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An Analysis
of the

Delta Mills Bridge

A Thesis Submitted to

The Faculty of
MICHIGAN STATE COLLEGE
of

AGRICULTURE AND APPLIED SCIENCE

by

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Charles BE Gates

A Candidate for the Degree of

Bachelor of Science

June 1943

THESIS

 

 

 

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This thesis is respectfully dedicated to my
parents, without whose help and consideration it

would never have been written.

Part No.

II

III

IV

VI

VII

VIII

IX

XVII

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TABLE OF CONTENTS

 

Title
Introduction
Check or Stringer Size and Spacing
Check of Floor Beams
Live Load on the Trusses
Dead Loads on the Trusses
Dead Load Stresses
Live Load Stresses
Design of Truss Members
Check of Lateral Bars and Spacing
Lateral Forces
TOp Chord Pins
Top Chord Pin Plates
Pin Design in the Bottom.Chords
Hanger Pin Plates
Upper Transverse Connection Plates
Diagonal Connection Plates
Check of Pedestals and Shoes
Conclusion
Bibliography

General Drawing of Structure

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INTRODUCTION

 

The bridge to be analyzed in this report is an old highway bridge
of the through truss type. It was put up in 1891, by the R. L.
Wheaton.Company of Chicago, Illinois. It is located at Delta Mills,
Michigan, which is a small community located about six miles to the
Northwest of Lansing. The greater part of the load travelling over
the bridge is composed of passenger vehicle traffic, for which the
structure was undoubtedly designed. However, since the World War II,
an increasing amount of heavier traffic has been using it. Gravel
and equipment trucks have been re-routed over this road, and the

structure gives evidence of failing.

It is a two—Span bridge, and has a span length of 120 feet. The
span is divided into eight panels of fifteen feet each, and the height
is also fifteen feet. The roadway is twelve feet wide, and the
total structure is fourteen feet wide. It is a pin—connected truss,
and is supported on rock-masonary piers founded in the Grand River.
The bridge has not been adequately protected against the weathering
effects, and is rusting and Scaling badly.

The object of this thesis is to determine whether—or—not the
bridge will safely carry a loading as prescribed by the American
Association of State Highway Officials as a H-lO loading. This load-
ing is caused by a train of trucks of the 10 ton type or an equiva-
lent loading. It will be determined just what loading is allowable
if this is not safe. Recommendations will also be made for contin—

uing the useful life of the structure, to prevent further failure.

 

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Part I

CHECK OF_STRINGER SIZE AND SPACING

In this part of the check analysis, it is to be determined
whether or not the stringers are of adequate size and spacing to
resist the forces set up by the K—10 loading as prescribed by the

American Association of State Highway Officials' Specifications.

1. Wheel Load. The stringers are 7 x 3—3/L @ 17.5# per foot sec—
tions, and are Spaced 2 feet center-to-center. The magnitude of the
wheel load is equal to 8,000# per foot of stringer(20,000# x 0.4),
which is reduced by the factor 2/6 because of the distribution to
adjacent stringers. The wheel load is than 2/6 x 8,000 or 2670# per
foot.

2. Impact. The formula that is used to determine the impact ratio
is:

Impact ratio = .50
L - 125

 

where (L) is the value ofdthe length of the stringer, in feet. The
impact ratio is never taken less than 30%(AASHO). Substituting the
value of 15 feet for (L), the ratio comes out 35.714%.

3. Dead Load. The dead load consists of the weight of the stringer
plus the weight of 3 inches of concrete roadway slab. A weight of
50# per foot is arbitrarily assumed as the weight of the stringer,
and from this the needed size of stringer will be determined, which
will later be corrected for the actual weight. The 3 inches of con—
crete is equivalent to a load of 37.5# per square foot(150 x %,
assuming that the concrete weighs 150# per cubic feet). This makes

the total dead load on the stringer 125# per foot.

4. Moments. The bending moment due to the concentrated wheel load

is:

B.M. flit—2.x 7.5

2
10,000 Foot—pounds

The bending moment due to impact equals 10,000 x 0.35714, or 3571.4

foot-pounds. The bending moment due to the dead load is 125 x'% x 7.52

which equals 3515.625 foot—pounds. The stringer must be designed to
resist the total of these three moments, which is 17,087 foot—pounds.

5. Section Modulus. The formula used for determining the modulus
is:

Total.Bending Moment
Allowable stress in Steel (18,000#/in27

17,087 x 12
18,000

Section modulus

 

11.38 Inches3 .
The actual section modulus of the 7 I 17.5 is 11.1 inchesB, which

is not sufficient for the assumed conditions. However, by inserting
the correct weight of 17.5# per foot in place of the assumed 50# per
foot, the following results are obtained:

Dead load moment = 2601.56 Foot-pounds
Total moment = 16,172.963 Foot—pounds
Section modulus = 10.78 Inches3

Since the actual section modulus exceeds the modulus required, the
stringer is satisfactory at this point.

6. Web Area. The compression flange is supported laterally by the
concrete so that the full unit stress is considered to be developed.
The maximum shear developed(live load and impact) is equal to

( 8,000 . 2,000 x 1%— ) x .31. x 1.357

The dead load shear is equal to 694 pounds = (37.5 x 2 + 17.5 ) x 7%.
Since the web area necessary is found by dividing the total shear by
the allowable shearing stress(ll,000#/in2), the required web area is
found to be 0.401 square inches. The available web area is 5-3/8 x
0.345 or 1.854 square inches, which is more than adequate for the
loading.

7. Summary. In summary, the stringers are of sufficient size and
spacing to withstand the loading of a 10 ton truck or the equivalent

Part II
CHECK 0F FLOOR.BEAMS
The floor beams are built—up members constructed as shown in
the sketch. They must be checked for section modulus, web area,

and adequacy of connections and Splice plates.

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1. Depth. The minimum depth to be used is equal to l/25 the length
of the floor beam or 1/25 x 14 x 12 = 6.72 inches (AASHO).
2. Loading. The maximum concentration on the floor beams at the

middle stringer comes frem the 10 ton truck when one 8,000# wheel

load is directly over the floor beam and a 2,000# load 14 feet away.

The concentration equals 8,000# plus 2,000# x 1/15 or 8, 133.33 pOunds.
3. Bending Moment. The maximum bending moment occurs at this point

‘Ihen similar loads are placed as near it as the axle Spacing of 6

feet and a clearance of 3 feet will permit. The result is a central

loading, with only one train of trucks on the span.

633.333”3 6’ _ 8/33.333#

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The bending moment at the critical section is the same as if the
wheel loads were applied to the floor beams directly, and they need
not be brought down through the stringers. The bending moment due
to the live load is equal to 32,533.33 foot—pounds(8133.33 x 4).
The bending moment due to impact is obtained by using the following

formula: _(_2000 _ L.)

(1600 +1..)
of the floor beam. The ratio is never taken less than 30%(AASHO),

Impact ratio =._%_.x ; (L) being the length
and equals 34.562%. The impact moment is 32,533.33 x 0.34562 or
11,244.170 foot—pounds.

Assuming that the floor beam weighs 200# per foot( to determine the
approximate size of floor beam needed ), the other portion of the
dead load on the floor beam is due to the 3 inches of concrete road-
way. The equivalent dead load is

(l%<3___+ 8.75) x15 1. 200

= 893.75 Pounds

The bending moment due to the dead load is found as followsz

Bending moment = 7(—§—9—3—é—Zi—)x l4 - ( 893.73 ) x g.
- 21,896.875 Foot—pounds
The floor beam must be designed to resist the total bending moment
of 65,674.377 foot—pounds.

