A SOLUTION QF A NUMBER
OF TYPICAL INDETERMINATE

STRUCTURAL PROBLEMS

Thesis for the Degree of M. S.
MICHIGAN STATE COLLEGE

John Emést Meyer
I992 ‘

 

THESIS

IIJIuI 4

 

A SOLUTION CF A 11113:? OF
TYPICAL INDETEREIHATE STRUCTURAL

PROBIEIS

by

John Ernest reigr

A THLS IS

Submitted to the Graduate School of Michigan
State College of Agriculture and Applied
Science in partial fulfilment of the
requirements for the degree of

118m 03:? SC Bil-TE

Department of Civil Engineering
1942

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Table of

Title ”hect

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Table of

Introduov

Solirtions
Frizz-Cd r”trueturee in Grace
Icflcctione of Structuree
01:0;1

Icthode of
Continuous and long

“F71

L\ -A

Contents
ion

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of“.

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‘

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I ' q ." O n
2111;. .L "I 5’ l c»

Ioder Jethous for Liflo 1 awe
Ar,hee and Closed Pings
TXflz mation of Synbols
Iiitliogttqfigr
lumbers in lower right co 1: we are

5 ~55 v I. - n
ukhf rcic2enoee.

("I

Totes on Contents
Iiecue ion oi the Theor' of Thr
an; Area omcznt lro o: ition

Contents

\

Pridges

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-19
39-51
51-102
103- 1

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1:4;- lxjfl;

155-171
17L-173
174

Introduction
he fresentation of the su hiCct nutter concerned With thL
'lculation of the stresses irwlitrwwi structures has been
'flie princij l concerw1cfi‘2uu1'11rh<2 it iCs in the fieli.of
Civil Engineerinf.ln hnnr instances t:ese ren huve been

.1."

limited to the Cytlication of particular theories in tie

solution of conflicntei structur: TrohlerC.ThCy all recog
nizcd trust the Trorei'tng to att: CITIYII’ indettrmfijugte st11 -
3.
.-E.i~

tural froclems is to build a nodel,;nd tien to actually
the Carticulur stresses.fnou§h work Ins been done in this
directicni to allCWi'u c dCVEICWTWHlt and :qfnlicaticricrf the

theo~iCs of structural stress e to the solution offarticuler

fro‘ol ens. I’ovrever there is still a lz‘.r,r“e Car-iount of correlation

to IC eccomrliChrd in the matter of the afflicttion of the

classical theories to the solution of thee CO- C; lled, rrscticu

inde,ttrrn2ntte SIJNICtHI€L_ Crotflmrus.

iDCcn I; .Crinter,of the Illinois Institute of iccrnolCCnu

0-! A t \r n W 5"; 1’ 4 ’ I r v —-° a
aclvui ltrlgri..untooi..tg 221 tnC‘ n ljs‘ie 01 strum: 1
e.1s. Ininis todk "”trCs es in :r ”Cd ”tructureC“ volume
r .1. ' . - .. . ° 4. .. ‘I ,
C thcJ-;IIDDlCKS.CrlFI€1 212s
n CC 1v the cl:ssioal

1:,"
2—313

Ir} 1.).
H m

0
he hes Tresented a nu ber of True
homn, tCCt in Hair instunoes, C c;.

T€th0d€ OI cJLIPM1lC to t“e solution of these V”OIlE\W.Tn the
'freC eni;,2t2n1coi.rgt 1;"CCC‘rteci..e1"i itil thtw=c FHLiCCTtthHlS In: a

in nan? cas ,hCen CLIIlEd out.
The anflic tion of sone of the siPClCr classical met303
to the solution oi problrfis concerned Witl res tiained beam

.L

I? U)

and frames results in such an involved frooeoure thLL a more
unique solution of such froblens seened Worth WhilC.This has

1

been done in this tFCsis in such cases Vnere it was felt thtt
this method Cf Lttsc A sinjlified tht Co lution of the froolem.

*‘

larticular attention is called to tle Ljylicution of tne

Theorem of Three Ionents and to the frorosition of [LIFE Jon»

ents to cases There tht oriinerf classical Ct od of uttack
r7011]d not yield enough equations to render tLC problem det-

Cr.*nate.-necc theorene,in combination with the benoing non-

ent oiC Cram by :‘C.I‘tS,i’I;:,V€' rude 310s: i‘olC unirue solutions of

ot7CW‘ isC 10T@"EJV1 cowrijcxttefi nrolflrvws.ln thft‘filites on
(jontCnts" :hi'ttis thesis 217”ci<n1 is ncdo Cd"U7is:ncthod of
attack ior the several classes of liolleru outlined in the
"Table of Contents".

The field of Civil ingincerin1 is 011 enough at *l_e fre—
CCnt time so tint little rcru ins in,tke "1;*<If new theorems
to be develored for the solution of structure froblens.ft
is feltgix: rve1,11n1.t tkei -;7et re1xj2m32n1ch "orvz'ms oe done
in the intercorr Clution of the cl assiccl theorers concezned
Pith £22,21ch structures,e.nd the :“l1cat one. of these inter-
c01rCl--tion' of t1:_Cle- cl;-ssio:;l t e01 ens to the solution of
_ __ oblem {The 1“.11.1"}“ose‘ of t1:is thesis
:es been to Ircsent a these of tnCse intercorrelutions.

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Problem.1,Page 16

Determine whether the structure shown is statically det-
erminate,statica11y indeterminate or unstable.

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Problem 2 , Page 16

Determine whether the structure Shown 1‘ _ '
erminate,statically indeterminate or unstaglgfatically dEt

 

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Problem 5,Page 16.
Determine whether the structure shown is statically det-
erminate,statically indeterminate or unstable.

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Determine whether the structure shown is statically det-
erminate,statically indeterminate or unstable.

 

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Problem 5,Page 16.

Study the layout of a typical four-legged tower and
decide how it may be braced and made statically determinate.
A three legged tower with battered legs is suggested.

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—10-

Problem 6,Page 16.

The dirigible frame shown is made up of decagonal rings.
Allow for six reaction components and show that it's stress-
es could be calculated by the equations of statics.

 

 

 

 

 

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Problem.7,Page 30.

Analyze for the
note that although
start the analysis

stresses in the framed pedestal shown.
there are more than six reactions,you can
quite simply by observing that at each

upper chord Joint all members except one,lie in a plane.~

 

 

 

 

 

 

 

 

 

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Problem.8,Page SO
Explain in detai1,why the analysis of this pedestal shown
would be more difficult than the one of Problem.7.Can you
write enough simultaneous equations from.the equilibrium of
Joints E,F,G and H to obtain the value of any bar stress?
Can any stress be determined from.1ess than 12 simultane-
ous equations for any combination of Joints?
(Tote) These questions form.a written discussion by Prof.
Grinter in Proceedings A.S.C.E. for September 1934,
pages 1086-1091.

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Problem.9,Page 50.

Analyze for the wind stresses in the 4 post water tower
shown.This is a standard tower with tank of 100.000 gal.
capacity.Wind pressure is obtained from.20 pounds per square
foot on the dimetral plane of the tank and 200 pounds per
vertical foot on the tower.Loads are given for the entire

structure.Select rods to serve as diagonals at a working
stress of.18000 pounds per square inch.

