moment area theorems: theorem 1moment area theorems: the vertical distance between any point on the...
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1
Moment Area Theorems:
When a beam is subjected toexternal loading, it under goesdeformation. Then the intersectionangle between tangents drawn atany two points on the elastic curveis given by the area of bendingmoment diagram divided by itsflexural rigidity.
Theorem 1:
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2
Moment Area Theorems:
The vertical distance between anypoint on the elastic curve andintersection of a vertical linethrough that point and tangentdrawn at some other point on theelastic curve is given by themoment of area of bendingmoment diagram between twopoints taken about first pointdivided by flexural rigidity.
Theorem 2:
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Fixed end moment due to a point load at the mid span:
(1) ----4
WL - MM
0EI
L2
MML
4WL
21
BA
BA
AB
>=+∴
=
++
××=θ
)2(WL83
M2M
EI
2L
L4
WL21
L32
LM21
L31
LM21
AA
BA
BA1
>−−−−−=+
××+
××+
××=
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Both moments are negative andhence they produce hoggingbending moment.
8WL
8WL
4WL
M 4
WLM
8WL
M
get we(2) and (1) From
BA
B
−=
−−−=−−=∴
−=
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Stiffness coefficients
a) When far end is simply supported
EI
MomentBB
Babout B &A between BMD of area of '=
EI3ML
EI
L 32
L M 21
2
=
×
××=
EI3ML
LBB 2
A I =θ=∴
A LEI3
M θ=∴
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b) When far end is fixed
EI
LM21
- LM21
EI
BMD of area
BA
A
==θ
( ) ( )1 -------- 2
M - B >=
EI
LM A
( ) 2--------- 2M M
0EI
32
LM21
3L
LM21
EI Aabout B & A between BMD of Areathe of Moment
AA
BA
BA
1
>+=∴
==
−
=
=
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Substituting in (1)
( )
( )
AB
ABB
ABA
L
2EI M
EI2
LM - M2
EI2
LMM
θ=∴
θ=+
θ=−
ABA L
4EI M 2 M )2(From θ==
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Fixed end moments due to yielding of support.
( )
M M Hence
0)MM(.ie
L2
MM
0EI
B and A between BMD of area
BA
BA
BA
AB
−=
=+−
+−=
==θ
( )
+=
×+××−=
==−
2
1
L 6
2 -
32
21
321
EI
Babout B andA b/n BMD of area ofMoment
EI
MM
EI
LLML
LM
BB
AB
AB
δ
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( )
δ
δ
26
BM hogging Hence 2
6
6
2
26
2-
L
EIMMNow
L
EIM
EI
LM
MMSinceLEI
MM
AB
A
A
BAAA
−=−=
=∴
−=
−=
+−=
Hence sagging BM
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Fixed end moment for various types of loading
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Assumptions made in slope deflection method:
1) All joints of the frame are rigid
2) Distortions due to axial loads, shear stresses being smallare neglected.
3) When beams or frames are deflected the rigid joints areconsidered to rotate as a whole.
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Sign conventions:
Moments: All the clockwise moments at the ends of membersare taken as positive.
Rotations: Clockwise rotations of a tangent drawn on to anelastic curve at any joint is taken as positive.
Sinking of support: When right support sinks with respect toleft support, the end moments will be anticlockwise and aretaken as negative.
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Development of Slope Deflection Equation
Span AB after deformation
Effect of loading
Effect of rotation at A
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Effect of rotation at B
Effect of yielding of support B
−++=−++=
−++=−++=
LL
EIF
L
EI
L
EI
L
EIFMSimilarly
LL
EIF
L
EI
L
EI
L
EIFM
BABAABBABA
BAABBAABAB
δθθδθθ
δθθδθθ
32
2624
32
22
624Hence
2
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Slope Deflection Equations
δ−θ+θ+=
δ−θ+θ+=
L3
2LEI2
FM
L3
2LEI2
FM
BABABA
BAABAB
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Example: Analyze the propped cantilever shown by using slope
deflection method. Then draw Bending moment and shear force
diagram.
