Download Theory of Indeterminate Structures_Syed Ali Rizwan...
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A Publication of Pakistan Engineering Congress, Lahore
TI1[ORY
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INDrTI:I!~IN4TI:
8TRUCTURI:S Second Edition
Syed Ali Rizwan . Professor of Civil Engineering . University of Engineering & Technology (UET), Lahore, Pakistan.
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PREFACE TO THE FIRST EDITION This book is a compilation of lectures delivered by the author to tp.e Civil Engineering students wherein internationally recognized books on the topic were followed with an entirely changed approach thus making it easy to understand the procedure for the solution of indeterminate structures. The book is useful for the undergraduate students taking a course on classic.al indeterminate structural analysis. No lengthy derivations are given and emphasis is on application. Method of Rotation. Contribution, which is an extension of , moment distribution method, has been added. All forces in parallel chord trusses have been. determined by the "Method of moments and Shears" which has been developed as an extension of method of sections in the first chapter. The emphasis is on explaining the basic concepts and their comprehension by the students and therefore only very few typically . select~d problems have been solved in steps in most of the cases to establish that structura1 anslysis follows a routine set of steps. No extra unsolved problems have been given in this edition of book, which may be added in subsequent editions if demanded by the readers. It is hoped that the effort shall be appreciated and more people would like to write books .on structural analysis. Readers are especially requested to forward their suggestions for future improvements to the publisher.
PREFACE TO THE SECOND EDITION The first edition of this book has been a huge success. After its carefuI.:.study, the top Civil Engineering students as \vell as professional engineers gave some valuable Gomments. These included upgrading the entire text on MS-Equation editor, possible inclusion of three more chapters on the topics of statically determinate arches, influence lines and the three momerit equation. They also suggested to include unsolved problems alongwith answers at the end of the book. All these suggestions have been fully incorporated in this revised, updated and eniarged Second Edition of the book. It was a monumental and a huge task which has been accomplished by the . grace of God almighty. The author is really indebted to the Pakistan Engineering Congress for bearing all the expenses incurred on the publication of this book. Without this financial support, it would not have been possible for the author to bring out this edition. It is hoped that the book would be liked by the students of Engineering Universities and Technical Colleges of the country and would be the text book for undergraduate Civil Engineering and architec.tural engineering students 111 the subject of structural analysis. Readers are requested to forward corrections and suggestions at the following E-mail address for further improvements. '",,:
[email protected]
..
.i I
ACKNOWLEDGEMENT Pakistan Engineering Congress is the oldest and the most prestigious and multidisciplinary organization of qualified engineers. It was established in 1912 and has been contributing. for the spread of engineering knowledge and know-how in the shape of Seminars, Symposia, congresses and Annual Conventions. The proceeding J.. '.~'
of Engineering Congress are kept for record and reference for engineering community. Now the Pakistan Engineering Congress lJas taken a very bold step to publish the /
books of some prominent authors on engineering issues. where the authors due to . . financial constraints are not in a position to publish the same from their own expenses. Accordingly Pakistan Engineering Congress paid partially for the publication of 15t Edition of this Book and now 2nd Edition is being published for which all the expenses have been borne by PEe. The Pakistan Engineering Congress is . thankful to the author, Syed Ali Rizwan, Professor of Civil Engineering Department, UET Lahore who has worked very hard to produce this excellent book which will be particularly very useful for the students in addition to practicing engmeers.
Engr. Ch. Ghulam Hussain Secretary PEe
I
.. . ~y BfTOV[O fATH[1! Syed Karamat Ali Shah (Late)
May God Almighty bless his soul with eternal peace and May Holy Prophet Mohammad (Salallaho Alahay Waaalehi-Wassallam) grant him His kind Shifaat (with the permission of God Almightyy on the Day of Judgement. Ameen.
TABLE OF CONTENTS .. DES.CRIPTION
Page #
Chapter 'One - Stability,.Determinacy o(Structures and Consistent Deformation Method ............... ,...................................... ,................................... 1
..
-}
Stability of Structures, Stable Structures, Articulated Structures, Continuous Frame, Determinacy .............. :.......................................................................... ;... :................................ 2
-}
;External Indeterminacy ...... ~ .....................................................................................: .................. 2
-}
Internal Indeterminacy ............ :............... ;............... :...................... ;.;· ............................................ 3
-}
Total Indeterminacy ................................................................................................................... 5
-}
Examples ........................................................................................................ :................... 6 -- 15
-}
Methods for finding deflections and rotations, Moment area Theorem No. (1) ............ ;............ 15
-}
Moment area Theorem No. (2) .................................................'. ............. :........... ·...................... 16
-}
Sign Convention for deflections, bending moment diagram by parts ......................................... 17.·
-}
Sign Convention for Shear force and bending moment ............................................................. 18
-}
First and Second theorem of Conjugate heam method .............................................................. 22
-}
. Types of Elastic Strain Energy .............................................................................................. :~. 24
.-}
Castigliano's theorems.·........................................... , ...; ............................................................. 25
-}
Consistent Deformation Method ............................ :..................................................... 25
-}
Propped cantilever Analysis ..... ·................................................................................................... 28·
~.
Alternate solution of propped cantilever .............................................. , ................................... 32
-}
Analysis of Fixed ended beam ..................................................... :............................................ 33
-}
Inversion of a Matrix ................................................................................................................ 40
-}
Second degree externally indeterminate beam ............................ ;............................................. 41
-}
~hird degree externally indeterminate frame ............................................................................ 47
-}
Analysis of Externally Indeterminate Trusses ........................................................................... 56
-}
Method of moments and shears ................................................................................... 56
-}
Solution Of Truss by moment an·d Shear Method ........ , ............................................................. 58 .
,:.:"-'
--}
First degree externally redundant Trusses ................................................................................ 59
--}
Second degree externally redundant Trusses ....................................................................... :.... 63
--}
Solution of 3rd degree externally indeterminate trusses ......................................................... :.. 68
--}
Analysis of 3rd degree indeterminate frames ............................................................................ 74
--}
Analysis of 3rd degree c;xternally indeterminate continuous healn .. :....................... :.................. 86
Chapter Two - Method of Least work ................................................................. 91 --}
Solution of Propped cantilever by least work ........................................................................... 92
--}
Analysis of Second degree externally redundant beams .......................................................... 100
--}
Internal Indeterminacy of Structures ........................................ : ............................................. 102
--}
Analysis of First degree internally redundant truss by least work ....... :.................................... 105
--}
Steps for solution of internally indeterminate Truss by least work .......................................... 114
--}
Analysis of Second degree indeterminate Trusses ...................................................... :........... 120
--}
Analysis of Second degree internally indeterminate Trusses carrying gravity and lateral loads ........................................................................................................................... 126
--}
Analysis of Second degree internally indeterminate Truss by Unii load Method ..................... 138
--}
More Examples of First and Second degree internally indeterminate Trusses .................. 143-154
Chapter Three - Introduction to two-hinged Arches ......................................... 155 --}
Types of Arches, Linear Arch ................................................................................................ 158
--}
Analysis of two-hinged segmental arches ............................................................................... 159
--}
Analysis of two-hinged segmental arches earring gravity and lateral loads ............................. 161
--}
Analysis of two hinged circular Arches .................................................................... :............. 169
--}
Arches with secant Variation of inertia .................................................................................. 174
--}
Analysis of two hinged Parabolic Arches ............................................................................... 182
--}
Eddy's theorem ....................................................................... ,.............................................. 185
Chapter Four - Slope-Deflection Method ............................................. ~ ............ 199 --}
Analysis of externally redundant beam due to gravity loads ................................................... 203
--}
Analysis of indeterminate beam due to support settlements ... ~ ........................... ~ ................... 216
--}
Analysis of Frames - without sidesway ....................................... :........................................... 233
--}
Analysis of Frames - with sidesway ....................................................................................... 240
--}
An~lysis
--}
Double-Storeyed frames with sidesway .................................................................................. 249
of two-bay frames with unequal column heights ........................................................ 246
Chapter Five - The Moment Distribution Method ............................................ 258 {>
Stiffness Factor, Carryover factor and Distribution Factor ..................................................... 260
{>
Steps involved in moment - distribution Method - Application to beams ............................... 262
{>
Application to frames - without sidesway .. :........................................................................... 274
{>
Analysis of Double-Storey frame carrying gravity and lateral loads: ....................................... 281
Chapter Six - Kani's lVlethod or Rotation Contribution Method ..................... 288 {>
Rules for Calculating Rotation Contributions - First and Onward Cycles .......................... :..... 290
{>
A typical Solution 01 continuous beam by Kani's Method ...................................................... 292
{>
A typical Solution of Frame without Sidesway ........................................................................ 294
{>
A typical Solution of Frame with Sidesway ............................................................................ 295
{>
Analysis of double-storeyed Frame carrying gravity and Lateral Loads .................................. 298
{>
Storey shear, storey moment .................................................................................................. 299
Chapter Seven - Introduction to Column Analogy Method ............................. 306 {>
Introduction, basic rules and sign conventions ....................................................................... 307
{>
Fixed ended beam by column analogy ...................... .- ............................................................ 308
{>
Fixed ended beam with variable cross-section ......................................................................... 319
{>
Stiffness and carryover factor for straight members with constant section .............................. 328
{>
Frames with one-axis of symmetry .......................................................................................... 332
{>
Analysis of gable frame by column analogy ........................................................................... 340
. {>
Portal Frame carrying gravity and lateral loads ...................................................................... 343 .
{>
Analysis of non-Prismatic fixed ended beam .......................................................................... 346
Chapter Eight - Plastic Analysis of Steel Structures .................. :..................... 350 {>
Shape Factor and its calculation for variolls sections ............. :................................................ 353
{>
Collapse load of a structure, Assumptions made in plastic theory, Fundamental Theorems of Plastic collapse ........................................................................................ , ........ 356
{>
Analysis of continuous beam by Mechanism method .............................................................. 359
{>.
Types of Collapse .................... :............................................................................ ~ ................. 362 Po~talF£am.€}. s:;.fl-Iryi!1g,9rilYi}y ami lateral load ................... ,.:............... , ...... : ......................... ·~?+~::~~H~~l*~~;~~~~~f~;>~~~'~.~.~:~·~·:: ":;:~: : 20 This truss of fig. 1.10 isstable & internally indetenninate to 1st degree. 2·
6
10
13
Fig. l.!1.
b= 16 r = 3
j = 10 b + r = 2j 17+3=2x10 20 = 20 This truss is Unstable by inspection although the criterion equation is satisfied. The members in indicated square may get displaced and rotated due to gravity loads. Always inspect member positions. Insert one member in the encircled box or manage prevention of sliding by external supports to make it stable.
NOTE:- The difference between the internal and the external indetenninacy is only in the definition of 'r' 1.4.3.
TOT At: lI:NDETERMINACY The question of total indeterminacy is of little interest and \\'e have got diffcient equations for different types Qf structures. For example, the previous equation, i.e., b + r = 2 j can be used to check the total degree of indetenninacy of an articulated structure like truss by slightly modifying the definition of "r~' which should now' be considered as the "total 'number of reactive components available".
6
THEORY OF INDETERMINATE STRUCTURES
b + r where
= 2j
b = Total number of bars. r = Total number of reactive components available.
j = Total number of joints Example No.1: Determine the external and internal conditions of stability and determinateness for the following structures:-
·1
, I .~
Fig. 1.12 . (i)
External Stability And Determinacy:- . Number of reactive components available = 2 Number of equations of equilibrium available = 3 Unstable. (Visible also)
(ii)
Internal Stability And Determinacy b =9 r = 3 j = 6 b + r = 2j
9+3=2x6 12 = 12 Degree ofIndeterminacy = D = 12 - 12 = 0 Stable aIld Internally Determinate, if arrangement is impr
>1'
LOCATION OF POINTS OF CONTRA FLEXURE :- These are in Span BC. A as origin. MX I
Write moment expression and equate to zero. = .,.. 4.602 XI
+ 25.434 (XI - 3 )
- 4.602 XI + 25.434 XI - 76.302 = 0
XI
D as origin.
MX2
= 3.663 m from A.
Write moment expression and equate to zero. = - 2.096
X2
+ 21.264 (X2
. - 2.096 X 2 + 21.264 X 2 19.168 X 2 X2
-
106.32
0
Ii I I!
-
- 5) = 0 106.32
=0
=0
5.547 m.
These locations are marked above in BMD.
I
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
47
3 RD DEGREE INDETERMINACY:Exampie No.4:
1.15.
Solve the frame shown below by consistent defonnation method. 10KN B .----"-------, C 2m 2m 3m
outer sides
/
4m outer sides
20 KN
EI:::;Constant
-
A
outer sides
'"
/
D
3m
inner sides
-
B.M is +ve for Tension on inner sides
Fig. 2.32
1.15.1. SOLU1:I0N: Sign conventionJor S.F. and B.M. remains the same and are shown above as well. In this case, any force or'moment which creates tension on the inner side of a frame would be considered as a (+ve) B.M. Removing right hand support to get BDS. The loads create three defennations as shoWn. 10KN B
B
C
2m
C
2m 4m
'3
6m
+
20KN
l'
D .6.DH
2
3m
eo
A
8~dV
.6. DV A
1
0
x--7<
J:. ddh
4--
8ddh
Fig. 2.33 (a) M - Diagram Fig. 2.33 (b) mH-Diagram ADH:::; Deflection of point D in horizontal direction due to applied loads on BDS. LlDV = Deflection of point D in vertical direction due to applied loads on BDS. e D = Rotation of point D due to applied loads on BDS; B 'c 'B C
Note:
4m
4m
4m
4m·
6m
+ 6m
I
t mv_Diagram
OCddvl
D
X ex: dd A
) 1
)( ex:'ddh
me-Diagram
1\
,I
1\
I.\
Fig. 2.33c B.O.S. under unit vertical , redundant force at D
Fig. 2.33d B.D.S. under unit rotational redundant moment at D
48
THEORY OF INDETERMINATE STRUCTURES
Where (See mH diagram Fig. 2.33b) 8ddh = Deflection of point D due to unit load at D in horizontal direction acting on BDS. 8'ddv = Deflection of point D, in vertical direction due to unit load at D in horizontal direction. a'ddh= Rotation of point D, due to unit load in horizontal direction at D acting on BDS. (See mV diagram Fig: 2.33c) 8ddv = Deflection of point D due to unit load at D in vertical direction. 8'ddh = Deflection of point D (in horizontal direction) due to unit vertical load at D. a'ddv = Rotation of point D due to unit vertical load at D. (See me diagram Fig: 2.33d» a'ddh = Horizontal deflection of point D due to unit moment at D. a'ddv = Vertical deflection of point D due to unit moment at D. add = Rotation of point D due to unit moment at D.
Compatibility equations :L\DH + Ho x cSddh + Vox 8' ddv + Mo x oc'ddh = 0 (1) Compatibility in horizontal direction at D. L\Dv + Ho x cS'ddh + Vo x cSdd v + Mo x oc'dd v = 0 (2) Compatibility in vertical direction at D. eo + Ho x oc'ddh + Vo x oc'ddv + Mo x ocdd = 0 (3) Compatibility of rotation at D Now evaluate flexibility co-efficients used in above three equations .. We know that L\ or e = J ~I (Mmdx)
. There are 12 co-efficients to be evaluated in above three equations.
So
L\DH
=
JMxmH EI dx
(1)
cSddh
=
J(mH)2 dx EI
(2)
fmH EImv dx
(3)
JM x (mv) dx EI
(4)
cS'ddv =
J{mH x mv} dx EI
(5)
cS ddv =
rmv)2 dx EI
(6)
cS'ddh =
L\Dv
=
d oc'ddv = Jmvxme EI x
(7)
,:
'.' " \
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
=. EI1
f
(8)
J
(9)
f.
(10)
( M )( me) dx
I (mH)( me) dx cx:'.ddh = EI
1 cx:'ddv = EI
cx:dd
=
(mv) (.mB) dx
iI f (me l
(11)
dx
MUltiplying the corresponding moment expressions in above equations, we can evaluate above deformations. Draw M-diagram.
D
3m
i
80 KN-m
~
A
x
~T..-
r
20KN M- Diagram
10KN
M = 10 x 2 + 20 x 3 = + 80KN-m Fig. 2.34 B.D.S under applied loads M - Diagram by parts
! xl
10KN 20KN-m
c
2oKN-mG
20KN
3m
4m
10KN
E 3m
I
I
I
·1
80KN-~
t
20KN
~ A~
t
10KN
M=20 x 6-20 x 3 - BO = 20KN-m
49
50
THEORY OF INDETERMINATE STRUCTURES
B
.'
4m E
6m
D~1
D
MH- Diagram
'4-. Fig 2.34b
Fig.2.34a
~
~ 0
4
j1
I
F
i
C
,t,
1'1.
1 .
E
.\
E
~1
D
~1
I
D
~A
A
4~i1
lj,
1 me-diagram (by parts)
my-diagram (by parts)
Fig 2.34c
I\ l
Fig 2.34d
Moments expressions in vCirious members can now be written in a tabular fOnTI. Portion AE BE BF CF CD
Origin A B B C D
Limits 0-:-3
0-3 0-2 0-4 0-4
M 20X - 80 -20 lOX - 20 0 0
mH X-2
-X+4 4 4 X
Put these moment expressions, integrate and evaluate co-efficients
mv -4 -4 X-4 -X 0
Me' -1 -1 -1 -1 -1
i j 1
I
I
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
;:
~I
=
EI
1
I
M(mH) dX
[3f (20X - SO)(X - 2) dX + J3(-X+4)(-20) dX + J(lOX: 2 - 20) 4 dX + 0 + 0] o
=
0 0 ,
~I [t (20X2 ~ ~OX - 40X + 160) +
!
(20X - SO ?dX+! ( 40X - 80) dXJ -
3 3 2 40X2 1 [120X SOX2 4)(2 1 20X --, 3 -------+160X +---soxl + 1---SOX EI 3 2 2 0 2 0 2
= ........
3
~
14] 0
'J
1 [(20 X 3 " = EI -3 - - 40 x 3- - 20 x (3t + 160 x 3) + (10 x 9 - . SO" x 3) + (20 x 4 - SO x 2)
110 = - EI
8 ddh
=.l , EI
Ie
mH)2 _ dX
8ddh = 109.33 EI
8'ddV
=
~I
I( mH)( mv) dX
=~I[Jo (X-2){ -4) dX + J(-X+4)(-4) dX+J (4 )(X-4) dx+f 4 (-X) dX +oJ 0 0 0
1[3
3
,2
2
]
= EI !(-4X+S)dX+!(4X-16)dX+!(4X-16)dX+ !-4'XdX
1 EI
= -
[I
4X2 ' '4X2 3 4x2 2 3 + 8X I + I - -16X I + I 16X I + 2 2 02 0 0
- -
I -4X212] 20
51
52
THEORY OF INDETERMINATE STRUCTURES
= iI
[I -
2 x (3/ + 8 x 3 1 + (2
X
32
16 x 3) + ( 2
-
1 = iI
22 - 16 x 2) + (- 2
X
22)]
= -~I
8'ddV
cx:'ddh
X
f
(mH)( m8) dX
[3
3
2
2
4
]
1 =EI ! (-1 )(X-2)dX +! (-I)(-X+4)dX+! -4dX+! -4dX +! -XdX
1
= EI
[I -T+ X/ X2
2
3
o
= iI [
X2 3 2 2 + 1T-4X/ + 1-4X/ + 1-4x/ + 0
0
0
(-~ + 2 x 3) + (~- 4 x 3) + (- 4 x 2) + (- 4 x 2) + (_~2
I
u'ddh =
80
= iI
1
= EI
X2l4]
I-T
0
-
0)]
-~I
f
M ( m8 ) dX
[3J- (20X - 80) dX + J320 dX + J2(-lOX + 20) dX + 0 + 0] 0 0 0
=
1
EI
[I
2
20X
.. 3
3
'i
2
I
21 ]
10X
--2-+80X~ + 120x~ +1--2-+20X~
1
= iI [(-10 x 32 + 80 x 3) + (20 x 3) + (- 5 x 4 + 20 x 2)]
230
18o -_ EI 1
L1 Dy
= =
iT fM ( my ) dX 1[3 3 2 . ] EI.! (20X - 80)(-4) dX +! (-20) (-4) dX +! (lOX -20)(X ~ 4) dX + 0 + 0 1[3!(- 80X + 320) dX +!380 dX + ~2(IOX
= EI
2
-
1
20X - 40X + 80) dX..J
1
I j
I
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
=
iI
[(-40 x
9
+ 320 x 3) + (80 x 3) +
C30 x 8 - 30 x 4 + 80 x 2)J
l@v=~1 o'ddh
= =
iI I( iI[i
mH)( mv) dX
=
iI [I
0
= -EI1
iI
0
J
0
0
.
1 = EI
J- 4XdX] , 0
[I --+8X\ 4X2 4X2 4X2 4X 1J + !--16X! + !--16X\ +-, 2
3'
3
0,2
0
2
2
2,
020
[(-2 x 9 + 8 x 3) + (2 x 9 - 16 x 3) + (2 x 4 - 16 x 2) + (-2 x 4)]
IO'ddh
Oddv
0
(-4X + 8)dX + J(4X - 16) dX + (4X -16) dX +
o
=
+J 4 (X-4) dX + f- 4XdX+ 0]
(X-2) (-4) dX + J(-X + 4) (-4) dX
o
'J( mv2 ) dX '
=
-MI
.
=iI ~16dX+~ 16dX+~(X2:""8X+16)dX+~+X2dX 3
3
2
'
2
[
l[ "
,(8,
]
,
') (8)J
=EI (16x3)+(16x3)+ 3-4x4+16x2 + +3"
Oddv
=
Ii7.33 EI
53
54
THEORY OF INDETERMINATE STRUCTURES '~,'
o:'ddv
1 = .EI
f
1 = EI
[3f +4dX+f3+4 dx+f2(-X + 4) dx+f2XdX]
mv x me dX
o
cx::dd
0
0
0
.
=
~I [ /4X r+ 14X r+1 - ~2 + 4X r+ I ~21 :J
=
~I[(4X3)+(4X3)+(-2+4X2)+ (;~J
= ~I =
f(
me /
[3.
dX
3
. 2
2
4
]
1 EI f (-1)2dX+f (-l/dX+f (-1) 2 dX+f (-l/dX+f (-1/ dX. o
0
0 0 0
4 EI
~ o:dd
= -
Putting all values of evaluated co-efficients, equations 1,2 and 3 become . 110 10933 '56' .30 - EI + xHD - EI X V D - EI MD
-m
906.67
56
and
~'- EI x HD
and
EI - EI
230
30
117.33
32
=
0
+ --m-.x VD + EI MD =0
32 14 X HD + EI x V D + EI MD = 0
-110 + 10933 HD ..:. 56 VD - 30 MD = O· 906.61' -56 HD+ 117.33 YD + 32 MD = 0 230 - 30 HD + 32 YD + 14 MD = 0
(1)
(2)"
(3) Simplifying ~ ~ ~
(I) (2) (3)
"
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
From Eq (1) -110 + 109.33 Ho - 56 Vo . 30 . . = -3.67 + 3.64 Ho - 1.86 Vo
~
(4)
Putting in Eq (2) 906.67 - 56Ho + 117.33 Vo + 32 (-3.67 + 3.64 Ho - 1.86 V o) "" 0 906.67 - 56 Ho + 117..33 Vo - i 17.44 + 116.5 Ho -:- 59.52 V~ = 0 789.23 + 60.5 Ho+57.81 Vo = 0 Ho = -13.045 - 0.95 Vo
~
(5).
Putting the value ofHo in Eq (4) Mo = -3.67 + 3.64 (...,13.045 - 0.95 Vo) - 1.86 Vo Mo = -51.15 - 5.32 Vo
~
(6)
MD
-flutting the values ofMb & Ho in Eq (3)' 230 - 30 (-13.045 - 0.95 Vo) + 32 Vo + 14 (":'51.15 - 5.32 V D) = 0 230 +391.35 + 28.5 Vo + 32 Vo - 716.1 - 74.5 Vo =0 -14 V o "" 94.75. = 0 Vo == :-6.78 KN Putting in (5) & (6) , Ho = -6.61 KN,
Mo = -15.08 KN-m
From any equation above.. We get
.1 Vo .
.
.
=-12.478 KN
-
I
..
Apply the evaluated strucfural·a~tions in correct sense on the frame. The correctness of solution can be checked afterwards by equilibrium conditions. . 10KN Br----:::-_'"""-_-=-----, C .2m 2m 4m
20KN
.
--7
3m . Ma=1.8 KN . A
~a=13.39'KN
. va (
i
1S.08KN=m
D)~ 6~61KN 12.478 KN .
= 2.478 KN
Fig. 2.35 shows all reactions after Evaiuation
I
-'
.
55
56
'THEORY OF INDETERMINATE STRUCTURES
l:Fx = 0 20 - Ha - 6.61 = 0
I Ha
13.39 KN
=
l:Fy = 0 Va + 12.478 - 10 = 0
I Va =
I
,(asuming Va upwards) -
2.478 KN
I
o Ma+ 20 x 3 + 10 x 2 - 12.478 x 4 - 6.61 x 2 - 15.08 = 0 (assuming Ma clockwise)
I Ma = LMa = 0 1.16.,
~ 1.8 KN-m
I
12.478 x 4 + 15.08 + 6.61 x 2 + 1.8 - 20 x 3 - 10 x 2 = 0 Proved.
ANAJ;,YSIS OF STATICALLY EXTERNALLY INDETERMINATE TRUSSES:A tt:uss. may be statically indeterminate if all external reactive components and internal member , forces may not be evaluated simply by the help of equations of equilibrium available. The, indeterminacy oftbe trusses can be categorized as follows :(1)
Trusses containing excessive e~ternal reactive components than those actmilly required for external stability requirements.
(2)
Trusses containing excessive internal members th~n required for internal stability requirements giving lesser the number of equations of equilibrium obtained from various joints. .
(3)
A combination of both of the above categories i.e. excessive external re~ctions plus excessive internal members ..
