

 

 

 

Span lengths Block D B t Length
ft. No. in. in. in. in.
20-30 1 30 15 9 18
3o-uo 2 32 16 8 1/2 18
ho-55 3 no 20 6 1/2 18
55-75 u M9 at 1/2 7 18
75-100 5 60 3o 8 15
100.125 6 72 - 36 10 18
END BLOCKS
Span lengths Block D B Length To Be Used
ft. No. in. in. in. With Block
NO.
20- 0 18 30 15 12 l
30- 19 32 16 12 2
no-SS 20 H0 20 12 3
55-75 21 ”9 2“ 1/2 12 h
75-100 22 6o 30 ‘12 5
100-125 23 72 36 12 b

a E. n 1n \ EH... ‘21.:iwg

{I'll lxlifl Ila: 1‘lli .

 

 

 

 

,» “vapw—W.

 

The following tables give the dimensions of the blocks as a summary:

 

‘- _-1»r---‘-“-m--WQ w

 

 

"v—

INTERMEDIATE BLOCKS

v"

 

 

 

 

 

 

Span lengths Block D B t Length
ft. NO. in. in. in. in.
20-30 1 30 15 9 18
30-ho 2 32 16 8 1/2 18
ho—55 3 no 20 6 1/2 18
55-75 u M9 2b 1/2 7 18
75-100 5 6O 30 8 18
loo-125 6 72 36 10 18
END BLOCKS
Span lengths Block D B Length To Be Used
ft. No. in. in. in. With Block
N00
20- O 18 30 15 12 1
30— 19 32 l6 l2 2
ho-ss 20 Ho 20 12 3
55-75 21 M9 2h 1/2 12 a
75-100 22 60 3O '12 5
100-125 23 72 36 12 6

 

 

 

 

tin

 

M6

 

BLOCKS FOR BENDING POINTS OF THE CABLES

 

 

 

 

 

 

 

Span Block D B t Length eso To Be Used Point of
Length No. in. in. in. in. in. With Block the Beam
ft. No. to be

Used.At
20.30 7 30 15 9 15 u 1 1/8
30410 8 32 16 8 1/2 15 3 2 1/8
no 9 32 16 8 1/2 15 5 2 1/1+
55 10 no 20 6 1/2 15 6 3 1/8
no.55 11 no 20 6 1/2 15 10 . 3 II“
55-75 12 u9 2“ 1/2 7 15 7 1/2 4 1/8
55-75 13 M9 2h 1/2 7 15 12 1/2 u 1/h
75—100 1h 60 30 8 15 12 5 1/8
75-100 15 60 30 8 . 15 19 5 1/h
100.125 16 72 36 10 15 15 6 1/8
100.125 17 72 36 10 15 2h 6 1/u
CABLES TO BE USED
Span To Be Used Bending Cable No. to e Pt per Total
Length With Block Points Size Be Used in. Cable Pt
ft. No's: in. lbs. lbs.
20-30 1, 7 & 18 1/8 1 1/h u u 98000 392000
o—uo 2, 8, 9, & 19 1/u, 1/8 1 3/8 4 5 117500 M70000
55 3, 10, 11 & 2o 1/u, 1/8 1 7/16 u 10 119000 #76000
55-75 h, 12, 13 s 21 1/u, 1/8 1 5/16 6 12 1/2 109500 657000
75—100 5, 1a, 15 s 22 1/u, 1/8 1 1/2 6 19 1u3000 858000
100-125 6, 16, 17 s 23 1/u, 1/8 1 5/8 8 2n 156250 1250000

-'

By using the above four tables in the Summary, prestressed, simply

supported beams from 20 feet to 125 feet can be built.

This of course is very economical and practical and cuts down the

time of construction greatly.

 

 

or

As an example let us take a beam thirty feet long:

1 -

2 -

Use

Block #18
Block 4 1
Block # 7
Block 47' 1
Block # 7
Block # 1

Block #18

12“

= 10.0"

= 3I_Ou

1:_3«

19!_6N
= 1'-3'

3o_on

1'-O”

 

30'-0'

14.1 1/u" cables with 325000 lb. < P, < 392000 lb.

Block #19
Block O 2
Block # 8
Block # 2
Block # 8
Block # 2

Block #19

“-1 3/8” cables with

12'
36"
15"
23a"
15"
36"

12"

= 10_On

3'-O"

1l_30

190_6n

= 1|_3"

3I_0n

v-0"

 

30'-O"

“53000 1b. << Pt << M70000 1b.

 

“7

 

APPENDIX A

DETERMINING BEAM DEPTH FOR VARIOUS FLANGE AND

WEB THICKNESSES FOR MOXENT

 

 

 

 

For

n I '-_.'
1 . -
J.
v .. 4
: ’ . :3
.-'-
, 14'-

 

1+9
Twenty-five Foot Span
= _§.Q....=
Impact Load 125 + 25 30%

Influence Line for moment at 1/8 point of span:

2.7M

11.211 21.875' 4

For l/b point of span:

 

 

 

 

H.69
I
i
P
L6 25L!— 18.75' c!
For 3/8 point of span:
5°85

 

 

 

 

50

For 1/2 point of span:

6.25

L/\A|

r**

 

Maximum.Moments:
hooo x 1.3 = 5200 lb.

16000 x 1.3 = 20800 1b.

2.7a x 20800 + 0.986 x 20800 = 57000 + 20500 = 77500 ft-lb.
u.69 x 20800 + 1.19 x 20800 = 97500 + 2u750 = 122250 ft-lb.

5.85 x 20800 + 0.608 x 20800 = 121700 + 12670 = 134370 ft-lb.
Use a moment of 135000 ft-lb.

1. Assume:

t = YD and B = O.ND.

AO

(0.un)(2YD) + (YD)(D - 2YD)
= 0.8YD2 + 102 - 212112 = D2(1.8Y - 212)

I0 = ¥g(3.ur - 10.8Y2 + 15.2!3 — 8r“)

 

.2.— 1210 = 3.uY — 10.8Y2 + 15.2Y3 - 82“
-313- 034 "v“ w
EL= ’20 = 1

575

 

 

 

 

Q = flatgiim= 313.33

M. - n “db = 913112

i1.

5‘
Assume: M“ = QBD2 JE-
135000 x 12 = 313.33 x 0.u

D3 1520

: (Y - 3.175?2 + u.u7v3 — 2.3551“)

n3 (3.uv - 10.8Y2 + 15.213 — 8!“)
( DIN )

 

 

Y n A0
0.05 32.9 92
0.1 27.6 121.9
0.15 25.u 1 .1
0.2 2n.u 167.0
0.25 23.8 18h.5
0.3 23.6 200.5
o.h 23.u 215
0-5 23.3 217

 

2. Assume:
t = YD and B = 0.5D

A, = 211% - 12)

I0

1
¥§ (u? - 1212 + 16Y3 - 8!“)

o(_ 1210 _ MY - 1212 + 1613 — SY

u

 

BD3 0.5

_ 12 1
D3 - Y — 322 + hr} - 215

 

 

Ao

 

92
120.9
1h3.1
166.u
186.5
200.3
228.2
235-5

 

t=YD andB=O.6D

u
¥§ (n.6r - 13.2?2 + 16.8Y3 - SY

= 212! — 13.2!2 + 16.8Y3 - 81”

 

 

= Y - 2.865Y2 + 3.65Y3 - 1.7M

 

04
U

 

U'I

\J'l

U1

NNNNNN

SSOOHN-F'KO

00000000
I

uzwlumwwo

:mowmquowox

92
121
1h5.5

185.6
20u.o
235.0
250.0

 

  

. . ”’“
""“-" "1"" W ;, a;

 

 

 

53

Thirty Foot Span

= __JZL..==
Impact Load 125 + 30 30%

Influence line for moment at 1/8 point of span:

 

 

 

For 1/M point of span:

 

 

For 3/8 point of span:

 

 

7.02
L 11.25' 18.75' J

 

 

5%

For 1/2 point of span:

 

Maximum Moments:
#000 x 1.3 = 52000 lb.

16000 x 1.3 = 20800 lb.

3.28 x 20800 + 1.53 x 20800 = 68200 + 31850 = 100050

5.62 x 20800 + 2.25 x 20800 = #6800 + 117000 = 163800

7.02 x 20800 + 1.78 x 20800 = 37000 + 1u6ooo = 183000

7.5 x 20800 + 0.5 x 20800 + 0.5 x 5200 = 156000 + louoo + 2600
= 169000

Use a moment of 185000 ft-lb.

1. Assume:
t = YD and B = O.ND

o n2 (1.8Y - 212)

>
I!

u
I = 25 (3.uv - 10.8Y2 + 15.213 - 81“)

 

°<= 12152 = 3.uY - 10.8Y2 + 15.213 - 81“
so} o;h
e.= °20 = 1

SJ

 

i‘:
s

 

 

 

 

 

 

 

55

Q = 53325.31= 313.33

2 C<
A M = D __
ssume QB

11
185000 x 12 = 313.33 x 0,1153 EL”? - lo-gYi + 15.2173 - 8T 3

 

 

 

 

(Y - 3.17512 + 11.14715 - 2.335513)

Y D A,
0.05 36.5 113.2
0.1 30.5 1h8.8
0.15 28.2 179.0
0.2 27.1 205.5
0.25 26.5 228.5
0.3 26.2 2u7
0.h 26.0 266
0.5 25.9 268

Assume:

t=YDmm3=06D
A.) 21320: - Y2)

I

u
D .
0 f5 (hr - 1212 + 16Y5 - 81“)

_ 1216 _ MY - 12?2 + 1613 - 81“
x' W” 0.5

D3=___;110—_.____?
Y - 312 + uy5 - 2

 

 

 

 

 

 

 

 

 

Y D A0
0.05 3u.5 113.2
0.1 28.8 15h.5
0.15 26.5 179.0
0.2 25.h 206.5
0.25 2u.7 229.0
0.3 eu.u 250.0
0.4 2u.2 281.0
0.5 2h.1 290.0
'3.- Assume:
t = YD and B = 0.6D
1)” 2 3 l:

Io = ifigu.6v _ 13.2Y + 16.8Y - SY )

g, = l1.61 - 13.2?2 + 16.8Y3 - 81!“

D3 _ 1551

’ Y - 2.865Y2 + 3.65Y3 - 1.7ui“

10 = 2D2(1.1Y - 12)
Y D A0
0.05 32.8 113
0.1 27.u 150
0.15 2501 179.5
0.2 2u.0 207.5
0.25 23.3 231
0.3 23.0 25h
0.u 22.8 291
0.5 22.7 309

 

56

 

 

57

Thirty-five Foot Span

= ___59___.=
Impact Load 125 + 35 30%

Influence line for moment at 1/8 point of span:

3»83

/\
. I .0... |

For 1/h point of span:

 

6.56

1‘ 8.75' 26.25' ;J

[i

 

For 3/8 point of span:

8.2

L/\j

If V

 

 

 

 

   

58

For 1/2 point of span:

 

 

Maximum moments:
“000 x 1.3 = 5200 lb.

16000 x 1.3 = 20800 1b.

3.83 x 20800 + 2.08 x 20800 + 0.328 x 5200 = 79500 + #3200
+ 1705 = 12h405

6.56 x 20800 + 3.06 x 20800 = 63600 + 136500 = 200100

8.2 x 20800 + 2.86 x 20800 = 59u00 + 170800 = 230200

8.75 x 20800 1,1.75 x 20800 + 1.75 x 5200 = 36u00 + 9100
+ 182000 = 227500

Use a moment of 231000 ft-lb.

1. Assume:
t=YDandB=O.uD

‘6 = D?(l.8Y - 2Y2)

u ,
I ¥§(3.ur — 10.8Y2 + 15.2?5 - 81“)

O

= 1210 _ 3.nY - 10.8!2 + 15.213 - 81“
803 ‘ 0;u

 

o<

 

"Iv “a?“ «fi—u—L‘ "'1.

 

1'.) ...I L

 

 

 

 

 

e20
E: =
D72
Y‘f + f _
Q= ct tw - 313.33
__:‘___..__
_ 2 cx
“w-“hn-Q“ 7?
2 1x
Assume Mw=QBD ..é.

2 3 _
231000 x 12 = 313.33 x 0,1,93E3AY - 10.31.11; + 15.2? 81“;

260

D3 = L o
(Y - 3.175?‘ + 11.11713 - 2.3551”)

 

 

Y D A0
0.05 39.)4 132
0.1 33.0 1714.2
0-15 30.“ 208
0.2 29.1 237
0.25 28.5 261+
0'} 2802 286.5
0.14 28.0 308.
0-5 27.9 311

 

Assume :

t=YDandB=0.5D

 

 

A, = 211%? - v2)
11
Io = £12411! - 121?2 + 16373 - 311*)
o, = 1210 = MY - 121'2 + 161:3 - 81“
1303 0.5
3 _ 2210

D..——-—-————-—-—-—-—
Y - 312+ 1:13- 211+

 

 

...—i}. -_‘_ — fl.