 

 

 

 

 

 

 

 

 

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.Problem.12,Page 34.
Obtain an expression for the deflection under a concentr-
ated load placed at the center of this simple span beam.
Repeat for deflection under one concentrated load for a

simple beam.carrying concentrated loads at the one third
PO ints o

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.Problem 13,Page 54.

~21)-

Compute the deflection under the 300,000 pound load for
the truss illustrated.Each member has an area of 15 square

 

 

 

 

 

 

 

 

 

 

 

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Problem l4,Page 34.

Derive an expression for the internal work in a beam.
Make use of the sketch that represents a differential length
of beam over which the moment may be considered to be const-
ant. (hint) First determine the value ofA. with the aid of
the flexure formula.

 
   

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Problem 15 ,Page 54

Prove at the internal work in a beam of any type is
equal to 1 .1. times the statical moment of the moment dia-
gram about it‘s base. The loads are gradually applied.

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Problem 16,Page 35.
Wbrk Problem 12 by use of Castigliano's theorem.
Problem.12 is stated as follows.

Obtain an expression for the deflection under a
concentrated load placed at the center of a sim-
ple beam.Repeat the discussion for the deflection
under one concentrated load for a simple beam
carrying concentrated loads at the 1/3 points.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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'Problem.17,Page 55.
‘Work Problem.13 by use of Castiglian'os theorem.When set-

ting up the expression for internal work, let P equal the load.

Problem.15 is as follows.
Compute the deflection under the 300,000 pound load

for the truss illustrated.Each member has an area of
15 square inches.

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Eroblem.18,lage 35.
Find the deflection under the 10,000 pound load for the
cantilever beam illustrated.Let P equal the load when set-
ting up the expression for internal work.

 

 

 

 

 

 

 

 

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Problem 19 , Page I; C- .

Compute the vertical deflection of the point A of the
truss illustrated.Areas in square inches are marked on the

members .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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lroblem.20,Page 40.
Compute the angular rotation of the bar AB of the truss

of Problem 19 .

(Problem 19 on previous page)

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Problem 2:1,Page 40.
Compute the maximum upward deflection of the 12 inch,.’51.8

pound I-beam supported and loaded as shown.

 

 

 

 

 

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-29-

.Pr°8%§§u%2’¥%§eb$3aing and the shearing deflection at the
center of a 12 inch 50 pound I-beam acting as a simple span
of 40 inches.The beam carries a concentrated center load of
100,000 pounds.Assume that the beam is restrained against
lateral buckling and that the web carries all the shear.

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-39-

Problem 23,Page 41.

Find the maximum vertical deflection of the Pratt truss
shown as produced by a 20 degree increase in temperature of
the top chord and end posts.Take the coefficient of expan-
sion to be .0000065.Is the deflection up or down?

Explain Why the cross sectional areas of the members are
not needed.What would be the maximum.deflection if all the
members of the lower chord were fabricated % inch too long?

9? v‘ofi/ (190" 1", /en, #19

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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-32-

Problem 24,?age 41. .

A steel pipe is bent as shown to provide for expansion.
The pipe has a polar moment of inertia of 292 and a moment
of inertia about a diameter of 146.Compute the deflection
of the point A in the direction of the 1000 pound load by
by taking into account flexure and torsion.NOte that the
member CD is bent in a diagonal plane.

 

 

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-34-

Problem 25 , Page 41.

Find the absolute deflection in space of point A of the
pipe considered in the previous problem (Problem 24).

   

 

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-55..

Problem.26,Page 47.

Compute the end slope and the maximum deflection of a 36
inch,230 pound WQF. beam.for a simple span of 60 feet when

9
loaded to it's allowable stress of 18000 pounds per square
inch by a uniform load.Do not develop a formula.

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Iroblem 27,Page 47.

A 24 inch,79.9 pound I-beam extends as a cantilever 15
feet long and carries a concentrated load of 10000 pounds
at 10 feet from.the fixed end.Find the maximum.fiber stress
and the slope and deflection at the free end.Inc1ude the

effect of the dead load.

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-37-

Iroblem.28,Page 47.

A beam.fixed at both ends carries a total uniform.load of
1200 pounds per foot on a span of 20 feet.Compute the maxi-
mum deflection if the section is a 12 inch WkF. beam having
moment of inertia of 182.4.Compare with the maximum.deflect-
ion caused by a center load of 16,000 pounds,which produces
the same fiber stress.make a further comparison with the
maximum.def1ections of simple beams loaded in the same man-
ner to this fiber stress and having the same section.

 

 

 

 

 

 

 

 

 

 

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-39...

Problem 29,Page 47.

Find the maximum deflection of the beam shown due to the
concentrated loads alone.The beam is of wood,12 inches wide
and 50 inches deep.Take the modulus of elasticity at 600,000
pounds per square inch.

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-45-

Problem 30,Page 47.

Since flexure formula for re-inforced concrete beams,
f¢3 2M/(kjbd') ,must be equivalent to f=Mc/I,and since c-kd,
we can readily show that I=Jfiifly2. Show the derivation of
this formula and than determine the theoretical deflection
of a concrete beam 12 inches wide and 18 inches deep to the
stee1,which carries 4-3/4 inch square bars,if the simple

span is 20 feet and the total load is 1000 pounds per foot.
Let E 30,000,000

 

 

 

 

 

 

 

 

 

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-41..

Problem 31,Page 48.

Check the deflection of Problem 30 by computing a moment
of inertia from the formula fl, H/pjbd . Explain why the act-
ual deflection of such a concrete beam could not be expected
to be as great as the theoretical deflection.

 

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-42-

Problem.32,Page 48.

Graphically determine the elastic curve and the.maximum.
deflection for a simply supported beam of 50 foot span load-
Ed by a uniformly increasing load from zero at one end to
600 pounds per foot at the other end. The beam is a 10 inch
35 pound standard I-beam.Neglect the weight of the beam and
compare the deflection obtained with the maximum allowable
deflection of 1/360 times the span.

   

 

 

 

 

 

 

 

 

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- 42A-

 

-43-

Prob.35, Page 48 '

Determine the elastic curve and the maximum deflection
for the beam illustrated. Kote that the funicular polygon
must be drawn so that there will be a zero deflection at A
and C. This may be accomplished by correcting the base line.
Treat the uniform.load as a series of concentrations.

 

 

 

 

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lrob. 34, Page 48

Determine the % reduction in maximum deflection in the
beam illustrated as produced by the increased depth near
the ends. This is a rather difficult problem. You should
take the simple beam.moment diagram, divide it by E.I. and
plot the result. Then shift the base line to produce equal
positive and negative areas. The fixed beam has no change
in slope between ends. This true h/EI curve placed as a

load on the conjugate beam is used to calculate the center
deflection of the haunched beam.