INTERNAL INDETERMINACY:b+r.:;:: 12j I . There are two equations of equilibrium per joint where b = number of bars or members. r = minimum number of external reactive components required for external stability (usually 3). j = number of joints. '" . The above formula can also be used to check the total indeterminacy of a truss if we define 'r' as the total number of reactive components which can be pr~vided by a typical support system. . METHOD OF MOMENTS AND SHEARS: ' . 1,17. A simple method is presented to evaluate axial member forces in parallel chord trusses. For other types of trusses method of joints, method of sections or MaxwelI's diagram may be used. For determining forces in members of trusses, this method has been used throughout this text. To develop the method, consider the truss loaded as shown below:
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
2P
3P
F
E
57
G
1 1
7 RA=T P
t
3@
>t
a
RD=
~
P
Fig. 2.16 A typical Truss under loads
1
C-onsider the equilibriwn ofL.H.S. of the section. Take 'D' as the moment centre: we find Ra Ra x 3a = 2P x 2a + 3 P x a
II
7Pa Ra = 3a
:1
I
j
or
7P
="3
Mc = o. and assuming all internal member forces to be tensile initially, we have Ra x 2a - 2P x a + SFG, x h = 0 (considering forces on LHS of section) '_' (RaX2a-2pa) . SFG - - , h
The ( -ve ) sign indicates a compressive force. , SFG
=
(Ra x
2~ -
2 pa)
= ~c
Or
.where numerator is Mc. Therefore
1
The force in any ch9rd member is a function of bending moment. "To fmd out the axilll force in any chord member, the moment centre will be that point where other '; two members completiig'the same triangle meet and the force will be obtained by taking moments about '( that point and dividing itby the height of truss. The signs of the chord members are established in the very j beginning by using an analogy that the truss behaves as a deep beam. Under downward loads, all upper \ chord members are in compression while all lower chord members are in tension.
" 1;\
,1
Similarly, Sac =
~
(using the guide line given in the above para)
Consider the, equilibrium of left hand side of the section and IFy = 0 Ra ~ 2P .;.. SFC Cos e
=
Ra - 2P) SFC .= (,.,Cos e
,
0
,where Ra - 2P is equal to shear force V due to applied loads at the section. So in general the force in any inclined member is a function of shear force.
V SFC - cose The,general formula is : S =
± (V) . . ± (Cos e).
58
THEORY OF INDETERMINATE STRUCTiJRES
Where V is the S.F. at the section passing through the middle of inclined member and '8' is the angle measured from "the inclined member to the vertical" at one of its ends. Use (+ve) sign as a pre. multiplier with the Cos8 if this angle is clockwise and (-ve) sign if e is anticlockwise. Take appropriate sign with the S.F also. This treatment is only valid for parallel chord trusses. The force in the vertical members is determined by inspection or by considering the equilibrium of forces acting at the relevant joints. To illustrate the method follow the example below.
1..I7.1: EXAMPLE:SOLUTION:-
Analyze the following truss by the method of moment & shear. Determine reactions and Draw SFD and BMD.
P
P
L
.
l'
h
.J, 8@a
Given Truss under loads.
1.S P
ol~______+_______~ ___
·+o_.S_P__________
~ ~ __
~--"-------------'
o S.F.D.
1.SP
~ .
3P
1.SP
.
,
4.SP SPa
4:SP
.
.
.
3 Pa
+
1.SPa
.
.
.
o~----------------------------------~
8.M.D.
Fig. 2.37
TOP CHORD MEMBERS. Considering the beam analogy of truss, all top chord members are in compression. Picking bending moment, at appropriate moment centers, from BMD and dividing by height of Truss.
SI'J'
3 Pa =-T
Sjk
=-T 5 Pa =-T 5 Pa =-T 3 Pa =-T 3 Pa =-T
Stl Slm
Smn Sno
3 Pa
Negative sign means compression.
,1 :
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
59
BOTTOM CHORD MEMBERS. All are in tension. Taking appropriate moment point and dividing by height of Truss. Sap Sbc Sde Sfg
=
1.5 Pa Spb = +-h-
. 4.5 Pa h 4.5 Pa = Sef= + - h = Scd =
+--
Sgb
+ l.5 Pa h
=
=
INCLINED MEMBERS.
.
.
+v
The force in these members has been computed by the formula. ±(Coser Follow.the guidelines. Sili
- --l.5P cose .
Sib
1.5 P +cose
Sbk Skd
. Length AI =
"a
0.5 P cose
Sdm
=--=--
Srnf
--+-Cos~ -
Sfo
=--=
Soh .
2
(if a and h are given, length and Cos e will have also late values) ,. h cose= ~~ +h
l.5P --cose 0.5 P -+cose - 0.5 P -cose ~ 1.5 P
-Vi + h
-l.5P
l.5P
-cose
cose
-l.5P +Cos9
VERTICAL MEMBERS. For all vertical members of trusses in this book,'member forces have been determined by Inspection or by Equilibrium of joints. So = Sbj =;= Sck = Sem = Sfn = Sgo = 0 Sip = - P ( If a and h values are given, all forces can be numerically evaluated) SId 1.18. EXTERNALLY REDUNDANT TRUSSES - FIRST DEGREE EXAMPLE 5:- Analyze the following truss by the force method. (consistent deformation method). The following data is given. . .
E=200 x 106 KN/m2 A=5x10-3m2 for inclineds and verticalS, A=4x 10-3 m2 for top chord members, .. A=6x 10-3 m2 for bottom chord members
60
THEORY OF INDETERMINATE STRUCTURES
SOLUTION:-
1 8m
4 @ 1.8m Fig. 2.38 Given Truss under loads
TOTAL INDETERMINACY :where r = total reactions which the supports are capable of providing.
b+r=2j 17+4 :;i:2xl0 21 :;i: 20 D = 21-20 =1
Indetenrunate to 1st degree. Apply check for Internal Indetenninacy :-
b + r = 2J
where r = Minimum number of external reactions required for stability.
17+3=2xlO 20 = 20 This truss is internally detenrunate and externally indetenninate ·to1st degree, therefore, we select reaction at point "C" as the redundant force. Remove support at C, the Compatibility equation is : .A C +
DCC x
Rc = 0
(Deflection at C due to loads pl~s due to redundant should be zero.)
or·
Rc ==
where
Ac =
Now we have to calculate Ac and F'UL
L AE --
DCC
to get Rc.
where F' = Force induced in members due to applied loads acting on BDS. U = Forces in members due to Unit load applied in direction· of applied loads, at external redundant support in BDS.
STABILITY. DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
61
72K
Fr-____~=-~~~H~--~~--~
Ism I(
) I
Fig 2.39a 8.0.S under applied Loads (F-Oiagram)
Fig 2.39b B.O.S under unit Vertical Redundant at C (U-Oiagram)
Analyze the given truss by the method of moments and shears as explained already for F' and U forces in . members.
1'----1
[email protected]·tRe=45.KN Ra=63 (F'-Oiagram) Fig 2.40 B.O.S under Loads
63 +
o
I
0'---------+----------, 4S
S.F.D" .. "
'-------4..,..,S=--'
.~113.4 162. . .
81
.' .
·O~OB.M.D.
.J
62
THEORY OF INDETERMINATE STRUCTURES
Detennine forces in all members of trusses loaded as shown in this question and enter the results in a tabular forin. (using method of moments and shears, F' and U values for members have been obtained). F
G
H
J
f
~ f3C1t) U=Diagram
+
S.F.D.
1.8
B.M.D. Fig 2.41 B.D.S under Unit redundant force at C
Member
U
F' (KN)
FG GH HI IJ AB BC CD DE AG GC CI IE AF BG HC .ID JE
0 -90 -90 0 +63 +63 +45 +45 - 89.1 +38.2 +63.64 -63.64 0 0 -72 0 0
0 - 1 - 1 0 +0.5 +0.5 +0.5 +0.5 -0.707 GC +0.707 -0.707 0 0 0 0 0
Ax 10-3 (mf 4 "
" 4 6
" "
L (m) 1.8
" " " 1.8
" " "
" "
2.55
5
"
" " " " "
" "
" "
1.8
"
" " "
.F'UL 10-3 AE x (m) 0 0.2025 0.2025 0 0.04725 0.04725 0.03375 0.03375 0.16063 0.06887 0.11473 0.1 1473 0 0 0 0 0
L F~~L = 1.02596 x 10-3
U2L -~ 10-3 AE
(m) 0 2.25 x 10-3 2.25 x 10-3 0 0.375 x 10-3 0.375 x 10-3 0.375 x 10-3 0.375 x 10-3 1.275 x 10-3 1.275 x 10-3 . : 1.275 x 10-3 1.275 x 10-3 0 0 0 0 0 U2L L AE =11.1 xlo-{i
Fi=Fi'RcxU 1 (KN)
0 +2.5 +2.5
0 +16.75 +16.75 - 1.25 - 1.25 -23.7 -27.2 - 1.76 +1.76 0 0 -72 0 0
,
,
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
, . F'UL ~ C = L ~ = 1.02596 ,
' 10-3 = 1025.96
X
10-6 m
2
occ=
Rc
U L'
X
63
L AE
=
1l.1 x 10-6 m
= ~ ~ C = -1025.96 X 10-
Putting these two in original compatibility equation
6
occ 1Ll x 10 6 Rc = - 92.5 KN. The (-ve) sign with Rc shows that the assumed direction of redundant is incorrect and Rc acts upward. IfFi is net internal force due to applied loading and the redundants, acting together, then member forces an calculated from Fi = Fi' - Rc x Ui~~, The final axial force in any particular member can be obtained by applying the principle of ·superposition , and is equal to the force in that particular member due to applied loading ( ± ) the force induced in the ' same member due to the redundant with actual signs. Apply the principle of superposition and insert the magnitude of redundant Rc with its sign which has been obtained by applying the compatibility condition to calculate member forces. ' 1.19. SOLUTION OF 2ND DEGREE EXTER.1~ALL Y INDETERMINATE TRUSSES:-
Example-6 : Solve the ,following truss by consistent deformation method use previous member propertIes. " 36KN
72KN
F~__~~__~~H__~~__-;J
lam
[email protected]>1 Fig 2.42 Given Truss
36KN
72KN
ram 6.C (F'-diagram)
o
145KN S.F.D.
63
I
I
45
'-------1-_0
o
'--___--->'\
162
o Fig 2.42a B.D.S under loads
+ L
B.M.D.
64
THEORY OF INDETERMINATE STRUCTURES "
i
1
(U diagram)
1 J'2 ~"____---,
I
1 2
i
+
---11 ~
1 - -_ _ _ ,~
,.,
1.8
.
o~o
S.FD.
2 B.MD.
+
(U 2 diagram) 0.251--_ _---,-,-_ _---,
0~1_~__L(+~)____~~--~o
(-) I
S.F.D. '--_---10.75 nAt:
o~
o~"oB.M.D.
"
Fig 2.42 e B.D.S under unit redundant at 0
"Compatibility equations are: t.C + Re. occ + R~ x ocd = 0 (1) Compatibility of defomations at C t.D + Rc . odc + Rd . odd = 0 (2) Compatibility of defonnations at D ocd = Odc by the law of reciprocal deflection. occ = deflection of point C due to unit load at C. odc = deflection of point D due to unit load at C. odd = deflection of point D due to unit load at D. ocd = deflection of point C due to unit load at D. " Flexibility coefficients of above two equations are evaluated in tabular fonn (Consult the attached table)
"t.C - ~F'UIL ="1026.2 x IO-6 m - "" AE W -
". occ
2:
F'U2L AE
"7
1
'l I
I,
579.82 x 1O-jj m
2
=2:
U L TE - I i.I x 1O-jj m
I
I
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
U,2L
bdd = LAB" bcd =
UJU,L
L~
= 9.3565 x 10
_6
m
6
= 6.291 x 10- m
Odc = L U~L = 6.291 x 10-6 m
Put these in eql!ations 1 and 2 ~
6
1026.2 x 10-{, + 11.1 X 10- Rc + 6.291 X 10-{, Rd = 0 579.82 X 1O-{; + 6.291 X 10-6 Rc + 9.3565 X 10-6 Rd = 0 Simplify 1026.2 + II.! Rc + 6.291 Rd = 0 579.82 + 6.291 Rc + 9.3565 Rd = 0 From (3), · = (-1026.2·~ 6.291 Rd) Rc 11.1 Put Rc in (4) & solve for Rd o.c;· (-1026.2 - 6.291 Rd) _ 57J.8L.+6.291 11.1 +9.3565Rd-0
. ~ ~ ~.
~
(1) (2) (3) (4)
. (5)
- 1.786 + 5.791 Rd = 0
IRd=
+ 0.308 KNI
. So,from (5), =:> Rc = (-1026.2 - 6.291 x 0.308) . 11.1 IRc ='-92.625 KN\ :. Rc;:: - 92.625 KN Rd = + 0.308 KN These signs indicate that reaction at C is upwards and reaction at D is downwards. By superposition, the member forces will be calculated as follows Fi = Fi + Rc x U I + Rd X U2 which becomes. Fi = Fi - Rc x U I + Rd X O2. It takes care of (-ve) sign with Rc. . . Equilibrium checks:.)
0.308 1.082
1.082
0.308
Joint D LFx=O LFy=O Equilibrium is satisfied. Only check at one joint has been applied. . satisfied at all joints.
In fact this check should be .
65
--.-------...
~---
-----
0'1 0'1
TABLE FOR EXAMPLENQ. 6 Mem ber.
Ax 10-3
(mi FG GH HI II
AB BC CD DE AG GC CI
IE AF BG HC ID
JE
4 4 4 4 6 6 6 6 5
5 5 5 5 5 5 5 5
L (m)
L . AE X 10-3 KN-m
. 1.8 2.25 x 10-3 1.8 . 1.8 .. 1.8 1.5 x 10-3 1.8 1.8 1.8 1.8 2.55 2.55 X" 10-3 2.55 2.55 " 2.55 " 1.8 ." 1.8 1.8 " 1.8 " _1.8 __
F' (KN)
.
. .. . ..
..
-
-
0 -90 -90 0 63 63 45 45 -83.1 38.2 63.64 -63.64 0 0 -72. 0 0
U1
0 -1 -1 0 +0.5 +0.5 +0.5 +0.5 -0.707 +0.707 +0.707 -0.707
0 0 0 0 0
U2
0 -0.5 -0.5 0 +0.25 +0.25 +0.75 +0.75 -0 ..35 +0.35 -0.35 -1.06 0 0 0 +1 0
F'U1L AE 3 X 10. (m)
U.'L AE 3 X 10(m) .
·U.UzL AE 3 X 10: (m)
0 +0.2025 +0.2025
0 0 2.25 x 10-3 . + l.I25 X 10-3
..
..
0 +0.4725 +0.4725 +0.0337 +0.0337 +0.161 +0.0689 +0.1147 +0.1147
0 0.375 xlO-J
0 +0.1875 X 10-3
0 0 0 0 0
0 0 0 0 0
...
,;
..
." 3
1.275 x 10-
2:1.0262' X 10-3 = 1026.2x 10- 6
. ..
.
Dl.l X 10-6
+0.5625 X 10-3 " +0.63 I x 10-3 " -0.63 X 10-3 + 1.91 X 10-3 0 0 0 0 0
L6.291 X 10-6
F'U2L AE 3 X 10(m) . 0 +0.10125
..
0 +0.023625
ReU.
+ RdU z (KN)
0 0.5625 x 10
3
.
0 0.09375 x 10
. +0.050625 .
.
0.844
X
10-3
" 3
+0.07952 +0.0341 -0.0568 +0.172 0 0 0 ,0
0.3 12 x 10-
.
.. 2.87
X
10-3
0 0 0 2.55 x 10 ·0
0
2:0.57982 10-3 =579.82
F=F'-
U/L AE 3 X 10. (m)
3
2:9.3565 X 10-6
.
X
10-6
3
0 +2.471 +2.471 0 +16.765 +16.765 -1.082 -1.082 -23.722 -27. I 78 -1.954 + 1.5 I 9 0 0 -72 +0.308
>-l
::r: tI1
o ~
~
~ •
I
_Jl .. ----.J
~
~
>-:l tI1
,CI.l
>-:l
~ (')
~
CI.l
l
~"",(",,; •.\:...-.. ~
-
---1'.·.'·Wi~;;'i1i.i;,ml-;m¥.·-~~l~~j"q,;:~;~""""~,~i'... ,.-~.,,..T'l"~-""O-""'------------------
CIl
~
E .:;! Cl
~
TABLE FOR EXAMPLE NO.7 Me m
ber FG GH HI IJ AB BC CD DE AG GC GO IE
AF BG HC lD IE
Ax 10-3
L(m)
L/AE
F'
UI
U2
U3
0 -90 -90 0 63 63 45 45 -89 38.2 63.6
0 -I -I 0 0.8 0.8 0.3 OJ -I
0 -I -I 0 0.5 0.5 0.5 0.5
0 -I -I 0 OJ OJ 0.8 0.8
.10-6
(m)'
4 4 4 4 6 6 6 6 5 5 5 5 5 5 5 5 5
L8 L8 L8 L8 " L8 L8 L8 L8 2.6 2.6 2.6 2.6
1.8 L8 L8 L8 1.8
2.95 2.95 2.95 2.95 L5 1.5 L5 1.5 2.55 2.55 2.55 2.55 2.55 2.55 2.55 2.55 2.55
-64 0 0
-72 0 0
-0 0.4 -0 0 I 0 0
0
-0.71
~O
0 0 0 0 0 0 0 0
0.4 -0 0.4 -0 0 0 0 0
F'UI' L/AE) I 0-6 0 101.25 IOL25 0 70.875 70.875 16.9 16.9 241 -341 56.8 56.8 0 0 0 0 0 391.65
F'U2L IAEx 10-6 0 202.5 202.5
0 47.25 47.25 33.7 33.7 161 68.9 114.7 114.7 0
0 0 0 0 1026.2
F'U3L1 AExl 0-6 0 IOL25 lOL25 0 23.625 23.625 §0.125 50.125 79.52 34.1 56.8 172 0 0 0 0 0 692.42
U12L1 AEx 10-6 0 0.5625 0.5625 0 0.844 0.844 0.0938 0.0938 2.875 0.312 0.312 0.312 0 2.55 0 0 0 9.3616
U22 LlAE
~
U32L1AE x10-6
UIU2L1 AExlO-6
x 10-6
0 0 2.25 0.5625 0.5625 2.25 0 0 0.09375 0.375 0.09375 00375 0.844 0.375 0.844 OJ75 0.312 1.275 1.275 0.312 L275 " 0.312 1.275 2.87 0 b 0 0 0 0 0 2.55 0 0 11.1 9.3565
0 LI25 1.125 0 0.625 0.625 0.188 0.188 1.91 -0.631 0.631 0.631
0 0 0 0 0 6.417
UIU3L1 AEx 10-6 0 0.5625 0.5625 0 0.281 0.281 0.281 0.281 0.946 -0.312 -0.312 0.946 0 0 0 0 0 3.517
U2U3 LlAE x10-6 0 1.125 LI25 0 0.1875 0.1875 0.5625 0.5625 0.631 0.631 -0.631 1.91 0 0 0 0 0 6.291
Z
>-< o"rl
Fi=Fi+Rb UI+RcU2 +RdU3 0 4.388 4.388 0 2.085 2.085 -2.047 -2.047 -45.329 -51.814 -3.828 2.999 0 32.797
()
CIl
I
~
()
~
:;.;:I
r:1
CIl
>Z o (")
~
CIl
Vi -l r:1
Z
-l Cl
-72
r:1
0.588 0
"rl
o
~
>-
::l
~
CIl
~
~
o
Cl
",
I
-.J
72
THEORY OF INDETERMINATE STRUCTURES
B'D'S under unit load atB for calculatingobb,ocb andodb
l~ '.
.
.
(U1 - diagram)
~----~~--~----~-----7
0.75
1.0 0.75
0.75
01
025' .
I
(+)
0
0.2S
1 0 .25
(-)
~ ~
S.F.D.
0.9
F
G
H
B.M.D.
J
I
~ .~
~.'
B.D.S under unit load ate E forcalculatingocc,obc andodc
.
~
1
U2-diagram
D
t o.5
to.5
0'05r-I S.. F D. L ====:::::+:!:-.-_-.....;.-_-_....,-+-__________--,,0 ' - - - - - - - - - ' 0.5 .
::: ~ 1.8
_B.M.D. U3 diagram for obd, oed and odd
~ I
0.25
i
i
1
0.25
0.75
(+)
.
1.3
(_)
" SDF
0.75
--~. ~
BMD
From the previous table we have the values of all flexibility eo-effieients as given below: . ~B=391.65 x 10-6 m ~C= 1026.2 x 10-6 m ~D=692.42 x 10-6 m obb= 9.3616 x 10":6 m , and ,aee = ILl x 10-6 m, odd = 9.3565 x 10-6 m obe = oeb =6.417 x io-{i m abd = odb = 3.517 x 10-6 m oed = ode = 6.291 x 1O-{i m
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMAnONS METHOD
Putting the values of flexibility co-efficients into compatibility equations we have. 391.65 x 10-6+9.3616 x 1O+Rb+6.292 x 10-6 Rc+3.517 x lO-6Rd= 0 ~ (1) 1026.2
X
10-6 +6.292
10-6 Rb+ 11.1 x 10-6 Rc + 6.291
X
X
10-6 Rd = 0 ~ (2)
579.82 x 10-6+3 .517 x 1O-6 Rb+6.291 x 10-6 Rc+9.3565 x lO-6Rd = 0 ~ (3)
Step No.4 Simplify equation (1), (2) and (3), we have 391.65 +9.3620 Rb+6.292 Rc+3.517Rd= 0
~
(4)
1026.2 + 6.292 Rb + 11.1 Rc + 6.291 Rd = 0
~
(5)
579.82 + 3.517 Rb + 6.291 Rc+9.357 Rd = 0
~
(6)
Multiply (4) by 6.291 & (5) by 3.517 & subtract (5) from (4) 391.65 x 6.291+9.362 x 6.291Rb+6.292 x 6.291 Rc+3.517 x 6.291Rd=0 1026.2 x 3.517+6.292 x 3.517 Rb+ 11.1 x 3.517 Rc+3.517 x 6.291Rd=0 - 1145.275 + 36.767 Rb + 0.544 Rc = 0 ~ (7) Multiply (5) by 9.357 & (6) by 6.291 & subtract (6) from (5) :. 1026.2 x 9.357+6.292 x 9.357 Rb+ 11.1 x 9.357 Rc+6.291 x 9.357Rd=0 579.82 x 6.291+3.517 x 6.291Rb+6.291 x 6.291 Rc+6.291 x 9.357Rd=0 ~.o
5954.506 + 36.749 Rb + 64.286 Rc:: 0 From (7),
Rb
= (1145.275 - 0.544 RC) 36.767 0
Put Rb in (8) & solve for Rc' (1145.275 _ 0.544 RC) 36.767 + 64.286 Rc = 0 5954.506 + 36.749 o
5954.506 + 1144.71 - 0.544 Rc + 64.286 Rc = 0 7099.22 + 63.742 Rc = 0
IRc
= - 111.374
KNI
Put this value in equation (7) and solve for Rb Rb = (1145.275- 0.544 x 11 1.374) 36.767
IRb = +32.797 KNI Put Rb and Rc values in equation (4) to get Rd. 391.65 + 9.362 x 32.797 + 6.292 x (111.374) +3.517 Rd = 0
IRd
= + 0.588 KNI
(8)
73
74
THEORY OF INDETERMINATE STRUCTURES
After reactions have been calculated, truss is statically detenninate and member forces can be easily calculated by'Fi = Fit + RbU J + RcU2 + RdU3 as given ill table. Apply checks on calculated member forces. . Step No.5: Equilibrium checks. Joint (C)
~172
~~
51.81
32.058 .
IFx
=
3.828
2.047
1111.374
0
- 2.047 - 32.058 - 3.828 x 0.707 + 51.814 x 0.707 = 0 -0.179:0 0=0
IFy:=;O
.
111.374 - 72 - 3.828 x 0.707 - 51.814 x 0.707 = 0 0.035 : 0 o= 0 (satisfied) Solution is alright. 1.21: ANALYSIS OF 3-DEGREE REDUNDANT FRAMES Example No.8: Analyze the following frame by consistent deformation method. 96KN 8 3m
~ F
3m
6m 31
c ,
.'-..:
36KN 21 --7 E '3m
I
7.5m
A
o SOLUTION :The given frame is statically indeterminate to the 3rd degree. So that three redundants have to be removed at support D or A. Consider H D, V D & MD as the redundants 96KN
8 3m
~ F
3m
6m 31
c
36KN 21 --7
3m
E
7.5m
A
o
r
I
. 75
STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD
I
i!
B 3m
m
C
17.5m
1 1
36KN
+--
396KN-m
1
\
Ai
OI)~H J.~«80 ~Ov
96KN
Fig. 2.45 BD.S under loads
+ Bi->"'..I.I-......;Ft--6_m __c =,
F
3m "E
E
1
~-
A
6m
3m 7.5m
1
+
"
6m
F
+
"1
DH
1.5m 1
-
1
OdVd~t9
C
C
3m
7.5m
3m
1.5
B 3m
Br-"_-I-_9_m_--,C
I
~Vd91
ocdh de
ocd9d9
Bdhdh
ocd9dh
mH-Diagram
mY-Diagram
m9-diagram
(BDS under redundants)
Compatibility Equations:~DH
+ HD x 8dh.dh + V D x 8dhdv + Mo x adhd8 =0 (l)
compatibility in horizontal direction at D.
~v
+ HD x 8dv.dh + Vox Odvdv + Mo x advd8 =0 (2)
compatibility in vertical direction at D.
So
+ Ho x adS.dh +Vo x ad8dv + Mo x ad8d8 =0 (3)
rotational compatibity at D.
We have to determine the following flexibility co-efficients. ilDH = Horizontal deflection of point D due fo applied loads.
Wv = Vertical deflection of point D due to applied loads. eo
= Rotation
of point D due to applied loads.
