surojit
TRANSCRIPT
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INTRODUCTION
Determinate structures are analyzed just by the use of basic equilibrium equations. By
this analysis, the unknown reactions are found for the further determination of stresses.
Redundant or indeterminate structures are not capable of being analysed by mere use of
basic equilibrium equations. Along with the basic equilibrium equations, some etra
conditions are required to be used like compatibility conditions of deformations etc to get
the unknown reactions for drawing bending moment and shear force diagrams.
!ample of determinate structures is" simply supported beams, cantile#er beams, single
and double o#erhanging beams, three hinged arches, etc.
!amples of indeterminate structures are" fied beams, continuous beams, fied arches,
two hinged arches, portals, multistoried frames, etc.
$pecial methods like strain energy method, slope deflection method, moment distribution
method, column analogy method, #irtual work method, matri methods, etc are used for
the analysis of redundant structures.
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What is Strain Energy :-
%hen an eternal load acts on a structure, the structure under goes deformation by the
eternal force & work done is takes place. 'his internal work stored as a energy in
structure & it helps the structure to spring back to its original shape & size when e#er the
eternal load is remo#ed pro#ided the material of the structure is still within the elastic
limit.
'his internal zone which is stored as a energy is due to the (straining energy).
%hen *+ -, *# -, * -, means in equilibrium is reached as per law of
conser#ation of energy the work done by the eternal force or load must be equal to thestrain energy stored.
'his concept of energy balance is utilized in structural analysis to de#elop a number
method to find out the deflection of the structure
/01 $tain energy 2 Real work method
/31 4irtual zone method
/51 6astigation7s method
8 Strain energy & complementary energy.
9et,
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:;i elementary stress
:ei elementary strain
:a c2s area of element
: 9ength of element
ere stress is gradually increased from zero to its final #alue (p) as stain increase from
zero to final #alue (e).
9et, ;i be the stress & strain
:ei& work done is takes place.
%ork done strain cur#e :#
?f stress > strain cur#e is 9inear the strain > energy of element
v δeρ ×××2
1
∴'he strain energy stored in hole structure
∫ ×××= dv ePU 21
×
2
1
stress strain d#
8 oa! "S De#ormation Relation
?f the Deformation (:@?)
9oad acting (=i)
∴ %ork done
∫ ∆ iδiρ
/in case of 9inear elastic problem1
$ Strain energy !%e to en!ing
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9et consider the is subject to pure being let area of the element :a & its distance from .
A. is () from the fleure theorem f
Y I
M.
+ bending moment
? moment of inertia of the section
f bending stress
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∫ ∫
=∴
L L
dx EI
Mudx
EI
IM0 0
2
2
2
22
$ 'ortal (rame :-
9et,
4a, 4d #ertical reaction at A & D
orizontal 'hrust at each support
w0&w3 eternal load.
'he portal frame is the empal
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%hen the frame is of uniformly rigidity !? the orizontal thrust can be gi#en by the
∑∫ == 0dsHδMδ
M
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E total strain energy.
u
3322112
1
2
1
2
1∆+∆+∆ ρρρ
FFFF./i1
9et1δρ
9oad Added after ;0, ;3, ;5 load applied.
Additional Deflection :@0 , :@3 , :@5 respecti#ely
ow du 2
1
:;0 :@0 G2
1
:;3 :@3 G2
1
:;5 :@5 FFFF.. /ii1
'he ecess strain > energy du
∴
u G du
∆+∆+∆+
∆+∆+∆
3322113322112
1
2
1
2
1
2
1
2
1
2
1δPδδPδδPδPδPP
FFFF. /iii1
ow,
:;0 G ;0 , ;3 , ;5 9oad to be applied & strainHenergy is gi#en by
$!
