Force method - vsb.cz
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Force method
Transcript of Force method - vsb.cz
Force method
Example 1
F = 50 kN
4
a b
2
E·I = const .E·I = const .
Example 1
F = 50 kNDegree of statical indeterminacy
=−−+=−−=
03)13(
3 ks pan
1
4
a b
2
=−−+= 03)13( 1
⇓⇓⇓⇓
Statical & deformationalequilibrium conditions
Have to be respected
Example 1 – release of redundant constraint
F = 50 kNMa
4
a b
2Ra
Rb
Ha
Example 1 – release of redundant constraint
F = 50 kNMa Basic statically determinate structure
4
a b
2Ra
Rb
Ha
Example 1 – release of redundant constraint
F = 50 kNMa
Statically indeterminate structure
4
a b
2Ra
Rb
Ha
M M0. loading state 1. loading state
a b
Ra
Ha
Ma F = 50 kN
a b
Ra
Ha
Ma
X1=Rb
1 1
Example 1 – deformational condition
F = 50 kN
Statically indeterminate structure
a b
Wb = 0
0. loading state 1. loading state
a
F = 50 kN
a
X1=1
δδδδ10
δδδδ11
1 1
F = 50 kN011110 =⋅+ Xδδ
Example 1 – deformational condition
Statically indeterminate structure
a b
Wb = 0
011110 =⋅+ Xδδ
0. loading state 1. loading state
a 1
F = 50 kN
a
X1=1
δδδδ10
δδδδ11
1
Example 1 – deformation δδδδ10
F = 50 kNReal loading Virtual unit loading
0. loading state
1a a
X1 = 1
1
Example 1 – deformation δδδδ10
F = 50 kN
0. loading state
Real loading Virtual unit loading
1a 1a
M0 M1
-200
X1 = 1
M0 M1
6
Example 1 – deformation δδδδ111. loading state
Real loading Virtual unit loading
1a 1a
M1 M1
X1 = 1X1=1
M1 M1
66
Example 1 – internal forces
F = 50 kNMa
4
a b
2Ra
Rb = 25,926 kN
Ha
Example 1 – internal forces
F = 50 kNMa= 44,44
4
a b
2Ra
Rb = 25,926 kN
Ha
24.074 -44.444
51.852-25.926
Example 2
a b
q = 10 kNm -1
6 3
E·I = const.
Example 2
a b
q = 10 kNm -1
6 3
Basic statically determinate structure
Example 2 – deformational condition
a b
q = 10 kNm -1
Statically indeterminate structureWb = 0
6 3
a 1δδδδ10
q = 10 kNm -1
0. loading state
a 1X1=1
δδδδ11
1. loading state
Example 2 – deformation δδδδ10
Real loading Virtual unit loading
q = 10 kNm -1
0. loading state
X1 = 1
a 1 a 1
Example 2 – deformation δδδδ10
q = 10 kNm -1
0. loading state
Real loading Virtual unit loading
X1 = 1
M0 M1
-225
-45
a 1 a 1
M0 M1
6
Example 2 – deformation δδδδ111. loading state
Real loading Virtual unit loading
X1 = 1
M1 M1
X1=1
a 1 a 1
M1 M1
66
Example 2 – internal forces
a b
q = 10 kNm -1
Ha
Ma
6 3
Rb = 41,25 kNRa
Ha
Example 2 – internal forces
a b
q = 10 kNm -1
Ha
Ma= 22,5
6 3
Rb = 41,25 kN
30.000
-45.000Ra
Ha
-11.250
22.500