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-45-

Erob. 35, Page 48
Compare by use of the f/c curves, the relative amounts of

the maximum deflections of 2 simply supported plate girders
of the same span and depth. One of the girders is of con-
stant section and the other is of constant strength. The
loading is uniform and the depth is constant from.end to
The value of the maximum.fiber stress is the same for

end.
the 2 girders. 25,4»;
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-46-

-rob. 36, Page 51

_ The actual changes in the lengths of the various truss
.nembers as caused by the load P are shown on the drawing .
of the truss. Obtain the shape of the deflected lower chord

by the method of angle weights. E=30,000,000

 

 

 

 

 

 

 

 

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-47..

Erob. 37, Page 51
Determine the deflection diagram for the truss shown as
caused by a dead load of 2000# per foot of the truss. The
.dead load stresses are given in Fig.126 of Vbl.l. Areas are
written on the members. Check the center deflection, obtain-

ed graphically, by a single calculation of virtual work.
E=30,000,000

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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-48...

Prob. 42, Page 59 .
A 10", 30-}? I beam acts as a simple span 25 feet long. Dem

'termine the maximum uniform load and the maximum fiber

stress if the limit on deflection is 1/560 times the span.

 

 

 

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6, 13, I6, I8. 19

Prob. 43, Page 59

A 30", 240# beam and a 36", 500# beam carry uniform

loads over the same spans and have equal maximum deflections
Determine the fiber stress in the 30" beam if the 56"

 

 

 

 

 

 

 

 

     

 

 

 

 

 

 

 

 

beam
is stressed to 18000# per square inch. Explain
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-50-

Prob. 44, Page 59

An I- beam.has a fixed deflection. A particular beam is
found to be 5Q% overstressed when it is deflected the re-
quired amount.‘What reduction in depth will produce the re-
quired stress. Why does the moment of inertia of the beam
not enter into this problems

P 44-3500" bet”

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-51..

Prob. 45, Page 59

The truss shown is to be erected by the cantilever meth-
od. If the dead load is 500# per foot of truss and if there
is a concentrated load of 4500# at Lsproduced by the erec-
tion equipment, find the proper length of the erection
tieback in order for L4to land on a seat at the same level
as L.. Areas of all members are shown on the figure. If the
lower chord members have excessive L/r values, how would
you stiffen them.to permit erection.

 

 

 

 

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-52-

Prob. 46, Page 64

Draw the final moment diagram for the two examples
given in Paragraph 47

2'1)? I‘K/emcal 0/ de/Offf
“on.

10' 7 30'

 

 

 

 

 

 

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-54..

Problem 4.7, Page 64
Repeat the analysis of the example of Paragraph 47 with-

out settlement where the ends of the beam are fixed against
rotation.

 

   
 

 

 

 

 

 

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Problem 48, Page 64
Obtain the moment diagram for a continuous three span
beam fixed at one extreme end and simply supported at the
other, where the spans are of equal lengths and the load-
ing is W'# per foot over the entire length. Such beams
commonly are designed for maximum.moments of WL/l2 for

interior spans and WL/lO for simply supported end spans.

 

 

 

 

 

 

 

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Problem.49, Page 70 .
Use least work method to obtain the moment diagram for

the beam shown.

       
 

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~58—

Iroblem 50, Page 70
Obtain a moment diagram for this bent with pin-end
columns restrained against lateral motion. All members

have the same cross-section.
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-59-

Problem 51, Page 70

Analyze the pin-connected bent for the stresses caused

by the 250 kip load.

Ifirone Menéev- 4r ’5 1‘0:

 

 

 

 

 

 

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Problem 52, Page 70 ‘
Analyze this mine head frame for the stresses caused by
the forces which are produced by the cable pull. Areas of
members are given on the figure. The joints are pin-connect-
ed. Take the stress in the sloping back brace as the re-
dundant stress S.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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-51-

Eroblem,53, Page 70

Analyze the trussed beam.for the stresses caused by a
uniform.load over the entire length of 1400# per foot.
How could the design be improved if the maximum fiber
stress is limited to 10,000# per sq.inch.

 

 

 

 

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Problem 54, Page 70

Calculate the maximum fiber stress in the trussed beam.
E:30,000,000# per sq.inch. Take Sxin the vertical strut as
redundant.

 

 

 

 

 

 

 

 

 

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Problem.55,

Page 70

-54-

Check the results obtained in Fig.62,page 68, by an-
other analysis using the left hand reaction and the stress
in the diagonal PC as the redundant forces.

 

 

 

 

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8/‘7/6

Iroblem 56, Page 74

The frame illustrated has all members of the same area.
Use a direct application of virtual work to obtain the
stress in the member in line with the loads.

[cl AC 5: J‘k #0100!ch

 

 

 

 

 

 

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-66.-

Problem 61,Page 76.
.Analyze for the stresses in the frame of Problem 56 by
use of the general method of indeterminate structures.

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Problem 62,Page '76.
Analyze for the stresses in the frame of Problem 52 by
the general method of indeterminate structures.

3 .

 

 

 

 

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-68..

Problem 65,Page '76.
Check the moments given in Problem 50.Use the general
method of indeterminate structures.

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Problem 64,Page 76.
Use the general method of indeterminate structures to ob-

tain the stress in the redundant strut of Problem 54.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Problem 65,Page '76.

Take reactions as redundants and solve for the moments at
B and C in the continuous beam of Fig.56,page 63.Neglect
settlement. (SlOpe Deflection Method)

 

 

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Problem.65,Page 76.

settlement.

\

Take reactions as redundants and solve for the moments at

B and C in the continuous beam of Fig.56,page 63. Eeglect

(Balancing Angles Method)

 

 

 

 

 

 

 

 

 

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Problem 65,Page 76.

Take reactions as redundants and solve for the moments at

B and C in the continuous beam of Fig.56,page 65.Neglect
settlement. (Balancing Mbments Pethod)

 

 

 

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IToblem.66,Page 76.

Find the stresses in the center diagonals of the truss
as illustrated.The single load is 70,000 pounds.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Problem 67 ,Page 81.

Obtain the moment diagrams for the loaded structure shown.
In this case check the results by statics and by drawing a
deflected structure to see that it is reasonable.

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-77-

Problem 68,Page 81.

. Obtain the moment diagrams for the loaded structure shown.
In this case check the results by statics and by drawing
deflected structure to see that it is reasonable.

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Problem 69,Page 81.

Obtain moment diagrams for the loaded structure shown.

In this case check the results by statics and by drawing a
deflected structure to see that it it reasonable
(Balanced Loments)

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-81-

Problem.70,Page 81.
Obtain moment diagrams for the loaded structure shown.
In this case check the results by statics and by drawing a
deflected structure to see that it is reasonable.
(Balanceding meents)

 

 

 

 

 

 

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-83-

Problem 7l,Page 81.

Obtain moment diagrams for the loaded structure shown.
In this case check the results by statics and by drawing a
deflected structure to see that it is reasonable.

 

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Problem 72,Page 8]..
Obtain the moment diagram for the beam caused by the
settlement of the support. C. E : 30,000,000

 

 

 

 

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lroblem 73, Page 81

Obtain the maximum moment for this beam caused by the

settlement at B. E is 50,000,000

 

 

 

 

 

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Iroblem 74,Page 81. -
Check the fixed end moments as given in.Figure 69 for the

beams (a),(b),(c) by use of the first two propositions of

area moments . P

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Problem.75,Page 81.