 

“' - .- '.-“ y,..-w “...-aw '5_Hm. - ._—-—w._w

_ _...-_,

"3' _.

 

 

3.

 

 

 

 

 

 

 

 

Y n 10
0.05 37.1 130.8
0.1 31.0 173
0.15 28.5 207
0.2 27.2 236.5
0.25 26.6 265.5
0.3 26.3 290
0.8 26.1 327
0.5 26.0 338
Assume:
t = YD and B = 0.6D
6
IO = ¥5(u.6v - 13.2!2 + 16.813 - 8!”)
cp = &;_g - 13.2?2 + 16.8Y3 - 8v”
.6
113 = 193—2 -
Y - 2.865Y2 + 3.65Y5 — 1.7m?E
Ab = 2D2(1.1Y - 12)
Y n 10
0.05 35.u 131.5
0.1 29.5 17%
0.15 27.1 209
0.2 25.8 239.5
0.25 25.1 268
0.3 2u.8 295
0.fl 2n.5 338.5
0.5 25.5 360

 

 

61

Forty Foot Span

- 0 =
Impact Load — i2§§;—EO 30%

Influence line for moment at 1/8 point of span:

h.37

aJ ,

For l/N point of span:

 

 

 

 

 

 

 

 

For 3/8 point of span:

 

15' 25' ,J

 

 

 

 

62

For 1/2 POint of span;

10

 

1L_'_

Maximum moments:

N000 x 1.3

5200 lb.

1600 x 1-3 20800 1b.

n.37 x 20800-+ 2.62 x 20800 + 0.87M x 5200 = 91000 + 5u500
+ #530 = 150030
7.5 x 20800 + u x 20800 + 0.5 x 5200 = 83200 + 156000 + 2600
= 2&1800
9.35 x 20800 + n.11 x 20800 + 0.622 x 5200 = 85500 + 195500
+ 3235 = 283235
10 x 20800 + 3 x 5200 + 20800 = 20800 + 62u00 + 15600 = 286000
Use a moment of 286000 ft-lb.

1. Assume:

‘ = YD and B 0.uD

‘6 = D2(1.8Y - 212)

I0 = ggl3-“Y - 10.8Y2 + 15,gy3 _ 8Yu)
CX: 8.19. = 3°uY - 10.8Y2 + 15,2!3 _ SYN»

 

BD3 0.h

 

 

L..‘..._,I -.

 

 

 

53

W2—

Q = r\fc‘t + ft! = 313.33

2
Mw-\M/I,,=QBD %
L‘—
6

Assume M' = QBDe

3 _ I.
286000 x 12 = 313.33 x 0.1193E3-W - 108:2 + 15.21 8‘! g
3 = 1.220
D (Y - 3.1751231- u.u7f5 - 2.3551'”)

 

*4

 

\11

U1

0
Wrwmwwwo
m

00000000
I. .

oooor-‘rumm u
00.0
OHmmmmzm

m

N

m

0

.n

\JJWWWMWW 42’

 

2. Assume:
t = YD and B = 0.5D

A, 2020! - '12)

I6 f—Eflr _ 121:2 + 1613 - 8!“)

q: 121, = u? - 1212 + 16Y3 — 811’

Bfi 06

D3= 2160
Y — 3! + hr} _ 2y

 

 

 

.4

U

 

U1

sees. a
U'IU'I

OOOOOOOO

L11:\JJNNHHO

Owko
I

NNN WWW
0100061008

. I I I .
ot—uumu mwxo

 

t = YD and B = 0.6D

I, = ¥g(u.6r - 13.2?2 + 16.813 - 81“)

 

c. = “.6! - 13.2226: 16.8Y3 - 88“
o ,

 

= 2380 -
Y - 2.865127+ 3.65Y3 - 1.71:3

 

'4

 

U1

noose
\J'IU"

c><3<3<3 c>k3<3<3
U1JrUJflJhJP‘hMO

38.3
31.6
29.0
27.8
27.0
26.7
26.5
26.h

 

 

 

 

 

 

 

 

 

 

 

 

 

Forty-five Foot Span

Impact Load = I§%9;rfi5-= 29.h%

Influence line for moment at 1/8 point of span:

11.92

/

For l/M point of span:

 

7

11.2 '

 

7—

For 3/8 point of span:

39-375'

33-75'

10.55

\\

1

JL____

 

L369

 

28.1'

 

 

"- ‘4‘" ' —~.ne- —

 

 

66

For 1/2 point of span:

11.25

/\
. ...... I

 

Maximum moments:

16000 x 1.29M = 20700

4.92 x 20700 + 3.56 x 20700 + 1.621 x 5175 = 73600 + 105500
+ 6u70 = 185570

8.u2 x 20700 + “.93 x 20700 + 1.h35 x 5175 = 102000 + 176100
+ 7&20 = 283520

10.55 x 20700 + 5.29 x 20700 + 1.81 x 5175 = 109200 + 218000
+ 9370 = 336570

11.25 x 20700 + n.25 x 20700 + u.25 x 5175 = 233000 + 88000
1 22000 = 343000

Use a moment of 343000 ft-lb.

1. Assume:

t=YDandB=0J+D

A0 = 02(1.8Y - 2Y2)-
I, = ¥;(3.AY - 10.8Y2 + 15.2?3 - SYM)

0,: 1210 = 3.61 _ 10.8Y2 + 15.2!3 — 81'1+
BD3 0.h

 

 

67

8,: 620 = 1

572

Q = fléliu = 313.33

M1v - Rudb = Q1392

5.
8
Assume “w = QBD2 :-
6 3.111 - 10.81: + 15.2Y3 - 815)

311.3000 x 12 = 313.33 x 0.1m3 E

 

 

 

 

D3 = 3860
(Y - 3.175172 + 11.1.1713 - 2.3559”)
Y D A,
0.05 115 172.2
0.1 37.7 227.5
0.15 35.1 277
0.2 33.1: 312.5
0.25 32.6 316-5
003 3203 375-5
0.1: 32.1 1109
0.5 32.0 1:10
Assume:

t = YD and s = 0.5D
1 = 202(1 .. 12)
1 = 11.32461 .. 121:2 + 16Y3 .81“)

121, = 111' - 121'2 + 1633- 8?"
DD3 0.5

3: 3280
D Y-3Y2+1+Y3-2Y

cg:

 

 

 

 

L___.. .1...1_

 

 

 

 

 

 

 

 

Y._ D A,
0.05 h2.5 171.5
0.1 35.5 227
0.15 32.7 272.5
0.2 31.2 311.5
0.25 30.u 3M6
0.3 30.1 380
o.u 29.9 u28
0.5 29.8 nun
3. Assume:
t = YD and B = 0.6D
I, = 22(u.6r - 13.2?2 + 16.8Y3 - 8v“)
ax: 14.6Y - 13.212 + 16.813 .. 818*
0.6'
Y - 2.865Y2‘+ 3.65!3 - 1.7uiE
A, = 2D9(1.1Y - 12)
Y D 1b
0.05 60.5 172.5
0.1 33.5 22h.5
0.15 30.9 262
0.2 29.5 313
0025 28'? 350
0.3 28.“ 387
0. 28.2
0.5 28.1 3?3

 

68

 

69

Fifty Foot Span

1 1:1. d=—5_°__=28.5
mpac oa 125+50 .%

Influence line for moment at 1/8 point of span:

 

 

For 1/1+ point of span:

 

 

 

1;

L12» - 37.5'

For 3/8 point of spam

 

 

70

For 1].? point of span:

 

Maximum moments:
8000 x 1.285 = 51h0 1b.
16000 x 1.285 = 20550 1b.

5.u7 x 20550 + 3.72 x 20550 + 1.97 x 51h0 = 76uoo + 112h00
+ 10120 = 198920
9.38 x 20550 + 5.88 x 20550 + 2.375 x 51u0 = 120900 + 192800
4 12200 = 325900
11.71 x 20550 + 6.87 x 20550 + 2.97 x 51u0 = 290500 + 133000
+ 15270 = 388770
12.5 x 20550 + 5.5 x 20550 + 5.5 x 51ho = 256500 + 113000
+ 28300 = 397800
Use a moment of 398000 ft-lb.
1 . Assume:
t = YD and B = o.hn
10 = 02(1. SY - 2??)
0:25 (3. u! - 10. 8Y2 + 15. 2Y3 - 8!“)

c< g 1210 = 3.ur - 10.822 + 15.223 - 81“
BD o.h

 

 

 

 

 

 

 

71

5 = 620 :1

575

Q = nfct + ftw = 313.33

M, - vwdb = Q3112 3%..
Assume M. = QBDZ 3(5—

 

_ . (my - 10.8Y2 + 15.213 -43)
2 - 313.33 x 0~”D3 ( 0.9 )

 

 

 

 

398000 J: 1
(Y - 3.175?2 + 9.97Y3 _ 2.3551”)
Y n 10
0.05 97.1 188.5
0.1 39.9 298.5
0.15 36.9 298
0.2 39.8 339
0.25 39.1 378
0.3 33.8 911
0. 33.6 996
0-5 33. ‘ 999
Assume:

t = YD and B = 0.50
A0 = 202(1 — Ya)

1;
IO = ¥§(9Y - 1272 f 1673 - 8?“)

= 1210 = 9? - 12!2 + 16Y3 _ 88“

303 0-5

810
D3 - Y - 3Y2 + 973 - 2;E

OK

 

 

 

 

 

 

3-

 

72

 

 

 

 

 

 

Y D 10
0.05 99.5 188.5
0.1 37.2 299
0.15 39.2 ‘ 298
0.2 32.7 .392
0.25 31-8 379.5
0.3 31-5 417
0.9 31.3 970
0.5 31.2 986
Assume:
t = YD and B = 0.6D
9
IO = ¥E(9.6Y - 13.2Y2 + 16.8Y3 - 8Y“)
°‘_ 9.6Y - 13.2Y2 + 16.8Y3 - 8Y“
— .6
83 = ————713——-——--:' 2°
Y - 2.865Y + 3.65Y3 — 1.79Y
A0 = 2D?(1.1Y - Y2)
Y D 10
0.05 92.5 189.5
0.1 35.9 251
0.15 32.9 299
0.2 31.0 396
0.25 30.1 385
0.3 29.8 926
0.9 29.6 991
0.5 29.5 522

 

 

 

 

 

73

Fifty-five Foot Span

Impact Load = = 27.8%

JXL____
125+55

Influence line for moment at 1/8 point of span:

Ch
I

31

6.875' I - - 98.125' _J

For 1/9 point of span:

 

 

 

For 3/8 point of span:

12.9

 

L: 20.6' 3 39.9'

 

 

 

 

 

79

For 1/2 point of span:

13-75

L 27.5' 27.5' J

 

 

 

Maximum moments:
"-000 x 1.278 = 5110 113.

16000 x 1.278 = 2011.50 11).

6.01 x 20950 + 9.26 x 20950 + 2.515 x 5110 = 87000 + 123000
+ 12850 = 222850

10.3 x 20950 + 6.8 x 20950 + 3.305 x 5110 = 210500 + 139000
+ 16900 = 366900

12.9 x 20950 + 7.69 x 20950 + 9.13 x 5110 = 263500 -+ 156000
+ 21150 = 990650

13.75 x 20950 + 6.75 x 20950 + 6.75 x 5110 = 281000 + 138000
+ 34500 = 453500

Use a moment of 959000 ft-lb.

1- - Assume:
t=YDendD=0.9D

10 = D2(1.8Y - 212)

 

1;
I0 = ¥§(3.9Y - 10.8Y2 + 15.2Y3 - 81'“)
7x = 1210 = 3.9Y - 10.8Y2 + 15.213 - 81'“

 

 

 

 

 

75

53”
ll

E:

H

I

0,: V1fcté+ ft! = 313.33

 

 

 

 

Mr -'W.Mdb = QBD2.%%—
2 04
A M = ...—...
ssume , QED E:
959000 x 12 = 313.33 x 0.9DBE3LuY -19685? + 15-2Y3 - BY“;
3= _ 5100
D (Y - 3.175?“ 9.975 - 2355?“)
Y D 10
0.05 99.2 205.5
0.1 91.3 266
0.15 38 325
0.2 36.5 372.5
0-25 35.7 919
0.3 35.9 951
0. 35.2 990
0.5 35.1 993

 

2 - Assume:

t=YDandB=0.5D

 

A0 = 2D2(Y — Y2)
u
= D _ 2 — 3 _ 9
o i§(9Y 12Y + 16Y 8Y )
(X: 121., = (9r .. 12Y2 + 16Y3 - 81'“)
BS3 ( 0.5 3
9300
D3 = u

 

Y - 3Y2 + 9Y3 - 2?