[ ] [ ] ( ) ( ) ( ) ( )33322211112
1
2
1
2
1∆+∆+∆+∆+∆+∆+ δPδPδPδP
FFFF. /i#1
!quation /iii1 & /i#1 ze set
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2
1
;0@0 G2
1
;3@3 G2
1
;5@5 G2
1
:;0:@0 G2
1
:;3:@3 G2
1
:;5:@5 2
1
;0@0 G2
1
;0:@0 G
2
1
:;0@0 G2
1
:;0:@0 G2
1
;3@3 G2
1
;3:@3 G2
1
;5@5 G2
1
;5:@5
2
1
;0:@02
1
;3:@3 G2
1
;5:@5 2
1
:;0@0
2
1
du H4
1
d;0 d@0 2
1
:;0 @0
2
1
:u 2
1
:;0 @0
1
1
∆= pδ
uδ
ere
114
1∆δPδ
smallI
Piδ
uδi =∆∴
JJJJJJJJJJJJJJJJJJJJJ
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PROBLEM NO - 07
BE=2.08 m & EC=2.08
Total Load= (2.08×2×30) KN
=125.284 KN
Hene !a= !d="2."42 KN
(#$ %$mmet$)
'oment 'em#e
'
H
*an+
e
H- B /$ 0 to"
!a/"H/
152
BE /" 0 to
2.08
H-C /- 0 to
"
!a/"H/EC /" 0 to
2.0
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1528
Member AB and CD
No
∑∫ = dy H M M H U δ δ δ δ
o =2
∫ −−K
-211// dy EI Y HY
=
dyY EI H ∫ K
-
323
=
( )K-52
53 Y EI H
=
( )5K.25
3 EI H
= 144HEFor AB & CD
NOW SPAN BE & EC
∑∫ = dx H
M M
H
U
δ
δ
δ
δ
∫ = -L.3
-3
H
M M
H
U
δ
δ
δ
δ
=
( ) ( )∫ −−−-L.3
-
3 .2KK0MKN3.K33 dx EI H x x
= 2E6/"2.4"2 "
]∫ ∫ ∫ ++ -L.3
-
-L.3
-
3-L.3
-5KK0M dx H dx x x Xdx
=
[ ] H EI
LL.ONPK.3KPL053 ++−
HENCE,
-=∑ H
M
δ
δ
144E7(4.88H/543.04)(2)E=0
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=144H7149."H/108".08=0
H=3."9KN
Finding bending momenFor !"an AB
'=/H- -=0
'=0 at
'=/H-
-="
'=/22.182 KN/m at B
For !"an CD
'=/H-
'=0 at
'=/H-
-="
'=/22.182 KN/m at C
F0R SPAN BE
'= ( ) H x x K0MKN3.K3
3
−−
:;en =0
'=/"H
=/22.182KN/m at B
:;en, =2.08m
'=
( ) H x x K0MKN3.K3 3 −−
=130.295/22.182/"4.89"
=43.21 KN/m at E.
FOR SPAN CE
'=
( ) H x x K0MKN3.K3 3 −−
:;en =0
'=/"H
=/22.182KN/m at C
:;en, =2.08m
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'=
( H x x K0MKN3.K3 3 −−
=130.295/22.182/"4.89"
=43.21 KN/m at E.
Finding Poin o# $onra#e%e're
"2."42 152 "(3."9)= 0
⇒2 4.2 7 1.5= 0
=
( ) ( ) ( )
03
M.010/N3.N3.N 3
×
−−±−−
= (74.2n+ (/)%>+n
Ca%$'%aion #or !(ear #or$e,
o BC
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Ta>n+ moment a#o?t C
!B (4)/22.182/22.182/30(4)(4)2 =0
!B =1.1KN
!C=120/1.10
=48.90KN
FOR SPAN AB
Ta>n+ moment a#o?t B
"H =/22.182
H=/3."9 KN
FOR SPAN CD
Ta>n+ moment a#o?t C
"H/22.182=0
@, H= 3."9 KN
22.18
B
H 22.18
C
H
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∴
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PROBLEM ) *
%Aan B ,$ d>tane om .
'$=/H$
Hδ
Mδ
=/$ 60 to "D
o Aan BC, d>tane om B.