4. Section Modulus. The section modulus required to resist the
above moment is 43.783 inchesB, which requires a l2x5§ I @ 40.8#

per foot section. Since the actual beam is a built up section, the

section modulus is calculated according to the formula:

 

Section modulus = i 3 where (I) is the moment of inertia
of the cross section, and (c) is the distance from the neutral axis
to the extreme fiber of the member in inches. The moment of inertia
of the web plate is equal to 1/12 bd3 and is 182.6 inches4. The mom—
ent of inertia of the flange angles is obtained by use of the transfer
formula with the distance between axes being 8.01 inches, and is equal
to 309.802 incheSA. The distance (0) is equal to 9 inches. The
section modulus available then equals (182.25 + 309.802) e 9 or
54.672 inchesB, which is greater than the required modulus. The
floor beam is thus safely designed at the center. Since the bending

moment becomes smaller as the end of the beam is approached, the

section may be reduced and still be safely designed.
5. Rivet Spacing. The first check will be made of the rivets in

the flange angles. The object is to determine the pitches of 7/8"
rivets in a built—up floor beam with the following loadings: a uni-
form load of 729.291# per foot, two truck wheel loads of 8,133.33#,

and a 36% impact. The wheel loads are spaced 6 feet apart.

The value of one rivet in bearing in the 3/8 inch plate is 7875#,

(7/8 x 3/8 x 24,000-assuming an allowable stress of 24000#/in2).

The distance between lines of rivets is 11.0 inches for the first
three feet of the floor beam. The decrease in the dead load shear

per panel is

 

729.291 x i; = 670 Pounds

The decrease in the live load shear per panel is

8,133.33 x 1.36 x 2 x 11
14 x 12

= 1450 Pounds

The vertical component of the live load is

81133033 X 1.36 x 2 XL;— x 623.7:
14 12 12

- 850 Pounds

 

The gross moment of inertia of the beam at a point 3 feet out from
the end of the beam is equal to the moment of inertia of the web
plate plus that of the flange angles. The moment of inertia of the
web is 1/12 bd3, which equals 95.270 inchesA, and that of the flange

is
2( 1.1 + 1.19 x 6.262 ) . 95.466 InchesA, making the total

moment of inertia 190.736 inches4.
The horizontal shear is equal to 493.2# and is obtained from the

formulas

Bendingqfloment(tan a + tan b)

Shear increment ' Total shear ‘ Distance between Rivet Lines

where (a) and (b) are the angles of inclination of the bottom and top

flanges with the horizontal. The value of the horizontal shear is

Shear =__§hear X Flange Moment of inertia
Dist. between Rivet Lines x Gross I

(umw—zwjx9i@6
11 x 190.736

493.2 Pounds

The resultant shear is equal to the sum of the squares of the com—
ponent parts and equals 976 pounds. The pitch then required is

found by dividing the allowable bearing by this resultant shear and
equals 8 inches. The maximum pitch used is 4% inches for this type
of member, and since the actual Spacing is 3% inches, it is more than
adequate.

The spacing is investigated at two other points, at 4% feet out from

the end and at the mid-point, and the results are

Distance from Required Actual

End of Floor Beam Spacing Spacing
3 Feet 4.5 Inches 3.5 Inches

4 " 4.5 " 4-0 "

7 " 4-5 " 4-5 "

The Spacing in the bottom flanges is the same as that in the top
flange, and both are sufficient for the loading.

6. Web Splice Plate. The plate is used at the mid-point of the
floor beam to transmit the load from one web section to the other.

The dimensions and rivet spacings are as shown in the sketch below.

 

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The gross moment of inertia of the plate is 95.5 inches4 as found
by using the formula 1/12 bd3. The moment of inertia of the holes

in the plate is equal to 53.5 incheSA. The moment resisted by the web

plate is equal to fI/c, where (f) is the allowable stress(16,000#/in2),
(I) is the net moment of inertia, and (c) is the distance from the

neutral axis to the extreme fiber.

 

Resisting moment = 16-003 2553.5

- 180,000 Inch—pounds
The eccentricity is equal to the resisting moment divided by the
summation of the vertical and horizontal distances to the rivets,
squared. The summation of the vertical distances squared is 224,
and of the horizontal distances 14 x 12 or 14. The eccentricity is
then 180,000 9 238 or 756. The horizontal component of the stress
in the critical rivet is 756 x 6, which equals 4,536 pounds. The

vertical component is

 

756 X 4.5 + 1150207 a 3753 Pounds, and the resultant
stress is 5900 pounds, which is the actual stress that the most
critically stressed rivet must resist. The allowable stress that the
rivet is capable of withstanding is 7,031 pounds, assuming an al-

lowable stress of 30,000# per square inch in bearing for double

shear.

The web Splice is of sufficient design to withstand the forces
set up,

7. End Connection. The end connection is with the use of two
3 x 3 x % angle section, 12 inches long, which contain five rivets,

Spaced as shown; 2% inches between rivets and 1% inches for edge

margin. The check is made to determine the Spacing needed for the

floor beam to resist the end reaction, using unit stresses of 11,000
in shear and 24,000 bearing(double shear), and using 7/8 rivets.

9.

The maximum end reaction occurs when the truck load is placed as

close to the left reaction or left end as is possib1e(one foot).

 

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To obtain the magnitude of the reaction, take moments about the
right end or

Left reaction = 16,480 Pounds

8395.8 x 13 + 262.5(11 +91 + 8395.8 x 7 + 262.5 x3 + 662.6398
14

The effective bearing length of one pair of angles is equal to 2 x 3
or 6 inches.. The required thickness is found by dividing the end
reaction by the product of the bearing length and the allowable
stress in bearing, and is equal to 0.115 inches. The actual flange
or angle thickness is 0.25 inches or more than double the required
0.115 inches. ~The number of rivets required in the web legs of the
angles is equal to the end reaction divided by the allowable stress
in double shear, which is 5400 pounds, making a total of 3.03 needed
in each angle. The actual number present is 5, which is an excess
of about 2 rivets per angle.

The number of rivets required in the stiffener plate, as determined
by bearing in the 3/8 inch plate, is 3000 x 3/8 x '7/8 or 7875#
divided into the end reaction. This number is 2.14, and Since 3.0

are used, the plate is safe.

10.

This concludes the check analysis of the floor beams, and the struct—
ure is so far capable of withstanding the loading with a fairly high

factor of safety.

Part III
LIVE LOAD ON THE TRUSSES

The live load for the trusses per lane is 320 pounds per linear
foot of lane plus a single concentrated load of 13,000 pounds(AASHO).

l. Lanes. The number of lanes is 1.33 or 1 + 3/9.

2. Load per Panel. The live load per panel per truss is composed

of the uniform loading and the concentrated load. The uniform load is

160 x 1.33 x 15_ = 3192 Pounds
2

The concentrated load produces a load per panel per truss of
13,000 x 1.33/2 or 8645 pounds.

3. Impact Load. The impact ratio for the chord members is equal

to

 

1 2000 — 120 x 1.33/2 _
3 3‘(1600 + 1200 x 1.33/2 “ 26'7%

Part IV

 

DEAD LOAD ON THE TRUSSES

The dead load brought down from each floor beam is equal to
( 693.75 + 22) x 7 = 5012 Pounds

11.

The weight of each truss, including bracing, per foot is approxe
imately equal to the length in feet plus one-ninth the total load

per foot from the floor system. The weight per panel is

5012 J, M x 1.267
8

2
l 0 x 15 + 9

 

— 2550 Pounds.

This makes the total dead load per panel 2550 t 5012 or 7562 pounds.

ar _
DEAD LOAD STRESSES
In this section, the computations will be made for a few members
to illustrate the method of procedure, and the remaining values will
be placed in tabular form.

1. Load Distribution. The dead load is assumed to be distributed
between the top and bottom chords at a one-to-two ratio. The top
chord is considered to take one-third of the total load per panel,
and the bottom chord takes two-thirds. The magnitudes are 2520#
per panel in the top, and 5040# per panel in the bottom chord.