 

 

 

 

3o

 

 

 

 

 

 

 

Y D .10
0.05 96.5 205.5
0.1 38.8 271
0.15 35.6 323
0.2 39.1 376
0.25 33.1 911
0.3 32.8 452
0.9 32.6 511
0-5 32.5 532

Assume:

*- = YD and B = 0.61)

10 = £_:(9.6Y - 13.2Y2 + 16.8Y3 - 8Y”)

Q( = 9. Y - 13.212 + 16.8Y3 - 81V"

0.6
D3 = 3180
Y - 2.865Y2 + 3.65Y3 - 1.79Y“

Y D 10
0.05 99.9 207
0.1 37.0 279
0.15 33-9 327.5
0-2 333-3 375-5
0.25 31.9 919
0.3 31.1 969
0. 30.9 539
0.5 30.8 569

 

 

76

 

Sixty-five Foot Span
- O _ '
Impact Load 125 + 65 - 26.3%

Influence line for moment at 1/8 point of span:

 

8.125' I 56.875' _1

For 1/9 point of span:

 

V7

 

16.2 ' 1+8. '

For 3/8 point of span:

14.0.6 '

 

77

 

78

For 1/2 point of span:

.25

 

Maximum Moment:
4000 x 1.263 = 50501b.

16000 x 1.263 = 202001b.

7.1 x 20200 + 5.35 x 20200 + 3.61 x 5050 = 108000 + 193500
+ 18250 = 209750

12.2 x 20200 + 8.68 x 20200 + 5.18 x 5050 = 246500 + 175500
+ 26200 = “#8200

15.2 x 20200 + 9.98 x 20200 + 6.96 x 5050 = 307000 + 201500
+ 32650 = 591150

16.25 x 20200 + 9.25 x 20200 + 9.25 x 5050 = 328000 + 187000
+ 96750 = 561750

Use a moment of 562000 ft-lb.

1 . Assume:
t = YD and B = 0.4D

1° = 02(1.8Y - 212)

I0 = 12;!3.9Y — 10.8Y2 + 15.2Y3 _ 8Y”)

07: 1210 = 3.9Y - 10.8Y2 + 15.2Y3 - 817“
BD3 0.9

 

 

 

 

 

79

- 920 =
EL- 575' 1

Q = 3%1121 = 313.33

 

 

 

 

 

MI ' Y1 “db = Q3132 35"
2 ax
Assume I": QED :-
: 3(339Y - 10.8Y2 + 15.2Y3 - 8Y“)
562000 J: 12 313.33 x 0.11-D ( '1‘ j
D3 = 6310
(Y - 3.175Y2 + 9.97Y3 - 2.355Y“)
Y D Ab
0-05 53 239
0.1 99.9 315
0.15 90.8 379.5
0.2 39.2 930
0.25 38.3 976
0.3 38.0 520
0.9 37.8 566
0.5 37.7 569
Assume:

t = YD and B = 0.5D
A0 = 2D2(Y - Y2)

H
II

25(9Y - 12Y2 + 16Y3 - BY“)

_ 1§gg_= (9Y - 12Y2 + 16Y5 _ 8Y“)
BD3 ( 0.5 —)

 

93 = .......Ji޴l..._
Y - 3Y2 + 9Y3 -758:

 

 

 

 

 

80

 

 

 

 

 

 

Y D Ab
0.05 50 237
0.1 91.6 311
0.15 38.2 372
0.2 36.6 928
0.25 35.6 975
0.3 35.3 529
0.9 35-1 591
0.5 35.0 612
3. Assume:
t = YD and B = 0.6D
Io = ¥;(9.6Y - 13.2Y2 + 16.8Y3 - 8Y“)
= 9.6Y — 1 .2Y2 + 16.8Y3 - 8Y“
6" m%“""m‘
3 ___V_ 9680 _
D ’ Y - 2.865Y2 + 3.65Y3 - 1.79Yu
10 = 2D2(1.1Y - Y2)
Y D 10 ‘1
0.05 97.7 239
0 1 39.7 315
0.15 36.9 378
0.2 35.0 991
0.25 33.8 986
0.3 33.5 539
0.9 33.3 621
0.5 33.2 661

 

 

 

 

 

 

 

 

 

 

81

Seventy-five Foot Span

=l~=
Impact Load 125 + 75 25%

Influence line for moment at 1/8 point of span:

8.2

/\
_J 1

For 1]“ point of span:

 

 

19.08

 

 

-L____

L8.75' A 56.25'

For 3/8 point of span:

 

 

. |l| liIlrJIIIil

 

 

 

 

 

For 1/2 point of span:

1

8.75

82

/\

L 37.5'

Maximum moment:
L1000 x 1.25 = 5000 111.

16000 x 1.25 = 20000 111.

 

37.5' J

8.2 x 20000 + 6.05 x 20000 + “.7 x 5000 = 160000 + 129000 + 23500

= 316500

19.08 x 20000 + 10.58 x 20000 + 7.05 x 5000 = 281500 + 211500

+ 35300 = 528300

17.58 x 20000 + 12.31 x 20000 + 8.81 x 5000 = 351500 + 296500

+ 99100 = 632100

18.75 x 20000 + 11.75 x 20000 + 11.75 x 5000 = 375000 + 235000

Use a moment of 669000 ft-lb.
1 . Assume:

t = YD and D = 0.9D

.10 = D2(1.8Y - 2Y2)

+ 53750 = 603750

IO = ¥g(3th - 10.8Y2 + 15.2?3 _ SYM)
I 2 3 9
o<= 1216 = 3.91? - 10.8Y + 15.2Y - 811

 

BD3 0.9

 

33

_°20_
6-175—1

Q = flgtgiia= 313.33

”I ' nfldb = QBD2 i.
8
Assume M" = QED? %

669000 x 12 = 313.33 x 0.9D3E319Y - 10.?2 + 15.2Y3 - 8Y“;
D3 = 7500
(Y - 3-175Y2 + 9.97173 - 2.35517”)

 

 

Y D A0
0-05 56 266.5
0.1 97 353.5
0.15 93.2 920
0.2 91.9 981
0.3 90.3 589
005 140.0 6%

 

2. Assume:
t=YDandB=o.5D
‘o = 2D3(Y - Y?)

11
{la-(91! - 121!2 + 1623 . 3y“)

I0

0‘: 1210 = (9r - 121‘2 + 16?} -57”)
BD3 ( 0.5 “)

_ 60
D3‘Y-3?+9Y3-2?3

 

 

 

 

 

 

 

 

 

 

*4

A

 

U1

oeeoee
mu:

OOOOOOOO
WFWIUNHHO

269.5
3N9
918
”79
533
669
689

 

t = YD and B = 0.6D
10 = $9.6Y - 13.2Y2 + 16.8Y3 - 81:")

(x: 9.6Y - 13.2Y2 + 16.8Y3 .. 817"
0 6

 

3 = 5560
D Y - 2.865Y’2-4- 3.65Y3 - 1.791174

10 = 2D2(1.1Y - Y

 

*4

 

U1

U1

\J'l

.OPOOOOOO

U1.t:‘\N|\)l\)l—'HO
I

NWKJ’IQMO‘J—‘kfi

WWWWWM

 

 

35

One Hundred Foot Span

=_fl.—=2 I
Impact Load 125 + 100 2 a

Influence line for moment at 1/8 point of span:

10.93

/\

For 1]”: point of span:

 

1L.—

18.75

/
L29 ‘ 75'

IV

/

 

L

For 3/8 point of span:

23.95

/

L/

 

 

 

 

 

 

 

 

86

For 1 [2 point of span:

25

/

l 50' 50.

l ‘ V

 

 

 

 

Maximum moment:
9000 x 1.222 = 9890 lb.

16000 x 1.222 = 19550 1b.

10.93 x 19550 + 9.18 x 19550 + 7.93 x 9890 = 219000 + 179500
+ 36300 = 929800
18.75 x 19550 + 15.25 x 19550 + 11.75 x 9890 = 366000 + 298000
+ 57500 = 721500
23.95 x 19550 + 18.2 x 19550 + 19.7 x 9890 = 959000 + 355500
+ 71900 = 886900
25 x 19550 + 18 x 19550 + 18 x 9890 = 988000 + 352000
+ 88100 = 928100
Use a moment of 928000 ft-lb.
1‘ Assume:
t = YD and B = 0.9D
10 = 02(1.8Y - 2Y2)
Io = ¥;(3.9Y - 10.8Y2 + 15.2Y2 — 8Y“)

0.: 1210 = 3.9Y - 10.8Y2 + 15.2Y3 - SY“
BD3 0.9

 

2.

_. 620 _
£51 -—-— D 2 1
Q = r”Lfct + ftw = 313, 33
2
MW — “Mm, = QBD

(DIR 048

A3 stuns M. = QED2

 

928000 J: 12 = 313.33 x 0,1.D3E3.9Y _ 10.832; 15,213 - 3Y3;

 

 

 

 

10900

3 _

D — (Y _ 3.175172 + 9.9723 - 2.35m“) .
Y D A6

0.05 62.5 3}?-
OJ 52.3 937
0.15 48.2 52”
0,2 96.3 600
0.25 95.2 66“
0.1+ 1419.7 7%
0.5 11111.6 796

Jlssmmme:

t=YDandB=0.5D

A0 = 2D?-(Y .. Y2)
I = 9.091! - 12Y2 + 16Y3 - SYH)
0 12

121a = (9Y - 12Y2 + 16Y3 - SY“)
393 ( 0-5

3 8780
D =Y-3Y2+9Y3-2Y’*

 

 

57

 

 

 

 

 

 

 

 

 

 

 

 

 

Y D 10
0.05 58.8 328
0.1 99.2 936
0.15 95.1 519
0.2 93.1 599
0.25 92 661
0.3 91.7 731
0.9 91.5 827
0.5 91.9 857
3. As sums:
1: =YDandB=0.6D
I0 = $19.6? - 13.2Y2 + 16.8Y3 — 8Y”)
d: 9.6Y - 13.2Y2 + 16.8Y3 .. 817“
3 = 7720
D Y - 2.86512 + 3.65Y3 - 1.791!”
2
A0 = 2D (1.1Y - Y2)
Y D 10
0.05 56.3 332.5
0.1 96.8 938
0.2 91.1 608
0.25 39.8 679
0.3 39-5 799
o. 39.3 865

 

88

 

 

 

 

 

 

 

39

One Hundred Twenty-five Foot Span

20%

= _50_..._.=
Impact Load 125 + 125

Influence line for moment at 1/8 point of span:

11

7.

1.6' 7 * 109.9!

 

J;

For 1 /1+ point of span:

23.95

L31.25' 1 93.75' 7!

 

F01- 3/8 point of span:

29.3

/

 

 

For 1 [2 point of span:

31.25

L 62.5' 62.5' A

w—— '1

 

 

Maximum Moment:
9000 x 1.2 = 9800 lb.

16000 x 1.2 - 19200 1b.

13.68 x 19200 + 11.9 x 19200 + 10.15 x 9800 = 262000 + 228500
+ 98700 = 539200
23.95 x 19200 + 22.95 x 19200 + 18.95 x 9800 = 950000 + 931000
+ 91000 = 972000
29.3 x 19200 + 29.05 x 19200 + 20.6 x 9800 = 562000 + 961000
+ 98900 = 1121900
31.25 x 19200 + 29.25 x 19200 + 29.25 x 9800 = 600000 + 965000
+ 116500 = 1181500
Use a moment of 1180000 ft-lb.
1 - Assume:
t = YD and B = 0.9D
A0 = D2(1.8Y — 2Y2)

H
0
II

¥;(3.9Y - 10.8Y2 + 15.2Y3 - SY“)

O<= 1219 = 3.9y - 10.8Y2 + 15.2Y3 - 8Y“
31) 3 0.1+

 

2.

Q = M: 313.33

“w - nude = Q3132 €-

Assume M' = Q302 28‘.—

 

91

1180000 1: 12 = 313.33 x 0.99ng ' 10'8Y2 “" 15:27} ' 8i)

 

 

 

 

0.9 )
D3 ___ 13210
(Y - 3.1751?- + 9.9716 — 2.35590

Y D A0
0.05 67.5 387
0.1 56.75 515
0.15 52.25 619
0.2 50 700
0.25 99.1 783
0.3 98.8 856
0.9 98.6 938
0.5 1#85 9 1

 

Assume:
t=YD andB=O.5D
A0 = 202(Y - Y2)

IO = 13.2491! - 121:2 + 16Y3 - 81:“)
o<

D3 = 11180
Y--3Y‘;-'+1+Y3-2Y1+

= 1210 = (9Y - 1211'2 + 16Y3 -8Y“)
DD3 ( 0.5 )

 

 

 

. . INN 11‘7“: I .-
.ln. 2.». . r
i hunlfi‘qfio‘:

. . . III": 18"

 

 

‘11: 1:7,. .171 IO. .