8dhdh = Horizontal deflection of point D due to unit horizontal redundant force at D
76
THEORY OF INDETERMINATE STRUCTURES
8dhdv = Horizontal deflection of point D due to unit vertical redundant force at D ad8dh = angular' deflection of point D due to unit angular redundant force at D 8dvdh = Vertical deflection of point D due to. unit horizontal redundant force at D 8dvdv = Vertical deflection of point D due to unit vertical redundant force at D ad8dv = Rotation deflection of point D due to unit vertical redundant force at D adhd8 = Horizontal rotation of point D due to unit rotation at pt D advdO = Vertical rotation of point D due to unit rotation at pt D ad8d8 = Rotation rotation of point D due to unit rotation at pt D 8dvdh = 8dhdv
(reciprocal deformations)
arl8dh = adhd8
(reciprocal deformations)
ad8dv = advd8
(reciprocal deformations)
Now these flexibility co-efficients can be evaluated by following formulae.
WH
=
JM xE1rnH dX
t.Dv
=
JM xEImV dX
8n
= JM~1m8
8dhdh
= J~RC t
1
>1
. ~ U2 f---71~ U2 BDS UNDER LOADS AND REDUNDANTS Choosing C as origin, Set-up moment expressions in different parts of this beam. 'wX 2 . Mbe = Re.X - -2-
.Mab '
( L)
L 2' < X < L. Write strain energy expression for entire
WX2
= Rc.x+Rb X-2' --2-
structure.
1
U2[ Re.X - -2WX2 J2 I L[ ( L ) WX2 J1 dX + 2EI Lz Re.X + Rb X - 2' - -2- dX
U = 2EI ~
Partially differentiate it W.r.t. redundant Re first. Use eastiglianos theorem and boundary conditions.
au
aRc
=
1
U2[ Rc.X -""'2 WX2 J I L [ ( L ) WX2 J [X]dX + EI L2 Re.x + Rb X - 2' - -2-· [XJ dX
6.e = 0 = EI ~
__L L'2[
0- EI f
o
2
Re.X -
wX 2
3
3J dX + .l.EI f?L[Re.X2 + Rb.X u_
.
4 U2
X . WX J o = EI1 [ Re.} - -8- 0
1r + EI
X
Z
-
Rb.LX wX 2 - 2
3J dX . Integrate It..
3 RbLX 2 WX 4JL X + Rb. } - --4-' - -g- U2'
3
LRc. }
Insert limits and simplify.
o=
Re.L3
5Rb.L 3
wL 4
-3-+~--g-
--+
(I)
METHOD OF LEAST WORK
101
' 1
Now partially differentiate strain energy w.r.t. Rb. Use Castiglianos theorem and boundary conditions. uU 1 l..i wX1 L [ wXL :=> .L ) aRb =Llb = 0 = EI Rc.x - -2-]132 F - Diagram (Truss under loads and redundant) NOTE: Only the rectangle of members containing redundant X contains forces in tenus of X as has been seen earlier. Now analyze the Truss.by method of joints to get Fi forces. JOINTLO
e
loU 1
--~---'~loL1
16KN
2:Fy = 0 LoU, cose + 16 = 0
- 16 LoU, = cose
-16 0.8
2:FX = 0 LoLl + LoU I Sine =0 LoLl + (-20) x 0.6 =0 LoLl - 12 = 0
ILoLl = 12 KNI· Joint U I
2:FX = 0 U I U 2 + X Sine + 20 Sine = 0
116
THEORY OF INDETERMINATE STRUCTURES
U, U 2 + X x 0.6 + 20 x 0.6 = 0
2:Fy = 0 - U, L, - X cose + 20 cose = 0
- U, L, - X x 0.8 + 20 x 0.8 = 0
U, L, = - 0.8 X + 16
IU,L, = -
(0.8 X - 16)/
Joint L, :O.8X -16
2:Fy = 0 - (0.8X - 16) + L, U 2 cose = 0 L,U 2 x 0.8=0.8X-16 :1
2:FX = 0 L,L2 + LlU2 Sine - 12 = 0
Put value of L, Uz.
I
L,L 2 + (X - 20) x 0.6 - 12 = 0
1
L,L2 + 0.6 X - 12 - 12 = 0
.\
IL, L2 = - (0.6X - 24)/ JointU2 (O.6X+12)
(X-20)
2:FX=O (0.6 X + 12) + U 2L3 Sine - (X - 20) Sine = 0 0.6 X + 12 + U 2LJ x 0.6 - (X - 20) x 0.6 = 0
'I
1
l
METHOD OF LEAST WORK
117
0.6 X + 12 + O.6U2L3 - 0.6 X + 12 = 0 U 2L 3=
-24
OT
L:Fy =0
- U 2L2 - (X - 20) cose - U 2L3 cose = 0
- U2L2 - (X - 20) x 0.8...:. (- 40) x 0.8 = 0 - U2L2 - 0.8 X -+- 16 + 32 = 0 - 0.8 X + 48 = U2L2
IU L
2 2
= -
(O.8X - 48)
I
Joint Lz:-
0.8 X- 48
x
e
0.6 X-24
48 L:FX =0
LzL3 + 0.6 X - 24 - X Sine = 0 L2L3= - 0.6 X + 24 + 0.6 X
IL2 L
3
= 24
KNI
L:Fy =0
-(0.8X...:.48)-48+XCose =0· -O.8X + 48 --: 48 + O.8X = 0 0=0 (Check) Joint L3:-· At this joint, all forces have already been calculated. Apply.checks for corretpess. 40
.
24"
e
32
118
THEORY OF INDETERMINATE STRUCTURES
~FX = 0 40 Sine - 24 = 0 40 x 0.6 - 24 = 0 24 - 24= 0
o= 0
O.K.
~Fy = 0 - 40 cose + 32 = 0 - 40 x 0.8 + 32 = 0 O.K. Checks have been satisfied. - 32 + 32 = 0 0=0 This means forces have been calculated correctly. We know that strain energy stored in entire ... "Fi2L Truss IS U =
C/.l ,...,
~ ~
+ 12
+\0.84 + 24 - 20 + 21.94
+ 1.94 -40 - 1.55
+ 30.45
I\~
THEORY OF INDETERMINATE STRUCTURES
120
EXAMPLE NO. 9:- By the force method analyze the truss shown in fig. below.' By using the forces in members L,U2 and L2U 3 as the redundants. Check the solution by using two different members as the redundants. E = 200 x 10 6 KN/m2 SOLUTION:-
~~~-r.~~~~~~~~~~4. I6m 48KN
J
96KN
r
72KN
48x4.5 + 96x9 18 18
48+96+72114 102KN «-+ 72x13.5 ' 18 F'- Diagram
=
f
6r
=114KN
B.o.S. Under applied loads only. Or F-Diagram
14 114KN
[email protected] ------~
+
'
102KNOL____
I ,. 0 -1=5=4=K=N==C=====r-~~1 42KN 459 KN-m
-
S.F.o.
1114KN
702KN-m
~+ 5133KKNN--lTm
o
~
~
+
, U1 0.6
U2
0
B.M.o.
U3
.~ ~. 6t
~~.
0
~
L1
U1
0
0'
U
L3 U3
I
o
~l
U,-DiagramforredundantX,
~,
i
6m
U2-diagram for redundant X 2
1 , ,Compatibility equations are: LlX,L + LlX,R, + LlX,R2 = 0
Here
R,
;=
X, ,
R2 == Xi
-+
(1) Change in length in member I due to loads and two redundants should be zero. -+ (2) Change in length in member 2 due to loads and two rcdundants should be zero. '
1.! 1
r I
!
METHOD OF LEAST WORK
121
if:
l .. prod· liced'In mem ber (1) due to app I'Ie d I·oads. Where LlXIL = l:.F'U L = DeflectIon
llXIR I = Deflection produced in member (1) due to redundant RI =
l:~L). XI·
LlxIRz = Deflection produced in member (1) due to redundant R z = l:(!l.::1L) . X 2 LlxzL = Deflection produced in member (2) due to loads = l: F;iL LlxlRI = Deflection produced in member (2) due to redundant RI =
l:~.::1L) . XI
llX2RZ = Deflection produced in member (2) due to redu.ndant R 2 =
l:(!lli). X2
From table attached, the above evaluated summations are picked up and fmal member forces can be seen in the same table. All member forces due to applied loads (Fi' diagram) have been detennined by the method of moments and shears and by method of joints for Uland U 2 diagrams. Evaluation of member forces in verticals ofF' - Diagram:Forces in verticals are determined from mothod of joints for different trusses shown above. (Joint L I)
76.5
76.5
.48
. l: Fy = 0 UIL I -48 =0
(Joint U z)
117
l: Fy = 0 - U 2Lz + 52.5 .CosS = 0 - U2Lz + 52.5 x 0.8 = 0 UzLz = 52.5 x 0.8
85.5
122
THEORY OF INDETERMINATE STRUCTURES
1.:Fy= 0
- U3L3 + 142.5 cose = 0 U3L3 = 142.5 x 0.8
Evaluation of forces in verticals of U 1 (Joint L I )
1.: FX = 0
1.:Fy = 0
UILI + 1 Cos
(Joint U I
).
1.:FX = 0
e=
0
-
Diagram:-
123
METHOD OF LEAST WORK
LFy
=
0
+0.8-UI~COSe
0.8 =
Ul~
X
= 0
0.8
U lL 2 = 1
so
U l U2 + 1 x 0.6 = 0
Putting value ofU lL2 in L FX.
U l Uz = - 0.6
Now from the table, the following values are taken. AXIL = - 0.671 x 10-3 AXIRI = 125.7 x 1O-6X1 = 0.1257 x 1O·3Xl AXIRz = 32 x 10-6 X2 = 0.032 x 1O-3X2 AX2L = - 6.77 x 10.3 AX2Rl = 0.032 x 10-3 Xl AX2R2 = 125.6 x 1O-6X2 = 0.1256 x 1O-3X2 Putting these in compatibility equations, we have. ~ 0.671 x 10-3+0.1257 x 1O-3Xl+0.032 x 1O-3X2 = 0
~ (1)
- 6;77 x W-3+O.032 x 10-3 Xl+0.1256 x 10-3)(2 = 0
~ (2)
dividing by 10-3 - 0.67l+0.1257X1 + 0.032X2 = 0 - 6.77 + 0.032Xl + 0.1256X2= 0 From(l), Xl
0.671 - O.032X2 0.1257
.~
(3)
Put Xl in (2) & solve for X2 - 6.77 + 0.032 [
0.671 - 0.032X2] 0.1257 + 0.1256X2 = 0
- 6.77 + 0.171 - 8.146 x 1O-3X2 + O.l256X2 = 0 - 6.599 + 0.1174X2 = 0 0.1l74X2 = 6.599 IX2 = 56.19 From (3)
Xl
KNI
. 0.671 - 0.032 x 56.19 0.1257
Xl =- 8.96 KN After redundants have been evaluated, final member forces can be calculated by using the formula shown in last column of table. Apply checks on these member forces.
'!.
124
THEORY OF INDETERMINATE STRUCTURES
CHECKS:(Joint La) 127.5
e iE----t>-76.5 102
LFX = 0 76.5 - 127.5 Sine = 0 76.5 - 127.5 x 0.6 = 0 0=0
LFy';" 0 102 - 127.5 Case = 0 102 -
12is x 0.8 = 0
0=0 The results are O.K. Follow same procedure if some other two members are considered redundant. See example ~o. 12.
I o
'"rj
l' tr:I
>-
en ....,
TABLE FOR EXAMPLE NO.9
~
Mem ber
Ax 10-3 (mZ)
L (m)
F' (KN)
, VI
Vz
F'VjL 10-3 AE x
V,lL AE (m)
(m)
X
10-3
V,V1L -3 AE XIO
F'VzL
(m)
'0-3
AE
x
!!l!:. 10-3 AE x (m)
1.5 1.5 1.5 1.5 1.8 1.8 0.9 0.6 0.9 2.4 1.2 1.2 1.2 1.2 2.4
4.5 4.5 4.5 4.5 4.5 4.5 6.0 6.0 6.0 7.5 7.5 7.5 7.5 , 7.5 7.5
+76.5 +76.5 +117 +25.5 - II7 - 85.5 +48 +42 +II4 -127.5 +67.5
o-
-52.5 0 -142.5
0 -0.6 0 0 -0.6 0 -0.8 -0.8 0 0 +1 +1 0
o·
0
0 0 -0.6 0 0 -0.6 0 -0.80 -0.80 0 0 0 +1 +1 0
0 -0.688 0 0 + 0.877 0 - 1.28 - 1.68 0 0 +2.1 0 0 0 0 2:-0.671 ~10-~3 ___
.
0 +5.4 x 10-3 0 0 +4.5 x 10 3 0 + 21.3 x 10 3 + 32 x 10-3 0 0 +31.25 x 10-3 +31.25 X 10-3 0 0 0 D25.7 X 10-6 ---
- --
----
--
---
0 0 0 '0 0 0 0 +32 X 10-3 0 0 Q 0 0 0 0 2:32 x 10 -
,---
6
------
~
V\X\+ VzX z (KN)
(m)
LoL, L,Ll L2 L3 L3 L4 U IU2 U2U3 UtL, U2L2 U3L3 LoU I U I L2 LtU2 U2 L3 L2U3 U3L4
o
F""F +
0 0 -" 1.05
O. 0 +0.641 0 -1.68 -3.04 0 0 0 -1.64 0 0 2:-6.77 ~1O-3
_ _-'
0 0 +5.4 x 10-3 0 0 +45 x 10-3 0 +32 x 10-3 +21.3 x 10-3 0 0 0 +31.2 x 10-3 +31.2 x 10-3 0 2: 125.6 6 X 10-
----
+76.5 +81.88 +83.28 +85.5 -111.62 -119.2 + 55.17 +4.22 + 69.05 - 127.5 + 58.54 - 8.96 + 3.69 + 56.19 - 142.5
I
.-----~---------
.....
N
VI
126
THEORY OF INDETERMINATE STRUCTURES
2.9. SIMULTANEIOUS INTERNAL AND EXTERNAL TRUSS REDUNDANCY EXAMPLE NO. 10: Determine all reactions and member forces of the following truss by using castiglianos theorem or method of least work. Consider it as: (i) internally redundant; (ii) internally and externally redundant. Nos. in ( ) are areas in x 1O-3m". E = 200
X
10 6 KN/m 2
20KN ,
\ !
8m
8m
SOLUTION: DEGREE OF INDETERMINACY:D = (m + r)
~
2j
= (10 + 4) - 2 x 6 = 2
Therefore, the truss is internally statically indeterminate to the 2nd degree. There can be two approaches, viz, considering two suitable members as redundants and secondly taking one member and one reaction as redundants for which the basic determinate structure can be obtained by cutting the diagonal CE and replacing it by a pair of forces XI - XI and replacing the hinge at F by a roller support with a horizontal redundant reaction HF = X2 • Applying the first approach and treating inclineds of both storeys sloping down to right as redundants. (I) WHEN THE TRUSS IS CONSIDERED AS INTERNALLY REDUNDANT:20KN
8m
8m
ApplYIng method of joints for calculating member forces.
127
METHOD OF LEAST WORK
Consider Joint (C) and all unknown forces are assumed to be in tension to begin with, acting away from the joint. Length AE= 10 m, cos e = 0.6 , sin e = 0.8 Joint (C) 20KN
3KN ----~4h._--~SeD
X1
SBe
L: FX
= 0 Sed + 3 + Xl Cos e = 0 Sed = - (3 + 0.6 x Xl) L: Fy = 0 - Sbe - Xl Sin e - 20 = 0 Sbe = - (20 + 0.8 Xl ) Joint (D)
20KN
............. .
--~
SSD
!FX = 0 3 + 0.6Xl - Sao SBO=( 5 +X l )
L: Fy
x
0.6 = 0
= 0
- SOE - 20 - Sao Sine =. 0 - SDE - 20 - ( 5 + Xl ) x 0.80 = 0 SDE = - ( 24 + 0.8XI ) Joint (B) (20+O.8X1)
6KN ----~~+---~~BE
2
L: FX
= 0 SaE + (5+X 1) x 0.6 + Xz x 0.6 + 6 SBE = - ( 9 + 0.6 Xl + 0.6 X z)
L: Fy
=
0
=
0
THEORY OF INDETERMINATE STRUCTURES
128
- SAB -:- X 2 Sine - (20 + 0.8 X 1)+ (5+X 1) Sine - SAB - 0.8 X 2 -20 - 0.8 Xl + 4 + 0.8 Xl = 0 SAB = - (16 + 0.8 X 2 ) Joint (E)
=
0
(24 + 0.8 x 1)
X1
'"
9+0.6X 1 + 0.6X2
e e
\ LFX = 0 9 + 0.6 Xl + 0.6 X 2 - Xl 9 + 0.6 X 2 = SAE x 0.6 SAE = (15+X 2 )
X
0.6 - SAE x 0.6 = 0
LFy = 0 - SEF - 24 - 0.8 Xl + Xl x 0.8 - (15 + X 2 ) x 0.8 = 0 SEF = - 24 - 0.8 Xl + 0.8 Xl - 12 - 0.8 X 2 = 0 SEF =, - 36 - 0.8 X2 SEF = - (36 + 0.8 X2 ) Enter Forces in table. Now applying Catiglianos' theorem and taking values from table attached. as L .' L S . aX • AE = 0 = 485.6 + 65.64XI + 2.7X2 = 0 (1) I
and
as L s. aX
•
2
or
L AE
=0 =
748.3 + 2.7XI + 62.94 X2
=0
(2) ~
485.6 + 65.64 Xl + 2.7 X2 = 0 748.3 + 2.7 Xl + 62.94 X2 = 0
~
(1) (2)
From (1) X2 = - (
. . (2) 485.6 + 65.64 Xl) 2.7 ) puttmg m
748.3 + 2.7 Xl - 62.94 (485.6 ~~i·64 X') = 0
~
748.3+2.7X I -I1319.875 - 1530.141XI - 10571.575 - 1527.441 Xl IXI ='- 6.921 From (3)
/"
(2)
=0
~
(3)
KNI'
X = _ (485.6 - 65.64 x 6.921~ 2 2.7 ) IX2 = - 11.592
KNI
Now put values of Xl and X 2 in 5th column of S to get final number forces SF as given in last column of table. Apply equilibrium check to verify correctness of solution.
"-----
I
I
; ~
&; > ~
TABLE FOR EXAMPLE NO. 10
L
Mem ber
(m)
AB BC
8 8
Ax 10-3 (m 2) 5 .5
DE BE
S 8 6
5 5 4
CD
6
4
AE BF BD CE
10 10 10 10
2 2 2 2
EF
~
r\ote:
By entering Value column of SF'
L X AE 10-3 8 X 10-3 "
S
-as aX I
as -aX 2
as
L
-3
S. aX I AE x 10
as
L
3
S. aX AE x 102
SF (KN)
~
0 (l02.4+5.12 X 2)LO 3 . - 6.726 -(20+0.SXl)(-0.8) - 14.463 S x 10-3 0 =(I2S+S,12XJ 10-3 (153.6 + 5.12X l) 10-3 0 0 " - IS.463 -(24+0.SX t t -O.S 3 7.S x 10 3 0 0 + 5.12X (230.4 -(36+0.SX2) -O.S -26.726 z) 10" (40.5 + 2.7Xl+ 2.7X2) (40.5+2.7 Xl+2.7 Xz) 10-3 +4.26 -0.6 -0.6 -(9+0.6Xl x 10-3 +0.6X2) 3 25 X 100 + 1.153 0 -(3+0.6X 1)(-0.6) ~(3+0.6Xl) -0.6 7.5 x 10-3 (I5+X2) 0 + 1 " 0 + 3.408 (375 + 25X2) 10- 3 ." 0 + 1 X2 0 (25 X 2) x 10-3 lL592 + 1 " (S + Xl) 0 (125+25X l)10 3 0 - 1.921 25 X 10-3X l + 1 0 0 - 6.921 " Xl L485.6X 10-6 + 65.64Xl I(748.3+2.7X l+62.94 Xl) x 10-6 x 1O-6+2.7x 10-6)(2 of Xl = - 6.921 KN and X2 = -11.592 in column 5 for S, net member forces are calculated and given in last -(16+0.SXV -(20 +0.8X l)
0 -O.S
-O.S 0
.....
tv \0
THEORY OF INDETERMINATE STRUCTURES
130
EQUILIBRIUM CHECKS;Joint (A) 6.726KN 3.408KN
HA
4KN
IFX
=
0
3.408 cose - HA - 0
IFy=O -6.726 + 4 + 3.408 Sine = 0 o = 0 Check is OK.
Joint (F) 11.592KN
26.726KN
)----HF
- - - - 1 . . ......
36KN
IFX =0 -HF+ 11.592 cose
=
0
IHF = + 6.955 KNI IFy =0 36 - 27.726 - 11.592 x Sine = 0 o = 0 (check) It means solution is correct. Now calculate vertical reactions and show forces in diagram.
Dl
METHOD OF LEAST WORK
20KN
20KN
3K:..:.N_~'--!;>_ _...:t
8m 6KNi-~-3>lt-~_~
8m HF=6.955KN HA=2.045Kn ~ VA=
f+-
4KN
i
6m
i
VF =+36KN
~I"-"'::":':":-"I>!
k'\fAL YZED TRUSS
LMA = 0 VF
X
6 - 20 x 6 - 3 x 16 - 6 x 8 = 0
LFy = 0 VA + V F = 40 KN
EXAMPLE NO. 11: CASE II: When the Truss is considered as both externally & internally redundant. Taking SeE & HF as redundants. Now Truss is determinate and calculate vertical reactions. 20KN-
20KN L:Fy
=0
VA + VF = 40 L:MA = 0 VFx6 - 3x16-20x6-6x8=0
Cos9=0.6
IVF = 36KNI and VA::; 4KN
I
Sin 9 =0.8
j 4KN
j36Kn
~6m--"1
Fig. 2.51
I
132
THEORY OF INDETERMINATE STRUCTURES
Compatibility Equations are:
as L I S'aHF' AE
=
0
(1)
Partial differentiation of strain energy W.r.t. HF =.6.H = O. (Pin support)
as
L
I s. ax . AE
(2)
0
=
Partial differentiation of strain energy W.r.t. X = elongation of
member tE due to X = O. As before determine member forces Si in members by method of joints. Joint (A)
4
IFX= 0 SAE SAE
cose - (9 -
H F) = 0
x 0.6 - (9 - H F) = 0
SAE =
(9 0.6 -HF)
IFy = 0 4 + SAB + SAE Sine = 0 4+S AB +(15-1.670Hd x 0.8 = 0 4 + SAB + 12 - 1.33 HF ':" 0 SAB = -16,..1.33H F
ISAB
=-
(16 - 1.33 Hdl
Joint (F)
36
METHOD OF LEAST woRk
133
'LFX = 0 - H F- SSF cose = 0 -HF-0.6 SSF=O -HF = 0.6 SSF I SSF = - 1.67 HFI 'LFy = 0 36 + SEF + SSF Sine = 0 36+ SEF - 1.67 HF x 0.8 = 0
ISEF = -
(36 - 1.33 HF)I
Joint (E)
x . SBE_-t---'JIII'
(15-1.67HF)
'LFX = 0 - SSE -:- X cose - (15 ~ 1.67 H F) cose = 0 - SSE - 0.6X - ( 15 - 1.67 Hd x 0.6 = 0 - SSE - 0.6X - 9 + HF = 0 HF - 0.6X - 9 = SSE I SSE = (H F -0.6X-9)1
2::Fy = 0 SOE +36 - 1.33 H F+ X Sine - (15 - 1. 67HF) Sine = 0 by putting sine = O.OS SOE + 36 - 1.33 H F+ O.SX - 12 + 1.33 HF = 0 SOE =-0.8X-24 I SDE = - ( 24 + 0.8X)
I
Joint (C) 20KN
3KN -----I~...._--s CD
SSG
x
134
THEORY OF INDETERMINATE STRUCTURES
:LFX = 0 SeD + 3 + X Case = 0
ISeD =
- ( 3 + 0.6X)
I
:LFy = 0 - 20 - Sac - X Sin e = 0 - 20 - Sac - 0.8X = 0
ISse = -
(
20 + 0.8 X )
I
Joint (D) 20KN
(24+ o.ax)
:LFX"; 0 3 + 0.6X - Sao Case = 0 3 + 0.6X - 0.6 Sao = O.
ISao =
(5 + X)
I
:LFy = 0 - 20 + 24 + 0.8X ..:.. Sso Sine = 0 - 20 + 24+ 0.8X - ( 5 + X ) 0.8 = 0 -20+24+0.8X-4-0.8X
=
0
0= 0 (check)
Calculation ofHF & X_:From the attached table, picking up the values of summations, we have. as L . :L. S; aH ' AE = 0 = (-1247.03 + 175.24 HF - 4:5 x X) 10-6
F
METHOD OF LEAST WORK
as
135
L
L. s. ax· AE = 0 = (460.6 - 4.5 HF + 65.64X) 10-6 .
and
-1247.03 + 175.24 HF - 4.5X = 0 + 460.6 - 4.5 HF + 65.64X = 0
~
~
(1) (2)
~
(3)
From (1) = ( - 1247.03
X
+ 175.24 HF)
4.5
Put in (2) to get HF
460.6 - 4.5 HF + 65.64 ( -
1247.034~ 175.24 HF)
= 0
460.6 - 4.5 HF - 18190.01 + 2556.17 HF = 0 -17729.41 + 2551.67 HF = 0
IHF
=
= (-1247.03
X or
6.948
KNI
Put this value in 3 to get X.
+ 175.24 x 6.948)
(3)
4.5
IX = - 6.541 KNI Now calculate number Forces by putting th,e ,,;alues of X and ... HF in S expressions given in column 5 of the attached table. These final forces appear in last column for SF' .
2.052Kn
~
i
4kn
i
36KN
!E-6~
Fig 2.52 ANALYZED TRUSS
..... VJ 0\
TABLE FOR EXAMPLE NO. 11 !Hem her
L· (m)
A (mz)
AB
8
5
L
S
- x 10-3 AE 8
X
10-3
-as
-(l6-1.33HF)
as
aHF
ax
+ 1.33
0
BC
8
5
"
-(20 +O.8X)
0
-0.8
DE
8
5
"
. -(24+0.8X)
0
-0.8·
EF
8
5
"
+ 1.33
0
BE
TD AE
6
6 10
7.5
4
X
-(36-1.33HF)
10-3
"
4
25 x 10-3
2
+1
(HrO.6x-9)
0
-(3+0.6X) (15 - 1.67H F)
-1.67
-0.6
10
2
BD
10
2
CE
10
.. .'