2. Member (a). The vertical component of the stress in this
member is equal to the reaction at the left end, since there is no
other vertical member at this point. The reaction is equal to
7/2 x 5040 + 2520 or 26,460 pounds. Since the truss is one that

is the same height as panel length, the angle at the supports or
panel points is 45 degrees. The stress in (a) is then 26460 x 1.414,

which is 37,500 pounds, compression.

3. Member (b). The stress in (b) is equal to the horizontal
component of the stress in member (a), because these two members
are the only members acting at this point that are capable of taking
a horizontal loading. The stress then equals 24,460#, tension.

4. Member (c). Since the only load acting on the member (0) is
the panel load of 5040 pounds, the stress in this member is 5040
pounds, tension.

5. For the stresses in the remaining members, refer to the

accompanying table of values on the following page.

Notezr Member (m) is made up of two cylindrical rods with an
unbraced length of 15 feet, and thus it cannot take any compression
whatsoever. To take compression, the members (n) are provided and
are called counters. These members do not directly take the compress-
ion, however. The force that would produce compression in the member
(m) produces a tensile stress in the counters, and they are so
designed to take all of this stress so that none of it will act on

member (m).

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Table
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DEAD LOAD STRESSES

 

MAGNITUDE

0F

STRESS

37§500 Pounds

26,460

5,040
45,360
26,700
26,460
13,860
16,000
45,360
56,700
60,480

7,240

5,350
66,600

9,900
96,320

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TYPE
OF

STRESS
Compression
Tension
Tension
Compression
Tension
Tension
Compression
Tension
Tension
Compression
Compression
Compression
Tension
Tension
Compression

Compression

Part VI

LIVE LOAD STRESSES

The uniform live load of 3192 pounds per panel is treated the
same as the dead load in calculating the stresses caused by it.
One-third or 1064 pounds per panel is taken as acting on the top
chord, and two-thirds or 2128 pounds on the bottom chord. The
stresses caused by the moving live load of 8645 pounds will be
determined by the use of load influence lines.

1. Uniform Live Load. Since the method of analysis of the stresses
produced by this load is the same as for the dead load, it will not
be necessary to repeat the procedure. The values of the stresses
can be found in the table on the following page.

2. Moving Live Load. The method of influence load lines consists
of placing the load of unity at various panel points and determining
the stress resulting from this position in the member being invest-
igated. The stress is then multiplied by the actual value of the
moving load, 8645 pounds in our case, to determine the actual
maximumvstress.

The influence line for member (a) is a sloping line going from
zero at the right end to 12,224 pounds at the left end, assuming
that load enters from the right.

This procedure is continued for each member in the truss and
points of maximum stress and reversal of stress are noted. The
maximum stress is that stress that must be used to determine the
design stresses. The influence diagrams for the various members

are as indicated in the following sketches and the stresses are
indicated in the table.

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MAGNITUDE

OF

STRESS

15,800 Pounds

11,172
2,128
19,152
11,280
11,172
5,852
6,780
19,152
23,942
24,536
22,292
841
33,850
4,308
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TYPE
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Tension
Tension
Compression
Tension
Tension
Compression
Tension
Tension
Compression
Compression
Compression
Tension
Tension

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Table

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MOVING LIVE LOAD STRESSES

 

MEMBER HAXIKUM MAXIMUM

TENS LE COMPRESSIVE
STRESS STRESS

a -- 12,224 Pounds

b 8,645 Pounds --

c 8,645 " --

d -- 12,968 "

a 7,600 " 3,060 "

f 8,645 " --

8 2,160 " 5,300 "

h 9,150 " 1,530 "

1 13,000 " --

3 16,200 ” --

k -- 17,290 "

1 p-- 4,325 n

m 6, 100 ” --

n -- 22,800 "

0 17,290 ” a-

p 32,200 n 8,650 "

17.

Part VII

DESIGN OF TRUSS MEMBERS

. The total loads(dead, live, and moving live) for each individual
member are shown on the following page, and are the loads that the

members must be designed to resist.

l. Member (a). The member (a) is the end post of the truss and
must resist a load of 65,524 pounds, compression. The cross section

of the member is shown in the Sketch below, and is analyzed as follows:

 

L I! If

 

 

 

 

 

 

 

 

 

LLB—1 —_LL:J‘ a u
Coveer/c lax;-
l 1'2 # l'
834 aLI/As N 3?- Rfve)‘:
.Lac'
if ___’:q_a__ n
1412" l

 

 

 

The area of the section is 2 x 3.36 4 12 x % or 9.72 square inches,
and the eccentricity of the neutral axis from the centroidal axis
of the channels is ( 4 x 4.11 ) f 9.72 or 1.69 inches. The moment
of inertia about a horizontal axis through the center of the web
is
l c 13 2
32.3 x 2 + 0‘12 x 12 x 4 + 3 x 4.11 )
= 115.016 InchesA

The area multiplidd by the eccentricity squared equals 27.72 and
by subtracting this from the moment of inertia about the web axis,
we obtain the moment of inertia about the centroidal axis of the
section, which is 87.296 inchesA. The radius of gyration is equal

to the square root of the moment of inertia divided by the area, so

(r2) will equal 87.296 + 9.72 or 9.00 inches. According to the

18.

Table

of

 

TOTAL LOADS ON MEMBERS

 

MEMBER TENSILE COMPRESSIVE
LOAD LOAD
a -- 65,524 Pounds
b 46,277 Pounds --
c 15,793 " --
d f- 77,322 "
e 47,150 " 1,550 "
f 46,277 ” --
8 5,300 " 25,012 ”
h 50,580 " 7,600 ”
1 77,512 " --
J -- 97,932 ”
k -- 102,236 "
1 41,100 " 33,855 ”
m. 12,291 " --
n 100,450 ' 22,800 "
0 21,598 " 9,900 "
p 52,200 ” 153,864 ”

19.

Specifications the formula to be used to determine the allowable
stress in compression is governed by the slenderness ratio L/r,
where (L) is the unsupported length of the member in inches, and
(r) is the least radius of gyration. If the ratio is less than
120, the formula to be used is

Allowable stress = 15,000 - L2 , and if the ratio

4 r2
is greater than 120, the formula to use is

2
Allowable stress = 15,000 —.§L;2— .

In this case the ratio is 15 x 12 + 3 or 60, which calls for the

use of the first formula. The allowable stress is then 14,100

pounds per square inch. The safe load that the member can carry
is equal to the allowable stress per square inch times the area of

the member or
14,100 x 9.72 = 137,000 Pounds.

Since the allowable load is over twice the actual load, the member
is safe.

2. Member (b). The load on member (b) is a tensile load of

46,277 pounds. The member consists of two bars of rectangular
section 2" x l" x 15 feet, spaced 6 inches apart. The cross section-
al area is 2 x 2 x 1 - 4.0 square inches. Using an allowable

tensile stress of 18,000 pounds per square inch, the allowable load
is 72,000 pounds, and since the actual load is only 46,277 pounds,

the member is safe.

3. Member (c). The load on this member is a tensile load of
15,793 pounds. The member is composed of two 15/16 inch square

bars, Spaced 4 inches center—to-center. The area is then 2 x 0.878

or 1.756 square inches. The allowable load is 18,000 x 1.756, and

equals 31,600 pounds. The actual load is 15,793 pounds and is less
than the allowable, thus the member is safe.

4. Member (d). The member (d) is a top chord member and is
composed of the same section as the end post member. The allowable
compressive load on the member is 137,000 pounds, as found in the
analysis of the end post, which is greater than the actual load of
77,322 pounds, making the member adequate in design.

5. Member (e). This member has both a tensile and compressive
load acting on it, and is known as a reversal—of—stress member.
Specifications require that for members subject to live loads pro-
ducing alternate tensile and compressive stress: to the net total
compressive and tensile stresses add 50% of the smaller of the two,
and proportion the member to resist either of the increasing stresses
resulting therefrom. The net total tensile load is 47,130 pounds
and the net compressive load is 1,550 pounds. The design loads will
then be 47,895 pounds and 2,295 pounds, reSpectively.

