analysis of statically indeterminate...
Post on 04Jun2018
226 views
Embed Size (px)
TRANSCRIPT
Analysis of Statically Indeterminate Structures
Force Method of Analysis
Chapter 10
Methods of AnalysisTwo different Methods are available
Force method known as consistent deformation, unit load method,
flexibility method, or the superposition equations method.
The primary unknowns in this way of analysis are forces
Displacement method Known as stiffness method
The primary unknowns are displacements
2
Force method of analysisThe deflection or slope at any point on a structure as a result
of a number of forces, including the reactions, is equal to the algebraic sum of the deflections or slopes at this particular point as a result of these loads acting individually
3
Force Method of Analysis General Procedure
Indeterminate to the first degree
1 Compatibility equation is needed
Choosing one of the support reaction as a redundant
The structure become statically determinate and stable
Downward displacement B at B calculated (load action)
BB upward deflection per unit force at B
Compatibility equation
0 = B + ByBB Reaction By known
Now the structure is statically determinate
4
Force Method of Analysis General Procedure
Indeterminate to the first degree
1 Compatibility equation is needed
Choosing MA at A as a redundant
The structure become statically determinate & stable
Rotation A at A caused by load P is determined
AA rotation caused by unit couple moment applied at A
Compatibility equation
0 = A + MA AA Moment MA known
Now the structure is statically determinate
5
Force Method of Analysis General Procedure
Indeterminate to the 2nd degree
2 Compatibility equations needed
Redundant reaction B & C
Displacement B &C caused by load P1 & P2 are determined
6
Force Method of Analysis
General Procedure BB & BC Deflection per unit force
at B are determined
CC & CB Deflection per unit force at C are determined
Compatibility equations
0 = B + ByBB + CyBC0 = C + ByCB + CyCC
Reactions at B & C are known
Statically determinate structure
7
Maxwells Theorem The displacement of a point B on a
structure due to a unit load acting at a point C is equal to the displacement of point C when the unit load is acting at point B the is
fBC=fCB
The rotation of a point B on a structure due to a unit moment acting at a point C is equal to the rotation of point C when the unit moment is acting at point B the is
BC=CB
Force Method of Analysis Procedure for Analysis
Determine the degree of statically indeterminacy
Identify the redundants, whether its a force or a moment, that would be treated as unknown in order to form the structure statically determinate & stable
Calculate the displacements of the determinate structure at the points where the redundants have been removed
Calculate the displacements at these same points in the determinate structure due to the unit force or moment of each redundants individually
Workout the compatibility equation at each point where there is a redundant & solve for the unknown redundants
Knowing the value of the redundants, use equilibrium to determine the remaining reactions
9
Beam Deflections and Slopes
3 2
3 2
max max
36
3 2
Px Lx
EI
PL PLatx L
EI EI
20
2
0 0max max
2
2
Mx
EI
M L M Latx L
EI EI
Positive (+)
Beam Deflections and Slopes
4 3 2 2
4 3
max max
4 624
8 6
wx Lx L x
EI
wL wLatx L
EI EI
3 2
3
max
2
max
4 3 , 048 2
2 48
0 or 2 16
P Lv x L x x
EI
L PLatx
EI
L PLatx
EI
Positive (+)
Beam Deflections and Slopes
3 2 3
4
max
3
max
224
5
2 384
24
wxv x Lx L
EI
L wLatx
EI
wL
EI
2 2 2 , 06
6 6L R
Pbxv L b x x a
LEI
Pab L b Pab L a
LEI LEI
Positive (+)
Beam Deflections and Slopes
3 2 3
3 2 2 3
3 3
9 24 16 02384
8 24 17384
2
3 7
128 384L R
wx LL Lx x xEI
wLx Lx L x L
EI
L x L
wL wL
EI EI
2 20
2
0max
0 0
3 26
9 2
6 3L R
M xv x Lx L
EIL
M L
EI
M L M L
EI EI
Positive (+)
Example 1 Determine the reaction at B
Indeterminate to the 1st degree thus one additional equation needed
Lets take B as a redundant
Determine the deflection at point B in the absence of support B. Using the momentarea method
Determine the deflection caused by the unit load at point B
39000 .B
kN m
EI
14
506=
300kN.m
B.M.D
BA
50kN
1 6 2300 . 6 6
2 3B
EI
Example 1
Compatibility equation
0 = B + ByBB
By = 15.6kN
The reaction at B is known now so the structure is statically determinate & equilibrium equations can be applied to get the rest of the unknowns
3576BB
mf
EI
15
1 12 212 12
2 3BBf
EI
112=
12m
B.M.D
BA
19000 576
0 yBEI EI
B.M.D
Example 2 Determine the moment at A
Indeterminate to the 1st degree thus one additional equation needed
Lets take MA as a redundant
Determine the slope A at point A ignoring the fixation at A. Using the momentarea method
1
1 10 10 333.3320
2 3d
EI EI
A
20k.ft
B.M.D
B1 1 333.33.