2 ,
..
-(16-U3H F)x 1.33x8x10 =-170.24x1O-3+ 14.15 X 1O-3H F 0
0
"
(-1.67H F)
-1.67
0
"
(5+X)
0
+1
"
(X)
0
+1
SF (KN)
0
- 6.759
- (20+0.8x)( -0.8)8 x I 0-3 = 128x 1O-3 +5.12x 1O-3X - (24+0.8>"'",,~" ·,~·c~",,_ ~,.,!114KN B.D.S. Under applied load only.
Or F'-Diagram
~-~~~-~~-~~--~L4 114KN 154
42
I
S.F.D.
' - - - -.....·114
702 B.M.D. 0.6 l1-Diagram
.,
. ,,
139
METHOD OF LEAST WORK
Compatibility equations are: ~l +AX1RI +AX j R2 = O·
-7
(1)
Here Xl =Rl
X2 =R2 Deflection created by applied loads and redundants shall be zero. AX2L + AX2RI + AX2R2 = 0 --7 (2) -
.
F'UL
L~.xIL
=
L. ~ (Change in length of first redundant member by applied loads)
AXIRI
=
. (1T 2L) L\. ~
AXIR2 =
Xl
(Change in length in first redundant member due to first redundant force)
L\.~) AE . X2
. .In first redundant member due .to second redundant force)
(Change In length
(Change in second redundant member due to applied load.)
.
f1!UL:\
.&x2R j = L \.~) .. X I (Change in length of second redundant member due to first redundant force.) AX2R2
= L f1!/L) \."AE" . X2
..
(Change in length of second redundant member due to redundant force in it.)
Forces in chord members and inclineds are determined by the method of moments and shears as explained already~ while for verticals method of joints has been used. EvalUation of force in verticals ofF' - Diagram (Joint~)
67.5
52.5
76.5--__--4>-85.5
96
LFX = 0 85.5 - 76.5 + 52.5 Sine - 67.5 Sine = 0 85.5 - 76.5 + 52.5 x 0.6 - 67.5 x 0.6 = 0 0=0 (Check)
LFy+O U2L2 + 52.5 cose + 67.5 Cos e - 96 = 0 U2~ = - 52.5 x 0.8 - 675 x 0.8 + 96 = 0
-""'"
T \.BLE FOR EXAMPLE NO. 12 AxlO-J (m2)
Member
---
-
1.5 1.5 1.5 1.5
4.5 4.5 4.5 4.5
+ 76.5 +76.5 + 85.5 + 85.5
F'UIUAEx I03(m)
U2
UI
F' (KN)
L - ____
-
LoLl Ll L2 L2 L3 L3 L4
L(m)
- --- -
----
BOTT.
CHORD '
0 0 -0.6 0
0 -.688 0 0
TOP
o
U2 .LlAE x IOJ (m)
U.U 2L1AE x 10J(m)
F'U2L1AE x IO J(m)
U 2 x 6 U
;=
13 > 12 so i= 1 .
First degree internal indetenninancy.
F2L . . 2 AE Strain energy due to direct forces induced due to applied loads in a BDS Truss.
au ax -
of
F.
ax·
L AE = 0
144
THEORY OF INDETERMINATE STRUCTURES
Note:- We seleCt the redundants in such a way that the'stability of the structure is not effected. Selecting member EC as redundant. E
F
/t~jm
AA~ 5KNt
~Z£
~10KN
~:'.
F-diagram B.o.S. under the action of applied loads & redundant. S.F.D. due to applied load only.
B.M.D. due to applied load only.
Method of moments and shears has been used to find forces in BDS due to applied loads. A table has been made. Forces vertical in members in terms of redundant X may be determined by the method of joints as before. From table.
of
L
IF. Ox . AE
=0
= - 331.22 X 10-6 + 51.49 X 1O-6X or - 331.22 + 51.49X = 0
Ix =
+ 6.433 KNI
The final member forces are obtained as ~elow by putting value of X in column 5 of the table. Member
Force (KN)
AB
+5
BC
+5.45
CD
+ 10
EF
- 9.55
BE
+0.45
CF
+10.45
CE
+ 6.43
BF
-0.64
AE
-7.07
DF
- 14.14
.,
"
~
g o ~
~
TABLE FOR EXAMPLE NO. 13
t:Il >-j
F (KN)
of ax
11.25 X 10-3 1l.25 x 10-3
+5 + IO-O.70X
0 - 0.707
0 - 0.707(10- 0.707X) 11.25 x 10- 3 = - 79.54 X 10-3 + 5.62 X 1O-3X
+ io - 5 - 0.707X
0
4
11.25 x 10-3 5.625 X 10 3
- 0.707
4.5
2
11.25 x 10-3
+ 5 - 0.707X
- 0.707
CF
4.5
2
11.25 x 10-3
+ 15 - 0.707X
- 0.707
CE BF.
6.364
2
15.91 x 10-3
O+X
1
6.364
2
3
15.91 x 10-
-7.074X
I
AE
6.364 6.364 .
4 4
7.96 x 10-3 7.96 X 10-3
-7.07 -14.14
0 0
0 - 0.707(- 5 - 0.707X) 5.625 x 10 3 =+ 19.88x 10-3 +2.81 x 1O- 3X - 0.707(5 - 0.70X) 11.25 10-3 = - 39.77 x 10-3 + 5.62 xlO-3 X - 0.707 (15 - 0.707X) 11.25 x 10-3 = - 119.31 x 10-3 + 5.6 x 1O-3 X + 15.91 X 1O-3 X (-7.07 + X) 15.91 x 10 3 = - 112.48 X 1O~3 + 15.91 X 1O-3 X 0 0 2,> 331.22 x 10-6 + 51.49 ;( 1O-6X
Mcm ber
·L (m)
A x 10-3
AB
Be
4.5 4.5
2 2
CD
4.5
2
EF
4.5
BE
DF
(mi
L AE
X
10-3
FaF L ax AE
x 10-3
~
......
~ t.p.
146
THEORY OF INDETERMINATE STRUCTURES
CHECK. Joint A.
5
LFX = 0 5 - 7.07 cose = 0 5 - 7.07 x 0.707 = 0 0=0 LFy = 0 - 7.07 x 0.707 + 5 = 0 o = 0 Check is OK.
EXAMPLE NO. 14:Analyze the following symmetrically ·loaded second degree internally indeterminate truss by the method of least work. Areas in ( ) are 10-3m2 • The value of E can be taken as 200 x 106 KN/m 2
T
3 3m
1 1< Selecting member BO and Before as redundants.
o
F
E
T 3m
1 j7.SKN
k
BOS under loads j7.SKN and redundants. 2@3m------~)1
METHOD OF LEAST WORK
147
SOLUTION: Note:- By virtue of symmetry, we can expect to have same values for XI and X2. It is known before hand. 7.5
I
S.F.D.
+
~.--------r-------~
L-------'h.5 22.5
~B.M.D. SFD and BMD in BDS due to applied loads are shown above. As in previous case determine member forces in BDS due to applied loads by the method 'of moments shears while method of joints maycbe used to determine member forces due to redundants acting separately. Apply super position principal. Then these are entered in a table given.. Summation of relavant columns due to Xl and X2 gives two equations from which these can be calculated. Putting values from table and solving for Xl and X2. [-2.65 x 10-3 (7.5 - 0.707X I ) - 2.65 X 103 (- 0.707X I ) -3.53 X 10-3 (- 0.707X 1 ) -3.53 x 10-3(15 - 0.707X I - 0.707X 2 ) + 10.6 X 10~3 (-10.6+Xd + 10.6 x lO-3 (X 2 )] 10-3 :: 0 - 19.875 + 1.874X I + 1.874 XI + 2.450 XI - 52.45 + 2.50 XI + 2.5 X2- 11'2.36 + 10.6 XI + 10.6 XI == 0 29.898 XI + 2.50 X z - 185.185 = 0 ~(I) CE eolS) - 2.65 x 10-3(7.5-0.707 X 2) - 2.65 x 10-3 (- 0.707 X 2) - 3.53 x 10-3 (15-0.707 X I - 0.707 X 2) - 3.53 x 10-3 (~0.707 X2) + 10.6 x 10-3 (-10.6+X2) + 10.6 X 10-3 X 2 '7' 0 - 19.875+1.874 X2+1.874 X2-52.95+2.50 X I+2.50X2+2.450 X2-112.36+1 0.6X2+ 10.6 Xi = 0 2.50 XI + 29.898 X2 - 185.185 = 0 = (185.185 - 2.50 X2) From ( 1, ) XI 29.898
(2) => 2.50 (
~(2)
~
(2: eol9)
(3) Put in 2 above
185.185 - 2.50 X?~ 29.898 "j + 29.898X2 - 185.185 = 0
15.465 - 0.21 Xl + 29.898 Xl - 185.185 = 0 29.689 Xl - 169.7 = 0
·jX? =+5.716KNI Put X2 in equation 3 to get XI. The final member forces are given in last column .. These are obtained by putting values ofX j and Xz, whichever is applicable, in column 5 of the table.
.t:>oo
TABLE FOR EXAMPLE NO. 14 F (KN)
8F
8F
Oxl
Ox2
+ 7.5 - 0.707 XI
-0.707
0
"
+ 7.5 - 0.707 X2
O'
4
"
0-0.707 XI
Mem ber
L (m)
AB
3
4
Be
3
4
DE
3
L AE x 10-3 3.75 X 10-3
Ax 10-3
(mi
EF
3
4
"
AD
3
3
5x 10-3
BE
3
3
CF
3
3
"
AE
4.2426
2
10.6 3 X 10-
BD CE BF
4.2426 4.2426 4.2426
2 2 2
0-3 xl
Member Forces F(KN) + 3.459
F 8F..110-3 Ox2 AE x 0
- (7.5 - 0.707 XI) 0.707 x 3.75 x 10-3
-0.707
0
- 0.707 (7.5- 0.707Xz) 3.75 x 10-3
+ 3.459
-0.707
0
0
-4.04
0-0.707X2
0
- 0.707
- 0.707 (- 0.707X I) 3.75 x 10-3 0
- 0.707 (- 0.707X2) 3.75 X 10-3
-4.04
0- 0.707 XI
- 0.707
0
- 0.707 (- 0.707X I) 5 X 10-3
0
-4.04
+ 15 - 0.707 XI - 0.707 XI
- 0.707
-0.707
- 0.707 (15 - 0.707X 1 5 x 10-3 - 0.707X 2
- 0.707 (l5~ 0.707X r - 0.7Q7x2) x 5 x 10-3
+6.918
.0 - 0.707 X2
0
-0.707
0
- 0.707 (- 0.707X z) 5 x"10- 3
-4.04
~ o"I1
-10.60+ XI
+1
0
1O.6xlO-3 (1O.6+XI)
0
-4.884
~tI1
.
.
"
F 8F L Oxl AE
....,
::r: tI1 o
...., tI1
(0 + XI) -10.6+ Xl (0 + X2)
" " "
+1 0 0
0 +1 +1
10.6 x 10-3 (XI) 0 0
0 .1O.6x 10-3 (-10.6 +X z) 10.6xI0-3 (X2 )
l
+ 5.716 - 4.884 + 5.716
! CZl
29.898XI +2.5.oX2-185.185
L 2.50XI + 29.898 X2 - 185.185
~
>-i
~
CZl
"","
',' ~.--.
--"'--"...... - ~.~ ~,
c.-"",,-,,-__,"-,,-,,,~-,-,-,=",,,,,,,-,=",,,,,,,,-",,,,=,,,,,,=,,-,,,,,,,-,,=~,,,,-,,,,,",:,,,,,,",,~-,,,,-,",.,,.,,,,,.'-';;"~n._,",""""'_'''''''''·''''''''_'",",,,,".::,
,~~~~."< .. .'.,..,:;,,,
t~··.>.,··:-r_··;'i1>\5KN
C - -_ _ _ _
---',~
30
.~
B.M.DO~~O
150.
THEORY OF INDETERMINATE STRUCTURES
10KN
~
3@3m----?l>l'5KN
. 10.-----,
S.FD
01-1--+--+---------, -
L - -_ _ _ _
--'t~
30
~15
B.M.D/~~
0---------------0 Member Forces Due to Redundants Only. Please number that due to separate action ofredundants Xl and X 2 member forces will be induced only in the square whose incllneds are Xl and X 2 • There will be no reaction at supports. Joint D:DG
L: Fy
CD~
0 DG Sine - 0 =
IDG =
L: FX
01
= 0
DG Cos
e + CD
= 0
ICD =
01
10int G:-
FG:?e1 e· X2
L: FX
=
0
- FG - X 2 Cos e = 0
IFG = -
0.707
x~1
L: Fy = 0 - CG - X 2 Sin e = 0
ICG = -
0.707 X2
!
CG
'"
.1
'I
I ·1"
,~ .I 'I
151
METHOD OF LEAST WORK
Joint C:O.707X2
5J
Be L Fy
= 0 CF Sin e - 0.707 X z = 0 CF = 0.707 X 2 0.707
L FX
= 0 - BC - CF Cos
e= 0
Joint B.
LFX = 0 - 0.707 X z -AB + Xz Cos e - Xl Cos
lAB L Fy
=-O.707X I
I
= 0
Xl Sin e + Xz Sin e + BF = 0 I BF
= -0.707Xl - 0.707X2 1
Joint A.
L FX
= 0 - 0.707 Xl + AF Cos e = 0
1
J
e=0
·i
=1=52==============================T=H=E=O=R=Y=O=F=mD===E=T=ERM===IN=A=T=E=S=T=R=U=C=T=URE~1 ~
,;~\
,1,'
0
~~
AE + AF Sin e = 0
·1~
IFy
=
,~
Joint E.
L: FX
=
0
IFy =0 0.707 XI - 0.707 XI = 0
o=
0 (Check)
Entering the values of summations from attached table, we have.
'" aF L.. F. aX
I
L -{j -{j -6 . AE = 0 = - 229.443 x 10 +29.848 x 10 X I+2.45 x 10 X2
'" ~ • l_ L.. F. aX AE - 0 -_ -168.9 x 10-6 +2.45 x 10-6 X I+29.848 x 10 2
-{j
X2
Simplifying - 229.443 + 29.848 XI + 2.45 X2
=
- 168.9 + 2.45 XI + 29.848 X 2
0
=
0
~ ~
(1)
(2)
From (1) X
= ( - 2.45 X2 + 229.443) I, 29.848 Put in (2) & solve for X2 , (- 2.45 X2 + 229.443) - 168.9 + 2.45 29.848 + 29.848 X 2 = 0
- 168.9 - 0.20 I Xz -'- 18.833 -'- 29.848 Xl = 0 - 150.067 + 29.647 X2 = 0 X _ 150.067 2 - 29.647
IX
2
=
+ 5;062 KNI
~
(3)
•_ _
'--o_~
~ tTl , >-l
TABLE FOR EXAMPLE NO. 15 Mem ber
L (m)
AB
3
BC
3
Ax 10-3 (m)2 4
4
L AE x 10-3 3.75 x 10-3
F (KN)
aF
aF
aX I
aX2
+ 7.5 - 0.707 XI
- 0.707
0
::r:
F aF L aX2 AE
F aF L -3 aX I AE x 10
--
x 10-3
Member Forces F(KN)
...
0
- (7.5 - 0.707 XI), 0.707 x 3.75 x 10-3.
+ 3.459
+ 7.5 - 0.707 X2
0
- 0.707
0
- 0.707 (7.5- 0.707X 2) 3.75 x 10-3
0- 0.707 XI
- 0.707
0
0
-4.04
0- 0.707 X 2
0
- 0.707
- 0.707 (- 0.707XI) 3;75 x 10-3 0
- 0.707 (- 0.707X 2) 3.75 X 10-3
-4.04
+ 3.459' "
3
4
"
EF
3
4
"
AD
3
3
5x 10- 3
0- 0.707 XI
- 0.707
0
- 0.707 (- 0.707X I) 5 x 10- 3
0
-4.04
BE
3
3
"
+ 15 - 0.707 XI
- 0.707
- 0.707
- 0.707 (15 - 0.707 XI 5 x 10-3 - 0.707X 2
- 0.707 (15- 0.707X I ,- 0.707x2) X 5 x 10- 3
+ 6.918
0
- 0.707
0
- 0.707 (- 0.707X 2) ' 5 X 10- 3
-4.04
3
3
0- 0.707 X 2
"
> >-l
DE
CF
t-<
tIl
(/l
"
- 0.707 Xl
o t:l o >-rj
:E
o
~
{""
AE
BD CE BF
4.2426
4.2426 4.2426 4.2426
2
2 2 2
10.6 x 10-3
" " "
-
3
- 10.60+ XI
+ 1
0
10.6xlO- (10.6+X I)
(0 + XI) -10.6 + X 2 (0 + X2)
+ 1 0 0
0 +) +1
10.6 x 10-3 (XI) 0 0
0
- 4.884
'
-,
L L-
229.443 x 10-(' + 29.848 + 2.45
"
0 10.6x 10-\-10.6 +X 2) ) 0.6x) 0-3 (X 2)
--
.
+ 5.716 - 4.884 + 5.716
L X
X
10-6 XI 10-6X 2
L-
168.9x 10-6+2.45x2.45x 10-6 XI +29.848 X 10- 6 X2
...... u. V.l
154
THEORY OF INDETERMINATE
STRUCTURES,~
=================================="1'
So
Xl =
- 2.45 x 5.062 + 229.443. . 29.848 by puttmg value ofX2 m (3)
IXl = + 7.272 KNI EQUILIBRIUM CHECKS:5.141
8.579
t5KN Joint B:-
7'27~2 6.28 5.062 4.859
a .
a
6.421
15
l: FX
= 0 6.421 + 5.062 cose - 7.272 cose - 4.859 = 0 0=0
l: Fy
= 0 6.28 - 15 + 5.062 Sine + 7.272 Sine = 0 o = 0 The results are OK.
Joint C:2.008
1.421
·6.421~5 l: FX
= 0 5 + 2.008 cose - 6.421 = 0 0=0
IFy = 0 1.421 - 2.008 Sirte = 0 o = 0 Results are OK.
.
INTRODUCTION TO TWO-HINGED ARCHES
155
CHAPTER THREE INTRODUCTION TO TWO-HINGED ARCHES 3.0, TWO-HINGED ARCHES:The following issues should be settled first. Definition. Types. Basic Principle and RM. Linear Arch. Mathematical Generalized Expressions. Segmental Arches. Some infonnation is contained elsewhere where detenninate arches have been dealt.
3.1. DEFINITION OF AN ARCH. "An arch can be defined as a humped or curved beam sUbjected to transverse and other loads as well as the horizontal thrust at the supports." An efficient u~e of an arch can be made only if full horizontal restraint is developed at the supports. If either of the support allows some movement in the horizontal direction, it will tend to increase the B.M. to which an arch is subjected and arch would become simply a curved beam. . The 13.M., in arches due to the applied loads is reduced due to the. inward thrust. Analysis is .. carried out to find the horizontal thrust and also to find the RM., to which art arch is subjected. Beam action V s arch action:
kP M1P
lP
B
~
, One reaction at support only Simple beam subjecte to applied !1 Vb transverse loads. T
Support, abutments or springing.
~
D.
!1
Va IE--- x· --71 I Vb Arch carrying vertical loads & horizontal thrust
Two reactions at supports
The above beam and arch carry similar loadings. If Mo = RM. due to applied loads at a distaJljOe X41n the simple span of a simple beam where rise is y. then bending moment in the arch is, Mx = Mo ± Hy where Mx is the RM., in the arch at a distance x. H is the horizontal thrust at the springings & ) is the rize of the. arch at a distance. 'x' as shown in the diagram. The (± ) sign is to be used with care and" (-) sign will be used if the horizontal thrust is inwards or vice versa. In later case it will behave as a beam.
156
THEORY OF INDETERMINATE STRUCTURES p
Under transverse loads, the horizontal thrust at either of the springings abutments is equal. In the arch shown above, the degree of indeterminacy is one and let us consider the horizontal thrust at support B as the redundant. The above loaded arch can be considered equal to the following two diagrams wherein a BDS arch is under the action of loads plus the same BDS arch under the action of inward unit horizontal load at the springings. p
S'
A
S
tva
Vbt
6sC
B.D.S. under applied loads (loads try to flatten the arch) .1BL stands for displacement of point B due to applied loads in a BDS arch .. +
(Flattened arch recovers some of horizontal displacement at B due to unit horizontal loads and will recover fully if full horizontal thirst is applied at B.) (Arch flattens out uI1G.er the action of applied loads because freedom in the horizontal direction has been provided at point B.) and all due to full redundant value. This forces the basis of compatibility. LillR stands for displacement of point. B (in the direction of for.ce) due to unit horizontal redundant force at B. Remember that a horizontal reactive component cannot be realized at the roller support. However, we can always apply a horizontal force at the roller. -
3.2. Compatibility equation" .1BL - (LillR) H = 0 ( If unit load is applied in opposite sense so that it also produces flettening, +ve sign maybe used in the equation and the final sign with H will be self adjusting.)
INTRODUCTION TO TWO-HINGED ARCHES
LlliL orH= t.BR
157
displacement at B due to loads displacement at B due to unit horizontal redundant
We will be considering strain energy stored in bending only.The modified expression for that for _ curved structural members is as follows.
lu=f~1 Where ds is the elemental. length along the centre line of the arch and U is the strain energy stored in bending along centre-line of arch. The bending moment at a distance x from support is Mx = Mo - Hy (Horizontal thrust is inwards).
(1)
Where Mo = Simple span bending moment ( S.S.B.M.) in a similar loaded simple beam.
~ f~~~s
U
If H is chosen as redundant, then differentiating U w.r.t. H , we have
au aH
(.L
= t.BH = 0 = JEI . M.
au
JEI' (Mo - Hy)(-y) ds
_f
0ds _
EI
H/dS
J
EI
Put M= Mo - Hy and then differentiate.
(.L
aH = t.BH = 0 =
-f H l
(aM) aH ds
(Hi - Mo y) ds
Simlifying
EI
f Mo
y
by putting M from (1 )
ds = 0
EI
f
MO yds EI or
JMoy.ds H=
EI ? Jy-ds
EI
.
"
Applymg Castlghano's 2nd theorem, t.BL becomes =
and t.BR=
-f',2 ds ~ EI
fMOEIYds
158
THEORY OF INDETERMINATE STRUCTURES
The algebraic integration of the above integrals can also be performed in limited number of cases when EI is a suitable function of S ( total curved arch length), otherwise, go for numerical integration. For prismatic (same cross section) members which normally have EI constant, the above expression can be written as follows:
3.3. TYPES OF ARCHES :The arches' can be classified into a variety of ways depending mainly upon the material construction and the end conditions.
of
Q) Classification Of Arches Based On Material of Construction :-
The following arches fall in this particular category: a) Brick masonary arches. b) Reinforced concrete arches. c) Steel arches. Thespan of the arches which can be permitted increases as we approach steel arches from the brick masonary arches.
(2) Classification Of Arches Based On End Conditions :The following arches fall in this particular category: a) Three hinged arches. b) Two hinged arches. c) Fixed arGhes. In the ancient times, three hinged arches have been used to support wide spans roofs. However, their use is very rare in bridge construction since the discontinuity at the crown hinge is communicated to the main deck of the bridge. In three hinged arches, all reactive components are found by statical considerations without considering the deformations of the arch rib. Therefore, they are insensitive to foundation movements and temperature changes etc., and are statically determinate. These are covered as a separate chapter in this book. The Romans exploited the potential of arches to a. great extent. However, their emperical analysis approach became available in the ear.ly 18th century.
3.4. LINEAR ARCH:This is just a theoretical arch at every X-section of which the RM. is zero. M =Mo-Hy=O or
Mo = Hy.(The B.M. due to applied loads is balanced by Hy). therefore, y =
M If
INTRODUCTION TO TWO-HINGED ARCHES
l:
This is the equation for the centre line of a linear arch. With the change in position and the l1lunt of loads on the arch, the corresponding linear arch would also change as Mo keeps on changing. Therefol there are infinite number of such arches for every load pattern and position on the aCtual arch.
EXAMPLE NO.1: 3.5. ANALYSIS OF TWO - HINGED SEGMENTAL ARCHES We develop the method for indeterminate arches starting with the simplest cases of segmental arches. Solve 1 following segmental arch by using the basic principles of consistent deformation method and by treating horizon thrust at support D as the redundant. The segmental arches could be used in tunnels and in water ways. OKN/m
T· 4m
+t_2_m.;...f__
EI=Constant
;~ _ _--'-4'_2_m+!
40 kN
40 kN
T 4m
t 2mr
EI=Constant
4,2m~
4m
**-~-- 8m ----'---+J.
40 kN
40 kN
.... ... ~
(Ha will occur only point D is a hinge support) . ' M - Diagram. Due to applied loads. Similarly reactions due to supermetricalloading.
+
m - Diagram.
'
Due to unit redundant at D. (X is varied along length of members). Find cose and Sine. cos e = 0.447,2, sin e = 0.8944.
50
THEORY OF INDETERMINATE STRUCTURES
Sab sin e + 40 =0 so Sab= 0~~4 = - 44.722. Consider equilibrium of joint A and project forces y-direction. (M-diagram) Consider same diagram with roller at D. Now consider joint A and Project forces in X direction to 'aluate Ha. Sab cose + Ha = 0 or -44.722 x 0.4472 + Ha = 0 orHa=20KN ompatibilityequation 6.DL - 6.DR. H = 0 6.DL Or H = 6.DR
6.DL =
f
6.DR =
f
Horizontal displacement ofD due to loads Horizontal displacement ofD due to redundants
Mmdx ' EI Applying Unit load method concepts, m 2 dx
EI
Now we attempt the evaluation of these integrals in a tabular form. X is measured along member IS.
Mem ber AB
Origin. A
0-4.472
BC
B
0-4
CD
D
0-4.472
6.DL =
Limits.
