Structural Analysis 2
Lecture 4:
INFLUENCE LINES OF STATICALLY INDETERMINATE STRUCTURES BY
THE FORCE METHOD
Maxwell's theorem of reciprocal
displacements
The method is based on Maxwell's law of interchangeability which is often used in mechanics:
the interchangeability of internal force and external displacement.
◦ Betti’s law: by virtual force theorem
◦ 𝑊
=𝑊
,
, ,
+𝑊
=0
, ,
◦ 𝑊 , = 𝑊 , , +𝑊 , , = 0
◦ Internal complementary work by definition :
𝑊
, ,
=𝑊
, ,
=∫
𝑑𝑠 + ∫
𝑑𝑠 + ∑
ℓ thus 𝑊
, ,
=𝑊
, ,
◦ Maxwell’s law : the force-system P is only a single force and 𝑝 = 1, and similarly Q is a single force and
𝑞=1
◦ 𝑊
◦ 𝑒
, ,
= 𝑝𝑒
=𝑊
, ,
= 𝑞𝑒
=𝑒
The type of movement corresponding to the tested internal force is inserted into the examined
point and the vertical displacement diagram of the support is drawn.
Maxwell's theorem of reciprocal
displacements
Betti’s law:
Maxwell’s law:
Influence line diagram of a
displacement:
𝑥
𝑃
𝑄
𝑊
, ,
=𝑊
, ,
𝑝=1
𝑢
𝑞=1
𝑢
𝑢
=𝑢
𝑃
𝑄
𝑢
𝑝=1
𝑞=1
𝜂 𝑢
𝑃
𝑄
𝑢
= 𝑒
,
Influence line diagram of a displacement
What are we looking for?
What do we use?
The influence line diagram of the vertical
displacement at the cross-section K:
𝜂 𝑒
A unit vertical force at the cross-section
K: 𝐹 = 1
The influence line diagram of the
horizontal displacement at the crosssection K: 𝜂 𝑒
A unit horizontal force at the crosssection K: 𝐹 = 1
The influence line diagram of the
rotation at the cross-section K: 𝜂 𝜙
A unit positive moment at the crosssection K: 𝑀 = 1
The influence line diagram of the relative
displacement of points A and B: 𝜂 𝑢
A par of unit forces on the lines of
influence passing through the two points
: 𝐹 = ±1
The influence line diagram of the relative
rotation of joint C: 𝜂 𝜃
A par of unit moments on both sides of
the hinge: 𝑀 = ±1
What do we count?
How?
Diagram of the
vertical
displacement
(deflection) of the
structure
Small
displacements,
moment load
method, virtual
force theorem, or a
combination of
these
Force method – influence lines
Statically indeterminate structures
Force method – influence lines
As a result of the moving unit force, the load factors change continuously.
So, we need to define the effect diagrams of the load factors in order to get the influence line
diagrams of the connection forces.
The system of geometric (compatibility) conditions/equations based on principles of
superpositioning:
◦ ∑
𝑎 𝜂 𝑋 +𝜂 𝑎
= 0 , 𝑖 = 1. . 𝑛.
◦ Here the unknowns are the influence line diagrams 𝜂 𝑋 of the connection/redundant forces 𝑋 .
◦ The 𝑎 unit factors and the coefficient matrix formed from them are the same as for a fixed/unmoving
load.
◦ The effect diagrams of the load factors 𝜂 𝑎 represent the influence line diagrams of the
displacements occurring at the location of the redundant forces.
Statically indeterminate structures
Force method – influence lines
The influence line diagrams of the redundant forces are given by the solution of the conditional
equation system:
◦ 𝜂 𝑋 = −∑
𝑧 𝜂 𝑎
, 𝑖 = 1. . 𝑛.
◦ 𝑧 means the impact factors.
With the action diagrams
, any reaction force, internal force or displacement can be
determined at any cross section of the structure:
◦ 𝜂 𝐶 =𝜂 𝐶
+∑ 𝐶 𝜂 𝑋 .
◦ 𝜂 𝐶
is the influence line of a mechanical property of a cross-section K of the base structure,
◦ 𝐶 the volume of a mechanical property from the applied unit redundant force at the position/case of
j.
◦ 𝐶 may be replaced by any mechanical feature : 𝑇 , 𝑀 , 𝑆 , 𝑒 , 𝜑 , 𝐵, …
Force method – influence lines
Steps
1st step: primary/base structure determination.
The considerations set out for stationary loads are applied.
2nd step: preparation of the flexibility coefficient matrix.
(Just like with a stationary load.)
3rd step: preparation of influence line diagrams of load factors 𝜼 𝒂𝒊𝟎 .
◦ The load factors are the displacements at the location of the cuts.
◦ Based on Maxwell's interchangeability theorem, their effect diagrams are identical to the vertical displacement diagram
created by a (virtual) unit force work-compatible with the displacement : 𝜂 𝑎 = 𝜂 𝑒 .