10A
d
L EI
233.33 .A
k ft
EI
B
A Ad1
Elastic Curve
Example 2 Determine rotation caused by the unit
moment applied at A
Compatibility equation
0 = A +MAAA
MA = 10k.ft
The moment at A is known now so the structure is statically determinate
2
1 10 2 10 33.331
2 3d
EI EI
17
33.33 3.330 AM
EI EI
B
A AAd2
Elastic Curve
1 B.M.D
BA
12 1 33.33 3.33.10
AA
d ft
L EI EI
Example 3
Example 3
Example 3
Example 3
Neglect the axial load
The end moments at A&B will be considered as redundants
From Table inside the front cover
1(375)A
EI
1(291.7)B
EI
Example 4

3.33BA
EI
13.33
20(2/3)10
6.67
6.67AA
EI
6.67BB
EI
3.33AB
EI
375 6.67 3.330 A BM M
EI EI EI
291.7 3.33 6.670 A BM M
EI EI EI
45.8 .AM k ft 20.8 .BM k ft
Example 4
24Example 4
Example 5Determine the reaction at the support for the beam shown, EI is Constant.Choose the internal moment at internal support as the redundant
Compatibility Equations
Example 5
Example 5
Example 6Determine the support reactions on the frame shown EI is Constant.
Example 6
Example 6
Example (Additional)The frame, shown in the photo is used to support the bridge deck.
Assuming EI is constant, a drawing of it along with the dimensions and
loading is shown. Determine the support reactions.
Example 7Determine the moment at the fixed support A for the frame shown EI is
Constant.
Example 7
Example 7
Example 7
Problem 1
Problem 2
Problem 3
Force Method of Analysis: TrussExample8
Determine the force in member AC.
Assume EA is the same for all the members
44
Example 8
Force Method of Analysis: Truss
46
Example 9
Determine the force in each member if the turnbuckle on member AC is used to shorten the member by 0.5in. Each member has a crosssection area of 0.2 in2 & E=29(106)psi
47
Example 9
Example 9
Example 10The beam shown is supported by a pin at A and two pinconnected bar at B.
Determine the force in member BD. Take E=29(103), I=800 in4 for the beam
and A=3 in2 for each bar.
Example 10
Example 10
Example 11The simply supported beam shown in the photo is to be designed to support a
uniform load of 2 kN/m. Determine the force developed in member CE. Neglect
the thickness of the beam and assume the truss members are pin connected to the
beam. Also, neglect the effect of axial compression and shear in the beam. The
crosssectional area of each strut is 400 mm2, and for the beam I=20(106)mm4 .