'= !a/'#/152
= "0/"H/152
Hδ
Mδ
= /" 60 to 4D
'oment 'em#e
'H '*an+e
H- B /$ 0 0 to "
"0/"H/BC /" 0 to 4
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152
' H-C /- 0 0 to ".32
'em#e B
:e no F =
∫ s EI M
δ 3
3
No
∑∫ == 0HδMδ
MHδ
Uδ
o
[ ]∫ "
0&CD AbSpan
Hδ
MδM
( ) ( ) dy Y HY ∫ −−"
02
=
dy HY EI ∫ K
-
323
=
( )K-52
5
3Y EI H
=
( )5K.25
3 EI H
= 144HE
∫ =
4
0 Hδ
MδM
Hδ
Uδ
=
( ) ( )∫ −−−4
0
2.""15"0 dx H x x
= /"0 "
∫ ∫ ∫ ++ 4
0
4
0
24
03""15 dHdx x x dx
=
EI
0
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=
[ ]HMaEI
2181
+−
For S"a+ BC
∫ = 4
0 2
2
EI
dx Hδ
MδM
Hδ
Uδ
=
( ) ( )∫
−−−40
2 ""15
EI
dx H x X A
=[ ]HAEI 144192048
1
++−
∴
No #or S"a+ CD
∫ = 53.K
- 3
3
EI
dY H
M M
H
U δ
δ
δ
δ
=
∫ −−53.K
-
11//
EI
dY y Hy
=1"8.29 HE.
G H
U
δ
δ
=0
:e +et
1"8.29H7144H719207144H/48!=0
@ ! =4079.5HIIII(1)
For S"a+ AB
∫ ×= K
- 3
3
dY V
M
EI
M
V
U
δ
δ
δ
δ
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@ H =0."89 KN
!=4572.25(."89)
!=4".55KN!=!C=120/4".55
=3.45 KN
For Finding Momen
For !"an AB
'=/H- -=0,
'=0 at
'=/H- -=",
'=/(0."89) (")
=/4.134 KN/m at B.
For S"an BC
' =! / "H/152
:;en =0.
'= /"H
= /4.134 KN/m at B' =! / "H/152
=4
=4".55(4) (."89) (")/15(4) (4)
.-/ 1N-m a $
For !"an CD
'= /H- -=0
'=0 t .
'= /H- -=".32
=/(0."89) (".32)
= /4.354 KN/m at .
For S"an BC
' =! / "H/152 =2 =4".55 (2)/(."89)(")/15 (2) (2)
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=29 KN/m at m>d Aan
Finding Poin o# $onra#e%e're
4".55 152 4.134= 0
⇒15
2 4".55 7 4.134= 0
⇒2 3.10 7 0.28= 0
=
( ) ( ) ( )
03
3L.-10/N0.50.5 3
×
−−±−−
= (73.1n+ (/)%>+n.
Ca%$'%aion #or !(ear #or$e,o BC
Ta>n+ moment a#o?t .!B (4)/4.134/58/30(4)(4)2 =0
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!B =5.53 KN
!C=120/5.53
=44.4" KN
For S"an AB Ta>n+ moment a#o?t B
"H =/4.134
H=/."89 KN
For S"an CD
Ta>n+ moment a#o?t C
"H/4.354=0
@, H= 0.25 KN
4.134
B
H
4.354
C
" m
H
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'rolem * +
$pan AB y distance form A.
+y H y
D"06 !oY Hδ
Mδ−=
$pan B6 distance from B
+ 4a H +B6 > 0M3
K- > K > 0M3
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IK-DK to H
Mn−=
δ
δ
+or Bar +
Hδ
Mδ 9imit
Ab Hy H - to K
B6 K- > K > 0M3 HK - to N
6D H H - to K
%e know E
∫ s EI M
δ 3
3
ow∑∫ == 0HδMδMHδUδ
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H 3LL- G 0P3- G 0NN
H PK- G 0NN
∴ ∑∫ = 0HδMδ
M
⇒H PK- G 0NN G 0NN -
⇒3LL PK-
5.555 Q.
ow, 4A 4B K- Q
46 4D K- Q As per 6ondition
m or
At D
+ H K
H/5.5551 K
+ 0P.PL Q > m at B or 6
ow, 0M3 > K -
HK
H K /5.551
H0P.PL Q > m at B
+ K- > 0M3
> K N K- N > 0M/N13 > K
H K
H0P.PL Q > m at 6.