Solution:
12wL
F,12wL
F2
BA
2
AB +=−=
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Slope deflection equations
( )
)1(LEI2
12wL
2LEI2
FM
B
2
BAABAB
→θ+−=
θ+θ+=
( )
)2(LEI4
12wL
2LEI2
FM
B
2
ABBABA
→θ+=
θ+θ+=
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Boundary condition at BMBA=0
0LEI4
12wL
M B
2
BA =θ+= 48wL
EI3
B −=θ∴
Substituting in equations (1) and (2)
8wL
48wL
L2
12wL
M232
AB −=
−+−=
048wL
L4
12wL
M32
BA =
−++=
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Free body diagram
wL85
RA =0V =∑
wL85
wLRwLR AB −=−=
wL83=
2L
wL8
wLLR
2
A ×+=×0MB =∑
BR
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0wXwL83
SX =+−=
L83
X =∴
2max wL
1289
M =∴L
83
L83
2wL128
9
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Example: Analyze two span continuous beam ABC byslope deflection method. Then draw Bending moment &Shear force diagram. Take EI constant
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Solution:
KNM44.446
24100L
WabF
2
2
2
2
AB −=××−=−=
KNM89.886
24100L
bWaF
2
2
2
2
BA +=××+=+=
KNM67.4112
52012wL
F22
BC −=×−=−=
KNM67.4112
52012wL
F22
CB =×=+=
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Slope deflection equations
( )BAABAB 2LEI2
FM θ+θ+=
B6EI2
44.44 θ+−=
)1(EI31
44.44 B →−−θ+−=
( )ABBABA 2LEI2
FM θ+θ+=
62EI2
89.88 Bθ×++=
)2(EI32
89.88 B →−−−−θ+=
( )CBBCBC 2LEI2
FM θ+θ+=
( )CB25EI2
67.41 θ+θ+−=
)3(EI52
EI54
67.41 CB →θ+θ+−=
( )BCCBCB 2LEI2
FM θ+θ+=
( )BC25EI2
67.41 θ+θ++=
)4(EI52
5EI4
67.41 BC →θ+θ+=
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Boundary conditions
MBA+MBC=0
ii. MCB=0
CBBBCBA EI52
EI54
67.41EI32
89.88MM θ+θ+−θ+=+
)5(0EI52
EI1522
22.47 CB >−−−−=θ+θ+=
)6(0EI54
EI52
67.41M CBCB >−−−−−=θ+θ+=
83.20EI B −=θ 67.41EI C −=θSolving
i. -MBA-MBC=0
Now
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( ) KNM38.5183.2031
44.44 – MAB −=−+=
( ) KNM00.7583.2032
88.89 MBA +=−++=
( ) ( ) KNM00.7567.4152
83.2054
41.67 – MBC −=−+−+=
( ) ( ) 067.4154
83.2052
41.67 MCB =−+−++=
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Free body diagram
Span BC:
Span AB:
75+ 2
5×5×20 = 5×R 0 = M BCΣ
KN 65 = R B∴100KN = 5×20 = R+R 0=V CBΣ
KN 35 = 65-100 = RC
51.38-75+4×100 = ×6R 0 = M BAΣKN 70.60 = RB∴
100KN = +RR0 =V BAΣKN 29.40=70.60-100 = R A∴
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BM and SF diagram
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Example: Analyze continuous beam ABCD by slopedeflection method and then draw bending moment diagram.Take EI constant.