 

 

 

 

Y D A0
0.05 63.8 386
0.1 53.3 511
0.15 99 612
0.2 96.8 700
0.25 95.7 789
0.3 95.9 865
0.9 95.2 980
0.5 95.1 1017

 

= YD and B = 0.6D

= ¥g(9.6Y - 13.21:2 + 16.8Y3 - 8Y”)

9.6Y - 13.2Y2 + 16.8Y3 - SY“

 

190 = 2D2(1.1Y — Y2)

- Y - 2.865Y2 + 3.65Y3 - 1.7917)‘L

 

 

Y D 10
0.05 61 391
0.1 50.9 517
0.15 96.6 618
0.2 99.5 713
0.25 93.2 799
0.3 92.9 889
0.9 92.7 1020
0.5 92.6 1089

 

 

 

 

 

 

 

 

 

 

 

APPENDIX B

DETERMINING BEAM DEPTH FOR SHEAR

 

 

 

  

at
..VHI(D\. I
It. ...‘1117 . I

 

- . . n . .9171, 1: ...o . i. . 11111071171111!” 991.31.11.09 7 Ir... . u 4.41 - . I
.. -..... a 4 11---: 11.91.19 .... 1.1 v
...v a. 4. v. 1 f .

  

...I.Illlr.l!1l

 

 

96
3For block #2, “0 foot span at 1/8 point of beam:

208 0 208 0 52%}

__ 119! ____l__._.1

 

 

Influence Line

( 0 =

16000(1 + + 125 20850 1b.
( 0 _

9000 (1 + $5437E33 — 5213 lb.

Maximum shear = 20850(0.875 + 0.525) + 5213 x 0.175 = 29200 + 912

= 30112 1b.

In order to allow for dead load say

 

 

Vhax = 33000 1b.

Y D A0
0.10 51.1 968
0.15 39.8 909
0.20 35.2 396
0.25 32.6 386
0.27 31.0 380
0.30 29.7 370

 

‘dlere 10 = 2D2(Y - Y2)

 

APPENDIX C

DESIGN

 

 

 

 

 

 

 

 

1..-.

 

9) Crossing

crossing

For 100 foot span

1) Crossing

crossing

2) Crossing

crossing

3)

Crossing

crossing

9)

Crossing

crossing

119

 

 

the P 3‘15 ' (Mw n; ) = ' —"7€§§%EBES""_')
A0(3’11; ‘ few) 832( 12790 ' 200°)
= .000001118
the sac axis = Eg_.= 15.36.
810
length:
1 1 1
"h" P ”‘1‘ ‘ 07.“ . 93.3; ' .3252... . 11333300,
= .0000003985
the °so axis = - ___.= - 15.36.
the 1"; axis = -W—~T= -m6m
' ftt) 832( ‘12790 ' °)
= - .000001185
the eso axis = E§6-= 15.36.
1 0.8
the 1°: axis — ‘09;th + ftw)=53—2E.§ETL117W)BT;;
= .000000591
the eso axis = - E§_.= - 15.36.
10
the If, axis = - Wwv-T: .. fir...)
0(516 _ few w) 832(-T§?§6~—-- 2000)
= .00000929
the 0,0 axis = EQ..= 15.36.

810

 

 

 

 

 

 

116

Use eso = 19 in. (See page 115).

Equation 1 for 75 foot span:

(Hdb ) (22.... 0000 + 2000)
A + f t 832
Pt = 67"“516 c ) = ( 12790 ) = 2050000 = 917000 lb.

1 0 2.235
19:35:30.. b.3521}—

Equation 3 for 100 foot span:

(M ) (29190000 )
“(51"; + f“) = 8327 12790 ”0): 1

0

)

515000
n$1 + 7W: ° ) 0.85((l + 1355—1-31 1395

P = = 830000 1b.

Use 6 - 1 1/2 in. cables Pt = 858000 11>.

830000 <858000 <917000 0.8.

The known factors for block #6 are as follows:
f5 = 6000 psi
fct and fa, = 1/3 1’}, = 2000 psi
ftt and f" = 0 (Assumed)
n = 0.85
D = 72 in.
B = 0.5D = 36 in.
010 = e20 = 36 in.
Aomin = 598 in2.
1: = YD = 10 in.

Actua1A0=2x10x36+10x52=1290ina.
I =72x72x72xl6__52x52x52xgé
12 12

1120000 - 305000 = 815000 in”.

31.: = $2.. = :—:6 = 813000 = 22650 163.

 

 

 

 

 

 

117

The intersection points of the four curves with the two axes:

For 100 foot span length:

1)

2)

3)

9)

Crossing

crossing

Crossing

crossing

Crossing

crossing

Crossing

crossing

For 125 foot span

1)

Crossing

crossing

 

 

 

 

 

 

the %? “‘13 = (' 1 9.51 = ( 1 1 0000
1°(fct + 31°) 1290(2000 + 22650 )
= .000000283
2
the ego axis = - £2_.= - 63 = - 18.25.
810
1 l
the -- axis = - _
P M 0000
t ‘Ogsf—Z - 99112221“): 0;
= - .000000995
the °so axis = EI6.= 18.25.
1 0.85
the -- axis=
p 0 00000
t %sa_-+ f“0=12uoé 22 50 980;
= .000000509
the so axis — - £2_.= - 18.25. ,
e
10
1 O. 85
the — axis - -
Pt (M. (3o50000_
A0(31e ’ fcwo 12“(N 22650 2000)
= .000001099
_ re
the eso axis — o - 18.25.
e
10
length:
l l l
the —' axis = =
P M 02 0000
t (fct + 322% 129022000 + 22650
= .000000292
1.2
the °so axis = _ _°_= - 18.25.
e

10

 

 

 

 

 

 

 

 

119

 

2) Crossing the 1—- axis = - 1 = - .1.
r (M b ) (2250000 )
= - .000000605

crossing the e

1.2
so axis = _°_= 18.25.
°1

 

 

 

 

0.8
3) Crossing the %: axis = 969517;: ftp) = 1mg9§12§g§00 + 0;
= .00000035
crossing the ego axis = - g. = - 18.25.
9) Crossing the 1%,: axis = - (u n 3 = ‘ 121.029_mio_,___0000 - 2000)
“(sfi- few) ( 22650 )

= .00001712
r2

creasing the °so axis = _°— = 18.25.
e
10
Use eso = 29 in. (See page 118)

Equation 1 for 100 foot span:

(8 ) (193%0000 1 2000)
P. -.- ELM: 1290 23 ° = M: 1530000 11..

1 I To 657

Equation 3 for 125 foot span:

1: 99900000 )
A°ES'1";+ “a: _ law‘T‘ae o + °>= 285C100.

Pt= ( e s )' ( 2 x 6 1.975 =123000°1b'
”(1*‘1‘2’1'3 0-85<1+'65‘7‘li

Use 8 - 1 5/8 in. cables 1°t = 1250000 1b.

1230000 < 1250000 < 1530000 0.K.

 

APPENDIX D

CABLE POSITIONS

 

 

 

Tnirf

‘31
in

L3

 

 

Block #2

Thirty foot span, top and bottom fibers at 1/2 point of the span:

P l+70000 M 561000

_£_= = 2 _22_= _______= 2 2
r] P: = 0.85 x 1470000 = 2 u = 2631000 = 2
10 W" 999 lb/in s—Lle “2600 1010 lb/in

P
I 93‘; e10 = 1#70000 x 5 x 16 = 905 1b/in2

 

 

 

 

 

o 1‘3
nPt e30 910 =0.85 x M70000 1: 5 x 16: 0 1b 1 2
Ie 65110 77 / n
At the 3/8 point of the span:
Ii __ 970000 _ 2 nab _ 528000 _ 2
‘0 — uoo — 1175 113/in 51° 2600 203 1b/in
331: 0'85 x “70000 = 999 1b/1n2 EL: M= 1050 lb/ina
$0 $1, 2600
P e e 470000x5x16 ~
1'. so 10 = = 2
I 705140 905 1b/in
r1Pt eso e10 0.85 x M70000 x 5 x 16 _ 2
“—17,—— _ 4173—11? - 770 113/in
At the 1/H point of the span:
Pt ”70000 2 “db _ “21000 _ 2
'1 Pt _ 0.85 x 1170000 2 M, 2386000 2
T _ T- 999 1b/in fiz- W— 918 I‘D/in
Pt 630 910 _ “70000 x 5 x 16 _ 2
“1.0—“ .. 15% — - 905 lb/in
V\Pt °so 810 = 0.85 x “70000 x 5 x 16 = 770 1b/in2

 

I0 915%

 

122

At the 1/8 point of the span:

 

 

 

 

if = 51%? = 1175 1b/1n2 lsiff= 222280 = 911.3 lb/in2
n71}: 0.85 x0970000 = 999 1b“112 LL:= 1:206:00 = 556 1b/in2
Pt esgoem = h7ooogélc3) x 16 = 5,43 1b/in2

W = 0'85 3:330 I 3 x 16 = 962 1b/in2

Forty foot span, top and bottom fibers at 1/2 point of the span:

 

 

 

Pt _ l+7oooo _ 2 {4111 = L90” = 2
E - W - 1175 “/1“ 51° 2600 372 “/1“
n Pt _ 0.85 x 1470000 _ 2 ML _ 11399000 _ 2
To_ - T - 999 1b/in 519 - $070—- 1690 lb/in
Pt eso e10 = M70000 x 5 x 16 = 905 1b/in2
10 415116

VlPt ego em = 0.85 x M70000 x 5 x 16 = 2
____Io M15110 770 1b/in

At the 3/8 point of the span:
p 970000 M<11: 937000 .
_t.= = 2 .__.= _____.= 2
A0 ‘17077 1175 lb/in 519 2600 360 lblin
npt _ 0.85 x 1470000 - 2 M. = 14337000 = 2
Pt ego e10 = #70000 x 5 x 16 = 905 1b/in2

Io I+151+0

nPt eso em = 0.85 x 1470000 x 5 x 16
I0 415%

 

= 770 1b/in2 .

 

 

 

..1
m,
0

t

 

 

 

 

1.
.

 

 

At the 1/4 point of the span:

 

 

§§-= Elgg%9-= 1175 1b/in2 2%E-= Zgggg9-= 288 lb/in2
r1:‘ = BLEEEE‘EZEBQE = 999 1b/in2 :52": zgggg99.= 1h05 ib/in2
Ps esgoelo = 9700005:05 x 16 = 905 1b/in2
CL§£_;§2_219.= 0.85 31212000 x 5 x 16 = 770 1b/1n2

At the 1/8 point of the span:
£fi-= Elfigg9-= 1175 lb/in2 §f3-= Eggg%9-= 168 113/in2
Bil = W9: 999 lb/in2 :16: gag-32 = 860 1b/in‘2

Pt °so e10 = l+7003: x 3 x 16 = 51+} 1b/in2
nPt °so e10 = 0.85 x 1470000 x 3 x 16
I ”h15ho

O

= 962 1b/in2

 

When block #2 is used for span lengths from 30 feet to ’40 feet,
1 nclusive, the cables will be bent twice. 330 = 5 inches between 1/14
Points of the beams and 080 = 3 inches at the 1/8 points of the beams
as shown on page 123. The cables will reach the middle of the beams
at the support.
At the bending points of the cables, rectangular blocks will be
used as support to the cables. These blocks will be designed after

311ear is checked for all the span lengths.

 

 

 

 

 

Block #3

125

Forty foot span, top and bottom fibers at 1/2 point of the span:

_ 1476000 _ 2
_ .133. — 1092 1b/in

nPI = 0.85 x 976000
10 93

81;?

= 928 1b/in2

Mdb _ 1090000

51s 9228
M, _ 9520000
31, 9228

Pt e20 em = 976000 x 10 x 20 = 1126 1b/in2

1o 8 550

nPl e50 e10 = 0.85 x l+76000 x 10 x 20 = 956 1b/in2

§5§§6

I0

11: the 3/8 point of the span:

 

 

Pt _ 976000 _ 2 m ___ 1020000
To — W— 1092 1b/in $19 228
r11: _ 0.85 x 976000 _ 2 14!— : 9920000
To“ T’ 928 1““ sis T272?-
Pt 680 610 _ ”76000 x 10 X 20 _ 2
rip, :30 elo = 0.85 x 9:602: x 10 x 20 = 95b 1b/in2
O 5

At the 1/9 point of the span:
Pt _ 976000 _ 2 Mdb = 816000
V1 r, _0.85 x 976000 _ 2 1‘1. = 3716000
‘0 - '7—36 _ 928 lb/in $1. T—eas
Pt °so em = 1176000 x 10 x 20 = 1126 1b/in2

Io 89550

“Pt eso em = 0.85 x 976000 1 10 x 20 = 956 lb/ina

 

10 89350

= 258 1b/in2

= 1070 1b/ 1112

= 291 lb/ina

= 1095 1b/ is‘2

= 193 1b/in2

= 877 1b/in2

 

128

At the 1/9 point of the span:

1», _ 1476000 _ 2 9st _ 1035500 _' 2
To _ .735— - 1092 1b/in E; - “9'22?— - 2% lb/in

'M 5935500

P 0.85 x 976000 2
BJ- = = 2 _W- = = L}
———-—7(——— 928 1b/in $16 T—228 128 lb/in

A0 93

IO 8 550

“P1; 830 610 = 0.85 X l476000 I 10 X 20 = 95° 1b/in2

 

 

 

Io 89550
At the 1/8 point of the span:
P 976000 M 603900 2
_t = = 2 _dh = = 1}
10 .736— 1092 lb/in 519 “228 1 3 1b/in
n?, = 0.85 x 976000 = 2 M = 3278900 = 2
928 lb/in .1. T2'2—8— 775 lb/in

A0 936 $1 a

P, 8 em = 976000 x 6 x 20 = 672 1b/in2
ID 89550
r113, e,3° e10 _ 0.85 1: 976000 x 6 x 20

_ 2
I 813557 - 570 lb/in

 

 

o
When‘block #3 is used for span lengths from 14-0 feet to 55 feet,
inclusive, the cable will be bent twice. At the 1/8 and 1/# points of
the beams as it is shown on page 127. The cables will reach the middle
of the beam at the end of the beam.
At the bending points of the cables, rectangular blocks will be
used as support to the cables. These blocks will be designed after

shear is checked for all the span lengths.