The member (e) is composed of two bars of rectangular cross section
2" x 7/8", and the total area is 3.5 square inches. The area required
to resist the tensile load is 47,895 r 18,000 or 2.70 square inches.
This area is less than the actual, therefore the member will resist

the tensile stress.

The member is also safe in compression, because the small load of
1,530 pounds is more than taken by the area of 3.5 square inches, at

an allowable stress in compression of 11,720 pounds per square inch.

6. Member (f). This member is composed of two rectangular bars,
2" x 1" x 15 feet, with a center-to—center Spacing of 8 inches. The

total area is then 2 x 2 x l or 4.0 square inches. The load on the

21.

member is a tensile load of 46,277 pounds, which is less than the
allowable load of 72,000 pounds, making the member safe in tension.
7. Member (g). This member is also a member that has a reversal
of stress and will be analyzed the same as member (e). The design
loads are 27,242 pounds in compression, and 7,950 pounds in tension.

The cross section of the member is shown in the sketch below.

54 3122164/‘7
Lacin q
~—E——— -E~

 

 

 

 

 

The area is 4 x 1.19 or 4.76 square inches, and the moment of iner-
tia about the centroidal axis is

Moment of inertia - 1.56 t z. x 1.19 x 3.512
a 60.36 Inches4
The radius of gyration is equal to the square root of 60.36 + 4.76

or 3.57 inches. Since the length of the member is 15 feet, the ratio
of length to radius of gyration is 50.4, and the allowable compressive

stress is

15,000 _ (15 x 12)2
4 x 12.7

Allowable stress

14,364 Pounds per square inch

The allowable compressive load is then 14,364 x 4.76 or 68,500 pounds
which is greater than the actual load of 27,242 pounds.

The net area in tension is 4.76 — 4 x 1/4 x 7/8 or 3.885 square
inches. Using 18,000 as the allowable tensile stress, the allowable
load in tension is 70,000 pounds, which greatly exceeds the actual

load of 7,950 pounds. The member is safe for both loadings.

22.

8. Member (h). This member is a member carrying a reversal of
stress. The design loads are 33,443 pounds tension, and 12,973 pounds
compression. The member consists of two bars of square cross section,
1" x l" x 21.2 feet long, and has an area of 2 square inches in
tension. The allowable load in tension is 2 x 18,000 or 36,000 pounds
and the actual load is 33,443 pounds. The member is safe by a small
margin in tension.

The radius of gyration squared is equal to

r2 = 0.0835 x 2 L 2:; l x 6.25
2 x l

- 6.334 Inchesz.
The ratio of the length to the radius of gyration is less than 120,

so the allowable compressive stress is

 

Allowable stress - 15 000 - (21.2 X 12)
-'~ ’ 4 x 6.334

- 12,430 Pounds per square inch
The allowable compressive load is 12,430 x 2.00 or 24,860 pounds,
which is greater than the actual load of 12,973 pounds. The member

is safe for the tensile and compressive stresses.

9. Member (i). The load on member (i) is a tensile load and is

equal to 77,512 pounds. The member consists of two bars of rectang—
ular section, 2%" x 1" x 15 feet long. The area is equal to 5.00
square inches, and the allowable load in tension is 90,000 pounds.
This load exceeds the actual load and thus the member is safe.

10. Member (j). The member (3) is of the same cross section as

. the end post member, and has loads of 97,932 pounds compression,
and or 0 pounds tension. The allowable compressive load is 137,000
pounds, which exceeds the actual load of 97,932 pounds. The allow-

230

able tensile load is equal to the net area(9.72 - 4 x 1/4 x 7/8) or
8.845 square inches multiplied by the allowable stress 18,000 pounds
per square inch, and is greatly in excess of the actual load of
9,150 pounds. The member is adequate for the tensile and compressive
loads.

11. Member (k). The actual load on this member is 102,236 pounds,

and the member is of the same cross section as the end post. The
allowable compressive load that the member is capable of taking
is 137,000 pounds and is in excess of the actual load.

12. Member (1). This member is of the same cross section and
length as member (g), and carries a load of 33,855 pounds compression,

and 41,100 pounds tension. The allowable loads are 68,500 pounds
compression, and 70,000 pounds tension. The member is safe.
13. Member (m). The member (m) is stressed in tension only

because of its large slenderness ratio. The load is 12,291 pounds.

The area in tension is equal to 2 x 0.7854 square inches(for 1"
square bars) and is 1.5708 square inches. The allowable load in
tension is then 27,300 pounds, which is in excess of the actual load
of 12,291 pounds. The member is then safe in tension.

14. Member (n). The loads on the member (n) 118,850 pounds
tension, and 34,200 pounds compression. The member is composed of
two square bars, 1% x 1% x 21.2 feet long. The area in tension is
2 x 1.5 x 1.5 or 4.5 square inches. The allowable load in tension
is 4.5 x 18,000 and is 81,000 pounds. This allowable load is less

than the actual load of l , pounds, and the member would not be

safe in tension for the H—lO loading.

The allowable stress in compression is equal to 7,230 pounds per

24.

square inch, using a distance of 2% inches between the axes of the
bars. The allowable compressive load will then be 4.5 x 7,230 and
equals 52,500 pounds. The actual load on the member is 34,200 pounds,
which is greater than the allowable load, and consequently the member
is unsafe in compression as well as tension.

15. Member (0). The loads on the member are 9,900 pounds com—
pression, and 21,598 pounds tension. The section is composed of
two rectangular bars, 2 x 3/4 x 15 feet long, Spaced 4 inches center-

to—center. The radius of gyration squared is equal to

1.2: 2L1/12x2x314342x3/4x4)
3

a 4.75 Inches2

The allowable compressive stress is

2
Allowable stress = 15,000 -.—Lyililéfi_.
1. x 4.75

- 13,300 Pounds/in?

The allowable compressive load is then~13,300 x 3.0 or 39,900 pounds.
This member is safe in compression as the allowable load exceeds the
actual load.

The allowable tensile load is 3.0 x 18,000 or 54,000 pounds, and
exceeds the actual load of 21,598 pounds, making the member safe
in tension.

16. Member (p). The loads on the member are 170,000 pounds com—

pression and 48,300 pounds tension. The Section sketch is below.

214x 37; a 7.8“

  

 

 

 

25.

The area of the section is 2.87 x 4.0 = 11.48 square inches. The

moment of inertia about the centroidal axis is
Moment of inertia - 4(2.2 4 2.87 x 4.02)
= 192.5 Inches4'

The radius of gyration squared is equal to 192.5 + 11.48 = 16.72 inches
squared, and the allowable stress in compression is 14,515 pounds
per square inch. The allowable compressive load is 14,515 x 11.48
and is 167,000 pounds. This allowable load is less than the actual
load and the member is unsafe in compression.

The net area in tension is 4 x 2.87 - 4 x 7/8 x 7/16 or 9.95
square inches. The allowable tensile load is then 9.95 x 18,000 or
179,000 pounds, which is greater than the actual load. The member

is safe in tension but unsafe in compression for the H—lO loading.

This concludes the analysis of the truss members, and except for
the members (n) and (o), the truss will safely take the loading.
The members that are overstressed are visibly overstressed, for they

are warped out of position, and other signs of failure are evident.

26.

.Part VIII

CHECK 0F LATTICE BARS AND SPACING

1. Lacing in the Verticals. The verticals are compression members
and are governed by certain specifications as regards lacing members.
The lacing bars shall be so placed that the ratio of the length to
the least radius of gyration of the flange included between their
connections shall not be over three—fourths of that of the member
as a whole. For our vertical members, the slenderness ratio is 60
for the whole member and 10 for the lacing bars. This condition is
satisfactory.