Develop the expressions for fixed end moments for any
beam from (d) to (h) as given in.Figure 69.Use any method
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Re-analyze for the moments in the continuous beam of Fig-
ure 70,u31ng only two simultaneous equations.

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IToblem 77 ,Pag;e 84.
Analyze for the moments at B and C of the bent shown.Use
II-values.

 

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Iroblem 78,Page 84. T
Analyze for the moments in the bent illustrated.The col-
umns and girder are of the same section.

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lroblem 79, Page 84
Repeat the analysis of Problem 77 with the bent changed
to the condition of fixed end columns.

 

 

 

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-95-

Iroblem 80, Page 84
Repeat the analysis of Iroblem 78 with the bent changed

to the condition of fixed end columns.

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Problem 81, Page 84

Obtain a moment diagram for the two bay bent illustratei
Use equation (23) to simplify the work. Draw the deflected
structure by use of the final moments,as a check on the
reasonableness of the work.

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lroblem 82, Page 87

The building frame shown represents th qe 2nd, 3rd, and 4th
' stories of the Wilson-Laney bent. * Compute the wind mom-
ents and check with these values: Mfi8'90,700fi1:3n3100,3003
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-102-

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-103-

Iroblem 90, Page 136

The influence for center reaction is given for the
truss illustrated. Compare with the influence lines for a
girder on the basis of 3 assumptions.

(1 uniform I

(2 triangular variation of I

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- 104-

Iroblem,91, Page 156

The influence for end reaction is given for the truss

illustrated. Compare with the influence lines for a girder
on the basis of 3 assumptions.

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~106-

Problem 95, Page 157
Use the influence lines for reaction shown in Fig.9? to

sketch influence lines for stress in the bars man-o-p-q of
the cantilever truss. Assume that the center diagonals can

resist tension only.

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Iroblem.97, Page 137

The following dimensions apply to the three hinged
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obtain the maximum values of HASopand Sanfor a uniform
D.L. of 3000# per foot and a uniform.load(L.L.) or 5000#
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IToblem 98, Page 137

Derive equation (2) of Far. 90 by use of the general
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Problem 99, Page 137

Compare the influence ordinates as given for the Niagara
Gorge Arch with those given by equation (2) and with the
semi-cubical parabola having a maximum ordinate 0.2L/h.
Compute the hdrizontaltthruSt produced by a 60‘rise in
temperature and the corresponding unit stresses in the memw
bers whose areas are given. Rise is 124 feet.

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Problem 100,2Page 157.

Same as Problem 99 except for the Hell Gate Arch as
shown.Use a '72'rise in temperature and compute Labout the
center of gravity of ,the chord areas at the crown.Use the
rise as 220 feet.For additional data on this arch refer to
Transactions A.S .C .E . 1918, pp852-113'7 .

 

 

 

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-116-

Problem 101, Page 157 .

Assume that the suspension bridge shown for Problem 103
has a pin at the midpoint of the stiffening truss.Compute
and plot the maximum moments in the stiffening truss for
each l/Bth point of the span as produced by a uniform live
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Problem 102 ,Page 15?.

Same as Problem 101 except to take the stiffening truss
without a center hinge and of uniform moment of inertia.
Let equation (9) page 129 represent the influence line for
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Problem lO4,Page 157.

Obtain the value of the horizontal cable pull caused by
a temperature drop of 60°F. for the suspension bridge of
Problem.103.Use equation (10) page 129,based upon an
average value of I of 3,000,000 inches‘.Express the results
in terms of stress in the chords of the stiffening truss if
these chords have equal areas of 67 square inches,at the
center spaced 25 feet apart.YJhat would be the approximate
percentage reduction of the cable pull if the side spans

were suspended.
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Problem 105,Page 137.
The exact value of H for the Manhattan Suspension Bridge

is 6 600,000 pounds when the Live Load is placed on the
nthe center span.lt is 347,000 pounds when the Live Load

is placed on the side spans.Compute the maximum positive
and negative deflections of the center span as suggested

in Paragraph 99.

 

 

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Problem.lO6,Page 138.

Assume that the suspension bridge shown in Problem.103
has a stiffening truss of constant section where I is
3,500,000 inches4.Each.cable has an area of 150 square inch,
and carries a Live Load of 3000 pounds per foot.Gompute the
approximate maximum.deflections with and without suspended
side spans.The cable extends 100 feet into the abutment to
the point of anchorage.The sag in the suspended side span
is found from.the relation given in Paragraph 96.

 

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-122-

Problem lO'7,Page 138.

DevelOp a general expression for temperature deflection
of a two hinged suspension bridge with straight back-stays
and use it to find deflection of the Lanhattan Bridge when
t=60° F.The exposed length of the cable is 3320 feet.

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”Problem 1<‘)s,Page lob.
Compute the approximate temperature deflection for the
bridge of Hoblem lO3,when t:65°F.

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-124-

Problem.109,Page 147.

Analyze the continuous beam shown and draw the final
moment diagram.Calculate the moments under each concentrated
load and determine the values of the maximum.moments.Locate
points of contraflexure and sketch the elastic curve for the

beam. 9 -
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froblem llO,Iage 147. ~

Analyze the continuous beam shown and draw the final
noment diagrams.Calculate the moments under each concentra-
ted load and determine the values of the maximum.moments.
locate points of contraflexure and Sketch the elastic curve

for the beam.

 

 

 

 

 

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Analyze the continuous beam shown and draw the final
moment diagrams.Calculate the moments under each concentr-
ated load and determine the values of the maximum.moments.
Locate the points of contraflexure and sketch the elastic
curve for the beam.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Analyze the continuous beam shown and draw the final

moment diagrams.Calculate the moments under each concentrat-
ed load and determine the values of the maximum.moments.
Locate the points of contraflexure and sketch the elastic

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Problem.ll3,Page 149. .
Analyze for the joint moments and draW'moment diagrams

for the frame shown.There are no joint movements.Locate
points of contraflexure and sketch the deflected structure.

 

 

 

 

 

 

 

 

 

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IToblem ll4,Page 149.
Analyze for the joint moments and draw moment diagrams
for the frame shown.There are no joint movements.Locate
points of contraflexure and sketch the deflected structure.
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Analyze for the joint moments and draw the moment diagram

for the frame shown.There are no joint movements.Locate
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Assume that a landslide moves the joint D in.Figure 122
a distance of 2 inches down and 3 inches to the right.

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Froblem.118, Page 152

Determine the maximum fiber stresses in the frame of
Fig. 122 caused by a clockwise rotation of 0.2 of the foot-
ing at D. Sketch the shape of the deflected structure.

 

 

 

 

 

 

 

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Problem 119 Page 157
Locate the points of contraflecture in the columns of

this simple bent for 3 cases of relative stiffness,(a) K9:
2K5 (b) K,.k$ (c) K9.Ke/2. Sketch the general shape of the

deflected structure.

 

 

 

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-157-

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shown.