 

 

 

 

 

132

At the 1/4 point of the span:

 

 

 

 

 

P, 657000 2 N 3878000 _ 2
— = ' = 1120 lb/in 4111: = 535 lb/1n
10 588 S1e 72 '
3.31 = W = 950 1b/1n2 1‘1. = 10218000 = 1912 lb/in2
A0 588 $16 72
Pt eso e10 = 657000 x 1225 x 29.5 = 1135 1b/in2
10 177500
rth eso 610 = 0.85 x 657000 x 12.5 x 29.5 = 965 1b/in2
I, 177500
At the 1/8 point of the span:
P 657000 2 Md-b 2262500 _ 2
_l = = —— = —— - 1 b 1
A0 588 1120 lb/in $18 7290 3 3 1 / 11
01:? 0.85 x 657000 M 6062500 2
___I.= ___________..= 2 _!_.= =
P e e 657000 x 7.5 x 24.5 _
t so 10 _ = 2
Io - 177500 680 lb/in
Y\Pt eso e10 = 0.85 x 657000 x 7.5 x 29.5 = 578 1b/in2

 

Io 177500
When block #9 is used for span lengths from 55 feet to 75 feet,
inclusive, the cable will be bent twice. At the 1/8 and 1/9 points of
the beam as it is shown on page 131. The cables will reach the middle
of the beam at the end of the beam.
At the bending points of the cables, rectangular blocks will be
used as support to the cables. These blocks will be designed after

shear is checked for all the span lengths.

 

- 2.54 ‘2‘ ~~,

 

 

133

Block #5

Seventy—five foot span, top and bottom fibers at the 1/2 point of the

 

 

 

span:

Pt _ 858000 _ 2 udb _ 7320000 _ 2
z;-— 832 - 1025 lb/in $16 - 12790 - 573 1b/in
r\P = 0.85 x 858000 = 2 §!_.= 15390000 = 2
-I;1 ‘———§3§—————- 872 lb/in 51e ‘IEfiii” 1200 lb/in
P e 0 858000 x 19 x 30

1: so 10 _ = 2
--————IO — ——————————383500 1275 lb/in

nPt e30 010 = 0.85 x 858000 x 19 x 30 = 1082 lb/inz

I 38 3500

0

At the ‘3/8 point of the span:

 

 

 

. 0 ,
I: 8 658000 = 1025 mm? Lay: “—50 0° = 530 lb/ina
lo 332 516 12790
F 0.85 x 858000 2 M' 1u550000 2
IL_1.= _____________.= 2 __—.= .———————-= 1 b 1
‘0 832 87 lb/in 51° 12790 1N0 1 / n
Pt 630 010 858000 x 19 x 30 2
________._.= _______________.= 2 b 1
Io 383500 1 75 1 I n
rlpt eg9 019 = 0.85 x 858000 x 19 x 30 = 1082 1b“1&2
IO 383500
At the 1/u point of the span:
P 8 000 M usoooo
_t= 58 g 1025 115/1112 _dl= L.— = 1429 lb/in2
lo 832 318 12790
. 2
YlPt = 0 85 x 858000 = 372 1b/in2 ¥:_,= 11?.9999.= 925 1b/1n2
‘o 832 816 12790

Pt eso e10 858000 x 19 x 30

Io - _-'"7fiii§i3""’

 

 

= 1275 1b/in2

th 930 010 0.85 at 858000 x 19 x 30

I0 333500

= 1082 1b/1n2

 

 

 

 

 

 

 

 

 

 

 

136

At the l/h point of the span:

 

P 858000 M 9750000 .
—3.= = 2 b 1 2 ~92-= .~__.._= ' lb 1 2
10 832 10 5 1 / n 519 12790 703 / n
P .
'14 = 932.15.515.83‘19 = 872 lb/in2 I5!.— = 1.939902 = 11% 113/1:12
no 832 516 12790

Pt ego 810 = 858000 x 19 x 30
Io 383500

 

= 1275 lb/ 1:12

r\Pt eso e10 = 0.85 x 858000 x 19 x 30

.. = 1082 lb/in2
I 383500

 

 

0

At the 1/8 point of the span:

 

 

 

2. = 82:?" = 1025 lb/in2 21% = Egg-g9 = ln+5 Un/im2
I§Et.= 9:§§§§§§§§999 = 872 lb/in? Ef;-= lg;g;%39-= 8“8 lblin2
31.153219: = @OWOBESOX 3° = 801+ lb/inz

h Pt °so e10 = 0.85 x 858000 x 12 x 30 = 68M lb/ina

IO 383500

When block #5 is used for span lengths from 75 feet to 100 feet,
inclusive, the cable will be bent twice. At the 1/8 and l/N points of
the beams as it is shown on page 135. The cables will reach the middle
of the beam at the end of the beam.

At the bending points of the cables, rectangular blocks will be
used as support to the cables. These blocks will be designed after

shear is checked for all the span lengths.

 

 

 

 

 

 

137

Block #6

One hundred foot span, top and bottom fibers at the 1/2 point of the

 

 

 

 

 

 

 

 

 

 

span:
P 1250000 2 M 19350000
J»: =10081b1 .4-‘l= , =81+b 2
Ab 12KB / n 51° 22650 5 1 /1n
r1P 0.85 x 1250000 _ M, 30u90000
—l= w- = ' 2 -—-—=—— = ' 2
‘6 12h0 856 lb/in $16 2§3§5N~’ 13h5 1b/1n
P e 0 1250000 x 2M x 36
.1 9° l9_= .. = 1 lb 1 2
_ 10 815000 325 / n
P 0.8 1 0000 2a 6
r‘ t °3° °19 = 5 x 25 ‘22. x 3 = 1128 1b/1n2
ID 815000
At the 3/8 point of the span:
Pt 1250000 2 Mdb 18150000 2
_= =1008 lb in _—=———g———-=SOO 1b in
. 7 M 8 0
£131.: 9_§§.x {329999.= 856 1b/1n2 —!L.= §§.?OO O = 1270 lb/in2
A0 12E6' 818 22050
P e e 12 0000 x 2M x 6
t so 10 = 5 3 = 1325 1b/1n2
10 815000

 

. 2“
Ilfl_fsslfln.= 9155 ‘ 1250000_§_ x §§.= 1128 1b/in?
10 815000

At the 1]“ point of the span:

 

 

Pt 1250000 2 “db _ IMSZOOOO - - 2
... = .... = 1008 1b in .... — ”-..... — 681 lb/in
. - M 2 l 0000
£3.11: 0 85 x 135.9999: 850 ”/1132 ...L: 21,“: 1020 lb/in2
A0 1250 lo 22650
P! eso elQ = 1250000 x 2M x 3§_= 1325 1b/in2
10 815000

VIPs eso 610 0.85 x 1250000 x 2M x 36
Io "‘ 815000 ‘ ""

 

 

 

.v.

I r. .
"L ,v..l 7.. y _. xvi . l
1.10

1 .1. 1.91.0.1. J artist‘s!

 

 

 

 

At the 1/8 point of the span:

 

 

 

 

 

E. = 1.3—$03 = 1008 113/1112 24:1: {Egg-g?—

{1: 28%:01250000 = 856 1b/in2 2L = 12:22:00
0 1e

Pt 8:: 910 : 1250;026:015 x'36 = 827 lb/in2

“\Pt eso e10 = 0.85 x 1250000 x 15 x 36 = 703 lb/ine

ID '815000’

138

=~373 lb/in2

= 600 lb/in2

One hundred twenty-five foot span, top and bottom fibers at the 1/2

point of the span:

 

 

 

= 1333 lb/in2

= 1955 lb/in2

P 1250000 M 30200000
t 2 db -
—-= = 1008 lb/in —- ......—
A0 1256 318 22650

P 0.8 x 12 0000 , an 0000
LE" "—6110"; = 85° lb/in2 :‘L‘ 23'2“?"
P e e 1250000 x 20 x 36

t 3°."19.= _121 — = 1325 lb/in2

IO 815000
. - ’ 2% '

V1§:_eso 210»: O 85 x l?50000 x x 36 = 1128 1b/in2

 

 

I ‘“"”'815000

0

At the 3/8 point of the span:

 

 

 

 

P 1250000 2 u 28200000
.1: ._= OOSlbi —dh=__....—_.
A0 12E6 1 / n 516 22650
n ,
0P, = 0.85 3: 1250000 = 856 lb/in2 1‘11. = 1690000
‘0 12ho 816 22050
Pt 690 910‘: 1250000 X 2’4 X 36 = 1325 113/1:12
Io 815000

P e e 0.8 x 1- 0000 x an x 6

W: 5 P5 3 = 1128 1b/1n2

I 815000

0

1205 1b/1n2

= 18u0 1b/1n2

 

 

 

 

 

 

7
..‘18147.
-__;,.__r_-._._+.4
1,1 1";
i
A

...A

  

   
 

 

 

. 1 . g
1
d‘

s

1"

....

 

 

A
l

' '44 35224955 '

11

1, ’IT 11v

1;.
T

 

I

    

 

 

 

 

 

.1 .. I
. b I . i
t . 2’»:

I
;
I|/P"“

,2 2.
177773173.”—

zeiz'f
1

 

  

 

T

Cable Stress ‘

v

 

 

Y

:73”)?

IT! ‘10
1

........

   

3

Bottom Fibers

I

 

.........

.........

 

 

 

.....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

luo

At the 1/u point of the span:

 

 

 

P 1250000 M 22h00000 2
.11: --=10081b'2 43-h.=———-——-—-= 8 lb'
A0 12u0 /1n 51¢ 22650 9 7 lln
11 P 0.85 x 1250000 , 3&080000

t = -- = 856 lb/in2 M—L= --—--,—----- = 1505 lb/in2
Ab 12E0 816 22650

P e e 1250000 x 2M x 36
t 99 10 = - __.._= 2
10 815000 1325 lb/in

 

 

YIPt eso e10 _ 0.85 x 1250000 x 2M x 36
Io ‘“‘ "' 815000

 

 

= 1128 lb/1n2

At the 1/8 point of the span:

 

 

 

 

P 1250000 M 13211000

t _ _ 2 db _ _ 2
-----—-;—-—-—-10081bin - - 8 lbi
A0 1240 / 519 22650 5 3 / n

P 0.85 x 1250000 5 2 M 19671000 .
11.1.: = , .n_. ._2_,._..== , 2
Ab 12h0 85o lb/in $16 = 22650 866 lb/in
Pt eso 610 = 1250000 x 15 x‘36-= 827 lb/in2

Io 815000

P e e 0.8 x l 0000 x 15 x 6

7‘ t. 8° 10 = 5 25 3 = 703 lb/in2

 

 

10 815000

When block #6 is used for span lengths from 100 feet to 125 feet,
inclusive, the cable will be bent twice. At the 1/8 and l/M points of
the beam as it is shown on page 139. The cables will reach the middle
of the beam at the end of the beam.

At the bending points of the cables, rectangular blocks will be
used as support to the cables. These blocks will be designed after

shear is checked for all the span lengths.

 

 

 

 

CHECKING FOR SHEAR

 

 

1M2

Thirty Foot Span Length

B10 cl: #1

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

V: 8x11 XIO=59001b.

Live load shear at the 1/2 point of the beam:

 

 

 

 

 

 

21200 21200
1
.5 _
'053 EL
15! .4 15'
tq—u—~— 30'

 

 

Influence Line

1600 ( + 0 = ’ . 2 = 2 .
0(1 155; 1e000 x 1 3 5 21 00 1b

v = 21200(0.5‘t 0.033) = 11350 lb.