Solution:
M KN 44.44 - 6
24100
LWab
F2
2
2
2
AB =××−=−=
KNM 88.88 6
24100
LbWa
F2
2
2
2
BA +=××+=+=
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KNM 41.67- 12
52012wL
F22
BC =×−=−=
KNM 41.67 12
52012wL
F22
CB +=×+=+=
M KN 30 - 5.120FCD =×−=
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Slope deflection equations:
( ) ( )1--------- EI31
44.442LEI2
FM BBAABAB >θ+−=θ+θ+=
( ) ( )2--------- EI32
89.882LEI2
FM BABBABA >θ++=θ+θ+=
( ) ( )3-------- EI52
EI54
67.412LEI2
FM CBCBBCBC >θ+θ+−=θ+θ+=
( ) ( )4-------- EI52
EI54
67.412LEI2
FM BCBCCBCB >θ+θ++=θ+θ+=
KNM 30MCD −=
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Boundary conditions 0MM BCBA =+
0MM CDCB =+
CBBBCBA EI52
EI54
67.41EI32
89.88MM,Now θ+θ+−θ+=+
( )5-------- 0EI52
EI1522
22.47 CB >=θ+θ+=
30EI52
EI54
67.41MM,And BCCDCB −θ+θ++=+
( )6EI54
EI52
67.11 CB >−−−−−−θ+θ+=Solving
67.32EI B −=θ 75.1EI C +=θ
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Substituting( ) KNM 00.6167.32
21
44.44MAB −=−+−=
( ) KNM 11.6767.3232
89.88MBA +=−++=
( ) ( ) KNM 11.6775.152
67.3254
67.41MBC −=+=−+−=
( ) ( ) KNM 00.3067.3252
75.154
67.41MCB +=−+++=
KNM 30MCD −=
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Example: Analyse the continuous beam ABCD shown in figureby slope deflection method. The support B sinks by 15mm.Take 4625 m10120Iandm/KN10200E −×=×=
Solution:
KNM44.44L
WabF
2
2
AB −=−=
KNM89.88L
bWaF
2
2
BA +=+=
KNM67.418
wLF
2
BC −=−=
KNM67.418
wLF
2
CB +=+=
KNM305.120FCD −=×−=
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FEM due to yielding of support B
For span AB:
δ−==2baab LEI6
mm KNM61000
151012010
62006 652
−=×××××−= −
δ+==2cbbc LEI6
mm KNM64.81000
151012010
52006 65
2+=×××××+= −
For span BC:
![Page 38: Moment Area Theorems: Theorem 1Moment Area Theorems: The vertical distance between any point on the elastic curve and intersection of a vertical line through that point and tangent](https://reader030.vdocuments.us/reader030/viewer/2022040202/5e6d9daf20f5b750f062c2b9/html5/thumbnails/38.jpg)
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Slope deflection equation
( )
( )1---------- EI31
44.50
6EI31
44.44- LEI6
2LEI
FM
B
B2BAABAB
>θ+−=
−θ+=δ−θ+θ+=
( )2------------ EI32
89.82
6EI32
88.89 LEI6
)2(LEI2
FM
B
B2ABBABA
>θ++=
−θ++=δ−θ+θ+=
( )
( )3--------- EI 52
EI54
03.33
64.82EI52
41.67- LEI6
)2(LEI2
FM
CB
CB2CBBCBC
>θ+θ+−=
+θ+θ+=δ+θ+θ+=
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( )
( )4--------- EI 52
EI54
31.50
64.82EI52
41.67 LEI6
)2(LEI2
FM
BC
BC2BCCBCB
>θ+θ++=
+θ+θ++=δ+θ+θ+=
( )5--------- KNM 30MCD >−=
Boundary conditions
0MM
0MM
CDCB
BCBA
=+=+
0EI54
EI52
31.20MM
0EI52
EI1522
86.49MM
CBCDCB
CBBCBA
=θ+θ+=+
=θ+θ+=+Now
Solving 35.31EI B −=θ 71.9EI C −=θ
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Final moments
( ) KNM 89.6035.3131
44.50MAB −=−+−=
( ) KNM 99.6135.3132
89.82MBA +=−++=
( ) ( ) KNM 99.6171.952
35.3154
03.33MBC −=−+−+−=
( ) ( ) KNM 00.3035.3152
71.954
31.50MCB +=−+−++=
KNM 30MCD −=
![Page 41: Moment Area Theorems: Theorem 1Moment Area Theorems: The vertical distance between any point on the elastic curve and intersection of a vertical line through that point and tangent](https://reader030.vdocuments.us/reader030/viewer/2022040202/5e6d9daf20f5b750f062c2b9/html5/thumbnails/41.jpg)
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