 

 

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Problem 122, Page 157

Check the following values for the moments in the bent
with leping legs as caused by a 1 inch horizontal side
lurch where E=B0,000,000# per sq. in. Draw a moment diagram
and sketch the deflected structure.

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Problem 124, Page 157
Obtain joint moments and sketch the shape of the deflect-

ed structure for the 2 story bent shown. The members are all
of the same cross-section,ie., K values of girders are twice
the K'values of columns. Final column moments are given on

the figure.

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Iroblem 146, lace 204

-153-

Accurately determine the springing moment for the arch
uniform vertical load of 100$
per ft. over the entire span.

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Problem.160, rage £29

Determine the end momen+s for the fixed end beam of 30
foot span loaded with a uniform.load of 200# per foot over
the left half of the span.

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Problem.161, Iage 22

Compute the corner moments at B and D for the bent of
Fig. 171 and complete the moment diagram. Sketch the shape
of the deflected structure.

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-157-

Pro‘blem 162, Page .229

Obtain the moment diagram for the bent of Fig. 171 as
caused by a horizontal load of 1000# at the mid-height of
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-158-

Problem 165, Page 229

Obtain the moments in a circular ring of 10 inch diameter
supported on a point reaction and carrying a concentrated
load at the top of 1000#. If the ring is made up of a sir:
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Problem 164, Page 229

Check the moments obtained in Art.163 for the elliptical
arch by using a cantilever from the right end as a static
structure.

Hint. Slide rule accuracy will not be satisfactory here
because of the difference of large quantities involved in
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Problem 167, Page 229

The symmetrically deepened beam illustrated carries a
uniform vertical load of 450;}? per ft. Obtain the moment
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Problem.168, Page 230

Analyze the unsymmetrically deepened beam illustrated

for the effect of the concentrated load. Obtain the moment
diagram to scale. Choose a cantilever extending from the
left hand end as the static structure.

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Problem 169,Page 257
Obtain a final moment diagram for the bent of Fig. 176
and sketch the shape of the deflected structure.

 

 

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iAnalysis of an unsymmetrical bent by the conjugate beam
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Problem 170, Page 237
Make use of the column analogy and the eneral column

flemre equation to check the moments as g ven for the
ben of Problem 120. Place pins at B and C to produce Ms.

Hint. Look for a possible negative column load.

 

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-168-

Problem.172, Page 23?
Repeat the calculations of Fig. 177 for a bent of ident-

ical dimensions but where the moments of inertia of the
vertical legs are interchenged. Check the results by use

of principal axes.

 

 

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Problem 173, Page 238

Make use of the kern relationship in the column analogy
to determine a general expression for the end moment in a
fixed end beam carrying a concentrated load P at distances
a and b from.the two ends.

7" A.
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Problem 175, Page 2.42

A fixed end bent 14 ft. high and 12 ft. Wide is made up
of 10 in. 150-7}? I beams with their Webs placed in the plane
of the bent. Find the maximum fiber stress produced by
each of three movements of the left footings, namely x=lin.
yzl in. and 9:0.1'. Use the dimensions given as center line

dimensions. . ,
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17102191“; L.

lxplanction o; irnLols
deflection, usually c;us ed by the ma'n loads.
hanges of angles of a tr ienfle 0: used by strerc.
cl lange of t szg of a suspended cable.
change in length of a member.
deflect: on,ueually caused by a unit load.

change in length of a small fiber.

small change in . .

end slope of a membe.r of a rigid irame.

factor defining orange in stiffness due to end

I'FS bralnto

otation of a member
summation sign.
end anple change of a pin end member
short length,usually.a part of a span.

from joint translation.

cross-sectioxial area,1:articularly of the anal-
0 gen .3 c o lur..n n
areas of tne I and T diaerr1o.

numbers of bars in a St: ce frame .
extreme fiber distance for cross-
ing bending.

carry over factor used in balancing angles.

fixed end moment at A of the beam KP.

depth of a beam or of any cross-section.
eccentricity of a loadgcoefficient of temperature
change.-

modulus of elasticity

section resisr-

unit stres s;eag of a cable in a suspension brifige
shearing modu 11US 01 elasticity.
height of a truss or story;rise of an arch.

neutral toint fo 0 Ht rust,shear,moment.
C slope dei‘lection)

moments of inertia and product of
number of joints in a space frame.
polar moment of inertia.

defined by the relationr a kL

inertia.

stiffness or I/l; subs pts 0 'nd 5 refer to
column and girder.
length as of a truss bar or of an entire scan.

a structure.
of member AB (slope deflection ;

virtual moment at any point of

moment at end A
* about A3.

fixed end m ment,eee C

11.51286 0

kern moment.

for slope deflection

static and indeterminate moments.

total moment in a story,uind shear times story

height.
.oment of

forces about x-x and axes.
rionent of forcesin the xy or xz plane.
cesienction of ti e neutral 1oint.used as a
C‘u‘becriri’

v..-.r
If

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a.n?+1
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X V "

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Txplnnation of Cynbols
Continued

number of ejace joints Vlere all bars lie in our

J.

l

statical norent of the cross-section of the
5

'us of a moment center.

. ese of columns to girders.

total stress in a bar.

change in temperature in dgpyegg p,
torque moment caused by the virt a1 loading.
restraint factor in moment distribution.
total shear caused by the virtual loading.
unit load.

total load; Tds/EI load;tota vork.
coordinates from the neutral point.
distances referred to a centroid.

redundant stresses or reactions.

IEibliogrexlgr .
1,;nericen I.s itute of Cteel Con: t:uction-Ct el Constructior,
- 3'0, CR-C7lq -btericcrl lrstiinrte of CIteel C 1u3 -
ruct‘on,1e" Yo: 1:, 1C4l.
L. 1bbitt E.T. and Poland J.J.- “.ater Supply Tngineerin{,4€6,
TcGran-Yill,k W York, lCL9.
o. * hlehem Steel Com: nV-L-taloeue 3-39,184-EO4,Iethlehem
Steel Compan",-et lehem Ta.,l{;9.
..'ish p C.T.-Structura 1 Design, l-o-.-Z,Tokn Tiley & Con,
Ye” York,193€.
b.1cu--e" L.A.-keinforced Concrete,l€5-lCE,D.Van Yostrand,
l'er' York, 19236 .
6.»r1nter l.:. “es ifn of Todern Steel Ctruetures,.C» 595,
- o"illa-,-em York, lCél.
I.T Bleory 0:? Iod61n Steel Ctructurcs Vol.1,77-100,
lEE- lCC,~-€-.CA,Iac ille n,ZeW ‘fork,lC CE.
l .2

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LIL}

7.—rinter
.-Cleory of Loiern Steel Structures Vol.2,

c-1llan,1e" York, 1937.

9.5001 G.A. and Iiinne Y.S.-Cteel and Timber Structures,4OC-4;”,

' chr'W- Hill, eW Y101k,lC£4. '

10.Yool G.A. and Kinnc Y.C.- -Ctresses in Framed Ctructures,
4?6-445,448-46€,TcGraw—Iill,IeW Y1ork.lCL5.

ll.Laurson I.G.znd Cox T.J.- Iechanics of materie s,LOl-E.OC,
le- 0,279-u00,.Cl-.l.,.onn Yiley C Con,1eW York

F.vrinter

(“'ZC‘
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15.”eeltgrl?.i3.and Tnsigr1ITJ aAnalrtic'1.1cxfluiiics for
ne1neer 93-C9,99 -lOl,Job 0-1- -iley 8 Con,1 eW York,
753.