 

 

 

 

 

1%}

Live load shear at the 3/8 point of the beam:

 

 

 

 

 

21330 21200
1m
0625
\ e 375
11.25' _L‘ 18.15'
v-I—v
30'

 

 

 

Influence Line
v = 21200(o.625 + 0.158) = 16600 lb.

Live load shear at the 1/M point of the beam:

21200 21200
lu'

~75
.283L\\\\\\\\>

7.5'111 2?.5'

30'

 

 

 

 

 

 

 

 

Influence Line

v = 21200(0.75 + 0.283) = 21900 lb.

 

 

 

111

Live load shear at the 1/8 point of the beam:

21200 21200

.875
.u09

 

 

 

 

 

 

 

Influence Line
v = 21200(o.875 + o.h09) = 27300 lb.

Live load shear at the support:

 

 

 

 

32000 21 00 53 0
1M' 4 1uv 4-3
1 \\
~53} ,,
.0005 N
V 30' J

 

Influence Line

( 0 = =
hooo(1 + 155 hooo x 1.325 5300 lb.

V = 32000 + 21200 x 0.533 + 5300 x 0.0668 = 32000 + 11350 + 35“

= #370“ 1b.

 

O. . , I
l. in... . I . . é .
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1M6

The vertical component of the tension in the cables between the

support and the 1/8 point of the beam:

 

j."

 

 

 

 

 

 

 

 

 

e 1 e
3.75' 22.5' . 3.15'
Tan 9 = 33: 11.25 9 = su° 55"13"
Cos 9 = 0.08851;
Cos 9 x Pt = 0.0885“ x 392000 = 31+700 lb.
= vm H

8h 10 t
vmx = 31730 lb. (See page 1%)
H = 1582 in3 (See page 36)

11

IO = 32887 in (See page 21)

t = 9 in (See page 18)

- 31730 x 1582 - 2
fsh - 32887 1 9 — 170 1b/in 0.x.

 

 

 

 

1M7

Block #2

Thirty foot span Length

lb = ”00 in2 (See page‘OZ)

Dead load shear at the 1/2 point of the beam:
V = 0
Dead load shear at the support:

_ Moo x 1 x 1 0 _
v- _E‘EII'L’ 6250 lb.

Live load shear at the 1/2 point of the beam:

 

 

 

21200 91200
11“
‘k\\
.5 \
.033§J
\IS

 

 

 

 

 

Influence Line

V = 21200(O.5 + 0.033) = 11350 1b.

 

 

r ‘ l‘g‘ ~—

.. 1. .
”....._««»
7.. ....

 

 

 

 

 

 

151

The vertical components of the tension in the cables between the

support and the l/M point of the beams:

 

 

 

 

 

 

 

 

 

 

 

 

ea 9' f 4/225: 9.
[3.75' 3.75' 15' _.75' .zslj
Tan 9, = 5%: 22.5 e, = 87° 27' 19"
Cos 9, = 0.011010
Cos 6% x Pt = 0.0uuu x M70000 = 20900 lb.
Tan 9. = 321 = 15 9,: 86° 11' 9"
Cos a; = 0.06651
Cos 55 x Pt = 0.06651 x #70000 = 31250 lb.
fsh = vatH
Vmax = 25025 lb. (See page 150)
I, = 1115110 in“ (See page m)
t = 8.5 in (See page 18 )

H = 8.5 x 7.5 x 3.75 + 16 x 8.5 x 11.75 = 239 + 1600 = 1839 1:13

_ 25025 x 1832 = 2

 

 

.Io'l‘,

,, 1.13.11: , ) t 7 : let; 41:16:33.151. .

 

 

 

 

152

Forty Foot Span Length

Block #2

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

= #00 x 20 x 150 =
V Inn. 8330 lb.

Live load shear at the 1/2 point of the beam:

20850 20850
110

V
.5 K\\‘\\\\T154L\\\\

 

 

 

 

20' 20'

 

 

 

 

Influence Line

16000(1 + 16%; = 16000 x 1.303 20850 lb.

(

V'= 20850(0.5 + 0.15) 13580 1b.

 

 

 

 

 

 

 

 

 

153

Live load shear at the 3/8 point of the beam:

20850 20850
11.»

e625 \
0275'\ ,

\ .375
V 15! Jr: 25!

no:

 

 

 

 

 

 

 

 

Influence Line
v = 20850(0.625 + 0.275) = 18800 lb.
Live load shear at the l/h point of the beam:

20850 20850 5212.5
- 1h' .1, 1n.

‘\
.75 \‘
.h
.05 N
\ .25

10‘ .1: 30'

110'

 

 

 

 

 

 

 

 

 

Influence Line

( JED = —
uooo(1 + 165) uooo x 1.303 — 5212.5 lb.

v = 5212.5 x 0.05 + 20850(0.75 + 0.4) = 2h261 1b.

 

 

 

 

15h

Live load shear at the 1/8 point of the beam:

 

 

 

 

 

 

20853 20850 5212.5
1u| 1n.
1
.875 R \l
.,25
.125 L\\\\\
“‘\~.125
_ju__., . 35'
40'

 

 

 

Influence Line
V = 20850(0.875 + 0.525) + 5212.5 x 0.175 = 30113 1b.
Live load shear at the support:

320 20850 52125
18' 18'

1L\\\\\\\\\4
1 '65 \fi

#1

 

 

 

 

 

Influence Line

V = 20850(0.65) + 5212.5 x 0.3 + 32000 = “7115.1b.

 

 

 

 

 

 

 

 

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156

The vertical component of the tension in the cables between the

support and the l/N point of the beam:

 

 

 

 

 

 

 

 

 

 

 

e V 91

z e, l T 9-

5! 5' V 20' 5' 5' Jl
Tan 9' =§%=30 ' e,=ss° 5°27"

Cos e, = 0.0333
0.0333 x 1+7OOOO = 15650 1b.

Cos a x Pt
Tan 67- = £03. = 20 97.: 87° 8' 15"
Co: 91 = 0.01&991+

Cos 6:. 1 Pt = 0.01991; 1: l+70000 = 23500 11:.

 

fsh = vTTtH
Vm = 31916 (See page 155)
H = 1839 in3 (See page 151)
I0 = 1«#15140 in1+ (See page I02)
= 8.5 in. (See page 18 )

_ 1 1+ x 18 _ « 2
fsh — 15% x 8.5 - 16b.u 1b/1n 0.x.

 

I .1 ll
. I 0r- IV V
. .' (I .-

- - 2.2.03). (L. Low... N’3.i§.§3

x. .-.-t I7 I

 

 

157

Block #3
Forty Foot Span Length
‘0 = u36 in2 (See page106)

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

V = u 6 x $041 1 O = 9070 lb.

Live load shear at the 1/2 point of the beam:

 

 

 

 

 

20850 20250
114'
‘
.5
015\
.5
20. L 20.
T I
no'

 

 

 

Influence Line

V = 20850(0.5 + 0.15) = 13580 lb.

 

158

Live load shear at the 3/8 point of the beam:

20850

20850

11;-

 

.625 \

 

L\\\\‘\\\o

 

15'

 

\0375

25'

 

uo'

 

 

 

Influence Line

v = 20850(0.625 + 0.275) = 13300 lb.

Live load shear at the 1/4 point of the beam:

20850

52l2.5

 

 

20850
<4 Ly' 1h'
{\q
.u
.0

 

 

 

30'

 

no:

 

 

Influence Line

 

v = 20850(0.75 + o.u) + 5212.5 1 0.05 = euooo + 201 = 2u261 lb.

 

 

 

32000

 

159

Live load shear at the 1/8 point of the beam:

 

 

 

 

 

20850 20550 5212.5
1h' 1M' 7
V
.875
.525
~175
“*\\ .125

 

'7T—_—

 

 

Influence Line

Live load shear at the support:

114'

20850

A

V = 20850(0.875 + 0.525) f 5212.5 x 0.175 = 29200 + 913 = 30113 1b.

5212.5

 

\

.65

 

 

 

 

 

L\\\\\“‘\\,
J.

Influence Line

V = 32000 + 20850 x 0.b5 + 5212.5 x 0.3 = 32000 + 13580 + 1568

= h71u8 1b.

 

 

161

The vertical components of the tension in the cables between the

support and the 1]“ point of the beams:

 

16" 1°"

 

 

 

 

 

 

 

 

 

 

 

S 6
I. 9. I f . 9' a.
l 5‘ 5' A 20‘ 5' S'J
F‘ ‘ '1
Tan 6. =§%=15 e.=86°11' 9..

Cos a. = 0.06651
Cos e. x Pt = 0.06651 1: 1176000 = 31650

Tan 9:. = b—g- = 10 = SW 17' 21"

Cos 9:. 0.09951

C03 91 1 Pt = 0.09951 1: 1476000 = M7400 1b.

 

f g vum H
' sh 10 t
v,1m = 28796 1b. (See page 160)
I0 = 811550 in” (See page 106)
t = 6.5 in. (See page 18 )

H = 6.5 x 13.5 x 6.75 + 20 x 6.5 x 16.75 = 592 + 2180 = 2772 in}.

_ 28 6 x 2 2 _ 2
£311 - M— 1145.2 lb/in 0.x.

 

 

 

'11.,“ ll)

 

162

Fifty—five Foot Span Length

Block #3

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

v = u 6 112 ' I 1 — 12500 lb.

Live load shear at the 1/2 point of the beam:

20150 20h50
1h'

.5L\\\:;;EE4L\\\\\\\>

 

 

 

 

27-5' 27.8'

 

55'

 

 

 

Influence Line

1600051 + 188 = 16000 x 1.278.= 20u50 lb.

v = 20u50(0.5 + 0.2u55) = 15280 1b.

 

 

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:1
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l I", I ‘
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163

Live load shear at the 3/8 point of the beam:

20u50

1m

20h50 5112.5
1h'

J

 

.625

3 ’1

 

.37N

 

20.635'

\.

375

3h.3]5'

 

 

 

55'

 

 

Influence Line

( 0)
#000(1 + 180

= #000 x 1.278 = 5112.5 1b.

V'= 20u50(0.625 + 0.37) + 5112.5 x 0.116 = 2u000 + 53k = 2u53u lb.

Live load shear at the l/h point of the beam:

20u50

20u50 5112.5
1h'

 

 

 

 

 

 

 

55'

 

 

Influence Line

V’= 20u50(0.75 + 0.u95) + 5112.5 x 0.2u1 = 25u50 + 1232 = 26682 1b.

{aviatfli‘uitll‘ it‘ll]... . . . 1P ,.I,|.

 

 

 

 

_
_
w

‘s Vets—L

 

‘

 

16“

Live load shear at the 1/8 point of the beam:

20u50 20u5o 5112.5
h'

1\‘"

1 1m

 

 

 

 

 

 

.875 .62
.3652
“‘~4 .125
6.815' 7 1+8.125'
55'

 

 

 

Influence Line

V = 20u50(0.875 + 0.62) + 5112.5 x 0.3655 = 30600 + 1870 = 32U70 1b.

Live load shear at the support:

32000
IN

20350
I 11.1.

5112.5

 

\

.756

 

 

 

 

4

Influence Line

‘7 = 32000 + 20u5o x 0.7u6 + 5112.5 x 0.u91 = 32000 + 15280 + 2515

_. .--—..{v—y—M -H.H' _a- ,4

 

fivm_h'om~:‘a—5r'tfir‘ 5;;1‘b—g~\—.—- '. ...-y v.7 . ;_ ..--1 —

 

 

 

 

.. ......» ruR- ....

 

 

 

 

.
1| Ii... 41.I. full ,1. 7 ,1.
.
.
.

 

 

 

» 4
. _ .....
n t. 9 .
. . , h.
T0. 141111 (61+. 11 TI. 1 IL 11.1119
v . .1 +. 1 .

 

 

 

I
[:2
#3..
A..
1:
I.
‘.
I’

1

1

 

 

 

 

 

 

' ‘ ’for Block

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The vertical component of the tension in the cables between the

support and the 1/14 point of the beam:

 

 

 

 

 

 

 

 

6
92 9. I T l 9. L
2 .875: W667i P7.5' J: 875' 6875'
Tan 9, = 8125i = 20.625 9, = 87° 13' 27'
Cos 9‘ = 0.0118112
Cos e, 1: Pt = 0.011812 1: M76000 = 23000 lb.
Tan 92. = $6.21: 13.75 6‘ = 85° 50' 25!!
Cos 6,, = 0.07253

Cos 92. 1: Pt = 0.07253 1: l+76000 = 35800 1b.

H = 2772 11:3

IO = 811550 in“

t = 6.5 in.

.vmaxH

 

fSh ‘ I t

0

(See page 165)
(See page 161)
(See page 106)

(See page 18)

-18932x2772_ -' 2
rah — 81550 x 6.5 - 166.3 lb/in 0.x.