%hen 3
'hen + N-.-3 Q > m
(in!ing 'oint o# contra#ele,%re
K- > 0M3 > 0P.PL -
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⇒3 > N G 0.553 -
H/HN1
( ) ( ) ( )
12
332.1)1(444 2
×
−−±−−
-.5O m
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'rolem *
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+ember +oment
δ
Hδ
Hδ
Mδ
Mδ
Mδ range
AB +a H y - Hy 0 - to K
B6 4a 0M3 > K HK - - to N
ow member AB
0= δ
Mδ
∴ ∫ "
0
2
2dy
EI
M
∫
"
0 2
2
dy EI
δ
Mδ"
0D)(61
×− dy Hy MaEI
-
B6
0= δ
Mδ
∫ 4
0 2
2dx
EI δ
#δM
(∫
−−40
2"15
EI
dx x H x AX
( ) ( ) ( )
−− 4
0
24
0
44
0
3
2
"
4
15
3
1 x
H x x
EI
A
[ ]$aAEI
489"33.211
−−
∑ =∴ 0 δ
Mδ
set
30.55 4A > KN >NL G S -
30.55 4A > NL PK /01
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ow for $; AB
∫ .22
dy EI
M
∫ = "
0 2
2dy
EIHδ
MδM
( ) ( )∫
−−"0 EI
dy y Hy Ma
( )
+−∫ ∫
"
0
"
0
21 dy Y HYdy MaEI
( ) +− 32 )"(
3"
21 HMaEI
[ ]HMaEI
2181
=−
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∫ dy EIM
2
2
∫
"
0 2
2
EI
dy Mδ
MδM
( ) ( )∫
−"0
1
EI
dy x Hy Ma
−∫ ∫
"
0
"
0
1dy Y Hdy Ma
EI
×− 3"2
"1 H
MaEI
EI
HMa 18" −
&
∫ dx EIM
C %&SPa'o(
2
2
∫ 4
0 2
2dx
EIHδ
MδM
-
∴K+a > 0L -
+a 5
3
H +a G 03 0-K.KO > 3.O4A -
H5 G 03 G 0-K.KO > 3.O4A -
P > 3.O4A H0-K.KO FFFFFFFFFF./???1
30.55 4A > NL PK- FFFFF/?1
P > 3.O4A H 0-K.KO FFFFFFF../???1
> -.54A H 00.LM FFFFFF../?41
& FFFFF./#14 G ?4
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/-.NN4a > 1 G / > -.54A1 3- /H00.LM1
-.NN4a > > -.54A L.0M
-.0N4A L.0M
4A ML.30 Q
∴ 4B 03- > ML.30
K0.OP Q
-.5 4A H 00.LM
> -.5 /ML.301 H00.LM
H00.LM G 0O.NK5
h M.K05 Q
∴ +a 5
5 /M.K051
0K.L5P Q > m.
K > 0M3 -
ML.30 > K /M.K051 > 0M3 -
ML.30 > 3L.-KM > 0M
3
- 5.P > 0.LO0 > 3 -
3 G 0.LO0 > 5.P -
s3 > 5.P G 0.LO0 -
( ) ( ) ( ) ( )
12
81.1149.39.3 2
×
−−±−−
12
8.29.3
×
±+
-.M taking >#e sign. from B.
5.5M m from B taking > #e sign.
y
0K.L5P > S -
0K.L5P Q > m at A
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+ +a > y Km
0K.L5P > M.K05 /K1
H0K.L5P Q > m at B
ow K > 0M3
%hen -
+ H K
H K /M.K051
H55.KOL QHm at B
%hen Nm+ 4A > K > 0M3
ML.30/N1 > K /M.K051 > 0M/N13
HN-.L5L Q > m at 6
%hen 3m, + 33.ON3 Q > m
+D G N- .L5L -
K > +D G N-.L5L G -
K /M.K051 > N-.L5L +D
+D HO.0K QHm
+oment
%hen -
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+ +D
HO.0K Q > m at D
%hen N
+ +D >
HO.0K > /M.K051 /N1
H3P.K03 Q > m.
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'aking moment about T3
0K.L5P > K G 0K.L5P -
-
4B 4B, G 4B3
ML.M0 G -
ML.30 Q
'aking moment about D
N > 3P.K03 > O.0K -
P.0P5 Q
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'rolem *
+ember +oment
Hδ
Mδ
δ
Mδ x Mδ
Mδ x range
AB +a H y Hy - 0 -HK
B6 4a 0M3 > K /HK1 - -HN6D +D > Hy - 0 -HK
ow for AB
( )∫ ×
−=
"
0 22dy
EI
Hy Ma
Hδ
δ
( )
∫ −
"
0
,2
4
1dy
Hδ
MδHy Ma
EI
( ) ( )∫ −×−"
02
4
1dy y Hy Ma
EI
( ) ( )∫ −−"
0.