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13.9eeley F.B.- 1T;esistance of Tater1=ls,-€ -FF,L€€-£, ,.a.-.C€
507-317,John ‘iley C Con,le eW York, lCSB.
16.3hedd T.C.-Structura Eesign in Steel,L4 4-LSS,CCS-79€,
Joh Wiley 8 Son,Te eW York ,1954.
lC.Ckcdd T :0. end Vawter J.- Tneor" of Simple Ctructures,lEZ-l.
0-4;3,440- /FU,.0vn Iiley 8 Con,Tew York,lC4l.
l'. . ut-erla nd II.-nd IOWman T.l.-Ctructural Pesign,4—E4,Al-E7,
141-.Cfl7,JOIn1‘fileg'é‘f?on,3krr1ork,lCCJ3.
17.7utherlcnd 3.;nd Clifford T.Y.-Eeiniorced Concrete Design,
ECG-uso,.0nu Yiley C Con,TeW York,l92€.
lF.Timosh<nko 8. nd Iacfiullour “ C.T.—Tlements of Ctrength of
Tfiter1916,190-224,292- "0.,n.van estrand,.e”'Yor-,
1C36..
1C urruxart T.C.end C'1our-e C.I. Tlemente ry Structural
nrineerinr, -il€, I-cGraW--1ll,1eW York .1941.
29.Trguhart 1.0.and C'Zourke C.L.-Design of Steel Ctructures.
24-S.E,TCG aW— Till, Yew York, lCEO.
£1.Y0ung C.R.- .tructura l Iroolcms,-.l-sll John Wiley 6 Son,
16’1f‘101‘k,19.5

Tote: on Contents
Troblem 7,}age ll.
Because this structure is deter 1nate,t1e stresses in the
members can he found very easily by Writint the eeustions
of equilibiium for non—coplanar,ccncurrent forces at each
joint.Tege 13 shows the metrod of tefoulet ing these equat-
ions to expedite their solution as simultaneous equations.

Iroblem 9,pa5e 16.
This is intended to be considered only es 3 planer struct-
ure,and one ordinary methods of Tree Tociy Dia rams and the
equations of equilieriu11for ccicurrent forces can be
&:“rli€d.o

- "'"“fo_~r'*

11011101 12 ,pege lC: .
This could be solved by several methods,hut the Area
Toment lroposition is the most readily understood and
afflied.

Problem 3,:zee LC.
An engineering theorem Which follOWs from the conservation
01 ene15y,etates that the internal tork of deformation
equals the external Work of the arrlied loeda;eccordinslv
if there is but one concentrated loa d on the structui e, the
deflection of it‘s roint 01 application in the direction of
the loao. can be determined frIm the equation of real Work.

r1

F;’j 0" DJ
}—’ O o

Irohlem l€,fage

Castigliano al deals With internal Work as rroduced by
a gradually e_ ied load.Castigliano‘s law states "The
derivative of b total interlal Work With respect to any
one load is equal to the deflection of the load joint in
the direction of the load.

1J-

:1
CD

Troblem 19,3e5e £6.
The method of virtual Work as used in this problem is
Widel" accejtcd as the riost satisfactory method of deter-
zzinin5 the deflection 01 eny roint on a body,in a direction

caused by all :os .iole 12inds of internal distortions.

lroblem L4,:are 32.
Virtual Work methods are very easily affllEd to this tyre
of froblem because the steel fire has a comhination of

bending end torsion sin several different ylanes
Iroble3n29,1a e 39.
The Mr ea lement Iro_eosition method is the most erredient
11et1 od fo this tyne of rrohlen.
Treblem 31,.eze 45

Cince deflection mav 1e con uted as the nenoinr moment
in the conjugate beam,produced by it's L/EI losding,it is

 

a"

clear tint the elastic curve of the reel be3m 's ident-
ical Tith the moment diarrem for tbe conjugate beam.?cncv
ve can draw a force diefrzx.snd a runicular jolyfon for
the I/ll losds;s t33t the funicular Tolyfon Till re-
present the moment diegr3n for the conjugate been or thC
elastic curve of the re3l beam. '

Iroblem 3C,jsge El.
Tbe :f11CCIth of the brCC Toment Irorosition can be
(ItcnfiLed to the field. oi trus1s eeflCcticns,rhere it is
knOTn as the .metlod of elastic Teig hts or an €1€ Tcights.

1

Iroblem 46, page EL.
This t"pe of froolC21 best Torked by the Theorem of
Three 1oments.lt is C so cellCd. "Clcssical" method of
enelysis.The formulas ce.n be cev Clored by double
integration,conbined Titlz the Area 10Te23t Iroyosition.
It is yossible to Tork this by ct e1 met ods,out they
are less greeticul.

m

m

lroblem 41, ‘ age 4<-1ronlCm 43,3sge 49- oble1 44,:CTC
The irCs loner t lroros iti uon is tbe Tost Ir3ctic3l
solution for thee C tvres 01 yroblens.

10

Iroblem 47,935e 54-lroblCT 48, tCTe SE.
The Theorem of Three Zoments ,ordinarily arrlied to a
continuous beam,is here arplied to a ombinzztion of
continuous and res trained beams .“e can apply it to a
restr3inCd besm by consie ering the restrsint to be re-
noved 3nd reylaced vith a beam of zero lcngth and zero
10 ad.

Troblem 49,:uyc 57.
This problem is Torked by the fonen nt T istribution methce.
The moments 3t 33? SN? orts are found in the oruin:“f
”a? THQ then are distributed to cuch been in the ratio
05 their I ValU€3.-see bottom rare 5?.

iroblen Ml, CL 59.

This trusse is 1nteinullr inceterTinate to t"1ie first
de51CC31e1ce,*C rCrmve:lxgr to oth.i_n 'Uu: st3tic struct-
ure.The stress in each bar is xrr-ssed in terms of tha
erylied losd,3nd the red.unda nt stress X.A redundant be:
is a bar that is oCT11ncc to take only one kind of strr s
and is very often subjected to the cryosite kind.Under
these condit1313 it does not so 0t and hence is culled
inactive or re lui 3+.

‘_ "J g} _ :1 1L ' , '~ “Lu ‘1‘ C; :3 ' .
m;i3 133 i3 botd iiternally and externally i;15Cter-
minute to the first defree. eice,Te rCLnove 0110 reucti on
uni cut one bwr to obt3in the static st1 flucture. Le strC;
i ~.;3c“ 'b31 i“ 'T‘:Cg,_;7 551 t 1‘“; T7‘ tze L5““1jxw7 131.1.