 

 

 

 

 

 

167

Block #H

Fifty—five Foot Span Length

2

A0 = 588 in (See page 109)

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

.75
v = 88 x f x 1 0 = 16880 1b.

Live load shear at the 1/2 point of the beam:

20u5o 20u50
15'

 

 

\.s

27-‘1’ .1. 27.51

I

 

 

 

 

 

55

 

Influence Line

v = 20u50(0.5 + 0.2h55) = 15280 1b.

 

168

Live load shear at the 3/8 point of the beam:

 

 

 

 

 

20u50 20M50 5112.5
1),“ 1k!
.625
'37 .11 L\\\\
\ .375

 

‘ 20.625' pJ 38.375'
'1

 

 

55'

 

Influence Line

V = 20N50( 0.625 + 0.37) + 5112.5 x 0.116 = 25000 + 534 = 2N53H lb.

Live load shear at the l/U point of the beam:

 

 

 

 

20u50 20u50 5112.5
1u- 1h'
'75 u
° 9 .2u1
\.25

 

13.75' _1_ h1.25'

'T‘
55'

 

 

 

 

Influence Line

V = 20N50(0.75 + 0.h95) + 5112.5 1 0.2Hl = 25N50 + 1232 = 26682 1b. ‘

 

169

Live load shear at the 1/8 point of the beam:

20u50 20u5o 5112.5
lh' - 1n.

 

.875
-3655

 

 

 

 

6.875' ”8.125'

 

55'

 

 

 

Influence Line
v = 20u50(0.875 + 0.62) + 5112.5 x 0.3655 = 30600 + 1870 = 32k70

Live load shear at the support:

32000 20u5o 5112.5
2 1h' iu'

IL\\\\q
.7u6 'u91 L\\\\\\\‘\\\\\\\\>
" J

55'

 

 

 

 

 

Influence Line

V = 32000 + 20k50 x 0.7H6 + 5112.5 x 0.u9l = 32000 + 15280 + 2515

= 119795 lb.

 

:02. cut 0:0.

4 a : 2, use:

 

 

171

The vertical component of the tension in the cables between the

support and the l/N point of the beam:

 

 

9?.

 

9. 1

9.

 

 

 

6375' 6.815; 27 . 5'

 

6.875 6.870“

 

 

 

 

Tan 9‘ = 87%; 16.5

Cos e, = 0.0605

9. = 86° 31' 55"

Cos 9. x Pt = 0.0605 1: 657000 = 39750 1b.

Tan 9,, = 87$;- = 11 9‘: 814° 18 2o"

Cos a; = 0.09053

0o. 9. 1: Pt = 0.09053 1: 657000 = 59500 lb.

v s
f = flai—
Sh Io t
vm = 35122 11). (See page 170)
I0 = 177500 in)4 (See page109 )
t = 7 in. (See page 18 )

H = 7 x 17.5 x 8.75 + 7 x 2u.5 x 21 = 1071 + 3600 = 11671 1x13

_ 35122 1: 1+6 [1

2
fsh - 177500 x 7 = 132.1 lb/in

O.K.

 

172

Seventy-five Foot Span Length

Block #M

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

_ 8 x .5 x l 0 _
v — 5§_"'%%E"-__5"" 22950 lb.

Live load shear at the 1/2 point of the beam:

20000 20000 50C0
lu' 1n.

\\

'5 '3135 .12

\5

 

 

 

 

 

 

Influence Line

’ ( O) = =
16000(l + 265) 16000 x 1.25 20000 1b.
( 50; _ =
u000(1 + 200 - #000 x 1.25 500 lb.

V = 20000(0.5 + 0.3135) + 5000 x 0.127 = 16270 + 635 = 16905 1b.

 

 

- ~_-1 7.2-”? v—-

 

ill .1: nil . 11116111,.-.1.’

 

Q
"a“ f
.L

 

"J‘
”‘0‘
7. I.
3‘
, ‘
-
A
'-
‘
~
41
.0

Live load shear at the 3/8 point of the beam:

20000 20000 5000
11' V 11+!

.2515
-<<\\\\\\\\\\\V .375 §

28.125‘ 46.875'
75‘

 

 

 

 

 

 

 

 

 

 

Influence Line
V = 20000(0.b25 + 0.439) + 5000 x 0.2515 = 21300 + 1258 = 22558 lb.

Live load shear at the l/N point of the beam:

20000 20000 5000
A luv 1n:

1L\\\~\1L\\\\\J

.75 .56h 177 L\\\\\\\\\\\\‘

8.75' 56°25'

 

 

 

 

 

 

 

 

75'

 

Influence Line

v = 20000(0.75 + 0.564) + 5000 x 0.377 = 26300 + 1885 = 28185 lb.

 

 

179

Live load shear at the 1/8 point of the beam:

9.375'

 

 

 

 

 

 

 

20000 20000 5000
1).“ 11+!
1
L\\\\4L~
\‘r
.875 .689 .502 \

.125
65.625‘
75'

 

 

Influence Line

v = 20000(0.875 + 0.689) + 5000 x 0.502 = 31250 + 2510 = 33760 1b.

Live load shear at the support:

32000

20000

1h'

10'

 

 

.627

 

5000

 

 

75'

 

 

Influence Line

v = 32000 + 20000 x 0.815 + 5000 x 0.627 = 32000 + 16280 + 3135

= 51u15 1b.

 

 

 

 

 

 

 

  

 

 

 

      

 

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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181

The vertical component of the tension in the cables between the

support and the 1/11 point of the beam:

 

12» 19"

 

 

 

 

 

 

 

 

 

 

 

9; ¢ 9' f a. 9:
9375' 3315' 37 é 1 1 49.315: 49.315'
Tan 9. = £1.35. = 119.075 9, = 86° 21:1 25"
Cos 9, = 0.06209
Cos 9, 1: Pt = 0.06209 3: 858000 = 531400 lb.
Tan 91 = 12%}. = 9.37 9‘: 83° 511' 30"
Cos 9; = 0.10612
Cos 65 x Pt = 0.10602 I 858000 = 91000 lb.
fSh - V?axtH
o
Vfiax = h4fl35 1b. (See page 178)
I0 = 383,500 in“ (See page 113)
t = 8 in. (See page 18)
H = 3 x 22 x 11 + s x 30 x 26 = 1936 + 62uo = 8176 1n3

1‘

Q

111414 ‘
sh = 3853023? = 118.3 1b/1n2 0.x.

 

 

One Hundred Foot Span Length

Block #5

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

_ 532 x 50 x 150 _
v - 1hh - hjhoo 1b.

Live load shear at the 1/2 point of the beam:
19550 lgiio h§88
1M' 1M1

 

‘

w W
.skbx

 

 

\.5 V !
i

50' 50'

 

w W

 

100’

 

 

 

Influence Line

1600021 + 22? = 16000 x 1.222 = 19550 1b.

< o)== =
uooo(1 +-§g§) #000 x 1.222 #888 1b.

 

 

182

v = 1955o(o.5 + 0.36) + #888 x 0.22 = 15810 + 107M = 1788” lb.

 

 

-1155" "
. .' s..11;118at111c

Live load shear at the 3/8 point of the bemm:

19550 19550 4888
14' 1u"

 

'625 .u85[.3h5

<\\\\\\\\\\\\.375 ‘
i

37-R' 09-5'

J

 

 

 

 

 

Influence Line
v = 19550(0.625 + 0.485) + M888 x 0.3h5 = 21700 + 1688 = 23388 1b.

Live load shear at the 1/4 point of the beam:

19550 19550 #38
11+" 11+.

‘NN,

'75 .61 h7\\\\\\\\\\\\\\\\\

35‘ _L. 75‘

100'

 

 

 

 

 

1
..LJL 1._

 

Influence Line

 

..u'4:.' "a f
- I
if! I . 1!.11- .
I
1

v = 19550(0.75 + 0.61) + h888 x 0.47 = 26600 + 2300 = 28900 lb.

~ ”Wifgffi “. i- .‘ -..

J

   
 

   

18%

Live load shear at the 1/8 point of the beam:

19550 19550 “.888

 

 

 

 

1n. #1MLJ
1N !
.878
\ .125 , ‘
12.5' x 87.5:

 

100'

 

 

 

 

Influence Line
V = 19550(O.875 + 0.735) + #888 x 0.595 = 29500 + 2910 = 32Nl0 lb.

Live load shear at the support:

2000 0 “888
3 30958.

 

"\‘N

.86 .72

 

 

 

100' J

 

 

Influence Line

v = 32000 + 19550 x 0.86 + #888 x 0.72 = 32000 + 16800 + 3520

= 52320 1b.

 

 

 

 

 

 

 

 

 

 

 

   

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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186

The vertical component of the tension in the cables between the

support and the l/M point of the beam:

 

 

 

 

 

 

 

 

Tan 67 = 1?: 21.12857

 

Cos a, = 0.01I667

Cos 97 x P
Tan 6. =l§% = 125

vmax = 50600 11).

H = 8176 177:3

)4.
IO = 383500 in

t = 8 1n

fsh

 

I12" I?"
9 7 9 7
7 e. I /eF ‘
12.5' 12.5' 25' 12.5' 12.5'
f

 

 

e. = 87° 19' 31+-

070u667 x 858000 = ”0100 1b.

6. = 85° 25' 31+"

— 0.07976 1 858000 = 68500 1b.

 

(See page 185)
(See page 181)
(See page 113)

(See page 18)

= w_ 2
383500 I g - IBM-9 1b/1n 0.x.

 

187

Block #6

One Hundred Foot Span Length
‘0 = 12h0 in2 (See page 116)

Dead load shear at the 1]? point of the beam:

Dead load shear at the support:

v = £259.5igg_£—159 = 6u600 1b.

Live load shear at the 1/2 point of the beam:

19550 19550 Mess
~1m 11+-

 

 

 

 

 

 

 

 

Influence Line

V = 19580(0.5 + 0.36) + NSSS x 0.22 = 16810 + 107” = 1788“ lb.

-. M ...

 

 

188

Live load shear at the 3/8 point of the beam:

19550 19550 #888
1M' 1M'

1\'
.h85 .3u

37-5' .1. 62.5'
'T‘

 

.625

 

 

 

100'

-JL ,._!L-..fi__.-___

 

Influence Line
V = 19550(O.625 + O.H85) + #888 x 0.3M5 = 21700 + 1688 = 23388 lb.

Live load shear at the 1/N point of the beam:

19550 19 50 uses
1u{ lui,

 

V

 

 

25' 75'.
100'

 

 

 

 

 

Influence Line

V = 19550(O.75 + 0.61) + “888 x 0.“? = 26600 * 2300 = 28900 1b.

 

189

 

Live load shear at the 1/8 point of the beam:

19550 19550 h888

 

 

 

 

 

 

 

Alh' ‘lu'
.875
.735 .59?
\10125 . E
12 2. . .4._ 87.53 i
100‘

 

 

Influence Line
v = 19550(0.875 + 0.735) + #888 x 0.595 = 29500 + 2910 = 32h10 1b.

Live load shear at the support:

52000 19550 uses
Alu' filu'

RN e i

.86 .72

 

 

 

 

100'

 

..“JL.__

Influence Line

v = 32000 + 19550 x 0.86 + #888 x 0.72 = 32000 + 10800 + 3520 =

= 52320 lb.