1
1
.+ ::,
40 X cose .;~,., 'Jr=40X0.477= 17.88X 40(2+X)-10X2= 80 + 40 X-lO X2 17.88 X 1
4.472
MmdX E I =EIf
f
m
M
(17.88X)(+0.894X)dX+
o
+ I.XSine=+0.894X +4 + 0.894 X
4
f80+40X-I0X)(+4)dX
EI 0
4.472
+ EI f
(17.88 X)(+0.894 X) dX
o
2 4.472 1 4 = Elf (+15.985X2)dX+ f(+320+160X-40X 2)dX EI 0 o 3
_ +31.969 / X EI . 3
=
I
4.472
0
1../ " 160X? + EI + 320X + 2 -
40X 3
4
2 /
0
+10.656' 3 1 ( 40 ) (4.472 -0)+ EI +320x4+80x 16- 3 x 16 EI
Integrate and put limits
INTRODUCTION TO TWO-HINGED ARCHES
6,DL
= + 2659.72 EI 1
m2dX
t.DR =
f
4.472
EI = EI J
, 1 (+ 0.894Xt dX + EI
1
4
4.472
J 16 dX + EI f
0 0 .
.
0
(+ 0.894Xt dX
0
2 4.472 0 16 4 = EI J 0.799 X- dX + EI JdX o
_ 1.598 EI =
D.DR
I X3 I 3
0
4.472. o
0·~i3 [(4.472i -
l.2.
+ EI I X
0] +
I
~~ (4 -
4 0
0)
= 111.653
EI
D.DL H --D.DR 2659.72IEI 111.6531EI
i H = 23.82 KN I EXAMPLE NO. 2:-
Solve the following arch by using consistent deformation method.
T 4m EI-Constt
,\t 2m ,Ote .
4m
The above redundant I segmental arch can be replaced by the following similar arches carrying loac and redundant unit load. r20KN/m
tp;:;c:;::=:c:x:::coc:x::\C
T
X is varied along member lengths.
4m
~A .. ~
i
4m
2m
Ra=20KN
~
2m
.... 0 I
I
1
U
tt.D}
M-Diagram
BDS UNDER LOADS
Rd=60KN
THEORY OF INDETERMINATE STRUCTURES
62
I:Ma = 0; Rd x 8 = 20 x 4 x 4 + 40x 4 .. Rd=60KN so Ra=20KN
+
I I
c X is varied along member lengths.
1
k~ m-Diagram BDS UNDER UNIT REDUNDANT AT D Compatibility equation is LillL - ~DR.H = 0 Where
~DL =
Horizontal deflection ofD in BDS due to applied loads. = Horizontal deflection at D due to Unit redundant. H = Total Horizontal redundant.
~DR
Or
md
~DL
H-~DR
~DL
=
f
MmdX EI
~DR= ·fm dX . EI 2
Member AB
Origin A
Limits 0-4.472
BC
B
0-4
CD
L'lDL.=
D
0-4.472
M 20X Cose+40X Sine 20Xx0.447+40X x 0.894 = 44.72X 20(2+ X)+40 x 4 -I OX2 40+20X +160 - IOX2 = -IOX 2 + 20X + 200 60X Cose=60X.x 0.447 = 26.82 X
m XSin8=O.894X
EI Constt.
Constt. +4 0.894X
Constt.
MmdX 1 4.472 I 4 E I = ElI (+44.72X)(0.894X) dX + EI I (-IOX 2 +20X - 200) 4 dX
f
o
0
rI
INTRODUCTION TO TWO-HINGED ARCHES
+
~I t
163
n (26.82X ) (0.894X)dX
o
4 472 .
_ 2 x 23.9771 l.33X EI
o
.1.1-
X31 3 + EI
0
.1.
3
ADL = 63.97 [4.472 ] EI
L.l.
bDR =
f
3
m" dX
1
3 3
-
=
bDR
EI
+ EI
4.472
EI = EI ~
_ 1. 598 1 X 01
l OX) 20X" 14 3 + 2 + 200X
4 .472
o
[-10 403 10 4 2 200 0 4J = +4893.8 3 x + x + x El
, 1 4 1 (O.894Xt + EI ~ 16dX + EI
l.§. + EI I X
0.~:3 [(4.472)3 _ 0] + ~~
4.472
~
,
0 (O.894X)-
4
I
0
(4 - 0)
111.653 EI
bDL ..
H - bDR + 4893.8/EI 111.653/EI
So
I H = + 43.83 KN I
EXAMPLE NO. 3:- Determine the horizontal thrust for the for following loaded segmental arch. Take EI equal to constant.
p
P
G
5m
0
.
·'-1· ~.· !;· ·
:~r
THEORY OF INDETERMINATE STRUCTURES
164
I SOLUTION :-
x
p
p
G
~----4-------4
\1
I
0 X is varied along member length
t
'pI
p
Now consider a BDS under Loads and redundant separately for the same arch and evaluate integrals. An inspection of the arch indicates that it is symmetrical about point G and is indeterminate to the first degree choosing horizontal reaction at F as the redundant, we draw two basic determinate structures under the action of applied loads and the redundant horizontal thirst at support F.
p
p
c
4m
G
o
5m
4m
M-Oiagram (80S under loads)
C,r-_ _ _..:;:E
B.o.S. under unit horizontal redundant load at F. m-Oiagram
f
! fNTRODUCTION TO TWO-HfNGED- ARCHES
165
Because of symmetry, Moments and hence stram ener y IS compute d fior haIff:·rame. Origin
Limits
AB
A
0-5
PX cose = XO.6 PX
Be
B
0-5
P (3+0.8X)
CG
C
0- 2.5
P (7+ X) - PX = 7 P
Portion
AFL
=
u
2
I5 (0.6 PX)(0.8X) dX EI
0
2P [
=
b.FL
b.FR
2.5
o
0
0
EI
49 dX
]
5
5
o
2 P [0.48 3 ill .-3-·x 5 +
0
.
0
0.48 x 53 5 X 52 ] 3 + -2-+ 12 x 5 +49 x 2.5
(deflection of point F due to loads)
2 5 , 2 5 2 2.5 EI I (O.8XtdX + EI I (16 + 0.36X2 + 4.8X) dX + EI I 49dX
2 EI
0
[I
I
_ 1- [ EI
0
3
5
0.64X315 0.36X 4. 8X2 1 2.5 ] - 3 - + 16X+ -3-+ -2- + 149X! o
=
I2.5 49 P d
[I 0.438X3 1+ I0.483 X31 + 5 2X2 + 12X !+ 149X 2.5I ]
o
b.FR
+2
7
0
5
. 570 P
-
EI
0
0.8X 4+0.6X
I5 P(3+0.8X)(4+0.6X) dX
5
="EI =
m
ill f 0.48 X2 dX + I (0.48 X2+5X+12)dX + I
_ 2P - EI
=
+2
M
0
0
l
0.64 X 53 0.36 3 4.8 x 52 3 + 16 x 5 + 3 x 5 + 2 + 49 x 2.5 .J
608.33
EI 570 P 608.32
H---
So
I H = 0.937 P I
X
166
THEORY OF INDETERMINATE STRUCTURES
NOTE :-
Compatibility equation is ~FL - ~FR x H = 0 ML= ~FRxH H= ~FL
'roo.
~FR
We take compression on outer side & tension on inner side +ve in case ofM and m-diagram. EXAMPLE NO.4 :Determine the horizontal thrust provided that EI = Constt for the following loaded segmental arch.:
SOLUTION:
p
p
:1
Taking horizontal reaction at F as redundant. LMa=O Rf.19=P.12+P.7+4.P,So
IRf= 1.211 P I . and therefore Ra is, Ra = 2P - 1.211 P
IRa = + 0.789 P
i 0.789 P
I
M-Diagram
1.211 P
l
I
INTRODUCTION TO TWO-HINGED ARCHES
167
c
o
4m
5m
m-diagram (Unit redundant at
Portion AB
~FL
Origin A
M
Limits 0-5
BC
B
0-5
CD
C
0-5
DE
E
0-5
EF
F
0-5
~)
0.789 PX cose+px Sine = 0.4734 PX + 0.8 PX = 1.2734 PX 0.789 P(3 + XCose l ) +P(4 + XSin8 1) - PX Sine l = 0.6312 PX+6.367 P O.789P (7+X)+Px7-Px3.;..PX =-0.211 PX+9.523 P 1.211 P(3+X cose l ) = 3.633 P + 0.9688 PX 1.211 PX Cos 8= 0.7266 PX Determine
m 1 x XCos e =0.8X 1(4 + X SineJ =4+0.6X +7 1(4 + X Sine l ) =4+ 0.6X X Sin e = 0.8X Sines and Cosines ofe and el.
[5f (1.2734 PX)(O.8 X)dX + 5f (0.6312 PX + 6.367. P)
1 = EI
o
0
5
(4 + 0.6X) dX + f (- 0.211 PX + 9.523 P)(7)dX o 5 .
5
+ f (3.633 P + 0.9688 PX)(4 + 0.6X) + f (0."7266PX (0.8X) dX o
P = EI
0
[5J 1.0 1872X dX + 5f (2.5248X + 0.37872X + 25.468 + 3.8202 . X) dX 2
o
2
0
5
+
]
5
f (- 1.477X +
66.661) dX + J(14.532 + 2.1798X
o
0
5
+ 3.8752X + 0.58128X 2 ) dX + f O.58128X dX
Simplifying we get.
o
P
=
5
2
EI j (1.97872X + 11.50428X + 106.661) dX o 3
~FL
5
p I 1.97972"3 X + 11.50428"2 X2 + 106.661 X 1 0 = EI
168
THEORY OF INDETERMINATE STRUCTURES
P[
53
6.FL 6.FR
52
1.97872 x 3" + 11.50428 x"2 + 106.661 x 5
= E1
]
759.56 P EI 1 = EI J(0.8X)2dX + J(l6+0.36x 2+ 4.8X) dX
[5
.5
o
.
5
5
o
0
0
.
5
+ J49 dX + J(16+0.36X 2+-L8X) dX + J0.64 X2 dX
_ J...
- EI
0
[0.64 53 16 0.36 x 53 4.8 X 52 4· 9 5 3 x + x5+ 3 + 2. + x
0.36 • 4.8· , 0.64 3] x 5' + - x 50. + x5 + 16 x 5 + 3 2 3 uAFR
=
H -
]
608.33 C ompatl·b·l· ---m--.
I Ity
Simplifying
. remaInS . t h e same. p . va Iues 0 f·Integra Is, we have equation uttIng
6.FL 6.FR
= 759.56
EI
I: /
/608.33 EI
I H = 1.2486 P I Now all reactions are shown.
F J.2486P IO.789P
11.211P
ANALYZED SEGMENTAL ARCH Check:
IMc
=
0
0.789P x 7 - 0.2486 P x 7 - P x 3 + P x 5 + 1.2486 P x 7 - 1.211 P x 12 = 0 0=0 O.K.
INTRODUCTION TO TWO-HINGED ARCHES
169
3.6. ANALYSIS OF TWO HINGED CIRCULAR ARCHES :-
o The circular arches are infact a portion of the circle and are commonly used in bridge construction. From the kn6wledge of determinate circular arches, it is known that the maximum thrust and the vertical reactions occur at the springings. The~efore, logically there should be a greater moment of inertia near the springings rather than that near the mid-span of the arch. The approach is called the secant variation of inertia and is most economical. However, to establish the basic principles, we will first of all consider arches with constant EI. The. following points are normally required to be calculated in the analysis. (1) Horizontal thrust at the springings. . (2)
B.M. & the normal S.F. at any section of the arch. Usually, the span and the central rise is given and we have to determine; (i) . (ii)
the radius of the arch; the equation of centre line of the circular arch.
Two possible analysis are performed. (I)
Algebraic integration.
(2)
Numerical. integration.
After solving some problems, it will be amply demonstrated that algebraic integration is very laborious and time consuming for most of the cases. Therefore, more emphasis will be placed on numerical integratien which is not as exact but gives sufficiently reliable results. Some researches have shown that if arch is divided in sixteen portions, the results obtained are sufficiently accurate. In general, the accuracy increases with the increase or more in number of sub-divisions of the arch. We will be considering two triangles. 1- ~ADO
2-
~EFO
By considering ~ ADO
THEORY OF INDETERMINATE STRUCTURES
170
i
R2 = (R_yC)2 + (L /2 R2 = R2 _ 2Ryc + yc 2 + L2/4 o = yc ( yc - 2 R) + L2/4 yc ( yc - 2 R) = - L2/4 or - yc (yc - 2 R) = L2/4
Iyc (2R - yc) = ¥I
(1)
By considering /j. EFO OF 2 = OE 2 + EF2 , , 2 R- =(R-yc+yt+X R2_X2 =(R-yc+yi
s
? (2) The detailed derivation of this equation Gan be found in some other Chapter of this book. In this case, S = R ( 2
e ) where e is in radiains. S is the total length along centre line of the arch.
fMyds H = rIds
as before obtained By eliminating EI as we are considering EI = Constt
EXAMPLE NO. 5:A two- hinged circular arch carries a concentrated force of 50 KN at the centre. The span & the rise of the arch are 60m & 10m respectively. Find the horizontal thrust at the abutments. SOLUTION :- The arch span is divided in ten equal segments and ordinates are considered at the centre of each segment.
o
INTRODUCTION TO TWO-HINGED ARCHES
R
=
8~:
+
T.
171
where R = Redious, yc = Central rise and
L
= Span of arch ..
_ (60)2 lQ - 8 x 10 + 2
R
=
50m
30 . Sina = 50 = 0.6
a
. Now compute angle a is radians . , we know
= 36.87"
1trad = 180 0
1800 = 1t rad 0
1
So
1t
= 180 rad
36.870 = 1;0 x 36.87 radians 36.87 0 = 0.6435 rad = a a = 0.6435 rad S = R (2 a) = 50 (2 x 0.6435), Where S is length of arch along its ceptre-line For circular arches. X is varied from centre to abutments.
I S = 64.35 m I 50KN C
A ~~~--~~~----~ 25KN
1f'*'P-~.
o H _JMyds
- rids
where
M ~ Simple span ( 8.S ) B.M. in the arch due to applied loads only.
Mbc OE
= Mac = 25 ( 30 - X ) in two portions at a distance X from mid span. = RCose
")'r:';' ~j;';
172
THEORY OF INDETERMINATE STRUCTURES ···'1······'·····'"
OD y
=
R-yc
= OE -
=
I
50-10=40m
OD
,'i
[Since OC = OD + CD = 50 and CD = 10 = Yc]
y = RCos 8 -40 and
ds. = Rde X = R Sin 8 Evaluation of Numerator :Mx = 25 (30 - X), ds = Rd8, y = RCose - 40 u
J Myds = 2 J [25 (30 - R Sin8)] [R Cose-40] [Rd8], By putting X, y and ds from above. Also put o
value of a which is in radians. 0.6435
=
50 R J
(30 - R Sin8)(R Cos8 - 40) d8,
we know, 2Sin8 Cos8 = Sin 28.
o
0.6435
(30R Cos8 Cos8 - R 2Sin8 Cos8 - 1200 + 40R Sin8) d8
= 50R J o
= 50R
=
I
R2
30R Sin8 + 2
.
+C
28
. 12008 - 40R Cos8
I
0.6435
0
Put limits now
2500 502 l 50 x 50 [ 30x50xO.6+~ 0.2S-1200xO.6435-40x50xO.S -4 xl+ 40 x 50 xIJ
= + 194500 J Myds = 194.5 x 103
Evaluation of Denominator :-
1 2 . Cos 8=2'{1 +Cos28)
We know and
Sin 8 = ~ (I - Cos28) 2
0.6435
(RCos8 - 40i (Rd8)
Jids = 2 J o
0.6435
= 2RJ
(R2 Cos 2 8 - SOR Cos8 + 1600) d8
o
0.6435 [R2
= 2R ~
2
R2 (
=
2R
J-
(1 + Cos 28) - SO R Cos 8 + 1600 d8
S' 28)
I2\8+T
-SORSin8+ 16008
I
0.6435 0
Integrate
Put limits now
INTRODUCTION TO TWO-HINGED ARCHES
= 2 x 50 [
50 T
2 (
173
0 9f\ . ] 0.6435 +2) - 80 x 50 x 0.6 + 1600 x 0.6435
= 3397.5
Jy2ds
H
=
+ 3.3975 x 103
194.5 X 103 = 3.3975 X 10"3
I H = 57.2 KN I EXAMPLE NO.5: BY NUMERICAL INTEGRATION:The values of X, y and M are detennined at the mid ordinates of the segments. The basic philosophy is that if we consider a very small arc length that would be regarded as a straight line and therefore we tend to average out these values. y = -.jR2_X2 - (R-yc) or
y = -.j50 2 _X2_ (50-l0)
or
y = -.j50 2 - X2 - (40)
For section (l) X I = 27, from (1),
YI=
(1)
See segments of Example 5 about 4 page before.
-.j50 2 -2i -(40)=2.08m
For section (2) X2=21 from (!),
M
= 25
( 30 - X )
= (750 -
2
Y2= -.j50 2 -21 -(40) =5.738 mandso on. 25X)
0 < X < 30 as before
Now do numerical integration in a tabular fonn as under.
Section, 1 2 3 4 5 6 7
S 9 10
X 27 21 15 9 3 3 9 15 21 . 27
S = 64.35 m and ds = 64.35 10 ds = 6.435 m
y. 2.08 5.380 7.69 9.18 9.91 9.91 9.18 7.69 5.380 2.08
M 75 225 375 525 675 675 525 375 225 75
My 156.00 1210.50 3883.75 4819.50 6689.25 6689.25 4819.50 2883.75 1210.50 156.00 2::31518
?
y-
4.33 28.94 '59.14 84.27 98.21 98.21 84.27 59.14 28.94 4.33 2::549.78
174
THEORY OF INDETERMINATE STRUCTURES
H _ fMyds _
L Myds
-
Lids
- f Ids
31518 x 6.435 549.78 x 6.435
(Note:- ds cancels out)
I H= 57.33 KN I A result similar to that already obtained from algebraic solution
3.7. ARCHES WIlH SECANT VARIATION OF INERTIA:If 10 is the sec arid moment of area of arch rib at the crown: Then secant variation of inertia means. I = 10 sec. a ds Cos a =dX
and
dy
~ .
s
0<
dx
Or
ds = dX Sec a Myds
H
f EI f&EI
If it is built of the same material, then E would cancel out:
Put I= 10 sec a
H
MY dX Sec a: 10 Seca
f H
Y2dX Sec a
f 10 Sec a
H _ fMydX
- fldX
INTRODUCTION TO TWO-HINGED ARCHES
175
Ifwe utilize the above expression for horizontal thrust, it may be kept in mind that integration can now take place in the Cartesian coordinate system instead of the polar coordinate system. 3.8. BY SECANT VARIATION USING ALGEBRAIC INTEGRATION:EXAMPLE NO.6: Analyze the arch in Example No.5:
o -l
E
EB 4 0 0 0 0 0
- 3.334 0
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- 6.667
+ 6.667
0
+14.444
-6.667
+6.667
+ 3.334
0
- 3.334
- 6.667 + 1:667 -5 +0.83
- 1l.l1 +2.778 8.33 +0.83
-14.44 4 5.555 +5.555 -0 0
+5.555
+ll.lll
- 5.555
-2.778 +8.333 -0.83
+6.667 -1.667 +5 -.83
r 3.334 - 3.334 0 0
0 0 0 0
- 3,334 + 3.334
0 0 0 0
0 0
~
to
0 DA 6 0 0 0
c: >-l
(5 Z ~ >-l
tTl
::r: o i
d
-
0 0 --,-
~-----
ochecks have been satisfied. Now SFD and BMD can be drawn.
tv -...j
\0
280
THEORY OF INDETERMINATE STRUCTURES·
B.M. & SHEAR FORCE DIAGRAMS :6.667 A
l
1
20KN
• 2m
j12.034KN
liB
.
7.966KN 12.034
Mx=12.034 x-6.667= 0 x=0.554m
14.444
4m
01
+ .
'l___
S.F.D. (KN)
-l+
-
IQ
7.966
Mx=7.966 x -14.444= 0 x=1.813 m '
, X
:.i B.M.D. (KN-m)
6.661 14.444 14.444
6.667
4m
B l,7.966KN
7.96~
i
1c
12.0S4KN
--+--.-+-f.---'I Q
1-1
1-.---1:
.
S.F.D. (KNJ
12.034
B.M.D. (KN-m)
LO
N
o
0
B
+ 6m ~ C"i
C'>.
LO
N
~
~
Mx=3.334 - 2.5 x=O x=1.334m
E
LO
N
0
r'-'=====~ THE MOMENT - DISTRIBUTION METHOD
Il
281
c
B
~
D
F E
Elastic Curve EXAMPLE NO. 7:- Analyze the following frame by Moment Distribution Method. SOLUTION:- This is a double story frame carrying gravity and lateral loads and hence would be able to sw~y both at upper and lower stories.
f2KN/m
3m
j
I I
3m
1
I
1 1
I 1 i j
j J
j
l
1
I
I !
Step 1: F.E.Ms Due to applied loads:3 X 32 Mfab = ~ = + 2.25 KN-m Mfba = - 2.25 KN"-m 3 x 32
Mfbc=~
= +2.25
Mfcb = - 2.25 KN-m. 2.5 2 Mfbe = Mfcd = IT = + 4.167 KN-m , Mfeb = Mfdc = -4.167 KN-m Mfde = Mfcd = 0 Mfef = Mffe = 0
282
THEORY OF INDETERMINATE STRUCTURES
Step 2: Relative Stiffness:I
L
I L
AB
2
3
~x
BC
2
3
DE
2
3
EF
2
3
Member
CD
5
BE
5
Step 3:
F~E.Ms.
Krel
15 3 2 - x 15 3 2 - x 15 3 2 - x 15 3 1 -x 15 5 1 -x·15 5
10 10 10 10
3 3
Due to side Sway of upper storey:-
D. 1
C
5m
1
F
A
0
Mfbc = Mfcb = + 6 1!:J. Mfde = Mfed =
=
+ 6E~il ) !:J. x 900 = + 1200 (Note: 900 value is an arbitrary multiplier)
+ 6 EI !:J. + 6 E(2 I) !:J. 3~
L2
•. x 900 = + 1200 (Because R IS clockwise)
Step 4: F.E.Ms. Due To Side Sway Of Lower Storey:C ...------=----''----, 5m 3m 21
3m 21
A
.i
r
283
THE MOMENT - DISTRIBUTION METHOD
Mfbc = Mfcb =Mfde = Mfed=
- 6E(2I) Ll
9
x 900 = - 1200 (R is counter clockwise so negative)
Mfab = Mfba =
Mfef= Mffe =
+ 6EI(2I) Ll 9
+ 6EI(2I) Ll 9
(R is clockwise, So positive)
x 900 = + 1200
x 900 = + 1200
(R is clockwise, So positive)
Determination Of Shear Co-efficients (Kh K z) for upper and lower stories :Upper Storey:
MCB ~
C
3m
3m
3KN/o/ B MBC ~~HB HB
=4.5+
Shear Conditions: L Upper story
Hb + He =0
2. Lower storey
Ha + Hf=
Lower Storey
o·
MBC+MCB
.
3
E
.
MED
~HE
~
H _ MED+MDE , E -:. 3
(I) where Hb and He values in terms of end moments are shown in the relavant diagram. (2)
MBA It:--
B
3KN/~
3m
HA~ A . ~M 'AB.
HA=
4.5+ MAB+MBA . 3
3m
~MFE ~
HF
HF = MFE+MEF
3
.
Where Ha and Hf values in terms of end moments are shown in the relavant diagram Now we attempt the problem in a tabular form. There would be three tables, one due to 10ads(Table-A), other due to FEMs of upper story (Table-B) and lower story (Table-C). . Insert these three tables here. Now end moment of a typical member would be the sum of moment due
THEORY OF INDETERMINATE STRUCTURES
284
to applied loads ± KI x same end moment due to sway of upper story± Kl x same end moment due to of lower story. Picking up the values from tables and inserting as follows we have.
SW~ly
Mab = 1.446 - K 1{l43.66) + K2 (1099.625). Mb'a = -3.833 - KI (369.4) of K2 (1035.46) Mbc = - 0.046 + KI (522.71) - K2 (956.21) Mcb = - 4.497 + KI (314.84) .w K2 (394.38). Mcd = + 4.497 - KI (314.84) + K2 (394.38) Mdc = - 3.511 - KI (314.84) + K2 (394.38) Mde = + 3.511 + KI (314.84) - K2 (394.38) Med= + 2.674 + KI (522.71) - K2 (956.29). Mef= + 1.335 - Kl (369.4) + K2 (1035.46) Mfe = + 0.616 - Kl (193.66) + K2 (1099.625). Mbe = + 3.878 - KI (153.32) - K2 (79:18) Meb = 4.0,09 - K] (153.32) - K2 (79.18) Put these expressions of moments in equations (1) & (2) & solve for Kl & K 2• - 0.046 + 522.71 Kl - 956.21 K2 - 4.497 + 314.84 Kl - 394.38 K2 +2.674+522.71 K]-956.29 K2+3.51l+314.84 K I -394.38 K 2= 13.5 1675.1 Kl - 2701.26 K2 - 11.858 = 0
. --+ (3)
1.446 - 143.66 Kl + 1099.625 K2 - 3.833 - 369.4 K] + 1035.46 K2 +0.646-193.66 K 1+ 1099.625 K2+ l.335-369.4KI+ 1035.46K2 = 40.5 - 1076.12 KI + 4270.17 K2 - 40.936 = 0
--+ (4)
From (3) K2
=(1675.10KJ-lI.858) 2701.26·
Put K2 in (4) & 'solve for K]
. _ 1675.10 KJ - 11.858) - 1076.12 KI + 4270.17 ( 2701.26 - 40.936 - 0
- 1076.12 KI + 2648 K] - 18.745 - 40.936 = 0 1571.88 KI - 59.68 = 0 KI =0.03797 From (5)
K _ 1675.1 (0.03797) - 11.858 22701.26 K2 = 0.01915
--+ (5)
THE MOMENT - DISTRIBUTION METHOD
Putting the values ofKl and K2 in above equations, the following end moments are obtained. FINAL END MOMENTS:Mab = 1.446 - 0.03797 x 143.66 + 0.01915 x 1099.625 = + 17.05KN-m Mba = + 1.97 KN-m Mbc = + 1.49 KN-m. Mcb = - 0.095 KN-m. Mcd = + 0.095 KN-m Mdc = -7.91 KN-m Mde=+7.91 KN-m Med = + 4.21 KN-m Mef= + 7.14 KN-m Mfe = + 14.32 KN-m Mbe =
-
3.46 KN-m
Meb = - 11.35 KN-m These values also satisfy equilibrium of end moments at joints. For simplicity see end moments at joints C and D.