2
2dy y Hy Ma
EI
[ ]HMaEI 2182
1
+−
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HMa 3"9 +−
ow
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[ ]HMaEI
842
1−
[ ]HMaEI
421
−=
(or /C
( )∫
−−=
4
0
22
2
"15dx
EI
H x x
Mδ
Uδ A
( )∫ ×
−−=
4
0
2
.2
"152dx
Mδ
Mδ
EI
H x x A
( ) ( )∫ −−4
0
2 .0"151 dx H x ax EI
-
∑ =∴ 0Mδ
Uδ
3+a > N -
+a 3
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[ ]HaEI
489"033.212
1−−
[ ]HaEI
24480"".101
−−
∑ =∴ 0 δ
Uδ
-G0-.KK4a > NL- > 3N -
0-.KK4a > 3N NL- FFFFFFF./??1
ow 3.3M NM FFFFF../#1
/i#1 G /#1
/H4a G 5P.M0 G 5.O1 G /4a > 3.3M1 - G NM
G5P.M0 G 0.NM NM
5.OL Qn
∴+a 5.OL 3 O.MK QHm
∴ 3.3M /5.OL1 NM.
4a M5.M-M Q
∴4D 5- N > M5.M-M
KK.OPM Q.
N4D
3N- G K/5.OL1 > N /KK.NPM1
3N- G 33.KL > 3KM.PL
H5.5 Q > m.
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+oment for 6D
+ +D > -
+ +D
H5.5 Q > m at > D
%hen K
+ +D >
H5.5 > /5.OL1 /K1
H3M.PL QHm at 6
0M3 > K
%hen -.m
+ M5.M-M - > 0M -3 > K/5.OL1
H 33.KL Q > m at B
%hen Nm.
+ 4A > 0M3 > K
M5.M-M/N1 > 0M/N1
3
> K /5.OL1 HNL.KK Q > m at 6
%hen 3m
+ M5.M-M/31 > 0M/313 > K /5.OL1
3N.55 QHm
%hen -
+ +a
O.MK Q > m at A.
%hen Km.
+ +a > y
O.MK > 5.OL K
H0M.03 Qn > m at B
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4A > 0M3 > K -
M5.M-M > 0M3 > K /5.OL1 -
M5.M-M H 0M3 > 33.KL -
5.MK > 3 > 0.M03 -
3 > 5.MK G 0.M03 -
( ) ( ) ( )
12
512.1145".3)5".3( 2
×
−−±−−
12
5.25".3
×
±+
-.NPM m.
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4B N > 33.KL > NL.KK > 3N- -
4B OO.L5M Q
46 03- > OO.L5MQ
N3.0KMQ
'aking moment about B
K G O.MK > 0M.03 -
0.3K Q
K > 5.5 > 3M.PL -
N.LLQ
taking moment about 6
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'rolem * 0
+oment +ember
δ
Mδ
Hδ
Mδ
Mδ
Mδ Range
+a > y AB - Hy 0 - to K
4A n > 0M3H
K
B6 HK - - to N
+ember AB
∫ = "
0
2
2dy
EI
M
δ
Uδ
∫ ×−"
0 2
)(2
δ
Mδ
EI
dy Hy Ma
-
B6 member
( )∫
−−=
4
0
2
2
"152
dx EI
δ
MδH x Ax
δ
δ
( ) ( ) ( )
−−
4
0
244
0
3
2
"
4
15
31
1 x
H x x
A
E
[ ]H E
A 489"033.211
1−−
-
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∑ =∴ 0 δ
Uδ
SG 30.554A H PK- H NL -
30.55 4A > PK- NL FFFFF/i1
-
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[ ]HMaEI
18"1
−
(or Spa1 /C
02
24
0=×= ∫ dx Mδ
Mδ
EI
M
Mδ
Uδ
∴ ∑ = 0MδUδ
( ) 018"1
=− HMaEI
+a 5
3.O4A -
V P > 3.O 4A H0-K.KOFFFFFFFF../iii1
30.55 4A > NL PK-
V -.NN4A > 3- FFFF../i#1
-.5 4a H 00.LM FFF../#1
4 G ?4
/ > -.54A1 G /-.NN4A > 1 3- G /H00.LM1
V -.0N4A L.0M
4A ML.30 Q
4B 03- > ML.30
K0.OPQ
-.54A 00.LM
V > -.5/ML.301 H 00.LM
V M.K05Q
∴ +a 5
0K.L5P Q > m
'oint o# contra #le,%re
4A > 0M3 > K -
-
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V ML.30 > 3L.-KM > 0M3 -
V3 G 0.LO0 > 5.P -
( ) ( ) ( ) ( )
12
81.1149.39.3 2
x
−−±−−
-.M taking > #e sign from B.