 

 

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y I . 0 0 . « 0% I I . L 10 o < . 0 0 - 0 . I a - I L I 01.4 0 L- 0 0 . .V .a . I I: . TekI‘ .4 Ifil-IIW’... 0 . p * . 4 -.I 0 Lf -0 4 A. ..f H O I a H
b. 0 0 0 fine... . 0 v + 0 . n H 4. 0 0 TLW-Iw u 0 . I c T .0 0 Q L 1 % .4 . u 0 0 0 o GIG-#144114. 0 0. L. H A. n . C. I .1 . ... .. n 9
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a Q- .91 . .0 n. . v o I g 0. L . . . . ..--0 . o + 0 0- 0 0 a 0 I - . o f n9 0 o 4 .0- . 1% f . 0 1*. 0 1.0 LT .1 0 0| . .W 0 I4 . . Ir- 4- 0 e 0. 1. q . v o .
IL—r 1 I L1 14 I I I.—1 I 4 _ .11 I . 1 1.1 1 L14 q 1111.10.11 4 1.1101110. .4
0 0-. . 4N. . . I . u . . h 0-. . . .f r . - — . I . I. 0 . - . A . I .. 0 . 0.’ 0..-. .4. P 0 . 0.71” I 0- .0. 1,4 ..I I. . L 0 I - .
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. I c . I . u . W . . . . L . 0 a . o 1 8 n . 0. 0 n w u . . + 0h II 71 0 I 0 0. a. a - 1f. . f I . .0 r1 p 0 I o
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L . . . . . . . I . + 4. . -.- . . I . . L. . . . .
e . I . v L . . . + I _ + I _ F 4 . I . .. o . o. I .
n I . u d n v 0 n o 0 L . + 0 0 . L - I v 0 w . -0. I .
Y 1 1.111'1101-"1¢_ - #1 II 0‘111101I4I-‘I‘IIL
v 0 c 4. I .v . I w 0 0. . u L 0 H e p 9 .0 v v 0. . 0 a 0 I y
. . .
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r 0 I I - I H n 0 I I 0 ... .- n I
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. . . . . P . H .. . I- . .. . - . ..fi . - .
.
a I . e I + W 0 I 0. a 0 . . w . I
l O I O * I * 0 l‘ ¢ ¢ . ‘ 0 I L .0 0. .10 .H I O I a
. . . + . L . I . . L. L . . . . i U ._ . . . . L
VII-011011.11! W. I 0 .I‘OIILVI‘I *1! 101 01 L 1110.1" 4 -0 1110.11.11.11- A
- L _. .. «.... .......L ...T -....L
I . I . o . I . . . L . o . . I . .. « 0 . . L . . . . . . . . .
. . . L . . . . . . . . . - . . . I . . ... I I . . - .. I . I . . .. . . .
. . 0 . o I ' I W o .0 0 I r 0 0 -0 . 0 0. Q 1 o 0 . - LY . 0 0 w I . I A
«k . 4 IF J 1. J1 1‘1 11 J. 4 1.1 I.“ r _ 4.]
. o 0 . 0 . 0 i 0 0 0.]. o .9 . 0. 01 .0 0 10 h 0 I I. . a. . . 0 ~ % I . 0
,0 l . .0 I. 1h 0 0110 0.1 0 ..- 0 W .9 f 0 n w... 0 .014 Lm 1' fiI 0 .0 o . .W. . . - 0
. 0 I 0 .v 0 a 9 0 #10. 0-, m0 T1011. w 011% I 0 .0 . 0 . o 0 .4 . .fl '- .0 .
. 101.0 7 .4. 4 0 V. 14 0.! g ‘I 01 0 Q. - V 0 k w- 0 L0 0 . .I. 0 o I 0 a o o o 0
.1111 1 W1 P 0 1. I» I .1 - 111+ I1 III II.1011.|10|.110101L
. 0 V 0 1. .7 ¢ ¢ 0 a 0 0 0 IL 10 0 v1 f 0 01.0 0 u 0 .- 0 0 . e . 0 1H . 0 o
. .0 - .. q .0 . 0 . . I -0 L7 0 .0 0 . + . 0 a -.. L .I . I o i . . k . .-
- .
- 0.0 . 0 I . o 0 o . a 0 0 . 0 4 . 0 0 0 w o . f . 0 . . 0 0 . I
0 0. . Lv - I I ... . 0 0 0 0 a c o w . . . o L I r1 . 0 0 o . a . v .
e L. r r

 

 

 

 

191

The vertical component of the tension in the cables between the

support and the 1/4 point of the beam:

 

 

 

 

 

 

 

 

. I 15" I2)“!
. - 1__......._
el\‘ ‘
7:%:;f;I I e. e;
. I ’
I}2.51‘12.5 50' 12.R' 12-‘LI
Tan 9. = L39= 16.6b7 9. = 86° 31.1
Cos e. = 0.05989
003 e. x Pt = 0.05989 3: 1250000 = 71900 11).
R
Tan 62 = li%= 10 ez= 814° 17' 21"

C03 at = 0.09951

Cos 6% x Pt = 0.09951 x 1250000 = 12N500 1b.

 

_ vmax H
fSh - I t
O
vm = 61200 1b. (See page 190)
u
Io = 815000 in (See page 116)
t = 10 in. (See page 18 )

H=10x26x13+10x36x31=33so+111so=1115601n3

= > _

815000 J: 10 0""

 

 

 

192

One Hundred Twenty-five Foot Span Length
Block #6

Dead load shear at the 1/2 point of the beam:

Dead load shear at the support:

Live load shear at the 1/2 point of the beam:

19200 19200 hsoo
11v 11v

L\\\L\\4
-5 o388 276

 

 

 

 

 

 

 

 

 

Influence Line

( 0 = =:

16000(1 + 5%6; 16000 x 1.2 19200 1b.
( O = :

u000(1 + 526; N000 x 1.2 hsoo 1b.

V = 19200(0.5 + 0.388) + 4800 x 0.276 = 17080 + 1325 = 18905 lb.

 

 

19h

Live load shear at the 1/8 point of the beam:

19200 13:00 usoo
1 ' 1hfl

 

\ 1

-875 .73 -58
.125

 

 

 

15-625" sh.375'

7F_—_—

125'

 

 

 

Influence Line
v = 19200(o.875 + 0.73) + usoo x 0.58% = 30800 + 2800 = 33000 1b.

Live load shear at the support:

32000 19200 usoo
1M' 1M'

 

\\

1 .888.77é

 

 

 

 

L; 125'
reef

_J

Influence Line

V = 32000 + 19200 x 0.888 + “800 x 0.776 = 32000 + 17070 + 3725

= 53795 1b-

     

 

 

 

 

 

 

   
      

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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A T . -. . v . .-. . . . . A A . 7v . . W. . . e r .. . .. #1.. A.. . _ v . u.
. .
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.<.. ..s.
J.
. . .
w...
. .7 .. H

 

 

 

 

.

 

A
.
-P._.eW_.
, . .
.
l
V
W
.

 

‘,‘;\y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mvuwxfl .w. Hummwwfl. ..H...m. .mWJ
9.... . .. .v.. ....n -ox..Mv..... .
..n . . T. . . v... ..- VA . v w; .-.
‘l v .. 7' 1 vA
. .. . .. ... .....t... .. VH.. .... ... .v.
. . .. o. . . . e. .4. .... ... ...¢. . . o. . L...Y a

 

 

 

 

 

 

 

 

 

196

The vertical component of the tension in the cables between the

support and the I/H point of the beam:

 

 

 

 

 

 

 

 

 

 

 

 

 

1.. 12“"
9 .'
L 9| 1 f 9' 9;
15.62.5' 1555' 52.5' 1562.5” «5.1.25'
- "1
Tan 9. = €741: 20.833 e. = 87° 15' 7“
C03 9| = 0.0‘+79o
Cos a, x Pt = 0.014796 1: 1250000 = 60000 lb.
Tan 62 = gig—‘5- = 12.5 6.: 85° 25' 31+"
003 5; = 0.07976
C03 92 x Pt = 0.0797e 1: 1250000 = 99600 1b.
f = vmax H
“h Io t
Vm = 68330 lb. (See page 195)
H = lMSbO in3 (See page 191)
I o = 815000 in“ (See page 116)
t = 10 in. (See page 18 )

f =68510x11+560= 2 ..
311 515000 I 10 122 lb/in 0 K

 

 

 

 

APPENDIX F

.1311 S IONS

1

BLOCK DI

 

 

1"

10

Block #2

 

3 Vb." ‘ 1 ..

 

0)

$5
SI
Mai

A

 

 

 

 

18“

 

 

 

 

 

 

 

 

 

 

 

01
A”;
Block #3
Ch
20" P‘ r*__;u2L_.1
\
l i '3
'—-—"__—W
6%: 672." 6% 4:
' V O
ru :
“i

 

 

 

 

 

 

‘

 

 

 

 

 

 

 

 

uZ/I 9

 

19a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2” 1 2' Block #14 18" .
l
I 7 . ’
‘4
1 A
gig/1+" 7n 8 3L"
'51 135'
Block #5
l 30“ l 18"
L J '
°§
L 11 ‘8" 11"
E

 

 

 

 

 

 

 

 

 

 

 

199

 

200

Block #6

36"

18"

 

 

 

 

 

 

3..

 

Ho:

mm:

Hos

 

 

 

 

 

 

 

 

 

 

 

L13" 10" 13'

 

 

 

 

 

 

 

201

Blocks for the bending points of the cables:

For Block #2 from 30 feet to “0 feet:

16"

F——‘|

 

H
\D
5 H14“ '8 llf' 3 3/‘4 \rs

_—

 

 

u3;T 3

 

Block“
I 16" .
£2
3 32“" 8 1/2m 33M =
_-
\N
N

 

 

 

 

Block #9

15*

f‘fl

 

Bound to

 

u27114 ___-

+—

prevent
chipping

 

 

 

1/8 point of beam

 

fly—*1

 

Round to
prevent
chipping

 

 

 

 

1]“ point of beam

 

 

 

Block #10

Block #Ll

 

_J____ Round to
Prevent
- Chipping
E. -
m
3

1/8 Point of the beam

 

 

15"

Round to
"Prevent

: Chipping

l/h point of the beam

 

 

 

 

202

 

 

 

 

203

For Block #M from 55 feet to 75 feet:

 

 

 

 

 

 

 

 

 

 

 

 

 

an 172" | ’ 15" _
KN
n1
8
8%; r9836 35
L Round to
‘———1 Prevent
w Chipping
O.
--%r
‘1
3
1/8 point of the beam
Block #12

2h 172" 1 15"
_ I I

 

 

MM

u6h

 

 

 

 

 

Chipping

 

 

 

w I I!
eyzg_7 __8§a \2
'{ [ L Round to
-——-J . ti Prevent

 

l/h point of the beam

 

Block # '3

 

 

 

 

 

For Black #5 from 75 feet to 100 feet:

 

 

 

 

 

 

 

 

 

 

 

20h
r___3‘i"___.‘ r._15_"__l
4
t
A)
8
S
11"J 8"JA 11" a
Bound to
...—4% .4494 Prevent.
H Chipping
__103
3

 

 

 

 

 

Block #1u

" 30" |

 

lan‘ 8"}: ll"

 

 

 

 

:06?!

8.

 

Block #15

ll _1
Wing.

 

 

1/8 point of the beam

r—‘ifl

 

Round to

 

 

Prevent

 

Chipping

 

 

l/H point of the beam

 

 

 

 

 

For Block #0 from 100 feet to 125 feet:
.9 35"

 

205

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

_. 15" I
A
an
H
I
N
13' .1410. L 13" “3
Round to
‘ Prevent
H Chipping
H
I
i
A
H
‘2
1/8 point of beam
Block #16
36" _l 15"
g.
3
N
n1
3
1 ".4: 10“! 13"
"3’
:1 c—JL g”
5:: Round to
Prevent

 

 

 

 

Block #17

Chipping
1/h point of beam

 

For Block #0 from 100 feet to 125 feet:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

205

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.9 36" n 15" l
A
\J'l
..I
8
u n N
13" _L10 1} r3
Round to
____ir_ \ Prevent
H Chipping
...:
I
H
o
8
1/8 point of beam
Block #16
.9 36" A [.991
g.
N
m
3
l ”9 10"L 1'5"
m
a
.:l 1:9r.
E}; Round to
Prevent
Chipping
l/H point of beam
Block #17

 

 

we Meoflm gee: eeee en es

206

maa aooam

 

15 ilk"

 

 

END BLOCKS

 

 

..Lb ‘5/’+"

 

 

 

 

1. 21 111+"

 

 

 

 

 

 

__10"

u

-5

 

how

20 u

20"

 

 

 

 

5 1116" holes

 

 

 

 

20"

10",:.10"

w

 

 

 

 

_ l
...—“W.___ C: ‘ +
'/' l

 

 

 

 

.

6+4.

 

 

 

N L‘ 10014
‘L‘W F

 

 

GHQ
—5, -+—

 

 

 

 

 

Block #20

 

207

To be used with Block #3

 

 

 

 

:* Moon gawk comm on 08

Hm.‘ 03on

1+ 5/8" holes

 

 

2n 1/2'

 

 

_
19
r.

.

2n 1/2"

 

 

 

W

 

 

 

 

 

 

 

 

7...... ......

 

   

209

mi Mooam A»: cows 09 on.

mm... 988

  

 

28 3/‘4-

 

 

 

9 31 If“

 

 

 

     

 

 

 

 

._ 37 1/h" . 31+ 1W -
I I
973'; 972; eye"
4%”
414'

 

 

 

("II II 1' I
II], 7M
ELM/.17, I: I. l'

 

 

112"

Block #23

To‘be used with Block #6

   

 

 

BIBLIOGRAPHY

Abeles, P.W.T11e Irincip_le_s and Pragtiva g: {gesggessed Concrete,
‘flew York: Frederick.Ungar Publishing C0., 112 pages.
John A. Roebling' 3 Sons Co. Rgehllnf gtrgnd and Eittings §9_r Pre

stressed Concngtg, Catalog No. T-9lb.
”agnel, G. Pr “est'_e eased Concrete, London: Publications Limited,
1050. 300 page

”Proceedings of the First United States Conference on Prestressed
Concrete", Hassachusetts Institute of Technology, Cambridge, Mass.,
Aug. 1“ to 10, 1951.

 

 

\l .I
11111)}

. J.Il\l.l’\|l/\Il|l\|j||l \ll \‘

 

 

 

 

A

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