Space for notes:
285
TABLE A OF
Joint Members K Cycle D.F 1. FEM. BAL. COM. 2. BAL. 3. COM. BAL End Moments. (change) near end - 1/2 (change) fat' end
I
e reI. l_ K
AB 10 0 +2.25 0 -0,906 0 +0.402 0 + 1.446 - 0.804 +0.792 -0.012 + 0.001
BA 10 0.435 .-2.25 - 1.813 0 +0.203
+ 0.271 + 0.060
0 + 0.027
- 0.086 + 0.024 +0.027 + 0.008 + 3.878 . - 0.046 - 0.289 - 2.296 - 0.079 + 1.124 - 1.172 - 0.368 + 0.123 +0.117
- 3.833 - 1.5.83 + 0.402 - 1.181 +0.118
EX~MPLE
NO.7 (Table A) .
BC
10 0.435 +2.25 -1.813 - 0.737 +0.203
CB 10 0.769 -2.25 - 1.474 -0.906 +0.049 +0.102 -0.018 -4.497 -2.247 + 1.148 - 1.099 +0.110
N
D
C
B BE 3 0.130 +4.167 - 0.542
A
CD 3 0.231 +4.167 - 0.443 -0.842 + 0.015 - 0.079 - 0.005 +4.497 + 0.330 - 0.328 - 1.002 '-0.00
DC 3 .. 0.231 -4.167. +0.963
E EB 3
DE 10 0.769· O' + 3.204 - 0.222 · + (>:906 -0.158 · - 0.526 + 0.008 - 0.290 +0.065 + 0.217
ED 10 0.435 0 + 1.813 + 1.602 - 0.57? - 0.263 +0.101
-0.271 - 0.173 + 0.030 + 0.030
+ 3.511 - 3.511 +0.656 +3.511 -0.165 - 1.337 + 0.491 · +;2.174 -0.164 - 0.217
+2.674 +2.674 - 1.755 +0.919
-4.009 +0.158 + 0.145 + 0.303
- 0.092
- 0.101
O~130
-.4.167 +0.542
EF 10 0.435 0 + 1.813
0 - 0.579 - 0.290 0 0 + 0.101 + 1.335 +0.616 + 1.335 +0.616 - 0.308 . -0.667 + 1.027 - 0.051 - 0.103 + 0.005
TABLE B OF EXAMPLE NO. 7' (Tabl~ B)
Joint Members K Cycle D.F FEM. 1. BAL. 2. COM. BAL. 3. . COM. BAL End Moments. (change) near end - 1/2 (change) far end
I I
A
AB 10 0 0 0 - 261 0 + 117.34 0 - 143.66 - 143.66 + 184.7 + 41.04 - 41.04
BA 10 0.435 0 - 522 0 + 234.68 0 - 82.08 - 369.4 - 369.4 + 71.83 - 297.57 +29.76
B BE 3 0.130 0 - 156 -78 + 70.14 + 35.07 - 24.53 - 153.32 - 153.32 + 76.66 -76.66 + 25.55
P
C
. BC 10 0.435 + 1200 - 522 -46.15 + 234.68 :1-.153.61 - 82.08 + 522.71 - 677.29 + 442.58 - 234.71 + 23.47
CB 10.
0.769 + 1200 - 923 -261 + 307.22 + 117.34 - 125.72 +314.84 ::;=: 885.)6 ~+" 338.65 '-:'546:si :cp 54.65. ,-"'
8 reI. _ K , ~::
DC' CD . 3 3 . 0.231 Q.231 '0 ' 0 . - 277 -277 -138 .. 5 - 138.5 + 92.28 +92.28 + 46.14 +46.14 - 37.76 - 37.76 -314.84 -·314.84 -314.84 -314.84 + 157.42 + 157.42 - 1?7.42 - 157.42 + 52.47 of 52.47
DE . 1'0 0.769 + 1200 - 923 - 261 +307.22 + 117.34 - 125.72
ED 10 0.435 + 1200 - 522 - 46.15 + 234.68 + 153.61 + 82.08
+ 314.84. - 885.16 + 338.65 - 546.51 + 54.65
+522.71. -' 677.29 + 442.58 -234.71 + 23.47
E EB EF '3 10 0.130. 0.435 0 0 - 156 , -522 0 -78 +70.14 + 234.68 + 35.07 0 - 24.53 - 82.08 :.. 153.32 + 369.4 - 153.32 - 369.4 + 76.66 + 96.83 -76.66 - 272.57 + 25.55 + 27.25
00
F FE 10 0 0 0 +0.906 0
F FE
0\
I
..,::r: fj
10
::tI
0 0 0
o 'Tj
- 261 0 + 117.34 0
- 193.66 - 193.66 + 184.7 - 8.96 + 0.96
~
Z
tJ t11
....j
t11
~
~
t11 !Zl
..,~ (')
~
!Zl
·
"-'~'''- .. --~'..-----:- ... ~- ..--.----- .. -~.."~.-,
~
g:j ~
~
~ 1
tJ .......
CIl ~
TABLE C OF EXAMPLE NO.7 (Table C)
Joint Members K Cycle D.F FEM. 1. BAL. COM. 2. BAL. 3. COM. BAL End Moments. (change) near end
A AB 10 0 + 1200 0 0 0 - 100.375 0 + 1099.675
- 100.375
- 112 (change) far end
I
- 18.11 + 1.81
I
e reI: _ K I
_
+ 82.27
... ___
._ _ . _ _
B BE
D
C BC
CB
CD
DC
DE
ED
10
10
10
10
0.435 - 1200 + 300 + 46.15 -- 200.75
3 0.231
0 0 0 - 60.0
-1200 +923 0 - 106.50
3 0.231 0 + 277
-30.0 + 10.82
- 53.25 + 36.21
- 100.375
- 164.54 + 50.19
-79.18 -79.18 + 39.59
-- 956.29 + 243.71
- 114.35
-- 39.59
-- 394.38 + 605,62 - 121.86 + 683.76
+ 11.44
13.20
BA
10 0.435 + 1200 0 0 -200.75 0 + 36.21 + 1035.46
3 0.130
- 402.81 - 159. I + 15.91
0.769
+ 89.49
- 68.38
+ 138.5 -32.0
0 +277 + 138.5 - 32.0
0.769 - 1200 + 923
O· - 106.50 - 100.375 + 89.49
- 16.0 + 26.88 + 394.38 + 394.39 -197.19 +197.19
- 16.0 + 26.88 + 394 ..38 + 394.:38
+ 805.62
- 197.19 +197.19
- 121.86 + 683.76
~
- 65.73
- 68.38
65.73
- 394 .. 388
0.435 -1200 0 + 461.5 - 200.75 - 53.85 36.21 -- 956.29 + 243.71 - 402.31
E EB 3 0.130 0
0 0 - 60 - 30 + 10.82
EF 10 0.435 + 1200 0
C
~
>-I
10 I
Q-,--
~
::r: o
0
0
tJ
0 - 100.3
~ ~
-I- 1200
0 - 200.75 0 +36.21 + 1035.46
~
F FE
75
0 1099.605
-79.18 -79.18 + 39.59
- 164.54 + 50.19
-- 100.375
- 159. I
- 39.59
-11·U5
- IS. II
I
+ 15.91
+ 13.20
+ 11.44
+ 1.81
I
-i-
+ 82.27 I
1
t~)
00
--l
288
THEORY OF INDETERMINATE STRUCTURES
CHAPTER SIX. 6. KANIS METHOD OR ROTATION CONTRIBUTION METHOD OF FRAME ANALYSIS This method may be considered as a further simplification of moment distribution method wherein the problems involving sway were attempted in a, tabular form thrice (for double story frames) and two shear co-efficients had to be determined which when inserted in end moments gave us the final end . moments. All this effort can be cut short very considerably by using this method. ~ Frame analysis is carried out by solving the slope - deflection equations by successive approximations. Useful in case of side sway as well. ~ Operation is simple, as it is' carried out in a specific direction. If some error is committed, it will be eliminated in subsequent cycles if the restraining moments and distribution factors have been determined correctly. Please note that the method does not give realist~c results in cases of columns of unequal heights within a storey and for pin ended columns both of these cases are in fact extremely rare even in actual practice: Even codes suggest that RC columns framing into footings or members above may be considered more or less as fixed for analysis and design purposes.
Case 1. No side sway and therefore no translation of joints derivation. Consider a typical member AB loaded as shown below:
Mab
Mba
rA~~~Bl >1
~---""""'L
A GENERAL BEAM ELEMENT UNDER END MOMENTS AND LOADS
General Slope deflection equations are. 2EI Mab .= MFab + L ( - 29a - 9b ) Mba
2EI
= MFba +T( -
. 9a-29b)
equation (1) can be re-written as Mab = MFab + 2 M'ab + M'ba . and
~
(1)
-~(2)
~
(3)
M'ab = rotation contribution ofI).ear end A of member AB = 2EI9a =--L- =-2Ekl 9a
M'ba = rotation contribution of for end B of member AB. So
where MFab = fixed end moment at A due to applied loads.
M'ba = _ 2 EI 9b
L
= _ 2Ekl .9b
~
(5)
~ (29a)
T I I I
[
289
KANIS METHOD OF FRAME ANALYSIS
Now consider a generalized joint A in a frame where members AB, AC, AD ......... meet. It carries a . . momentM.
I
B
o For equilibrium of joint A, 2:Ma = 0 or
Mab + Mac + Mad + Mae ..................= 0
or
2:MF (ab, ac, ad) + 2 2:M' (ab, ac, ad) + IM' (ba, ca, da) = 0
Putting these end moments in form of eqn. (3)
Let IMF (ab, ac, ad) = MFa (net FEM at A) So
MFa + 2 2:M' (ab, ac, ad) + 2:M' (ba, ca, da) = 0
From (6), 2:M' (ab, ac, ad) =
-"21 [(MFa + IM' Cba, ca, da)]
~ (6) ~ (7)
From (4), 2:M' (ab, ac, ad) = - 2Ekl ea - 2 Ek2 ea - 2 Ek3 ea + .............. .
=or.
2 Eea (Ik), ( sum of the member stiffnesses framing in at joint A)
Sa - LM' (ab, ac, ad) -2E(Ik)
~
(8)
From (4);M'ab = - 2 Ekl 8a. Put 8a from (8), we have [2:M' (ab, ac, ad)] _ .h , , , _. M ab - - 2E kJ 2E (Ik) - 2:k [2:M (ab, ac, ad)] From (7), Put 2:M' (ab, ac, ad) So
M'ab
=K[-t(MFa+ IM' (ba, ca, da»]
290
THEORY OF INDETERMINATE STRUCTURES
or
M'ab = -
on similar lines
.
and
I
Mad
1k 2"Th [MFa
M'ac = -
+ 2:M' (ba, ca, da)]
k2 2"1 2:k [ MFa + 2:M' (ba, ca, da)]
1 k3
= - 2" Ik [MFa -I;' 2:M' (ba, ca, da)] 1.:::.
1.!
sum of the rotations contributions of far ends of members meeting at A.
rotation contribution of near t:nd of member ad.
Sum of rotation factors at near end of members ab, ac, ad is
= -
t'
[sum of rotation factors of different members meeting at a
.. Is.equa '. I to:-2"1 ] JOInt Therefore, if net fixed end moment at any joint along with sum of the far end contribution of members meeting at that joint are known then near end moment contribution can bede&rmined. If far end contr.ibutions are approximate, near end contributions will also be approximate. When Far end contributions are not known (as in the first cycle), they c~n be assumed to be zero.
6.1.. ,RULES FOR. . CALCULATING RO'rA TION .CONTRIBUTIONS :- Case-I: Without sides way. . . Definition: "Restrained moment at a joint is the meeting at that joint."
1.
algebrai~
sum of FE.M's of different members
Sum ofthe restrained moment Ofajoiilt and all rotation contributions bfthe far ends of members meeting atthat joi~t is multiplied by respe9tive rotation factors to get the require4 near end rotation contribution. For the first cycle when far ~nd contributions taken as zero (1st approximation). are not known, they may
be
2.
By repeated application, of this calculation procedure and proceeding froin joint to joint in an arbitrary sequence but in a specific direction, all rotation contributions are known. The process is usually stopped when end moment values converge. This normally happens after three orJour cycles. But values after 2nd cycle may also be acceptable for academic.
6.2. Case 2:- With side sway Goint translations) In this case in addition to rotation contribution, linear displacement contributions ( Sway contributions) of columns of a particular storey are calCulated after every cycle as follows:
- - - - - - - - - - - _.._ - - - - -
TI
KANIS METHOD OF FRAME ANALYSIS
291
I
I I I
I
6.2.1. For the first cycle. (A) --» Linear Displacement ContriOution ( LDC) of a column = Linear displacement factor (LDF) of a particular column of a story multiplied by [storey moment + contributions at the ends of columns of that story] 3 Linear displacement factor (LDF) for columns of a storey = - 2"
I ,
. . 3 k Lmear dIsplacement factor of a colunm = - 2" Ik
6.2.2. (B) --»
t
Where k=stiffuess of the column being considered and Ik rs the sum of stiffness of all columns 15
2kN/m 4m
no B.M.O
A
31~8
o
l'
10 10
10
7.5
15
5
Ms-diagram
15.5 Ms diagram
El
.-.~------
-----
....
_----._------------------------
345
COLUMN ANALOGY METHOD
Properties Of Analogous Column Section :Sketch analogous column section and show loads on it. BMD along column AB is split into a rectangle and other second degree curve.
A=
y_
G
x 4) x 2 +
5
1.63 m From line Be
GY
3x Ix= 12 . =
3
(1) 0 [°.51; 4 + (0.5 x 4) x (0.37)2] + 3"x3 x (1.63)- +2 12
8.55 ril4
(1)
[4 120.5 +(4x0.5)x(1.5t~J 3
Iy= 3" x(3) 3 +2 =
x 3) = 5 m"
(3x5)x(~)+2[GX4)x2J
-
Y=
G
x
9.83 m4
P,
y 1.0
0.5
l~
B
1.63 m
4m
P2", x i"- 1
x
0 37 .
P 3
",
1
[\. '---
2.37 m
11 i
'--- D
~
~
y.
IE
y.
y 3m
>{
1
3'46
THEORY OF INDETERMINATE STRUCTURES
Total load on top of analogous column section acting at the centroid. P ~ 3.75 + 30 + 10.67 = 44.42 KN upward. 1 Pl= -2x 1.5x5=3.75, P2=7.5x4=30,
,
4 x 7.5 P 3 = - - = 10
2+ 1
4
X' = 4" = 1 meters for A.
Mx = - 3.75 x 1.63 + 30 x 0.37 + 10.67 x 1.37 = 19.61 KN-m clockwise. My = 10.67 x 1.5 + 30 x 1.5 + 3.75 x 1 = 64.76 clockwise. Applying the general formulae in a tabular form for all points offrame. 'Ma = (Ms- Mi)a
+ M, y + My X ( M 1.) a = 1: A - Ix Iy Point
A
B C D
Ms
PIA (1)
- 31 -15 0 0
-8.88 - 8.88 - 8.88 - 8.88
.!,
,
'I
Mx
h' y C2) .::. 5.44 + 3.74 + 3.74 - 5.44
Mz Iy . X (3) - 9.88 - 9.88 + 9.88 + 9.88
Mi (l )+(2) + (3)
M Ms-Mi
-24.2 -15.02 + 4.74 -4.44
-6.8 + 0.02. -4.74 +4.44
EXAMPLE NO. 14:- Analyze the following beam by column analogy method. SOLUTION :~hoosing
B.D.S as cantilever supported at B
A1T ~
~B
f"L~~~~~~~rv____~____~~
""
2m
.,
2m"
4m
...'
Ms-diagram due to u.d.l. only
96 4m
2m
2m
1['
J
-----------
-_._._------_.
------------------
COLUMN ANALOGY METHOD
347
~
40 Ms diagram due to concentrated load only Slectch analogous column section and determine its proteins P, 2.14
d
b
a
--
-
-
-
24 .
72 3.21m Ms-diagram EI due to u.d.l
P3=18.. 67
P4=80
~
40
~f
diagram due to point load.
Slectch analogous column section and determine its properties. P3
Yo
1/3#:
I I
4.78
K
3.22
3.21
1.33
Yo
>!
24 x 4 48 x 4 PI = - 3 - + - 2 - + 24 x 4 == 224 KN.
!
I
:
1/1.5 1 column ~~=:~~~~:t:::::::::J Analogous section
Corresponding to full Ms diagram, due to u.d.!.
Location of PI from B 224 x X = 96 x 1.33 + 96 x 2 + 32 x 5 X = 2.14 meters .
P4 = Note:
1
2x
4 x 40 = 80 KN,
Corresponding to full Ms diagram due to point load.
Area of32 and its location ofMs diagram due to u.d.!. has been calculate d by formula e used in moment - area Theorems. 2
0
area (abc) == JMxdX == J-1.5X- dX = o
J(M x) X dX =
J2 -:-1.5X3dX =
1-
o
-
-6
X =_4
1_ 1.53 X312 =-4 0
1.5 4
X412 = - 6 . 0
= l.5m from 4
A
area (bcdc) = J (Mx) dX == J - 1.5X 2dX o
2
Jo
1.5 X2 dX
348
THEORY OF INDETERMINATE STRUCTURES.
=
I
- 1.5 4
J(Mx)X dx = J-
X314
3"
0 -
i
I
X312
1.5 3"
-
0
=-
.
28 ;
~
1.5 X3dX '- j - 1.5 )(3 dX = - 90
o
-90 X=-28 . =
3.21 meters from A (centroid of area bcde)
P4= 80 KN
Total concentric load on analogous column section.
= - 224
+ 1.33 + 18.67 - 80
284 KN (upward)
= -
Total applied moment = M = - 224 x 1.68- SO x 1.89 - 18.67 x 1.57 - 1.33 x 33 x 3.28 = - 426.79 KN-m(It means counter clockwise) T~is
total load P and M will now act at centroid of analogous column section.
Properties of Analogous Column Section. ,
1 3
1 1.5
A=-x2+--x2+1x4=6
.
. ( 1)
(1 x 4) x 2 + 2 x 1:5 ?< 5 +
_
X= =
(1). 3' 2 7 x
x
6 3.22 from B.
. 1 x 4"
?
(-1.51) x?3-
Iyoyo = 12' + (1 x 4)(1.22t +::
I:'
T
12
~
+
(1
)
IS x 2
, (l.78t
ri
i
349
COLUMN ANALOGY METHOD
G
3
2) 1 ? 12 +(3X2)C3.78t X
+ =
12 25.70 m 4
.) . P (M! a= A
Me
±-r-
- 284 426.79 x 4~ 78 =-6- +. 25.7 =
+ 32.05 KN-m
(Ms)a = 0 Ma = (Ms - Mi)a = 0 - 32.05
I Ma = (Mi)b =
32.05 KN-m
I
R. _ Me A r
. - 284 =-6-
426.79 x 3.22 25.7
=-100.81 (Ms)b = - 72 - 40 = - 112 Mb = (Ms - Mi)b
=-112+100.81
I Mb = -
11.19 KN-m
I
The beam has been analyzed. It is now statically determin,ate.
350
THEORY OF INDETERMINATE STRUCTURES
CHAPTER EIGHT 8. PLASTIC ANALYSIS OF STEEL STRUCTURES 8.1. Introduction: Although the terms Plastic analysis aJid design normally apply to such procedures for steel structures within the yield flow region, at almost constant stress. however the Idea may also be applied to reinforced concrete structures which are designed to behave elastically in a ductile fashion at ultimate loads near yielding of reinforcement. The true stress-strain curve for a low grade structural steel is shown in fig. 1 while an idealized one is shown in fig. 2 which forms the basis of Plastic Analysis and Design. E (B,C) Plastic
f f Stress
AB-Elastic BC-Yeild points CD-Plastic Strain fiow DE-Strainhardening EF-Failure
i
A
0
Stress
--7 Strain E
Fig 1:
i A
----7
Strain E
Fig 2:.
8.2. Advantages of Plastic Analysis 1. Relatively simpler procedures are involved. Ultimate loads for structures and their components may be determined: Sequence and final mode of failure may be known and the capacity at relevant stages may be determined. 8.3. Assumptions in Plastic bending 1. The material is homogeneous and isotropic. 2. Member Cross-section is symmetrical about the axis at right angles to the axis of bending. 3. Cross-section which were plane before bending remain plane (lJter bending. 4. The value of modulus of Elasticity of the material remains the same in tension as well as in compression. 5. Effects of temperature, fatigue, shear and axial force are neglected. 6. Idealized bi-linear stress-strairi curve applies. 8.4. NUlnber of Plastic Hinges "The number of Plastic Hinges required to convert a structure or a member into a mechanism is one more than the degree of indeterminacy in terms of redundant moments usually. Thus a determinate structure requires only one more plastic hinge to become a mechanism, a stage where it deflects and rotates continuously at constant load and acquires final collapse. So Mathematically N = n+l where N = Total number of Plastic hinges required to convert a structure into a mechanism. n = degree of indeterminacy of structure in terms of unknown redundant moments. and 2. 3.
I \
PLASTIC ANALYSIS METHOD
351
8.S. Plastic Hinge. It is that cross-section of a member where bending stresses are equal to yield stresses cr=cry=fy. It has finite dimensions. MpC M£ so Hp = Zp cry or cry From bending equation cr = ~ or cry = I Zp I M crI or M where = Z y Y So M = crZ and Z is elastic section modules and is equal to the first moment of area about N. A Z = fA ydA.
y
From elastic bending
8.6. Plastic moment of a rectangular section. Consider a simple rectangular beam subject to increasing bending moment at the centre. Various stress-strain stages are encountered as shown below. B
a< 0" Y E < E Y O"=O"y E
=EY
O"=O"Y E
>E y
O"=O"Y
E»Ey
C D
12 2
T
'--_--', case A: M E y case 0
0"=0",
Various Stress-strain distributions Case A - Stresses and strains are within elastic range. Case B - Stresses and strains at yield levels only at extreme fibers Case C - Ingress of yielding within depth of section. Case D - Full plastification of section. On the onset of yielding cr = cry and M = My = cry.Z. On full plastification cr = cry and M = Mp = cry.Zp. fJr Zp = fA yda (First moment of area about equal area axis). All compact sections as defined in AISC manual will develop full plastification under increasing loads realizing Mp. However local buckling of the compression flange before yielding has to be avoided by providing adequate. lateral support and by applying width / thickness checks as was done during the coverage of subject of steel structures design. Case B. Stresses and Strains at yield at extreme fibres only. Consult corresponding stress and strain blocks. M = Total compression x la = Area x cr x la Area in compression (from stress block). where . Area· cr Average compression stress. la = Lever arm i.e. distance b/w total compressive and tensile forces. So
M
(B~)
(cry 2+
0) . ~ D
352
THEORY OF INDETERMINATE STRUCTURES
In general M = Cjd or Tjd , where C and T are total compressive and tensile forces respectively which' have to be equal for internal force equilibrium. BD2
6
or
My
=
cry
So
My
=
cry.Z.
BD!"[' but 6 =Z "Z =Elastic Section modules
I
=C
Case D: Full plastitication, cr = cry upto equal area axis. M
CIa
=
(B. ~)
(cry)~
or
Zp
BD2
cry. 4 or Mp and y1
A
or Zp = 2 [y1
= cry. Zp
+ y2
where la is lever arm BD!
,
= 4 ' where ZP = Plastic section Modules.
+ y2] (first moment of areas about equal area axis)
(distance from equal area axis to the centroids of two portions of area.)
=D/2
Case C: Moment Capacity in Elasto - Plastic range. Extreme fibres have yielded and the yielding ingresses in the section as shown by the stress - distribution.
C1
D
where la1 la2 C1 C2
"2
z
=.lever axis b/w C1 andri =lever axis blw C2 arid T2 =Av.stress X area of element No.1 =Av-stress acting on element; N()f,2; x area of element 2.
D
"2
z
L
i
"
cr
Ie
Y
1
case C : Stress-Distribution
c, . 1.,] (A) • 1. 1
M
--
(cry 2+
---~---'--------------
0) Z. B =
[ Z+
Z2]
Z
2 ] ' 4 laz = [ '3 x Z x 2 = '3 x Z
=(crY)B(~-Z) Cz
~ ~ ~ ~~ + 2
cry
2ZB "and so, putting values Of C. , C2
lal and laz in equation A above.
... 353
PLASTIC ANALYSIS METHOD
4
x - Z 3
M M
Simplifying
BZ"
Mr
where Mr is moment of resistance.
Mp
=Mr=cry.B (
3D\; 4Z',\)
For rectangular section.
Calculating on similar lines, Plastic moment for various shapes can be calculated.
8.7. Shape Factor(y) It is the ratio of full plastic moment Mp to the yield moment My. It depends on the shape of Cross-section for a given material. . ME. cry. Zp Shape Factor = y = = cry. Zor y = (Ratio of Plastic section modulus to My Elastic Section Modulus). 8.8. Calculation of Shape Factor for different Sections.
¥-
B
.AU
J'
;t
2
(1) •
y1
D
So
For rectangular section. BD) I = 12 ' C = BD) x 2 Z 12 x D = A Zp = "2 [YI + Y2]
dy
1y
y2
(2) •
8.8.1
B
D
Z ,
C
= "2
BD2
6 -
-
BD
2
l-D4 + DJ 4
or alternatively, Zp
= fA ydA. DI2
=
2
f
y. Bdy
0
y
~
Z
=
BD2x 6 4 X BD'
D/2
6 4
=
1.5
2B
f
ydy ..
0
y
1.5
so [Mp is 1.5 times My)
or Zp
=
BD' 4
THEORY OF INDETERMINATE STRUCTURES
354
8.8.2
For Circular Cross-section 1tD4 D1 A = 2!. I = 64 , 4 1tD4 1tD3 2 I Z = C = 64 x D = 32 , A
+ y1]
= 2 [YI
Zp
(a) Cross-Section
1tD~ [2D
3n + 2DJ 31t '
8 Zp
=
D3 6
(b) Strain Distribution
(c) Stress at full plastification Distribution
4xD 4r D -2D , yl = 2 31t = 31t x 2 = 31t D3 x 32 ' 32 ~ 1.7 Y = Z 6 X 1tD' = 61t 1.7, [Mp is 1.7 times My] Y r
=
8.8.3 Hollow Circular Section
.NI§.