The lacing bars shall be so proportioned to resist a shearing stress
normal to the axis of the member equal to 2% of the total compressive
stress in the member. The allowable compression per lacing bar is equal
to 0.02 x 157,424 or 3150 pounds. The allowable tension per member
is 0.02 x 61,980 or 1240 pounds. Specifications require that the
minimum width of a member in which % inch rivets are used is 1% inches,
which is the width of the members actually used. The sketch below

indicates the size of members and spacings of the lattice bars.

W/dfh /21 u

. F— RivefsEI-"Dia.
Thickness -

 

 

The minimum thickness of single lattice bars is 1/40 of the length
from center-to-center of end holes and is 0.242 inches which is

slightly less than the actual thickness of 0.25 inches. ”The compress»
_ 2
ive strength is equal to (15,000 - L2/4r ) x the area of the member,

27.

and equals
( 15,000 - ——;QEZ——) x 1% x-% , where (r) for a
4 r
rectangular bar in terms of the thickness is t2/12. The compressive
strength is then equal to 3830 pounds. This quantity represents
the load that the bar is capable of withstanding, and exceeds the
actual stress of 3,150 pounds, making the member safe in compression.

The tensile strength of the lacing bar is 18,000(l% — 5/8) or
3,940 pounds, which exceeds the actual stress of 1240 pounds, so that
the member is also safe in tension.

2. Lacing the End Posts and T0p Chords. These members are also
compression members and are governed by the same principles as the
lacing in the verticals.

The actual compression per lattice bar is 0.02 x 102,236 or 2,044
pounds, and the actual tension per bar is 0.02 x 16,120 or 322,40
pounds. The minimum width is 2 inches when 5/8 inch rivets are used.

The sketch below indicates the Spacing and size of members.

[—

Wl'dfh 2 ” Rivcfs {’b/a.

II

77: ickncss g-

 

 

The minimum thickness is found the same as in the previous analysis
and is 0.375 or 3/8 of an inch, which is the actual size of member
used.

The allowable cumpresSive strength in the members is

 

1 000 — 225
( 5, 4 x 1/12 x 9/64 ) x 2 x 3/8

7,560 Pounds

Compression

28.

The allowable tensile strength is

Strength ( :2 — 0.625 ) x 3/8 x 18,000

9 g 300 Pounds
These allowable loads exceed the actual loads and indicate that the

lacing was properly designed.

iPart Ll
LATERAL FORCES

The American Association of State Highway Officials Specifications

are quite explicit and inclusive.

The wind force on the structure shall be assumed as a moving hor—
izontal load equal to 30 pounds per square foot on one and one-half
times the area of the structure as seen in elevation, including the
floor system and railings, and on one-half the area of all trusses or
girders in excess of two in the span.

The lateral force due to the moving live load and the wind pressure
against it shall be considered as acting 6 feet above the roadway
and-shall be as follows: 200 pounds per linear foot for highway
bridges.

The total assumed wind force shall not be less than 300 pounds
per linear foot in the plane of the loaded chord, and 150 pounds

per linear foot in the plane of the unloaded chord on truss spans.
The wind.pressure of 50 pounds per square foot on the unloaded

structure, applied as in the first paragraph, shall be used if it

290

produces greater stresses than the combined wind and lateral forces
of the first two paragraphs.

1. Wind load. The area of the truss as seen in elevation is

21.6 x 1 c 53.2 Square inches End posts
90 x 10/12 = 75 Square inches T0p chord
(15 x l)2 = 30 Square inches Panel braces
7/12 x 120 = 70 Square inches Stringers

9/12 x 15 x 4 = 45 Square inches Web members
3/12 x 120 = 30 Square inches Floor slab

2/12 x 2 x 110 x 1.75 = 64 Square
inches Railings

l x 8 = 8 Square inches

 

Total = 375.2 Square Inches
The wind load obtained by this method is 375.2 x 30 x 1.5 e 90 or
188 pounds per linear foot. Since this is less than the 300 pound
minimum, the latter is used.
The loads to be used then are
300 - 200 — 500 pound per foot on loaded chord
150 pounds per foot on the unloaded chord.
This makes a panel load of 7500 pounds for the loaded chord and
2250 pounds for the unloaded chord.
2. Stresses in the Top Bracing. The shear in the panel (ab) is

equal to 7,500 x 5/2, which is 18,750 pounds. The shear in panel (hfl

 

 

 

 

 

 

 

 

 

 

a b C d
8 c o E 14
a! bl C! d’

649/: = 90‘

30.

is equal to 2 x 18,750 — 7500( 2 + 3 + 4 + 5 + 6)/6, or 12,500 pounds.
The shear in panel (cd) is simply the panel load or 7,500 pounds. The
angle that the diagonals make with the chords is 46.7 degrees,jand
the sine of the angle is 0.728. The stresses are found by multi-
plying the shear in the panel by the Sine of the angle. The stresses

are shown in the following table:

PANEL ‘ SHEAR MEMBER STRESS
ab 18,750 ab'; a'b ‘£ 13,640
be 12,500 bc'; b'c 1 9,120
cd 7,500 cd'; c'd £ 5,450

The above stresses are calculated on the assumption that both
diagonals in any panel are in action at the same time, one in compress—
ion and the other in tension.

The stresses in the transverse struts equal oneehalf of the joint
load or 3,750 pounds.

3. Check of Design. The members are one inch circular bars, 21.1
feet long. The allowable load in tension is l x 0.785 x 18,000 or
14,100 pounds. The allowable load in compression is l x 0.785 x the

allowable stress.’ The allowable stress is

15,000 __ 121.]. X 12‘)2
4 x 6.4

Allowable stress

12,500 Pounds per Square inch
The allowable load is 12.500 x .785 or 9,800 pounds in compression.
iSince the allowable loads exceed the actual ones, the members are safe.

It is not generally assumed that the kind of stress changes during

31.

the passage of the live load; thus the members may be designed as
simple tension or compression members, which are not subject to a
reversal of stress. This is the opinion expressed in the book
Steel and Timber Structures, by H001 and Kinne; Second Edition, 1942;
page 313. Using this assumption, the laterals will not be investi-
gated for reversal of stress.

4. Check of Transverse Struts. The loads on the transverse struts
consist of 3,720 pounds, compression. The radius of gyration of the

strut cross section is

1.10 L 1.19(3.01)2
2 x 1.19

2.66 Inches2

ll

The allowable stress is

15,000 __114 x 12)2

ll bl
A owa e stress 4 x 7.08

14,000 Pounds per in?

The allowable compressive load is then 14,000 x 2.38 or 33,200

pounds, whiCh is greater than the actual load of 3,720 pounds.

The assumed loading is satisfactory because the Substitution of
the 50 pounds per square foot load on the unloaded structure results
in stresses about one-half of those now present.) .

4. Stresses in Bottom Bracing. The bottom bracing is arranged as

shown in the sketch.

(1 b C d g

 

 

 

 

 

 

 

 

 

 

a’ ’ c’ ecu/5:126

32.

The load in the bottom bracing is 150 x 15 or 2250 pounds per panel.
The shear in panel (ab) is 2250 x 7/2, and is equal to 7880 pounds.
The shear in panel (be) is 2250( 2 + 3 + 4 + 5 + 6 + 7 + 8)/8, which
is 9850 pounds subtracted from the shear in panel (ab). The result
is 5,900 pounds.

The Shear in panel (cd) is equal to 6 x 2250 — 2250 x 33/8, and is
equal to 4,220 pounds.

The Shear in panel (ef) is equal to the load or 2,250 pounds. The
angle that the diagonals make with the chords is 46.7 degrees, and
the sine of the angle is 0.728. The stresses in the members are ob-
tained by multiplying the panel shears by the sine of the angle, and

are given in the following table.