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ions on a oiairs1 oi the structure.
fi-Co: FVItC 5. tii nttion 1?;c tors, ;t 1117 jo
d thFSQ vglurs inside a bot at Each 30 oint.
nt “fiir =1os at 1&r5x371n1wefil nc(‘1‘fio cc71t,
e =n i7gjznomcnt 1r0"11;V=r71"ttIoution
-1 b lunc1n” hOHrzit un5ntctutu th-
corrcsro on: i 7 En actent.1»11nninL moments arc of
of os ‘te s to ,o to unoalgnce5 joint momcnt.5“1” a
hor1207t;1 ins U-Q€3 ouch 13s13ncinr mowcnt,to snow

7,. . . .1. .~ .43. ' . 1 11,7 , ., :7
tfiaut 11:5 ruvwc;1us TIbQVT: bin}; 1 :79 c151- o11117cc:1 urn; Dotti

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3 5rticle.
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4- MC IV‘ tux. 0; EH! OXEJ.-IOKKN1to ‘.t 7116 111 ermhs 0; too
fi'“*cr" Tlico tag cart“ ovor fioficn s in a onlnwn unloi-

the fix :1 end. r.1o:1e~nts r the b-,l..:1cinj' 7o ’16th exietmi
t‘L:r%?.':1 11 11‘r3r 07": 10‘1e11t in: of? 51:2. segue: ingri ;.s tTva
oc7um:Sforr113“1111113171: 1K) (n5, mat CL. one;31111' it's
vslte.T1cy JHJt be cons degei us eliitionel fixed end
‘zo1ents it the jelnts Trage- th e are reoor-ie'l.
o-Tuccess1vely bulznce (so; oth:i joint of the structure,
141211“ c-1ne to iCCOifii a lsnciIr rfionents L911 ,urry ovcr‘
£1ctor3 ”1th rroner sifn.Lote tflut the c:r“y over ”o:-
ents unlmjgfin3 ,‘-eviousl;'lyfihnice5 joizte,1.11 tF1t sue?

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liorfiizorrtml. 1.1Jle? .1: : caniri.-e1¢,1 ;-111113

below tle
tge frocess o rebulgncizig.
r571 .. rm '.~,. 4 ° .11 11' 11 1'.’ -
6-11e lTOCgoS o belen011_ dCl rice is cwg71e1 0:1 in mJJ
' - .1 ~ 1, .. 1. . ..-,-. .1. . '7 -.
deitire l s 171e-'rce,17.n l i11e “:13: cine; tie 1:1ts oeCCLie

*3

ti
neg ifioly s1ull,'“ich snows tiat Lie joints are
sult will be reuched after 6101 JOlnt

rw-vfifi

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1x11;nice;3 -.-.;is 17e
171.5 seen seleneei about three tir1es.-_ie tine required

should not £713.65 tcn :T-ii‘ttiiteas :te1 S3‘un for a. continuous
bea3,even Then t f513 is usfle b7r an inexgerienceg

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opfrator.

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117

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9811»o _!_

Troblen 113,;19e 118.
Th's is the ernlicet io n o: t"e Tuluncine “omen
to a; fIKZ:1€,Iin?VtC££i of‘:1 (Nlt_.JIO” beten.Zfiie .o
xsct1y tin :ner::1s xilsined in txe frevious e“t

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lroblen 117,3sfe 132.

Ch s is 1 sinrle case of joint tr s11tion,”h101 me?
occur or Locieent or 5ee75n.1oreve1 tie re e.'ular metlod

01 o1luicin" n ne:nts,c oujle5 with a sinnle ejjlicetion
of t1e Are; lonents cwn be used.

lroblen 1:4,;w3e 39.
Cne cunnot fully underetsni hi method of successive
correcti ns,u71ess he can form a mentul ficture of the
thys1 icel be uvior of tie structure,the corresronds to
eu.h step of tie process.It is desirable. ,to consid er
hat joint rotation end joint trs.c Ml tion as seyernte
uovements to be permitted eltern t(ld.in. structure is
allowed to move with joint 3 fi: {01 against rotetion.Then,
the joints ng sllofei to re tete, out the structure is
res rained a
1:11.165 evs FOE
def lectei ruo
moveJ..
Since esch Joint is to be bulsnced only once before
additional fi"ed er M5 moments ere introduced into the
colunn,it is known eheud of t fie,tbst such correction
moments rill be rositive.TbJ olloming rules should be

C inst trftnsletion.Tnis is permitted as meny
sible, ntil tiie St ructure reaches it's trur
sitionu,witn all 1m157qerr restruints re-

helpful in tlis kind of solut .

l.-The structure is allowed t deflect unier tie 1015:
”Wile tie joints ere fixei a? inst r05; tion an? the
:fizcfi ( 11;; :ents turf c 1cu31 ted.

«fle'm n-

‘1 y-r
.\ C-

“,7.

¢.-The fixed end monents ere be]_1nced ground the joints
initil each joint is in CCu1-_11111°tutt is the
summation of moments at ee,h are equal to zero.
DurinC thi.s :rocess, no t11nslutions of the joints
are sllo1'ed. 7113 see: is the sex-1e as the Cross
met 01 of onlsncing fixei end moments.

3. -1T 1 the bslznccd monents of step 2,the colux1n she rs
Lirer<ietr1fi11ne 1,1r1:1:?ro:1 the Liiiierencems bei'"e€*1 t11I
onlan£.UJ-L story s7we1rs axul 'U1e true storfirss1eurs , We
artificial restraininf forces at the joints are fc 1d.

4.-'.L".‘1C artii'icigl joint resfrt ;:.intsfo*.1n:1 in s'ep 3 or
CliI in ted by srfilyin; CC uwl end orgositc joint
force . -

5.-The e ru ture is nllovcd to ocflec With fixed joints
under tn action of this new force sysf n,;ixcd «PL
:xxnents re 0 lcnlctco,ioirflziwvrcnts ere belunced,ci.'
1 second set of joint restr'ints is determinCi.It
will be found tflbt t cond set of joint res-
traints is smaller c, less ingortance t1sn the
first se t.

e.-$n’ frocess isd re11ectei until a final set of joint

roctrriit is rter rmir ed that is small enough to

neglect entirely.

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7. -fhe iinal monents are obtained. b: so 1n? slnebrnic-
filly, tie several sets of oils cei nonents obtained
iro:a the or: ginsl loads,;nd tie several sets of
forces that bclwnce the erti:iciel joint restrsintC.

