1. The document provides examples of constructing influence lines for statically determinate beams and trusses. It defines influence lines and shows how to determine the influence line for reactions, shear, and bending moment at various points.
2. Example problems are worked out step-by-step to show how to construct influence lines for a simple beam and a beam with a hinge support. The influence lines provide the response of the structure due to a moving unit load.
3. Equilibrium equations are also used to determine influence lines by relating reactions, shears and moments. General expressions for shear and moment are developed for a beam with multiple spans.
Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
Aircraft Structures for Engineering Students 5th Edition Megson Solutions ManualRigeler
Full donwload : http://alibabadownload.com/product/aircraft-structures-for-engineering-students-5th-edition-megson-solutions-manual/ Aircraft Structures for Engineering Students 5th Edition Megson Solutions Manual
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Este ejercicio consiste en la resolución de una estructura articulada o truss en inglés, utilizando las ecuaciones del equilibrio y el método de los nodos
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
Este ejercicio consiste en la resolución de una estructura articulada o truss en inglés, utilizando las ecuaciones del equilibrio y el método de los nodos
Diseno en ingenieria mecanica de Shigley - 8th ---HDes
descarga el contenido completo de aqui http://paralafakyoumecanismos.blogspot.com.ar/2014/08/libro-para-mecanismos-y-elementos-de.html
check it out: http://goo.gl/vqNk7m
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Analysis of indeterminate beam by slopeand deflection methodnawalesantosh35
Slope-deflection method ,Slope-deflection equations, equilibrium equation of
method, application to beams with and without joint translation and rotation, Sinking or yielding of support,
ANALYSIS OF CONTINUOUS BEAM USING STIFFNESS METHODkasirekha
THIS METHOD IS USEFUL TO ANALYZE THE INDETERMINATE STRUCTURES USING MATRIX APPROACH. THIS METHOD IS ALSO CALLED DISPLACEMENT METHOD AS IT DEALS WITH UNKNOWN ROTATIONS.
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Influence linebeams (ce 311)
1. 1
! Influence Lines for Beams
! Influence Lines for Floor Girders
! Influence Lines for Trusses
! Maximum Influence at a Point Due to a Series
of Concentrated Loads
! Absolute Maximum Shear and Moment
INFLUENCE LINES FOR STATICALLY
DETERMINATE STRUCTURES
3. 3
Example 6-1
Construct the influence line for
a) reaction at A and B
b) shear at point C
c) bending moment at point C
d) shear before and after support B
e) moment at point B
of the beam in the figure below.
B
A
C
4 m 4 m 4 m
4. 4
SOLUTION
• Reaction at A
+ ΣMB = 0: ,0)8(1)8( =−+− xAy xAy
8
1
1−=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 m
Ay By
1
x
0
4
8
12
x
1
0.5
0
-0.5
Ay
5. 5
A
C
4 m
Ay By
8 m
• Reaction at B
+ ΣMA = 0: ,01)8( =− xBy xBy
8
1
=
1
x
4 m 8 m 12 m
By
x
1.5
1
0.5
0
4
8
12
x
0
0.5
1
1.5
By
6. 6
• Shear at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
01
8
1
1 =−−− CVx
xVC
8
1
−=
ΣFy = 0:+
0
8
1
1 =−− CVx
xVC
8
1
1−=
ΣFy = 0:+
VC
MC
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
8. 8
• Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
VC
MC
0)4)(
8
1
1()4(1 =−−−+ xxMC
xMC
2
1
=
VC
MC
0)4)(
8
1
1( =−− xMC
+ ΣMC = 0:
+ ΣMC = 0:
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
xMC
2
1
4 −=
10. 10
Or using equilibrium conditions:
• Reaction at A
+ ΣMB = 0: ,0)8(1)8( =−+− xAy xAy
8
1
1−=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 m
Ay By
1
x
11. 11
A
C
4 m
Ay By
8 m
• Reaction at B
1
x
4 m 8 m 12 m
By
x
1.5
1
0.5
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
01=−+ yy BA
yy AB −=1
ΣFy = 0:+
yy AB −=1
12. 12
• Shear at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
01 =−− Cy VA
1−= yC AV
ΣFy = 0:+
0=− Cy VA
yC AV =
ΣFy = 0:+
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
VC
MC