PANEL SHEAR MEMBER STRESS
ab 7,880- ab' ; a'b 1 5,720
be 5,900 bc‘ ; b'c i 4,290
cd 4,220 cd' ; c'd 2 3,070
de 2,250 ‘ de' ; d'e i 1,630

The stresses of 1125 pounds are taken by the floor beams, dispens—
ing with the use of other transverse struts. The floor beams have
adequate compression areas to handle this load.

The members are the same as the top laterals, that is, one—one inch
round rods. The allowable load is 14,100 pounds in tension and 9,800
pounds in compression. These values are well in excess of the max-

imum of 5,720 pounds, and the member is safe.

33.

5. Overturning Stresses. The wind blowing against the side of the
bridge has a tendency to overturn the structure about the supports
on the leeward side, and thus introduces stresses in those members.
This load is taken by the end posts and lower chord members, and
the web members are not stressed. The overturning moment is equal
to Ph, where (P) is the magnitude of the force brought to each portal
strut, and (h) is the depth of the truss. The resisting moment is
Re, where (R) is the value of the reaction, and (c) is the width of

the truss. The sketch of this condition is shown below.

p 2 t”!
a? {f .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

flf Leeward Truss Tfl/l
C (C
b
11‘? fight
19% C :J'R ‘2’: l Windward Truss 1%);

_ The magnitude of the force brought to each portal strut is 7,500 x
2 + 3,750, or 18,750 pounds, and the equivalent loading at the portals
is Ph/c or 18,750 x 15/14 + 20,100 pounds.

The stress in the end post is 20,100 x 1.414, or 28,400 pounds,
compression on the leeward side and tension on the windward Side.
By combining this load with the load due to dead and live stresses,
the total load on the end post is 79,724 pounds compression, and
42,600 pounds tension, by use of the reversal of stress theory.
The member is capable of taking 137,000 pounds compression and 176,000
pounds tension, and is thus safe for the above loads.

The load in the bottom lateral (b) is 56,327 pounds tension, and

34.

30,150 pounds compression. The allowable tensile load is 72,000 pounds

and the allowable compressive load is

 

2
Allowable load =(15,000 - 4¥?2: ) x Area
-(15,000 _ (15 x 12)2 ) x 4

 

4 x (1/12 x 1“} 8 + 4 x 2.25)/4
- 10,400 pounds/in 2 x 4
- 41,600 Pounds.
These loads are greater than the actual loads and the member is thus
safe.
The load in member (f) is the same as for (b), and Since they are
of the same section, this member is also safe.
The loads on member (i) are now 85,540 pounds tension, and 30,150
pounds compression. The allowable load in tension is 90,000 pounds,
and in compression is

Allowable load

2
(15,000 _ Apr ) x Area

2
1 000 _ (15 x 12) .
( 5’ 4 x .4(l/12 x l x 2.55 + 2.5 x 1.52)

x 5.00

55,350 Pounds
The allowable loads exceed the actual loads, making the member safe
for the loading.

The member (0) has the following loads acting on it: 53,815 pounds
tension, and 34,308 pounds compression. The allowable tensile load
is 54,000 pounds, and the allowable compressive load is 39,900 pounds.
The tensile load requirement is barely sufficient, being only 185

pounds safe. .The compression requirement is adequate.

Since none of the members were over-stressed, the lateral system

35.

is sufficiently designed to withstand the loading.

6. Portals. The construction of the portals is indicated below.

I

/4-

 

 

 

 

 

 

Cl

 

0
77am:

  

7}u:3

 

ANA/embers 44:3»,1coa5"

 

 

 

 

The chief function of the portals is to transfer the reactions of the
upper lateral system to the supports of the bridge. In addition, it
stiffens the and posts against vibration. The point of inflection of
the end posts is (15 - 3.61)/2, or 5.70 feet from the base. The load-

ing is as Shown, and results in a direct stress in the end post of

 

 

 

 

 

@500“6 e“ ‘3 1": was-0*
n: n . d
f 9
h
P ~ r f ”i
OH? O L"
Inf/ecfl'on Us
K
,\4— 2 375'“ ,(dL— 9, 3 75 "'

(18,750 x 9.30) + 12, or 14,500 pounds. To determine the vertical
component of the stress in members (an) and (uh), take moments

about the point of inflection.

Stress in (mn) - 9-3.£9-3751-62 x 14.500

= 0 Pounds.

This indicates that members (am) and (nb) are unstressed.

36.

The vertical component of the load on (fme) and (eng) is 14,500 pounds
and is tension for the former and compression for the latter. The
horizontal component is 14,500 x 6 + 3.6, and is 24,100 pounds.

The stress in (ac) is equal to 2/3 x 18,750 — 9,375 + 24,100, or
27,225 pounds tension. The allowable load in tension is equal to
the area times the allowable stress, and/is

Allowable load

(2.48 — 2 x 7/8 x 1/4) x 18,000

36,800 Pounds

The net section in tension was obtained by subtracting the 7/8 inch

rivet holes from the area of the angle.

The stress in (eb) is equal to 1/3 x 18,750 + 24,100 — 9,375 or

20,975 pounds compression. The allowable compressive load is

(72)2
4 x 0.418

a 29,000 Pounds

Allowable load —(15,000 —

\

The members so far are safely designed for both tension and compress-
ion. A reversal of direction of wind load will result in a reversal
of the Sign of the stress in the members, and an allowance should

be made for the increased stress possibly resulting therefrom. The
margin of the allowable loads over the actual loading is sufficient
to make this allowance.

The loads on the member (fme) are equal to the Square root of the

sum of the squares of the components, and are 28,000 pounds. These
loads are tensile loads when the wind is from the north side of the
bridge, and compression when from the south. .The allowable loads

are 36,800 pounds in tension and 29,600 pounds in compression. Since
both are greater than the actual load, the member is safe.

The portals are of sufficient design to carry the wind loading.

37.

Part X

TOP CHORD PINS

The pins will be analyzed for the required thickness to resist
the moment on them, and the thickness of plate required for the

connections.

1. Pin at (B). The pin at this point is a 3 inch pin and is
loaded as shown. The thickness of bearing plate required for each

channel web of member (a) is 65,524 + (2 x 3 x 24,000), or 0.456

3 " Pin 0 I 77I 32; #

  

inches, which is less than the actual thickness of 0.625 inches.
The Value 24,000 used in the denominator of the expression is the
allowable bearing on pins, in pounds per square inch. The thick—
ness required for member (d) is 0.536, and the actual thickness is
0.625 inches. The thickness required for member (0) is 0.11 inches
and is less than the 0.94 inches available. The thickness required
for member (e) is 0.306 inches, which is less than the actual 0.875
inches.
The vertical loads on the pin are arranged as shown. The maximum
moment is at the center of the pin and equals 46,300 inch-pounds.

Bending Moment = 1.5 x 15,500 + 30,300 x 3 + 46,300 x 3.5

38.

The horizontal loads on the pin are spaced as below, and the
moment is equal to
Bending Moment : 46,100 x 3.5 + 31,300 x 3 4 77,400 x 3.5

: 16,100 Inch-pounds

 

 

 

 

 

 

 

é ‘
. i 3
"5'0 I5“ 5 1 vs
' - - :10 '
0 § 8 8
o "I R .
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J/er-f'ica/ Loads Horizonfa/ (gods

 

The total moment is equal to the sum of the squares of its com-
ponents and is 49,000 inch—pounds.
The minimum diameter of pin to be used is found by use of the
formula 2
Diameter = .753 x Cube root of the Moment.

This formula is derived in Steel and Timber Structures, by Hool

 

and Kinne, 1942, on page 350.

The diameter then required is 0.753 x 36.5, or 2.74 inches. The
actual diameter is 3.0 inches, and is thus safe as computed. The
member (c) is evidently a special eyebar, for there is no standard
eyebar of this size in use at the present time.