Take E=200 GPa
Example 11
Example 11
Example 11
Example 11
Example 11
Symmetric Structures
Antisymmetric Structures
Transformation of Loading
Influence Lines for Statically Indeterminate Beams
Reaction at A
1
Scale Factor 1
y DA
AA
AA
A ff
f
Reaction at A
Influence Lines for Statically Indeterminate Beams
1
Scale Factor 1
E DE
EE
EE
V ff
f
Shear at E
Influence Lines for Statically Indeterminate Beams
Moment at E
1
Scale Factor 1
E DE
EE
EE
M f
Influence Lines for Statically Indeterminate Beams
Example 11
Example 11
Example 11
Example 11
Influence Line of a continuous Beam
4m 8m
A B C
G D E F
Draw the Influence Line of1. Reaction at A, B and C2. Shear at G and E3. Moment at G and E
Influence Line of RA
Influence Line of RA
1
3
1 1
2
3
2 2
0 8
5.33
12
0 4
18.67 64
6
xA
xA
x
x xf
EI EI
x
x xf
EI EI EI
4m 8mA
B C
D
16/EI8/EI 4/EI
5.33/EI
x1y=x1/2x2
y=x2
18.67/EI
64/EI
Influence Line of RA
1
3
1 1
2
3
2 2
0 8
5.33
12
0 4
18.67 64
6
xA
xA
x
x xf
EI EI
x
x xf
EI EI EI
Point fxA fxA/fAA
A 64/EI 1.0
G 28/EI 0.4375
B 0 0
D 14/EI 0.2188
E 16/EI 0.25
F 10/EI 0.1562
C 0 0
Influence Line of RA
0
8 12 1 0
3
2 8
c
B A
AB
M
R R x
R xR
Point x RA RB
A 12 1.0 0
G 10 0.4375 0.5939
B 8 0 1
D 6 0.2188 1.078
E 4 0.25 0.875
F 2 0.1562 0.485
C 0 0 0
Influence Line of RBUsing equilibrium conditions for the influence line of RB
0
4 8 1 8 0
18 2
B
A C
AC
M
R R x
x RR
Point x RA RC
A 12 1.0 0
G 10 0.4375 0.0312
B 8 0 0
D 6 0.2188 0.1406
E 4 0.25 0.375
F 2 0.1562 0.6719
C 0 0 1
Influence Line of RCUsing equilibrium conditions for the influence line of RC
1
Check
0y
A B C
F
R R R
Influence Line of VG
Influence Line of VE
Influence Line of MG
Influence Line of ME
Influence Line of ME
Live Load Pattern in Continuous Beams
LL
DL
Support Reactions
Influence Line for positive reaction at support 1
Load pattern for maximum positive reaction at support 1
Span Positive Moment
Influence line for positive moment at 7
Load pattern for maximum positive moment at 7
Support Negative Moment
Influence line for negative moment at support 2
Load pattern for maximum negative moment at support 2
Internal Shear
Influence line for positive shear at 7
Load pattern for maximum positive shear at 7
Moment EnvelopesThe moment envelope curve defines the extreme boundary values of
bending moment along the beam due to critical placements of design
live loading.
Example of Three Span Continuous Beam
Moment Envelopes
Dead Load Only
BMD due to DL
Moment Envelopes
Live Load Arrangement
Max. Positive Moment at span 1&3(Min. Moment at span 2)
Max. Positive Moment at span 2
(Min. Moment at span 1 & 3)
Moment Envelopes
Live Load Arrangement
Max. negative Moment at support 2
Max. negative Moment at support 3
Moment Envelopes
Live Load Arrangement
Max. positive Moment at support 2
Max. positive Moment at support 3
Moment Envelopes
Max. +M at span 2Max. +M at spans 1&3
DL only
Max. M at support 2 Max. M at support 3
Max. +M at support 2 Max. +M at support 3
Summarize all critical cases
Moment EnvelopesMoment Diagram for All Cases
Solving each of these cases and combining the results with the dead
only case, results in six different moment diagrams.
The dead load only case has been left as a heavier line.
Moment EnvelopesMoment Diagram for All Cases