Td
.
D X--+--Ir-----r-t- 31t
'!2Q..
-----::-
'"
+ M3 (2e1» - M4 4>
+ 2M3 - M4 2 x 63
All are equal to respective Mp. Putting values.
(1)
+ 42
1.944
2. Sway Mechanism Fig B. (24),,) 68=. Ml (-8) + M2 (8) + M4 (-8) + M5(8) 144)"
=...;
1441..
Ml + M2 - M4 + M5
42+42+42+42
(2) or)..
=
1.166
3. First Combined Mechanism Fig C
=
+ M2
(24 t.) (6¢)
+
252 A
- MI + 2M3 - 2M4 + M5 294 ).. = 1.166 252
(36/.) (34))
Ml (-4»
(0) + M3 (24))
+ (3)
M4 (-24»
+
M5 (4))
368
THEORY OF INDETERMINATE STRUCTURES
4. Second Combined Mechanism Fig D
(36 A.)3~+24A. (8+~)6=MI (-8 --~)+M2 (8)+M3 (2~) + M4 (8 396 A. = - Ml + M2 + 2M3 - 2M4 + 2M5 396 A. = 2(42)' + 42 + 2(63) + 3 x 42 + 2 x 42 462 A. = 396 = 1.166 A.
+ 2~)
+ M5 (8
+ ~) ~ == 8
= 1.166.
Note: In overcomplete collapse,more than one mechanism give the same value of collapse load factor. Any or both of the collapse mechanisms can contain extra number of plastic hinges than those required for complete collapse. So in this case fig c and d mechanisms give the same value. This was the case of over complete collapse.
Space for notes:
(
,.
:'
+ 369
THE THREE MOMENT EQUATION
CHAPTER NINE 9. THE THREE MOl\1ENT EQUATION Most of the time" we are concerned with the classical analysis of statically determinate structures. In this chapter we shall consider the analysis of statically indeterminate (externally) beams due to applied loads and due to settlement of supports. It must be remembered that supports for beams may be walls or columns. As we know that for the analysis of statically indeterminate systems, compatibility of deformations is also essential requirements in addition to considerations of equilibrium and statics. By compatibility it is understood that deformations produced by applied loads should be equal to those produced by redundants. It has been already mentioned that reactions occur at supports in various directions if (i)
There is some action (applied load) in that direction.
(ii)
There is restraint offered by support in that directions
Action and reactions are equal iii magnitude but opposite in direction. In the structural analysis it is sometimes customery to think that rotations are generally associated with moments and deflections or translations are associated with loads. It must also be kept in mind that we never analyze actual structural systems or sub-systems, it is only the idealized ones which are analyzed" Representing beams and columns by just a straight line located on their centroidal axis is also a sort of idealization on the structural geometry. Reactions and loads are, therefore, also idealized and are shown by a sort of line loads acting on a point. The three-moment equation is a good classical analysis tool in which support moments produced by the loads as well as by the differential settlements can be easily calculated by using second-moment "area theorem which states that "The deviation of a point A on the elastic curve w.r.t any other point B on the elastic curve is 1 " equal to EI multiplied by the moment of area of B.M.D's between those two points." The moments of B.M.D's are taken about a line passing through the point of loaded beam where deviation is being measured. The method is essentially based on continuity (equality) of slopes on the either side ofa support by reducing an indeterminate system to its determinate equivalents as follows by using supperposition.
An indeterminate beam under applied loads and redundant moments is equated to corresponding detemrinate system carrying these two effects separately" Let-us derive the three-moment equation. Consider a generalized two-span beam element under the action of applied loads and redundant support moments acting on BDS.
370
THEORY OF INDETERMINATE STRUCTURES
• .A'·LI' . . .
.
A,
w
.. c'
,, ' .... , ....
.' I
I
...... ". .",. .,
",'"
'"
'"
~
",'"
I
.
I I I
A. rl______~~~~--~~~~~~~------------~__il
C
1 Fig (a)
).
. I
12
I
./
/
1~----------------~~7~----~------------~7
L1
L1
BMD due to applied loads on simple spans
Fig (b)
',I
-- ....
. Ma
Generalized redundant moment diagram
A3
:
A, ...... _ - - -
~I
....
A,
--
.
---
I
--1 Me _----I --All
fig (c)
~I
Fig(a) is .an indet~r.minate beam subjected to applied load (udl ill this case) which has shown seitlement ~uch that support i3 'is at a lower elevation than support at A and C and difference of elevation '\;V.r,t intermediate support B is h. and he. The angle aB on either side .of support B must be equal. Fig(b) is·RM.D. -due to applied load on silllple spans where AI is Area of B.M.D. on span LI and Az is area of B.M.D. on span Lz: al and a2 are the locations of centroids of B.M.D's on LI and Lz from left and right supports respect.iveiy. Sp invoking continuity of slopes and knowing that for small angels e = tane.
=
CCI Lz
THE THREE MOMENT EQUATION
371
Evaluate AAI by second Moment Area Method. We know that
ha - deviation of point AI on the elastic curve from the tangent drawn at point B on the elastic curve.
expressing A3 and A4in terms of moments 1 r LI m: AlaI + :3
.
AAI
ha -
1
x
2
1
2" MaLI + 3" LI X 2" MbLI
] .
divide by LI ha __1_ [AlaI + MaLI + MbLIJ LI Ell LI 6 3
(1)
CCI Nowevaluate Lz on similar lines. We have from geometry CCI
= =
CIC I
=
lr 2 EIt Azaz + As 3" Lz + A6
-
CC I
(deviation of point C from tangent at B) - he . X
X
L2]. :3. -
he
expressing As and A6 in terms of Moments 1 r 2 1 L2 1 ] EIt A2az + 3" L2 x 2" MbL2 + :3 x 2" MCL2
CCI
r
1 Ll L2Z] = EIt Azaz + Mb 3" + Me 6 - he
cel L2
_
I
rAzaz Mb Lz Me L2]·he L2 + 3 + 6-L2
-EIt
- he
divide by Lz
(2)
Equating (1) and (2), we have
1
rAzaz Mb L2 . Me L2] he L2 +-3-+-6- -Lz
Eit
372
THEORY OF INDETERMINATE STRUCTURES
Multiply by 6E and simplify. we have after re-arrangement
(it b)
(b)
6Alai A~a~ Eh. Ehe Ma - t) + 2Mb -+- +Mc =- -6- + 6-LI - +6-L2 ,II ,II 12 12 IILI hb The above equation is called three-momeilt equation.
~
9.1. Analysis of Continuous Beams by three-Moment Equation. We apply three moment equation to two spans ata time which gives us one equation. With the successive applications, the required member of equations are obtained and are solved simultaneously. EXAMPLE:
Analyze the continuous beam shown below by three-Moment equation. Take E = 20 X 106 KN/m~ and Ie = 40 X 10.6 m4.
12 KN
B
A
9.6 KN/m
32 KN C
3m
~
D
~"L~"~DO
~'1~4-----:-:----~~~14~~--:-:-----~~1~4--~--:~-C----+~ll~4~~_:_=_OC~~ Fig (a)
9.6x8' =76.8 8
A,
32x6 '=48 4
~ 409.6
=0
BMD Fig (b)
144
SOLUTION: When a fixed support at either end is encountered, an imaginary hinged span of length La and Interia 10 = 00 is added to conform to acted support conditons and to make the method applicable in similar situations. ' The same has already been dop.e in Fig(a). Fig (b) is the BMD's on simple spans, their Areas and its locations. Apply three-moment equation to spans AB and BC at a time. We have 6) . ( 6 Ma ( 2Ie + 2Mb 2Ic
8)
+ 41c + Me
( 8) 4Ie
=- 6 x 0-
6 x 409.6 x 4 4Ic x 8
Simplify and multiplying by Ie both sides of equation, we get. 3Ma + 10 Mb + 2 Mc = - 307.2 put Ma = - 24 KN-m 10 Mb + 2 Mc = - 235.2 divide by 10 Mb + 0.2 Mc = - 23.52 (1)
373
THE THREE MOMENT EQUATION
Now apply three-moment equation to spans BC and CD 8J ( 8 6J (6) 6 x 409.6 x 4 6 x 144 x 3 Mb ( 4lc) + 2 Mc 4Ie + 3Ie) + MD 3Ie = 4Ie x 8 - 3Ic x 6 Simplify and multiply by Ic, we have, 2 Mb + 8 Mc + 2 MD = - 307.2-144 = - 451.3 Mb + 4 Mc +, MD = - 225.625 Now apply three-moment equation to spans CD and DDo
6) . ( 6 Mc ( 3Ic + 2 MD 3Ic
LO)
+ -;- + Mdo
divide by 2 (2)
(LO) 6 x 144 x 3 -;- = - 3Ic 6 x
Simplify and multiply by Ie both sides of equation. 2 Mc + 4 MD = - 144 divide by 2 Mc + 2 MD = - 72 (3) We have obtained three equations from which three-Unknowns Mb, Mc and MD can be calculated. Subtract equation (2) from (1)
= - 23.52 Mb + 0.2 Mc Mb + 4 Mc + MD = -225.625 - 3.8 Mc - MD = ,202 ..105 Multiply equation (4) by (2) and add in equation (3) -7.6 Mc - 2MD = 404.21 Mc + 2 MD = -72 - 6.6 MC = 332.21 So
Mc
(4)
= - 50.3 KN-m
put Mc in equation (1), we get Mb = - 13.46 KN-m put Mc in (3), we get MD Finally Mb = - 13.46 KN-m' Mc = - 50.3 KN-m MD = - 10.85 KN-m .
=-
10.85 KN-m.
Checks: The above calculated values of moments are correct if they satisfy the continuity of slope requirements. Slopes at any intermediate support point can be calculated from the two adjacent spans by using conjugate beam method. While applying checks, it is assumed that reader is well conversant with the conjugate beam method. Before we could apply checks, it is necessary to plot reactant moment diagram (support-moments) to get their contribution in slope calculation. Here is the statement of conjugate beam theorem number one again. . "The shear force at any point on the conjugate beam loaded with
~i diagram is the slope at the
corresponding point in the actual beam carrying applied loads." In applying the conjugate -beam method, we must use the original sign convention for shear force as applied in strength of Materials subject. (i.e., . "left up, right-down, positive)
374
THEORY OF INDETERMINATE
6m
Sm
+
A4
6m
+
A6
AS
A B C
o
:
___________________
•~_t AS
24
STRUCTURE~
0
. . 0 '10.S5 13.~:__________________ .; ____. r~--t----------.-------A7
--
t
.A9
Fig (c)
BMD divided into convenient shapes.
50.3
Fig(c) is the reactant moment diagram The areas of positive BMD's act as loads in downward direction to which reactions are upwards. The areas of negative BMD's act as loads in upward direction to which support reactions are downwards. The direction of reaction is accounted for in the' signs appropriately.
A4
= 13.45 x 6 = SO.7
A7
= S(50.3 ~ 13.45) =
A5
= 6(24 -213 .45) = 31.65
AS
= 1O.S5 x 6 = 65.1
A6::::i 13.45 x S
146.2
- 6(50.3 - 1O.S5) ~ S' 35 A9 2 - 11. .
= 107.6
Checks. SPAN AB S.F at A= 9a'
9a
=
1 [A4 2 ] EI -2-'3 A5
30.725
= -EfC
=
1 [- 80.7 2 .' ] 2Elc -Z.--'3 x 31.65
(There is no check on this value as, it is nota coIitimiQussupport)
_1_[SO.7
9b
2EIc =
25.45 EIc
4
+
31.65J 3
.
Clockwise.
SPAN BC 1 [409.6 107.3 1 ] 4EIc -2-""-2--'3 x 147.5
9b 25.46
EIC
9b
9c
=
'.
Clockwise
1 [- A2 4EIc -2-
A6
2
+""2 + '3 A7
J'
=
1 [- 409.6 4EIc 2
107.3
2
]
+ -2- + '3 x 147.5
i
375
. THE THREE MOMENT EQUATION
I I
- 13.18 EIc
ec SPAN CO
ec
=-
13.16 Elc
eo
=
eo =
0
_1_ [144 3Elc - 2
+
65.1 2
+
118'.33J ,3
(Fixed end)
All slope values have been satisfied. This means calculated support moment values are correct. Now bea~ is statically determinate we can construct SFD and BMO very easily. We have seen that· numerical values of E and I are required in this case only if one is interested in absolute values of e. However, these values are required while attempting a support settlement case. Determine reactions and plot SFD and BMO. 12 KN
32 KN
l
A 2m
6
B 6m
t
C
~ t i 32.031 KN
13.76
9.6 KN/m
D
3m
~ i
6m
5.B06
69.203
26.194 KN
33.79 1.76 -+-
0
SFD
l
0
5.B06
12 43.009
BMD
.~--------------~--------~__-+----~.
50.30B
~
-I
-[
376
THEORY OF INDETERMINATE STRUCTURES
EXAMPLE~2:
Analyze the continuous beam shown below by three moment equation if support at B sinks by 12 mm. Take E = 20 x 106KN/m2; Ie = 40 X 10.6 m4.
c
B
A
D •••••••~DO
12mm Sm
8m
Sm
21c
41c
31c
Lo
.
"10 = ex:
B' "'"
.'
2:~
A,}.8
~
A·
A,'
,
C · · ·,·;·- .
A"'""
1....
.
o ----~--.-.-+---'00:::::-.-+---'-"0;:.....--"""'T"~-""'7",-' -.--;;.1.----------. 0 "'.......... ,#,,*,:,, ·A.
Fig (b)
.
.....
•••• •••• ~~.
"
A,
.
.
Reactant moment diagram A, to A~ are areas of adjusted BMD.
SOLUfION: As the extrerrle right support is fixed, an imaginary Hinged span of length Lo and Ic = 00 has already been added to make the method applicable and to conform to the support characteristic at D. Now it is a sort of continuous support. Only analysis due to differential settlement at B is required. Had there been some applied loads also, those could have been considered at the same time also. . . Now EI = 20 x 106 X 40 X 10-6 = 800 KN-m2 • we also know that Ma = 0 and MDo = 0 being extreme hinge supports. Spans AB and Be When we consider these spans and compare them with the derivation, we find that situation is similar so both ha and he terms are positive and equal to 12 mm using three-moment equation. 6) ( 6 8) ( 8) Ma ( 21e + 2Mb 2Ic + 4Ic + Mc' 41e put Ma
= 0,
6E x 12 x 10.3 . 6E x 12 x 106 + 8 .
simplify and multiply by Ie
2Mb (3+2) + Me (2) put EI
3
=
= EIc x
12 x 10-3 + 0.75 EIc x 12 x 10-3
= 800
10 Mb + 2 Me ~. 9.6 +7~2 Mb + 0.2 Me =_1.68
= 16.8
divide by 10 (1)
THE THREE MOMENT EQUATION
377
Spans BC and CD Comparing these two spans with the derivation, we notice that ha term is equal to - 12mm and he term·is zero. . 8) Ma ( 4Ic
(8 2Me 4Ic
+
+
6) 3Ic
+
(6) Md 3Ic
=
6E(-12x1O"3) 8
+
0
Simplify and multiply by Ie 2
Mp + Mb
+ 2 Md = - 7.2 + 4 Me + Md = - 3.6 8 Me
divide by 2 (2)
Spans CD and DDo I (
There is no load and settlement on these two spans so right handside of equation is zero
j Me
Ij
.We
(3~e) + 2Md (3~e + ~) + Mdo (~) = know that Mdo
Simplify
an~
= O·,
0
Lo
-00 = 0
multiply by Ie
+ 4 Md = 0 Mc + 2 Md = 0
divide by 2
I
2 Mc
!
Above three linear simultaneous equations which are solved. Subtract (2) from (l)
I I
I
Ij II I
1
Mb + 0.2 Mc Mb + 4 Mc ~ 3.8 Me - Md
(3)
=
+
Md
1.68
= - 3.6 = 5.26
(4)
Now multiply equation (4) by 2 and add to equation (3) -7.6 Mc - 2 Md Me + 2 Md - 6.6 Me Mc
=-
Md
= -T
Mb
= 2 KN-m
1.6 KN-m Me
=
+ 0.8
= 10.56
= 0 = 10.56
THEORY OF INDETERMINATE STRUCTURES
378
Plot end moment diagram. Add and subtract equal areas on spans BC and CD and apply . conjugate beam method.
1
.
Al
=
'2 x6x2
A2
=
'2 xSx2
A3
=
'2 x 6 x O.S
A4
=
'2 x S x
AS
=
'2 x 6 x
=6
1
=S
1
1
. I
= 2.4
1.6 = 6.4
1.
1.6 = 4.S
Compute slopes at supprts. ea
= Slope due to settlement (configuration) + due to end moments 3 3 12 x 10_1_ [AI] ~ 12 x 10- , _1_ -3 = 6 .' + 2EIc 3 6 + 1600 3 - 3.2S x 10 rad ..
[§.] _
Span AS eb
= =
3
J= 12
12
X
10-
-s
X
10-4 rad.
12
X
10-3
6
1 [2 + 2EIc -3 Al
X
6
6
10-
1 [2 ] + 1600 -3 x 6
Span BC 9b
=
s·
eb
= - S
X
=
12
X
ec ec
= - 1
X
8
1 [2 1 ] 12 X 10-3 1 + 4Eic 3 A2 - 3 A4 = S + 4 x SOD
[''32 x 8,'73. . t'x 6.4,.']
10-4 rad.
10-3
1 [1 . 2 ] + 4EIc -3 A2 + 3 A4
10-3 rad.
Span CD ec ec
ed
ad
1 [1
= 0 +3EIc
3 A3 '-23 AS]
1
= 3 x 800
[1
3 x 2.4 - 32 x 4.S]
-1 x 10-3 rad.
1 [2-3 A3 + 31AS] = 0 + 3 x 1.800 [2. -3 x 2.4 + 31x 4.S]
= 0 + 3EIc
o
(Fixed end)
379
THE THREE MOMENT EQUATION
Checks on slopes have been satisfied so computed moment values are correct. Now beam is determinate. SFD and BMD can be plotted. Resolve same problem, for a differential sinking of 12 mm at support C. we get the following equations.
= -
(1)
=
(2) (3) .
0.72 8.4 = -4.8
Mb + 0.2 Mc Mb + 4 Mc + Md Mc + 2 Md Solution gives Mc = + 3.49 Md = - 4.145 Mb = - 1.418
apply continuity checks and plot SFD and BMD. Unsolved Examples: Solve the following loaded beams by three-moment equations. 70KN
A~~_ _ _ _ _ _!_3m~B_ _ _ _ _ _ _ _ _ _~~C
zs: .
8m
12m
EI = Constt.
Final equations: Ma + 0.5 Mb Ma + 5 Mb + 1.5 Me Mb + 2 Mc
= - 90.312 = - 213.12
(1)
(2) (3)
=0
End Moment Values: Mc = i6.41 Mb = - 32.82 Ma = -73.91 SOKN
.
.
I
24 KN/m
1S KN/m. . ....
Sm
72 KN ..
A~ Sm 3Jc
+
Final Equati0t.ls: 2Ma+Mb 2 Ma + 6.4 Mb + 1.2 Me 1.2 Mb + 8.4 Mc
4m
C
B
14
!
12m
10lc
= - 216 = - 1555.2
=-
1495.2
+ (1) (2) (3)
24 KN
ZS
! Lo
E
o Sm 21c
+. 1.Sm
THEOl3-YOFn~DETERM"INATE STRUCTURES
380
End moment values: Ma = - 0.361 KN-m Mb =- 215.28 Kn-m Me "= - 147.25 Kn-m
~A
B C D
3-------ZS-,...--:--~Q'----~
!
.15mm
E = 200 x 10' KNlm2
Ie = 400 x 10" m' 31e
10le
6m
12m
Final Equations: 2Ma+Mb
+
= -600
6m "
(1) (2) (3)
= 1800 = - 600
2 Ma + 6.4 Mb + 1.2 Me 1.2 Mb + 8.4 Me
21e
End moment values:
Ma Mb Me
= - 537.69 KN-m = 475.38 = - 139.34 KN-m 15KN
3 KNlm
~
Q 5m 21
C
B
A
20KN
+
8m
+
8m
!
-~ 21
End moment values: Ma = - 75" KN-m Mb = 21.75 Me = - 60KN-m 12KN !2m
9.6 KNlm
B
A
I~
32KN
C
~ 6m 21
+
8m 41c
+
!
3m
0
"~
6m 31e
-I
381
. THE THREE MOMENT EQUATION
Final equations: 10 Mb + 2 Me 2 Mb + 8 Me
= - 235.2 = - 451.2
(1)
(2)
End moment values: Ma = ~24 KN-m Mb = - 12.88 Me = - 53.18 Md = 0 12 KN
9.S KN/m
tzm
~
z:;; I~ Final equations: 10 Mb + 2 Me 2 Mb + 8 Me + 2 MD 2 Me + 4 MD
32KN
C
B
A
Sm
8m
+
21
= - 235.2 = - 451.2 = -144
·He
+
!
0
3m
Sm
3Ic
~
,
(1)
(2) (3)
End moment values: Ma = -24KN-m Mb = - 13.455 . Me = -50.33 Md = - 10.835 B
A
C
!
3m
0
4.Smm
I.
2m
+
Final equations: 10 Mb + 2 Me 2 Mb + 8 Me + 2 Md 2 Me + 2 MD
~ Sm
21e
8m
+
41c
= 6.3
(1)
= -2.7
(2)
=0
(3)
+
6m
31c
~I
THEORY OF INDETERMINATE STRUCTURES
· 382
End moment values: Ma = 0
Mb = 0.7714 Me = - 0.707 Md =0.707 64KN .
A
~ Final equations: 2 Ma + Mb 2 Ma + 10 Mb Mb + 2 Me
3
~
3m
~
9m
EI = Constt.
= - 144.
+
C
B
3 Me = - 288 =0
(1) (2) (3)
End moment values: Mb = -19.2 Me = 9.6 Ma = - 62.4 B
A
4.Smm
I~ Final equations: Mb + 0.2 Me Mb + 4Me + MD Me +·2 MD End momentvalues: Ma = 0 Mb = 5.45
Me
= -0.27
MD = -5.86
6m·
8m
21e
4fe
=5.4 = -1.5 = -12
D
C
(1) (2) (3)
I I
.~
+
3mm E = 200 x10· KN/m' Ic=400 x 10" m~
Sm ·3Ie
~I
r
383
INFLUENCE LINES
CHAPTER TEN 10. INFLUENCE LINES This is also another very useful technique in classical structural analysis. Influence lines are plotted for various structural effects like axial forces, reactions, shear forces, moments and thrust etc. As structural members are designed for maximum effects, ILD's help engineer decide the regions to be loaded with live load to produce a maxima at a given section. " An influence line is a graphical representation of variation of a particular strucrural effect at a given section for all load positions on its span. " Two methods, viz, static method and virtual displacement method are used for the construction of ILD's. Mostly it is the later method which is prefered. All structures in general and Railway and Highway bridges in particular are frequently subjected to various types of moving loads. As influence lines describe variation at a particular section for all load positions on span, the effects of moving loads can be calculated very easily. It must be remembered that a system of moving loads moves as a unit. For Railway bridges standard cooper's E-60 and E-72 loadings are used whereas for highway bridges AASHTO lane loadings and truck loadings or sbmetimes tank loadings are used. When dealing with calculations regarding moving loads the problem is how to place the system so as to produce maximum effects at a given section. Sometimes mathematical criteria are used for the live load purpose and sometimes simple inspection is made. In each case influence lines help us simplify the things. 10.1. Statical Method of Constructing Influence Lines
In this method, a load may be placed at several positions within span/(s) and a mathematical expression for a particular structural effects at a section is set-up. By placing limits of X (the distance), the shape and ordinates of influence lines (called influence co-efficients also) can be determined. For example consider the cantilever loaded below and let moment at fixed end A be represented by its influence line. For a generalized load position as defined by distance X in the diagram, moment at A is. p
L
X •
A~:it-_ _t,,--4---:-- B 4
LL.D. for Ma
Ma
= - P (L -
X)
O
e"1
-
b L
=
1
a L
e"2
v P
M
RA
'. ' .
=1
RB
...
b
a
{-
iB
ae," = be,·
(
L
)
415
INFLUENCE LINES ,,
., . . .
b
a
We have obtained ILD for B.M at X in a simple beam Let us now consider the shown conjugate beam. (
2m
4m
>1
1.0 = p
c
A
o
•
B
7Q.
fRO Rak-______________~~---------s-m--------~JRb Sm
)(
<
4m
Sm
>I<
>1
Applying same concepts we get following ILD
0110
~~~
0.61
0/10
__~r.~____~~~~______________~~B 0.4
416
. THEORY OF INDETERMINATE STRUCTURES
Consider a propped canti·lever MB
P=1
x
(L-X)
A. B ~------------~--~--------------r--y
P If support at Ais removed, this will be deflected snape.
&aL
L-x B.M.D due to load on BDS as cantilever supported at 8.
Applying moment area thereon, deflection at part A due to loads is
AXL
=
~I [~(l - X? (I - ~ (I - X»)]
~I Ra =1
i
Now consider load under redundant Ra = 1
(deflected shape) of BDS
==:;Jl
8.M.D. for Ra
=1
\;:L=2/2
-==:;;._ _ _ _ _ _ _ (+_)
U3
Applying moment area thereon, deflection at A due to Ra
iI [~ (;9]
i~I
=1
JXX = (1)2 = .Equation for compatibility t!.ai - fXX Ra = 0 because A is a support. Net deflection should be zero.