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
13. 13
yC AV =1−= yC AV
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-0.5
VC
8 m 12 m
x
4 m
0.5
-0.5
B
A
C
4 m 4 m 4 m
14. 14
• Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40 <≤ x 124 ≤< x
0)4(1)4( =+−+ Cy MxA
)4(4 xAM yC −−=
+ ΣMC = 0:
VC
MC
xAy
8
1
1−=
A
Cx
4 m
124 ≤< x
VC
MC
A
C
1
x
4 m
40 ≤≤ x
xAy
8
1
1−=
0)4( =+− Cy MA+ ΣMC = 0:
yC AM 4=
15. 15
yC AM 4=)4(4 xAM yC −−=
2
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
4 m
8 m 12 m
MC
x
-2
B
A
C
4 m 4 m 4 m
16. 16
• Shear before support B
A
C
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-0.5
VB
- = AyVB
- = Ay-1
1
Ay
8 m
VB
-
MB
x
Ay
8 m
VB
-
MB
VB-
x
-1.0-0.5
17. 17
• Shear after support B
A
C
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
VB+
x
1
VB
+ = 0
4 m
VB
+
MB
1
VB
+ = 1
4 m
VB
+
MB
18. 18
• Moment at support B
A
C
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-4
MB
x
1
MB = 8Ay-(8-x) MB = 8Ay
1
Ay
8 m
VB
-
MB
x
Ay
8 m
VB
-
MB
19. 19
P = 1
'x
• Reaction
Influence Line for Beam
C
A B
P = 1
Ay By
δy = 1 'yδ
LL
s
y
B
1
==
δ
C
A B
L
0)0()(1)1( ' =+− yyy BA δ
'yyA δ=
20. 20
δy = 1
'yδC
A B
P = 1
Ay ByLL
s
y
A
1
==
δ
C
A B
L
P = 1
'x
'yyB δ=
0)1()(1)0( ' =+− yyy BA δ
23. 23
• Shear
CA B
P = 1
a b
L
L
sB
1
=
δy=1
δyL
δyR
'yδ
A B
VC
VC
P = 1
Ay By
δy=1
L
sA
1
=
0)0()(1)()()0( ' =+−++ yyyRCyLCy BVVA δδδ
')( yyRyLCV δδδ =+
'yCV δ=
BA ssslopes =:
26. 26
1=+= BA θθφ
• Bending Moment
a b
L
a
h
A =θ
'yδ
b
h
B =θ
1
A B
MC
MC
P = 1
Ay By
h
CA B
P = 1
0)0()(1)()()0( ' =++++ yyBCACy BMMA δθθ
')( yBACM δθθ =+
'yCM δ=
1)( =+
b
h
a
h
)(
,1
)(
ba
ab
h
ab
bah
+
==
+
31. 31
• General Bending Moment
θA = 3/4 θB = 1/4
φ = sA + sB = 1
θA = 1/2 θB = 1/2
φ = sA + sB = 1
θA = 1/4 θB = 3/4
φ = θA + θB = 1
x
MC 3L/16
x
MD
4L/16
x
ME 3L/16
A
C D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
33. 33
Example 6-2
Construct the influence line for
- the reaction at A, C and E
- the shear at D
- the moment at D
- shear before and after support C
- moment at point C
A B C D E
2 m 2 m 2 m 4 m
Hinge
36. 36
A B C D
E
2 m 2 m 2 m 4 m
RE
RE
x
-2/6
2/6
1
37. 37
A B C D E
2 m 2 m 2 m 4 m
VD
VD
VD
x
1
2/6
-1
1
• sE = sC
sE = 1/6
sC = 1/6
=
-2/6
=
4/6
38. 38
Or using equilibrium conditions:
VD = 1 -RE
1
VD
x
2/6
-2/6
4/6
RE
VD
MD
4 m
1
x
VD = -RE
RE
VD
MD
4 m
RE
x
-2/6
1
2/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
39. 39
A B C
D
E
2 m 2 m 2 m 4 m
4
-1.33
MD
x
(2)(4)/6 = 1.33
φD = θC+θE = 1
θC = 4/6
2
2/6 = θE
MD MD
40. 40
Or using equilibrium conditions:
MD = -(4-x)+4RE
1
RE
VD
MD
4 m
1
x
MD = 4RE
RE
VD
MD
4 m
RE
x
-2/6
1
2/6
MD
x
-8/6
8/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
44. 44
A B C D E
2 m 2 m 2 m 4 m
Or using equilibrium conditions:
1
VCR
x
0.333
1
0.667
RE
x
-2/6 = -0.333
1
2/6=0.33
VCR = -RE
RE
VCR
MC
VCR = 1 -RE
RE
VCR
MC
1
46. 46
A B C D E
2 m 2 m 2 m 4 m
MC
x
1
-2
Or using equilibrium conditions:
1
RE
x
-2/6 = -0.333
1
2/6=0.33
MC = 6RE
RE
VCR
MC
6 m
'6 xRM AC −=
6 m
'x
RE
VCR
MC
1
47. 47
Example 6-3
Construct the influence line for
- the reaction at A and C
- shear at D, E and F
- the moment at D, E and F
Hinge
A B CD E F
2 m 2 m 2 m 2 m2 m 2 m
49. 49
2 m 2 m 2 m 2 m
A B
CD E
2 m 2 m
F
RC x
RC
1
0.5
1.5
2
50. 50
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
VD
VD
VD
x
-1
0.5
-0.5
1 1
=
=
51. 51
A B CD E
2 m 2 m 2 m 2 m2 m 2 m
F
VE
VE
VE
x
1
-0.5
-1
0.5
-0.5
=
=
52. 52
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
VF
VF
VF
x
1
=
=
53. 53
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
MD
MD
MD
x
-2
θD = 1
-1
1
2
54. 54
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
φE = 1
ME
ME
ME
x
θC = 0.5θB = 0.5
(2)(2)/4 = 1
-2
-1
55. 55
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
ME
ME
MF
x
θF = 1
-2
56. 56
Example 6-4
Determine the maximum reaction at support B, the maximum shear at point C and
the maximum positive moment that can be developed
at point C on the beam shown due to
- a single concentrate live load of 8000 N
- a uniform live load of 3000 N/m
- a beam weight (dead load) of 1000 N/m
4 m 4 m 4 m
A BC