2. Pin at (C). This pin is a 3 inch pin also and is loaded about

t

39.

the same as the previous pin. The computations are carried on in the

 

 

 

 

 

 

 

 

 

same manner as before and are listed in the table. a
’ en
r .
o.
! 2 1.2 l ' I~°
' .5' 2.5
‘8 8 b h
\9 V ‘9 "1
_ 0. Q‘ I D k
N N
u
Verf/ca/ Load.) Horizon fol loads
Table of
REQUIRED BEARING THICKNESSES
MEMBER LOAD REQUIRED _ ACTUAL
BEARING BEARING
THICKNESS THICKNESS
d 77,322 Pounds 0.536 Inches 0.625 Inches
J 97,932 " 0.678 " 0.625 "
g 24,192 " 0.168 " 0.375 "
Table
of

REQUIRED PIN DIAMETER COMPUTATIONS

VERTICAL MOMENT 20,000 Inch-pounds
HORIZONTAL MOMENT ' 20,000 Inch—pounds
TOTAL MOMENT 29,200 Inch-pounds
REQUIRED DIAMETER 2.32 Inches
ACTUAL DIAMETER 3.00 Inches

The pin and Connecting plates are of sufficient cross section

to withstand the loading up to this point.

PIN AT (D)

 

 

 

 

 

no a"
(4029

Ver fire I L 0 ads

 

| .25'
275 3‘

q r.
0 0‘
a: m

 

 

Her/z ons‘a/ lo ads

 

Tgble of

REQUIRED BEARING THICKNESSES

MEMBER LOAD
j 41,232 Pounds
k 70,138 "
m 40,000 "
1 28,382 "

REQUIRED ACTUAL
BEARING BEARING
THICKNESS THICKNESS
0.343 Inches 0.625 Inches
0.584 " 0.625 "
0.333 " 0.940 "
0.236 " 0.375 "

Table of

REQUIRED PIN DIAMETER COMPUTATIONS

VERTICAL MOMENT
HORIZONTAL MOMENT
TOTAL MOMENT
REQUIRED DIAMETER

ACTUAL DIAMETER

28,382 Inch-Pounds

28,382 Inch—pounds
40,000 Inch-pounds
2.55 Inches

2.50 Inches

The pin is not safe for the loading as designated because the

required diameter is 0.05 of an inch larger than the actual diameter.

The connecting plates will transmit the stress safely, however.

Bart XI
TOP CHORD PIN PLATES

These plates are used to transmit the stress from the chord members

to the pins, and must be designed to resist the tensile loads thus

occuring.

1. Pin Plate at (B). The plate layout is shown below, and must be

 

 

D

 

 

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h u

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o o C ” a

25'

 

 

 

 

 

 

designed to carry a moment of 49,000 pound—inches.
The allowable load in Shear on one 7/8 inch rivet is 11,000 x 0.601
and is 6614 pounds. The number of rivets needed to resist the moment
is equal to 49,000 f (6614 x 12) or 7.4. Since 10 are used, the connect—
ion is more than adequate.
The net section occurs at the section ABECD, and is (7-1.2) x 3/8
or 2.18 square inches. The allowable load in tension per plate is
2.18 x 18,000 or 39,300 pounds, and the actual load is 38,661 pounds.
The plate is adequate for this joint.
2. Pin Plate at (C). The moment to be resisted is 29,200 inch—

pounds, and with the use of 1 inch rivets, the required number needed

is 3.37. The number of rivets is met, as 4 per plate were used.

The net section is a section straight across the member, through the

43.

axes of the holes. It is equal to (6 — 0.87) x 1, and is 5.13 Square

 

 

 

 

 

 

3 8 Thickness 33”
a!— O O 9,. ..
o v Rive {'8 / Did.
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inches. °The total allowable tensile load is 5.13 x 18,000 or 92,500
pounds per plate, as compared with an actual load of only 48,966 pounds
per plate.
The plate is then adequately designed for the above loading.
3. Pin Plate at (D). The tetal moment on this plate is 40,000
pound4inches per plate. With the use of 1 inch rivets, the allow-

able stress per rivet is 11,000 x 0.7854 or 8,620 pounds. The

number of rivets needed in the connection is 4.64 = 40,000 + 8620.
The actual number used is 5, so the requirement is met. The net

section.is equal to (7 - 1) x 3/8, or 2.25 square inches. The allow-

 

II

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able load is then 2.25 x 18,000 or 40,500 pounds per plate, while

the actual load is 35,069 pounds.

The member is safe.

4. Pin Plate at (E). The plate at (E) is the same as the plate
at (D), the allowable load being 40,500 pounds, which is greater

than the actual load of 35,069 pounds.

Part XII

PIN DESIGN IN THE BOTTOM CHORDS

The procedure followed here will be the same as that followed
in the analysis of the t0p chord pins, and reference may be made to

that section for the method of procedure. -

PIN AT (A)

 

 

 

46 300* 4
4c, 300

Var/veal and HorizonI-a/ loads

VERTICAL MOMENT 66,300 Inch-pounds
HORIZONTAL MOMENT 66,300 Inch-pounds
TOTAL MOMENT 94,000 Inch—pounds
REQUIRED DIAMETER 3.49 Inches

ACTUAL DIAMETER 3.50 Inches

The pin is barely safe and is very close to being overstressed.

r V
I .
‘V.’ .

 

 

 

 

M, 400‘

 

49460‘*

E

o Horizonf'o/ leads

V.‘ 11311101". 11 ifiilfs'LIITJT 57, 700 Inc h~£JOuudS

HORIZORTAL MOMENT 19,985 Inch—pounds

TOTAL MQMTNT 61,000 Inch-pounds
ngfllfifld DIAMETER 2.96 Inches,
ACTUAL DiAdETfiR 4.00 Inches

C
\'
ls)

‘- . o w 'I ' "" " ‘ ' 0" 'v ”r - 'I" ‘14 a ‘4" n1 ~ 0: u- . '3'!»
The member is bdlu by a edict; Letter 01 scout 1.3, whlbh

adequate fur the connection.

‘91:: 4T @1

 

 

 

 

 

Q
V

 

Verfica/ loads Horizonfd/ loads

VERTICAL MOMENT 59,755 Inch—pounds
RORIZOEThL hangs? 60,547 Inchupounds
TOTAL MJMENT 85,103 Inch-pounds

iil‘l,UI.”.ff"I‘ iFIr“-J’.”_"'I‘:TF’. :3- 31 Inch-es

46.

ACTUAL DIAMETER 4.00 Inches

The pin itself is more than adequate to

resist the load placed

upon it, and the plates are also safe by a fairly wide margin.

 

 

 

 

 

 

 

 

 

PIN AT (n)
4 é
. , J t , ,
7% It‘s L540 [dis ("‘6 a" “k, /%
| a v 8 | n V 3 g.
. ~ M
Q :1 '5‘; {Q N lo l"
Vorft‘ca/ (oads Horizon/'a/ Loads
VERTICAL MOMENT 103,500 Inch—pounds
HORIZONTAL MOMENT 16,000 Inch—pounds
TOTAL MOMENT 104,800 Inch—pounds
REQUIRED DIAMETER 3.54 Inches
ACTUAL DIAMETER. 4.00 Inches
The required plate thicknesses are:
PLATE REQUIRED ACTUAL
THICKNESS THICKNESS
1 0.264 Inches 0.375 Inches
1 & o 0.280 " 1.00 "
h & n 0.201 "' 1.00 "

The pin and plate are both of the required thickness, and are

satisfactory.

 

47.

 

 

 

 

 

 

 

PIN AT (ML
a s.
j" n J 3 I" I. ‘ u l J
b t t
R 8 a 8 ‘2 e
w h N ‘1 m m
. R up ,3 ' N‘ b~ m m~
N N a.) N \l ‘0 ‘0
Vertical Loads AHorl'zon/"a/ loads

 

 

The thicknesses of the member plates are sufficient to withstand

the loading.