417
INFLUENCE LINES
Sal
Ra Rb
=
=
Ra
= fxx
P(l - X)2 (21 21 3
+ X)
after putting values of Sal and fxx
(equilibrium requirement)
1 - Ra
So we get We know
Ra x L - P (l - x). Put value of Ra and simplify
Mb
)1
pX(l2 - X 2 2[2
This expression will help in plotted lLD for Mb
ILD for Ra Mb
P=1
(l -X)
X
B
A
~l Ra
=
\\'hen \\'hen
f,
P(l - X? (21 213 X
=
0
X = 5
IlO for Ra
{
10m
Rb
(
+ X)
=> =>
Ra Ra
= =
1. 0
5
16
(put in above equation for Ra) (put in above equation for Ra)
1.0
L - - -__________J -__________
-=~=_
_____ O
Simplify lLD for Rb can be plotted as below: _ - - - - - - - , Rb flO for Rb
Putting boundary conditions in the Mb expression ILD for Mb is obtained.
1.0
418
THEORY OF INDETERMINATE STRUCTURES
3/161
_ PX Mb -
(I' - X2)
2t
ILD for Mb Ral - P(l - X) + Mb Mb = 1(/- X) - Ral
=
0.,
10.20. ILD for shear at Section mn:
IE Load to rightof ' mn. Vmn = Ra x a it mean ILD for Vmn will be same as ILD for Ra multiplieci by a for this portion
Mb
f
x m
A
B
_~{ ~
n
~l<
' a=4m 10m oE------:OO!,----_ ___ b_=6_m [., "l
m Mb
1.0
m
A
B
c
Vmn n Load on left of mn
Ral··..... /
1.0 Vmn =Rbxb for this portion, ILD for Vmn is same is ILD for Rbx b
' . I.~D. for Ra x a
---
10-----
~~.forRaxb ---------~- .......... /'
10.21. ILD for Mmn Consider a hi:e where ILD If ~~ment deSired B
Ii
IS
1.0
.... _- ...... _- .. _- --------C
Z
1~E______1_om________~__~~1~v~______6_m_______.~1
r iI
419
INFLUENCE LINES
i
~=1 z; ,7QO
ti
B
! ~
I
~c
-+----1 ~
I
R'lK---(_ L 1
_L2
~~ R,
"
I~_~~r' , Primary structure or BOS under load P = 1 and redundant Rb at B.
.,
1
_'_
~
~~IRb
.
~ ~
State-I
x
Compatibility equation at point B. Rb ebb- Py = 0
P= 1 Rb
=~
~~
.
~ l~smt~1I Rb
= 1.0
P
b
a
c
1.0
ebb
B" We know this is ILO for moment at B in a simple'beam.
420
THEORY OF INDETERMINATE STRUCTURES
f _ b2 X2) y = PbX 6EIl ( -
(X
=0 -
a)
PaX f 2 2 Y = 6EIl ( - a - X)
(X
=0-
b)
hX(f -Ii - X2)
Y =
6EIl
12X ([2 - Il - 1I 2)
8bb .-
6EIl 2
1t b
8bb
and
1
2
i
3Ell [
Rb = X ([2 _1I2 - X2)] 21I2 b
X
=
0 - It
with
Origin at A
2
Rb
=
X ([2 - 1I - X2) (2h 2 Ii)
X=o to h Origin at C
'i!
ri I
421
INFLUENCE LINES
Now assume same values of spans and re-calculate.
B
1(~
We know 11 + 12 L
=
_______
r;;;,.c
,r______
L_2_=_6_m____~1
L_1_=_1_0_m__________
A111l1'~ lp =1
B
~c
Compatibility at A
Ra Baa - Py Ra
=0 =
.(.L) oaa
.---------~----------~~~------------~c
1.0
B
I.L.D. for Ra
Rb
x ([2 (2
122 - X")
X ll2 /.
12)
X (16 2 - 6 2
2 x 102
-
X
X2)
6
by putting values of Ld j and I,
422
THEORY OF INDETERMINATE STRUCTURES
Rb
X
<
Ra
Rc
0
0
1
0.1825
2
0.36
3
0.5275
4
0.68
5
0.8125
6
0.92
7
,0.997
Calculate'
Calculate
8
1.04
yourself
yourself
9 10
,,'
0
Calculate
1
yourself
2
3 4
-
ILDfor Ra can be obtained from ILD for Rb. Taking moments about C is equality to zero. R31 SoRa and
+ Rb x 12 =
Rb . = .
P(l - X)
= 0
Pe~ X) _R~lz
(l;Zz ~)
(211 I - II X ...: X2)
423
INFLUENCE LINES
1
Ma
A~) Ra~(
Al;;,)
(~ (L-X)
X
Ma
Mb
p = 1.0
lp
Raj
B
4'Rb Mb
(~
B State-I
lRb
1.0
At fixed support, Sa = 0
B State-II
4EI
2EI -1-
1.
4EI
1
I~ (-) .
. ................•......' . " ..
~.
£lrD=D~~
~.~
'
2EI -1-
BDS under redundant moment.
424
THEORY OF INDETE.RMINATE STRUCTURES
CHAPTER ELEVEN .,1
11. THREE HINGED ARCHES These are Curved Structures which are. iIi use since ancient times. These were mostly used in buildings and the abatements used to be very thick. As our analysis capacity increased due to faster computers, it is now possible to understand behaviour of arches for various suppbrt. load and material conditions. These days arch bridges. either in Reinforced concrete or the pre-str~ssed concrete are .. becoming a common sight due to asthetics of curved surfaces. Arches when loaded by gravity loads, exhibit appreciable compressive stresses. At supports; horizontal reaction (thrust) is also developed which reduces the bending moment in the arch. Aches can be built in stone, masonry, reinforced concrete and steel. They can have a variety of end conditions like three hinged arches, two hinged arches and find arches. Considering the geometry these can be segmental, parabolic and circular. An arch under gravity loads generally exhibits. three structural actions at any cross-section within span including shear force, bending moment and axial compressive force. The slope of centerline of arch keeps on varying along span so above mentioned three structural actions also vary along span.
11.1. Eddy's theorem: The bending moment at any point on the arch is the difference between simple span bending moment and product Hy". Where H is the horizontal thrust at supports (springings), y is the rise of arch at a distance X from the origin. . . Shape of simple span bending moment diagram due to applied loads is also called linear arch. Hy may also be termed as equation of centerline of actual arch multiplied by a constant, (H). Consider the following arch carrying the loads PI, P2 and P3. The shaded area is the BMD. P2
Bending moment at X is Mx
= VaX -
Mx
= ).tX -
Hy - PI(X - a)
Hy. (Eddy's theorem)
425
THREE HINGED ARCHES
Where Ilz simple beam.
= Va x
x - PI(X - a) == Simple span bending moment considering the arch to be a
The inclined axial force (normal thrust) also contributes towards vertical shear force in addition to applied loads and reactions.
11.2. Three-hinged arch: If an arch contains three hinges such that two hinges are at the supports and the third one anywhere within span, it is called a three hinged arch. This type of arch is statically detenninate wherein reactions, horizontal thrust and all internal structural actions can be easily determined by using the laws of equilibrium and statics. If the third hinge is provided at the highest point, it is called crown of the arch. Consider a three hinged arch with third hinge at the crown, then
= J.lX -
Mx
Hy (1) becomes at center
SO H
= Vc
B
A
Mc = J.lc - Hyc = 0
H--"7
~
i
i
(2)
Cutting the arch as shown, and projecting forces along axis 1-1 and 2-2 and putting V = Va - PI we have.
P 2
v
. L.
I ....••.......•
= H Cos8 + VSin8 Q = H Sin 8 - Vcos 8 P
(3) along 1-1
(4) along 2 - 2
H--"7
i Va
11.3. Parabolic Arch If a three·hinged parabolic arch carries udl over its span, the arch will carry pure compression and no SF or BM. This is because the shape of linear arch (BMD due to loads) will be the same as shape of actual arch. For a parabolic arch having origin at either of springings, the equation of centre line of arch at a distance X from origin where rise is y will be. y
at X
Yc So
y
= C.X (L L
= 2" ' y = yc. L L
= C. 2".2" = 4L1c
•
(5) constant C will be evaluated from boundary conditions.
X) we get or X (L - X)
(6)
426
THEORY OF INDETERMINATE STRUCTURES !.§'
The slope ecan be calculated from ~ dX
= tan e
.~ L2 (L - 2X)
(7)
=
11.4. Circular Arch:
If arch is a part of Circle, it is convenient to have origin at the centre. ·c
.
Consider trjangle OEF
.................... : F
, ,
,
y ..
Or
R2
=
X2 + (R -yc
+ y?
,
(8)
: D
. . . .J. . . . . . . . . . . . . . . . . . . . .
(
... X
),
"
and we also have from triangle ADO "L2 ?? "4 + (R - yc)- = R-
.
"
,
'(8] , . ,
R "
L2
yc (2R - yc)
= 4'
,
(9)
'II.
= ..JR2 - X 2 -
'
"
.. .' ,
'
",
, " 'II
As span and central rise are usually known, Radius of ar{;h R can be calculated from (9) Equation (8) can bge written as y
,
,I
.. .'
#
~ ,t~ ~'
o (R - yc)
Now once the basic equations for parabolic and circular arches have been e&tablished, let us solve some numericals.
EXAMPLE NO.1 Analyze a three-hinged arch of span 20m and a central rise of 4m. It is loaded by udl of 50 KN/m over its left half. Calculate maximum positive and negative moments if (i)
The arch is parabolic
(ii)
The arch is circular
SOLUTION: 1. Arch is Parabolic
l:Ma = 0
50 KN/m
o:r:r::cr:o C
Vb x 20 = 50x 10 x 5 2500
Vb
= W = 125 KN
Va
+ Vb
'::So r;.
= 50 x 10 = 500 KN
Va= 500 - Vb
= 375 KN
20
= 500 - 125
Vb=125
THREE HINGED ARCHES
427
H:. 125 x 10 H = yc == 4 = 312.5 KN
and
= -{Va2 + H2
= -v3752
+ 312.5 2 Va
Vb
Tan8b =
II =
375 3.12.5 = 1.2
II =
Maximum positive Moment It is expected· in portion AC. Write generalize Mx expression. 50X" . Mx = 375X --2-· - 312.5y ~ . 4x4. 0 Now y = L2 (L - X) =202 X(20 - X) = 0.04 (20X - X-)
y = 0.8 - 0.04X 2 So Mx = 375X - 25X 2 - 312.5 [0.8X -,O.04X2]
== 375X - 25X 2 - 250X + 12.5X2 Mx
= 125X -
= Vx = 0 = X
Simplifying
12.5X2
125 - 25X
= 5m from A. Putting. Value of X in Mx expression above.
So M';'ax = 125 x 5 - 12.5
X
52
= 625 - 312.5
Mmax
= 312.5 KN-m
Maximum negative moment: . It would occur in portion BC at a distance x from B. Mx
=
125X - 312.5y
Putting equation of.y.
= 125X - 312.5 (0.8X - 0.04X2)
I
Mx = 125X - 250X Mx = - 125X
'1
dMx dX = V x = 0
II I
I
X
+
12.5X2
+ 12.5X2
!
+
+ 312.5"
97656.25
+ 9765.25
= -V 140625
8a == 50.19°
dMx d)(
H2=-V1252
Rb = -V 113281.25 = 336.57 KN
= -1238281.25 = 488.14 KN Tan8a ==
+
Rb = -V 15625
H = 312.5 KN R.a
Rh =-1Vb"
=-
125
+ 25X
= Sm from B.
125 312.5 = 0.4
428
THEORY OF INDETERMINATE STRUCTURES
So putting value of X in Mx expression above .
. Mm••
Ml1la~
= - 125 X 5 + 12.5(5i = - 625 + 312.5 = - 312.5 KN-m
SOLUTION: Considering Circular Arch EXAMPLE NO.2: Now or Solve the following loaded three hinged Circular Arch
50 KN/m
ccr:r:ccr:J
C
y
x
~----'~---------
va=375!
!Vb=125
Step 1. Reactions: As before reactions are same. Step 2. Equation of Circular Arch The general equation is (X - h)2
+ (y -
k?
= r2
hand k areco"ordinates at tbe centre and r is radius of Circle. There are three t,mknown in above equation, Viz, h, k and r and these can be determined from the following boundary conditions . Origin is at point A . · . . Boundary conditions l.
At X = 0,
y
=0
It gives
(_h)2 h2
2.
3.
At X=20,
At X=1O,
Y= 0
Y=4
Subtract (1) from (2) 400 - 40h = 0 Or
h = 10
It gives
It gives
we get
+ (_k)2
+ k2 =
r2
= r2 (1)
+ (_k2) = r2 400 + h2 - 40h + k 2 = r2 (10 -hf + (4-k? = r2 100 + h2 - 20h + 16 + k2 1 16 + h2 - 20h + k 2 - 8k = (20 - hf
(2)
8k r2
= r2 (3)
429
THREE HlNGED ARCHES
Put value of h in (1) and 3
or
100+k2=r2
(1)
116 + 100 - 200 + k" - 8k = r"
(3)
16 + k2 - Sk = r2 (3) 16 + k2 - Sk = 100 + k" (by putting Value ofr2 from 1)
=
8k
16 - 100 = - 84 -84
= -S-
k
= - 10.5
Putting k = - 10.5 in (3)we get r2
=
16 + (- 10.5)" + 8 x 10.5
= 16 So
+ 110.25 + S4
= 210.25
r = 14.5 meters. Putting Values of h, k and r in general equation, we get (X - 1O)~
=-
y
+(y +
y
= 14.5 2
+ -.j14.5 2 -
10.5
(y + 10.5)2
10.5)2
= 14.5 2 -
Simplify it, we get.
(X - 10)2
(X - 10)2
= - 10.5
+ -.j210.25 - X 2 - 100 + 20X
=-
+ -.j1l0.25 - X2 + 20X
10.5
(4)
L2
We know, . yc (2r - yc) =
"4
(5)
and
y
= ~ r2 -
J-
(~ - X
(6)
(r - yc)
These equations are same as were used in derivation earlier.
Alternatively to avoid evaluation of constants each time, equations (5) and (6) can be used. Equation (6) is the equation of Centre-line of Circular arch.
Step 3: Calculation of Maximum moment. Maximum positive moment occurs in span AC. Write Mx expression 50X2 Mx = 375X - -2- - 312.5 Y put y from (4) above.
Mx
= 375X -
25X2
= 375X -
25X 2 + 3281.25 - 312.5
-
312.5 [- 10.5 + -.jIlO.25 - X2 + 20X]
-J 110.25 -
X 2 + 20X
430
THEORY OF INDETERMINATE STRUCTURES
New maximum mement eccurs where shear ferce is zero.. So. dMx -d- = Vx = 375 - 50X x 375 - SOX =
312.5 (- 2X + 20)
2-..j 110.25 -
X2 + 20X
= 0
312.5 (-X + 10) -..jllO.25- X2 + 20X
6.25 (10 - X) 75- X = . -..j1l0.25 - X2 + 20X
divide by 50
multiply by - 1
We get
X_ 75 = 6.25 (X - 10) .. -..jUO)5 - X2 + 20X (X - 7.5) './ 110.25 - X2 + 20X = 6.25 (X - 10) (X - 7.5)2 (110.25 - X2 + 20X) = 6.252 (X (X2- 15X + 56.25) (110.25 - X2 + 20X) er
2
1l0.25X
-
4
X + 20X
3
-
3
1653.75X + 15X
-
square both sides
lW,
Simplify
= 39.0625 (X2 -
20X
+ 100)
300X == 39.0625X2 - 781.25X +'3906.25 2
+ 6201.56 - 56.25X2 + 1125X Simplifying - X 4 + 35X 3 - 285.0615X 2 + 252.5X + 2295.3125 =0 X 4 - 35X3 + 285.0625X 2 - 252.5X - 2295.3125
er
= 0
New it is censidered apprepriate to' selve this equatien by Medified Newtqn,- Raphsen iteratien selutiens which in general is ~n-1 = Xn +
f (Xn) (Xn)
f
(A)
3 So. . f (X) =, X4 - 35X + 285.0625X 2 - 252.5X -. 2295.3125 .
And differentiate,
fl
(X) = 4X3 - 105X2 + 570. 125X - 252.5
In general, it is recommended that first roet Xn sheuld be always taken at 1 because it cenverges very fast. Hewever, knowing that B. M will be maximum near the middle ef pertien AC, we take Xn = 2 (to' reduce number of iteratiens possibly) and selve in the fellowing tabular ferm. Evaluate f(X) and f/(Xn) expressiens. . Iteratien Number
Xn
f(Xn)
fl (Xn)
Xn + 1 frem A abeve
1
2
-1924.06
499.75
5.85
2
5.85
147.251
290.1629
5.3425
3
5.3425
- 30.3142
'406.3845
5.417
4
5.417
- 0.58794
390.546
5.418
r
431
THREE HINGED ARCHES
So we get Xn and Xn + 1 as same after 4th iteration. So X = 5.418 m
put this in MX expressions
Mma., = 375 (5.418) - 25 (5.418i + 3281.25 - 312.5 ...)110.25 - 5.418" + 20 x 5.418 = 280.066 KN-m
Maximum negative moment in the arch Let us assume that it occurs in portion BC at a distance X from A.( 10 < X < 20) Mx=125 (20 - X) - 312.5 (- 10.5 + ...)110.25 - X2 + 20X)
Simplify
= 2500 - 125X + 3281.25 - 312.5...) 110.25 - X2 + 20X or
Mx = 5781.25 - 125X - 312.5 ...) 110.25 - Xl + 20X
Maximum moment occurs where SF is zero, So differentiate Mx expression w.r.t. X.
°
_dM_x = = _ 125 _ --;3=12=.=5=(-=2=X=+=2=0::::)= dX·· 2..yllO.25 - Xl + 20X
or
o=
_ 125
+ 312.5(X - 10) ..y110:25- X2 + 20X
125 ...)110.25 - X2 + 20X
=
312.5 (X - 10) squaring both sides. We have,
15625 (110.25 - X2 + 20X) = 97656.25 (X 2 110.25 - X2
i
I
.1
"1
I
-
20X + 100) Simplify
+ 20X = 6.25 (X2 - 20X + 100) = 7.2SX2 - 145X + 5i4.75
dividing by 7.25
Solve this quadretic equation. I1
!
I
II
X-
20 ± .y400 - 284
2
20 ± 10.77 2
=
I
X =
,
So Mmax = 5781.25 - 125 x 15.385 - 312.5...)110.25 - (15.385)2
I
15.385!l1 from A
Put this v(lJue of X in Mx expression above.
1
~
i
i
I !
!
= 5781.25 -
1923.125 - 312.5..y181.257
=-
+ 20 (15.385)
349.1l5KN.m
43i
THEORY OF INDETERMINATE STRUCTURES
11.5. Derivation for center-line of a parabolic arch with supports at different levels.
yc
U2
U2
1~1
Me
= 9.6 KN-m.
r 447
APPENDIX - UNSOLVED EXERCISES j
I I
EXERCISE 23: . Solve the following beam by using slope deflection method.
[
I E
=
15m
~
200 GN/m 2
1
1
2OK
:'-i
40KN
B
::zs:: 7171TT
2m
)1
1<
~
3m
C
0
Dr
~
~
5m
4m
ANSWERS:
= =
MBA MCD
-23.676 KN-m ,
MBC
23.676 KN-m ,
MCB
=
-8.072 KN-m
8.072 KN-m
EXERCISE 24: Solve.the beam by using slope deflection method.
l.Sm
1
120
1
40KN
B
::zs: mm
mm
~
KN
3m
4m
1
20 KN
C
A mm I
o
Sm
ANSWERS: MBA = -16.21 KN-m
MBC
16.21 KN-m
-20.15 KN-m
MCD
20.15 KN-m
MCB EXERCISE 25:
Solve the following beam by using slope deflection method.
A B C 0 ~~,------3-m------~~-----4-m----·~~b~»~·----s-m-----~~
~2mm
"""
ANSWERS: MAB MCB
= =
0
MBA = 56.26 KN-m
-28.475 KN-m ,
MCD
=
28.475 KN-m
MBC = -56.26 KN-m MDC
=
0
448
THEORY OF INDETERMINATE STRUCTURES
EXERCISE 26: . Solve the beam by using slope deflection method.
c
B
A
4m
, 3111
,4
5m
771771
~mm ANSWERS: MBA
-10.251 KN-m,
MBC
MCD = -11.878 KN-m,
MAB
= =
10.251 KN-m
=
MDC
MCB.
=
11.878 KN-m
0
EXERCISE 27: Solve the following beam by using slope deflection method.
A
B
.is.. tmtf
C
,4
3m
4m
Tm1f
D
::zs:: 1fm1
~,
5m
lm~
l'mm ANSWERS: MAB = 0 MCB
MBA = 45.88 KN-m
-16.59 KN-m
MCD
=
MBc: = -45.88 KN-Ill ' MDt = 0
16.59 KN-m
EXERCISE 28: Solve the following beam by using slope deflection.
fA
,,1
36
~
KN
3 KNlm
6 KNlm
~
~
~ EI
=
constant
ANSWERS: MAB
24.324 KN-m ,
MBA
MCB = -13.716 KN-m ,
MCD
=
-18.852
MBC = 18.852 KN-m
13.716 KN-m
MDC
0
., i
APPENDIX - UNSOLVED EXERCISES
449
EXERCISE 29: Analyze the following frame due to settlement of 12mm at support D.
E
=
200 GN/m 2
EXERCISE30: Analyze theJrame by M.Distribution method.
10 K..";
I.5m
21 10K~
I.5m
D
ANSWERS: MAB
8.2745 KN-m
MBA
-2.651 KN-m ,
MBC
MBD
-1.362 KN-m
MCB
-1.9985 KN-m ,
MDB
=
4.013 KN-m 4.944 KN-m
450
THEORY OF INDETERMINATE STRUCTURES
EXERCISE 31: Analyze the frame by M. Distribution methoI
ANSWERS: MDA
=
4.87 KN-m
MBA = -8.56 KN-m
8.31 KN-m
MEB MFC
=
9.07 KN-m
MAD
=
4.45 KN-m
MAB
MBC
=
5.61 KN-m
MBE
MCB = -12:85 KN-m
MCF
= = =
-4.45 KN-m
2.95 KN-m 12.85 KN-m
APPENDIX-UNSQL VED EXERCISES
451
EXERCISE 33: Analyze the frame by Moment distribution method.
25
1in
31
3m
21
B
E 31
3m
21
A
r
F
1
I<
>1
ANSWERS: MBA = -1.294 KN-m
MBE
MBC = -2.715 KN-m
MCB = -3.841 KN-m
MCD
3.747 KN-m
MDC = -3.708 KN-m
MDE = 3.708 KN-m
MED
2.848 KN-m
MEB = -4.011 KN-m ,
MEF = 1.163 KN-m
MFE
EXERCISE 34: Solve the following frame by moment distribution method.
2~N/m
.. !
\
I
c~~~~~~~~
1 0 KN --7f--'--'--'--L-.JL-.J'---1--1
0
51
6m
21
Bv-~
50
24 KN/m
____
~
____
6m
21 ~
E
1 0 KN --7f--'--'--L....-JL-.J'---1'---1--1
!
I Ii
I
I !
I tI
= 4.025 KN-m
-0.524 KN-m
MAB
8m
21
21
A
6m
F
8m
>1
=
O.p72 KN-m
·452
THEORY OF INDETERMINATE STRUCTURES
ANSWERS:
=
MBA = -10.29 KN-m
MBL . =
MBC = -46.82 KN-m
MEB = -56.22 KN-m
MCD = 56.22 KN-in
MDC = -89.83 KN-m
MDE
MEB = -137.68 KN-m
MEF = 64.44 KN-m
MFE = 56.87 KN-m
MAB
8.52 KN-m
89.83 KN-m
~7.1l
KN-m
MED = 73.23 KN-m
EXERCISE 35: Find vertical and horizontal deflection of Point C Determine for exercise in member due to applied loads and then due to unit vertical and horizontal load at C. Use method of moments and shears valid for parallel chord truss and inspection. Make a table. Draw SFD & BMD. Number the members.
8@5m JE
Chord members
=
12 x 10-3 m 2
Inclined and vertical members
=
6 x 10-3 m2
E = 200 x 106 KN/m2 !:J.Vc == 18.213 mm
ANSWERS:
!:J.Hc
=
3.328 mm
EXERCISE 36: Analyze the truss by taking members force EF as redundant. Use consistent deformation method. E
=
200
X
106 KN/m 2 . ,
A
=
2x
E
1Q-3m 2
for all members. F
4.5 m
30KN
ANSWERS:
FEF
=
-27.16 KN
[email protected]
15KN
1
453
APPENDIX - UNSOLVED EXERCISES
EXERCISE 37: A three hinged parabolic arch is loaded as shown. Determine reactions. Determine moments. at 6, 12 and 18m for left support.
6m
31m
1<
I
I
)1
ANSWERS: Va M6
= 2005.96 KN = -27 KN-m
Vb
=
1853.54 KN
M12 = 61.44 Kn-m
H
=
M18
2639.41 KN
=
-467.71 Kn-m
EXERCISE 38: A three hinged circular arch is loaded as shown. Determine max +ve and negative moments in arch. Repeat the Exercise considering it as parabolic arch.
223 KN 3m
c 3m
A
21m
I(
)1
ANS\YERS:
Va
= 191.14 Kn
Vb
=
31.857 KN
.H
Positive moment under load = 402.84 KN-m Max. negative moment
~
I
=
-88 KN-m at 5.5 m from C in position BC.
111.50
iI I
Yl
THEORY OF INDETERMINATE STRUCTURES
454
For Parabolic Arch Same Reactions (M +ve) max under load = 410 KN-m (M
~ve)
max
=
-85.1 KN-m
at 5.24 m from center.
EXERCISE 39: A three hinged parabolic arch is loaded as shown. Calculate reactions and Max. moment at 15m. 27m from left springing. . 111 KN 334 KN
c Sm
37m
IE Sm
IE N] ~C
>1
1<
S.Sm
)1
ANSWERS: . Ha Vb
d.
>1
= =
82.79 KN 108.23 KN
= 128.179 KN . . ' 1 M15 = 165.1 KN-m Hb
Va ==' - ,"
... ' , '
',or
M27 = 77.75 KN-m
If arch is circular Va
=
132.86 KN
Vb
136.179
M9
Hb M27
=
= =
112.13
Ha = 74.83
216.25 KN
M15 =
36.57 KN-m
EXERCISE 40: Find plastic moment of the following loaded beam. \.5 TIm
.r! mT1T
i!:RS
136.76 i