(in!ing /en!ing moment #or 3/
+ +a > y
-
+ +a
0K.L5P QHm at A
(or Spa1 /C
+ 4A > 0M3
> K%hen -
+ K
H K /M.K051
H55.KOL Q > m at B.
(or 3/ Span
+ +a > y
Km
+ +a > K 0K.L5P > 55KOL
H0K.L5P QHm
(or /C
+ 4A > 0M3 > K
Nm
+ HN-.L5LQ >m of 6
3m
+ 33.ON3 QHm
but 6D.
+ -
Rollar supportI
At D
-
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Calc%lation S(
'aking moment about 6.
4B? N > 55.KOL > N-.L5L > 3N- -
4B0 N 3N- G55.KOL G N-.L5L
4B0 OL.K3P Q
∴460 03- > OL.K3P
KK.K3P Q
-
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H K G 55.KOL G 0K.L5P -
L.N3 Q
)#81.""
838.40=
-
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-
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'rolem : 4
+ember +oment
Mδ
Mδ
Hδ
Mδ Range Remarks
AB +a > y 0 Hy - to K
B6 4AH0M3 >
K
- HK - to N
N%mer 3/
( ) ( )∫
−=
"
0 2
12
EI
dy Hy Ma
Mδ
Uδ
[ ]HMaE
I18"
1−
N%mer /C
( )∫
−−= 4
0
2
2
0"152
EI
x dx H x x A
Mδ
Uδ
-
∴ ∑ = 0MδUδ
K+a > 0L -
V +a 5
-
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No2 #or 3/
( ) ( )∫
−−=
"
0 2
2
EI
d y #y #a
Hδ
Uδ
( ).218
1
HMaEI +−
(or /C
( ) ( )∫
−×−−=
4
0
2
2
""152
EI
dx H x Ax
Hδ
Uδ
( )HA 1441920482
1++−
∴ ∑ =0HδUδ
H0L+a G O3 > NL 4A G 0P3- G 0NN -
V H a G 03 G 0-K.KO > 3.O 4A -
4a
)# x
"02
30 4=
V H5 G 03 G 0-K.KO > 3.O /K-1 -
V P MM.55
K.0M Q
∴ +A 5 K.0M
0L.NMQ > m
∴ +a 0L.NM QHm
4a K-Q
K.0M Q
$imilarly
+D 0L.NM QHm
4D K-Q
(in!ing point o# contra#le,%re
4a 0M3 > K -
K- 0M3 > K /K.0M1 -
-
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V K- > 0M3 > 5K.P -
V N > n3 > 3.N5 -
V3 > N G 3.NK
( ) ( ) ( )
12
4".2141"4
×−±−−
2
48.24+
-.OK taQing >#e sign from B
/on!ing 5oment #or Spa1 3/
+ +a > y
%hen - at A.
+ a
0L.NMQHm.
+ +a > y
0L.NM > K.0M/K1 when y Km
H0L.NM QHm at b.
(or Spa1 /C
+ 4a 0M3 > K
-
+ HK
H5K.PQHm at T
+ 4A > 0M3 > K%hen Nm.
+ 4A > 0M3 > K
K- N > 0M/N13 > K /K.0M1
/3N- > 3N- > 5K.P1 Q > m
5K.P Q > m at 6.
(or /C Span
%hen n 3m+ 4A > 0M3 H K
-
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K- 3 > 0M N > K /K.0M1
03- > K- > 5K.P
35.0 QHm.
/en!ing moment #or CD.
%hen y -
'hen + 0L.NM Q > m at D
%hen y Km, + H0L.NM QHm at 6.
ta6ing moment ao%t C
4B N > 5K.P > 5K.P > 3N- -
4B OL.NMQ
46 03-HOL.NM N0.MMQ
-
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'aking moment about A
H K G 5K.P > 0L.NM -
5.-OM.
Ta6ing moment ao%t D.
H K > 0L.NM G 5K.P -
V 5.-OM Q
-
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