VERTICAL MOMENT 136,250 Inch—pounds
HORIZONTAL MOMENT 81,250 Inch-pounds
TOTAL MOMENT 159,000 Inch-pounds
REQUIRED DIAMETER 4.07 Inches
ACTUAL DIAMETER 4.50 Inches

The pins are safe, and will transmit the load without being over—

stressed.

This concludes the study of the pins in the bottom chord, and
there remains only the analysis of the plates that connect the
vertical hangers to the bottom chord pins; the plates that connect
the upper transverse members to the top chord; the pin plates that

hold the lower chord and upper chord laterals; and the pedestals.

48.

Part XIII

HANGER PIN PLATES

These plates are used to connect the truss hangers to the chords
at the top and bottom. They are all of the same size, and carry
maximum loads of 157,424 pounds compression, and 61,098 pounds tens-
ion. The stresses are assumed to be direct stresses and no reversal
of loading takes place.

1. Rivert Required. The above loads are the actual loads per
pair of plates, two being used at each connection. The allowable
shear in a one inch rivet(assuming an allowable stress of 11,000
pounds per square inch) is 11,000 x 0.7854, or 8,640 pounds. The
number of rivets required per plate is then 61,098 + 2 x 8,640, and
equals 3.55. The actual number used is 4 as shown, so the plate is

adequate at this point.

 

 

 

 

 

 

 

 

T3 '
O : O 777 ickness 3-
n Rive/s InD/‘a.
. O 0
°’ [N
z " .
h) 3 Pm
la’l‘gg'l 21"]: .

2. Tension Requirement. The net area of the plate is the smallest
when taken across the axis of the 3 inch pin, and is equal to
Net area = (8-3) x 3/8
= 1.875 Square inches

The allowable tension load is then 18,000 x 1.875 or 33,800 pounds,
which is less than the actual load of 30,549 pounds.

 

49.

3. Compression Requirement. The compression area is equal to the
gross area of the plate, which is 3.00 square inches. The allowable
load that the plate will take in compression is than 18,000 x 3.00
or 54,000 pounds. This is quite a bit less than the actual load of
78,712 pounds, which would be expected upon examining the plate
out at the bridge. The plates at the mid—panels are visibly failing,
and the ones at the panels next to them are beginning to buckle.

The design is definitely not sufficient for this loading.

PART XIV

UPPER TRANSVERSE CONNECTION PLATES

These plates are used to connect the upper transverse struts
to the chords, and are designed as shown below. Since the loads

on them are so light, 3,720 pounds, it is easily seen that the

 

 

 

 

 

 

I12"
E;—
I o o n
'7 Rivefs / 020.
° ° Thickness in
t 4
+ 7
, o (J o o
'W

 

design is excessively safe. The allowable load in tension is about
60,000 pounds, far in excess of the actual load. The stresses due
to the overturning moment are also so small that they will not

affect the strength of the connection.

50.

Part XV

DIAGONAL CONNECTION PLATES

These plates are the plates that hold the diagonal members of
the sway—bracing system to the transverse members, and are designed
as shown. The maximum load on the plate is 5,720 pounds, which is
the load in member (ab') at the middle panels.

The actual shear on the plate is equal to 5,720 x sine of the
angle the diagonal makes with the plane of the plate, or 5720 x 0.728
which is 4,160 pounds. The allowable shear is 4,860 pounds on a
3/4 inch rivet, which exceeds the actual load and thus is safe. The
plates are more than adequate for resisting pulling out or shearing

off of the diagonals.

Part XVI

CHECK 0F PEDESTALS AND SHOES

l. Bearing Plates. The end reaction is 66,300 pounds and assuming
an allowable bearing stress on the masonary piers of 200 pounds
per square inch, the required area of bearing plate is 66,300 + 200
or 331.5 square inches. The actual plate is 18" x 18", giving an
area of 324.0 square inches, which is not sufficient. The plate
is 1% inches thick which is sufficient, however.

2. Shoe Design. The shoe is constructed as shown and has a bear-

ing area of 3.5 x % or 1.75 square inches. Using an allowable

5]..

bearing stress of 18,000 pounds per square inch, the allowable

'

w x2 /”
éd‘ 4*:

@- '

 

 

 

 

 

In?

 

 

 

 

 

 

 

 

 

 

 

 

[I

 

43” /6'

au'
1
\

 

 

 

 

 

bearing load is 18,000 x 1.75 or 31,600 pounds per web of the shoe.
This value is less than the actual load of 33,150 pounds, and the
shoe is consequently overstressed. A

3. Bolt Shear. The connection to the end posts is accomplished
by the use of 3.5 inch pins, set in the shoe webs. The allowable
shearing stress to be used on the pin is 8,000 pounds, and the
area is 9.621 square inches. The allowable shearing load on the
pin is then 9.621 x 8,000 or 76,968 pounds, which is greater than

the actual load of 64,300 pounds. The pin is safe.

The shoe and pedestal system is slightly inadequate to withstand
the loading of the 10 ton truck or equivalent H—lO loading. The
connection between Spans is a 7 inch pin with double loading, and
results in the same proportionate loading in excess of the allowable

loading.

This concludes the analysis of the structure, and the summary
and conclusions as regards the results follows on the remaining

pages.

520

Part XVII

CONCLUSION

The bridge is, for the most part, safely designed for the H—lO
loading. It will not, however, withstand the loading without causing
failure in some of the members. The mid-panel truss members, and
supports, and hanger pin-plates need replacing or redesigning. The
lower chord pins at the mid-panels are also on the verge of being
overstressed, and should be considered in the strengthening of the
structure to safely carry the loading of a 10 ton truck.

It is not known just what loading was used in originally design-
ing the bridge, but it would seem that a loading similar to the one
used in this analysis was used. The members as originally designed
for the bridge would, I believe, resist the loading of the 10 ton truck,
but they were not adequately protected by painting or some other
method. The sections have been reduced from 10 to 20 per cent because
of rusting and scaling, and thus are not able to take the amount of
load that they were originally designed to take. To prevent further
deterioration, the bridge should be painted or rust—proofed as soon
as is possible.

If the structure is to be left as it is without any strengthening
or replacement of existing members, I would recommend that a load
limit of seven and one—half tons—a seven and one-half ton truck-be
placed on the structure to prevent any of the members being over-
stressed.

It wouhi not take a great deal of extra work and expense to make

the structure safe for the loading. The failing members could be

53.

removed from the truss and replaced by ones of suitable section with—
out affecting the other members to a very great extent. The replace-
ment of the supports will be a bit more difficult and may not be"'

necessary. The supports are very close to being adequate and possibly

they could be made satisfactory with the use of stiffeners.

In summary, the bridge will not be safe for the H—lO loading, and
a maximum of 7% tons should be allowed on it. It should be painted
to prevent further rusting and scaling, which is preceeding at an

alarming rate.

540

. .1 - . - “.13
.‘ I
i g _ v 4». 73:3,;

BIBLIOGRAPHY

 

The following books were used for reference to method of proced-

ure in the analysis of the preceding report.

The Civil Engineering Handbook by Leonard Church Urquhardt, C.E.,

Editor-in Chief; First Edition; 1934

Steel and timber Structures by George A. Hool, S.B. and W.S.

Kinne, Editors-in—Chief; Second Edition; 1942

Structural Design by Carlton Thomas Bishop, C.E.; 1938

 

steel Construction by the American Institute of Steel Construction,

New York, N.Y.; Fourth Edition; 1941

Structural Theory by Hale Sutherland and Harry Lake Bowman;

 

Third Edition; 1942

Mechanical Engineers Handbook by Lionel S. Marks; Fourth
Edition; 1941

 

 

 

 

 

DELTA MILLs HIGHWAY Final/6H 711241.93

 

 

 

Top CHOA o 3,

 

To 4' w

 

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