A brnt such as this one is best analyzed for Wind mom—
ents by the Central met“od of successive corrections.
1&3 5.;1tsridJLC jOlJTt, one rnnrt be afi;le tx><lisL1 ijr1te t7(
{iieoixg'rrcLau1ei7nnr the '“ind. forcens,o'en1f 17v: coljr1ns
fihen the joints are held efuinst otot101.Lnis distric-
ution is made according to t e stifiness ctio.
lroblem 1.6 ,;:;;e 53.
The le ng Mti of the arch ring is divided by triel with ;
1wir of dividers into eight s~ct- ons,mnion scale
or “TOD-Ci?" @13le 4.25 feet Cool-1.5110 net 01 the. divis-
ions ure t e-n locwtr1,t1( cooro inLtes x and y from the
left hand syringing,of {och center point being s oiled.
The arch is cut E.;.’.'.'L-.3f at the left abutment;t71us it be-
comes a cantilever from the right hsnd end.ine value
of ds/Tlg"fiich is constant for each division,vill be
t: ken as unity.31ncc it eyfe rs in e-s cn tern of the
Siymltgneous courtion tot will no "fritteIn,ds/.TII nay
be Cive n a unit 1elne,tfienever it is constant.

roblen,lffl,yere 154.
'fizis TTOEHETI is solvei.1vr tfie "Coljrn1.flnrlOFfi”‘:nethodJ

 

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t Concertion". This Icutr l loint Conccption ullofi:
b0 refrc $6-1t C€rtoin Arch rrorcrtics as zcro 9 ant—
ics,ii"U1€ ori‘“H1<3T coorfifrugtes,01'iflMT'roint o-
19111c1t101 of ti‘ relun4 nt regctions (f,‘f ‘) is

c pcr*cl to .ccnt,oid.of ‘U C {lastic '"‘ic1ts bflii
whcr- cit c -X or y-y axis is LU LTiG of syn1c-
tlf'. '191"‘01 :To1‘ a szttxmivzwi Sj1r1cinfic a:rc:1,191{ IT.T;
is 011 tnx '\C1 1cn1.:c‘is (n? sictfirtsz.

In the d? €10“: cnt oT 11c cola”n snLlony Tron the
lwn1tral o 1rt21ct od, tix1f.o llo“1nr {rut-tion '3 :JfiPiVGF
at; fi¢£s/¢}:ZE¥&QQQHL9'* sagij

This com-tion, "111101 163111936118 the indetc:_':’:1inatc nor-1-
{nt ”1CU~564 by the ncntrcl roint rcgction,corrcsronis

3(4-

141 T0121 to 1116 ccrr1on c*1u.tiorl for 11-161 ix1inr"ULL
Cures-:1 ill b3: 11 6006? d135i0c~l1 L'f lOCmiCd. 00111;.II’11—l ‘e-‘Ij’
(I
5= 7% r ”3f t”?!—

In tl‘ic co ”1321151 tc 11.6.1.1 r11{'-1t 1011, one visu;‘..lize.'l the ”06-1-11
LS lOLLO-(Td ”ltd t-1C :7th ‘-"-iC~:rC~1"l,:19-DC€ t1]. prg 911071131 bC
no (u:“ficulfi" in V4C11-l1"‘rui it 1mxre,¢¢ t1 logd.amrtin:
uron an unulocoua coln1n.11 such a column is visual-
ized,A must be it's cross scctional urcc,IX-x, Iy-y the
moments of inprtia about he X-X and y-y 8.118 through
the neutral point.fhis analogcus c011 1111 is loaded With
an imarinory distrioutcd load ”,1nich,bccanse of it's
ccccntri it" “ro4nces tiltinc m0“€nts Iy-x -:m. Ty-y
Ubout the ccntroidul UIis.Then t11e indc crminate
mon{nt L; is loun<:1—t any noint in t { ring as .16

a

stress CIi-.t1nc there in t.1is ccccntriCUlly loaded
colLRn.

-1. ‘3-

A Short Discussion of the Theory of Three homents
and Area K ment Pro3osition ,usin3 the Eendin3
Ioant D1C 3ram by Iarts A33lication.

The originator of the Theory O.L Three I.ments,:mile
Cla3eyron (1799-1864),was the consultant en3ineer on some
ver3 im3or tCCnt research 1or]: carried on in Iuss ia i1 the
JCar 1534.Ie tested t:1e stren t11 of Iron brid3es that were
encouz1tCer "1ile on a military m'ssion for tI 1e rench
Government.

In 1583,he set forth a thCory conccrning the eCuality
of internal work of stresses, to tie externCl vork of the
loads.This theory is the basic frinci3Cl of our mode rn
theory of indeterminate structures

About 1957,he 3resented the Theory of Three homents for

he case of a uniformly loaded C3an of uniform cross-section.
Due to Professor‘Cla3eyron's retiring nature,he never 3ub-
lished this theor3r.Tne importance and wide application of
the method has since been realised, and it is no*: given
ention as one of 5 e 11rortant t1C ories regarding t1e sol-
ution of beams and frames.

The met1od involves the Considering of any three succ-
Css ive s3ans of any continuous span beam,startin3 from the
left end,as one unit.ICuations are th en formed in terms of
the bendin3 moment over each su3301t,len3th of each of the
three ejans,and the s1atical moment of the bendin3 moment
diagram.This bending moment dia3ram is a result of whatever
system of loading is present on the s an.

The usual a331ication of t1is method has been to con-
tinuous beams,s'm31y su3301 ted at their extreme ends.The
beams can be of any number of spans of the same or differentS
lengths,and of the same or d fierent cross-sext ons.T.e 10
can be of any Kind,concent1ated,un11orm,o1 any any combination
of the two.

‘T1ile it is not: nenti oned in ter t books,it can ale 0 bC
a33lied very advanta3.eously to otIer struo urCs, uc1 as
four sided fre1es and restrained be3111s .This is quite a uni; e
1331ication of the L'hCory of Three loments,and on numerous
occasions in this Thesis,it has been used as the princiral
:1Cthod of solution.2he a331ication to res rained beams offers
a much more simple solution than any of the standard met I1ods.

To C33ly it to a restrained beam (one embedded in masonry
at both ends),re must remove the restraint from each Cnd,and
replace the same with a spa n of zero length,an zero load.

“C then have the equivalent or three spans,loaded as desired.
Cur equations,as mentioned at the be3innin; of this article,
can then be a331ied.

This cun still be a very com3liCCth 3rOCCdure, if we
I t 1e ordinary bending: :1ozz1C11t Ciarrav.1e.--o.*ever by usinr
C16 Area Ioment Iro3osition mCtAOu,.1tn he Bending oment

_iC3ram by “”rt1.111 Cad of a com3osite bending moment dia-
11am,we can sim3lify the taking of the statical - nts.
/ The bending mo 1Cnt d1CCrC1 by CC rts,as used in these

To1”fi " 9 “ ‘ "_’ t t T _-Ifir” t7' ;:fi' Torcnt

Troy Sition.It means,th:t no plot the oc1zlin§ Hoicnt din rsx

for each 1031 sererately, instead of drawi.n¢ tie conjosite
Lifrgm.-bis Wiles it much easier to find the distance fror

o roqent center to the centroid of the ’eniinr noment din—i
firm.

'ncse statical moments are thin inoorforated into the
fronOSWtion tt.'t The statical moment of the area of tLe

xcniing LOWLnt 51a 1am h ct"eLn any tro oint: L LnC T,
'ividtfl.1;r the :ztfihutt of I113.EZTC€H1U£T(V€ elas‘::«3itj,:xxi
T,t*c gortrrtcy? inertia,ic («nuil to the rLchxrtion of one
rm? tne 3 int: CRKT"L o: 7{3£V”'H‘C other. Fox-'Wi"tunCL,if TE
LIL tlLL FfL‘tiCLl ryoncnt of t’pc LIL; -T :nt tlic :0 nt T,t}:er~
'C ‘oulo find the dcflection of B ehotL or Eclon'A.

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