(𝜂 𝑎 = 𝑒 figure from 𝑋 = 1 .)
4th step: determination of the influence line diagrams of the redundant forces 𝜼 𝑿𝒋 .
◦ This step requires determining the inverse of the coefficient matrix.
◦ The elements of this are the so-called 𝑧 factors. The resulting 𝜂 𝑋 effect diagrams are curved line diagrams.
5th step: the preparation of an influence line diagram determining a mechanical feature 𝜼 𝑪𝑲 .
◦ We perform the principle of superposition in calculation for each ordinate. (Just like in case of a standing load.)
◦ The resulting figures are curved figures, that also show the characteristics of the figures of statically determinate structures:
unit jump in the shear force diagram, v-shaped fracture in the moment diagram, locations of zero points, etc..
Illustrative examples
E1: structure indeterminate to the first degree
𝜂 𝐶
𝑓
𝜂 𝑋
=−
𝜂 𝑎
𝑎
=𝜂 𝐶
+
𝐶 𝜂 𝑋
E2: structure indeterminate to the 2DOF
Inverse of a 2x2 matrix:
𝑎 𝑏
𝑑
=
𝑐 𝑑
−𝑐
−𝑏
𝑎
E2: structure indeterminate to the 2DOF
𝜂 𝐶
=𝜂 𝐶
+
𝐶 𝜂 𝑋
Example
E3: branched holder indeterminate to
the first degree
Determine the moment influence line diagram of the cross-sections 3 and 6 of the
beam,
and
, and the reaction force influence line diagram
at point
B!
kNm2,
1
,
2
𝜌
3
4
m/kN.
5
6
7
𝐸𝐼
𝐸𝐼
A
C
𝐸𝐼
B
16 m
8m
6m
E3: branched holder indeterminate to
the first degree
1st step: Free body diagram, primary/base structure determination.
1
2
3
5
6
7
𝐸𝐼
𝐸𝐼
A:
4
𝑋 =1
𝐶=2
𝐸𝐼
𝐵=3
16 m
8m
6m
E3: branched holder indeterminate to
the first degree
2nd step: preparation of a matrix containing the unit factors .
m3
Moment diagram from unit force 𝑀 :
1
A:
2
3 𝐸𝐼
𝑋 =1
4
5
6𝐸𝐼
𝐶=2
16 kNm
𝐸𝐼
𝐵=3
16 m
7
8m
6m
E3: branched holder indeterminate to
the first degree
3rd step: preparation of influence lines of load factors
𝒊𝟎
.
Based on Maxwell's interchangeability theorem, this is a diagram of the vertical displacements
caused by a unit load..
7
5 𝜗 6
1
2
3 𝜗 4
The moment figure from the unit load 𝑀 :
𝜗 =
𝜗 =
= 64 m2
= 128 m2
4
8
12
𝜑 = 𝜗 = 42,67 m2
8
16 kNm
B
𝜑 = 𝜗 = 21,33 m2
1
The vertical displacement figure from the unit
load 𝑓 = 𝜂 𝑎 :
C
2
3
4
5
6
𝜑 =𝜑
B
7
E3: branched holder indeterminate to
the first degree
4th step: creating the influence lines of the contact forces 𝜼 𝑿𝒋 .
Calculation of 𝒇𝒊𝒚 = 𝜼 𝒂𝟏𝟎
i
𝜼 𝒂𝟏𝟎
𝜼 𝑿𝟏 = −
𝜼 𝒂𝟏𝟎
𝒂𝟏𝟏
1
−42,67 16 − 128 16 2/3
-2048
0,911
2
−42,67 12 − 4 12 6 − 12 /2 8
-1376
0,612
3
−42,67 8 − 8 8 4 − 8 /2 8 2/3
-768
0,342
4
−42,67 4 − 12 4 2 − 4 /2 4 2/3
-288
0,128
5
0
0
0
6
+21,33 4 − 8 4/2 4/3
64
-0,028
7
0
0
0
1
2
3
4
5
6
7
5
6
7
𝜼 𝒂𝟏𝟎
1
2
3
4
𝜼 𝑿𝟏
E3: branched holder indeterminate to
the first degree
5th step: preparation of an influence line diagram determining a mechanical feature
.
◦ So three concerning influence line diagram must be created: 𝜂 𝑀 , 𝜂 𝑀
and 𝜂 𝐵 .
◦ The tree multipliers/weighting values per unit load must be determined : 𝑀 , 𝑀 and 𝐵 .
◦ Then the influence line diagrams must be prepared according to the formula.
𝑲
.
E3: branched holder indeterminate to
the first degree
𝜂 𝐶
1
2
3
4
5
6
1
-4
3
2
2
5
6
7
𝜂 𝑀
3
1
4
4
5
𝜂 𝑀
6
+
𝐶 𝜂 𝑋
7
𝜂 𝑋
-8
=𝜂 𝐶
7
𝜂 𝑀
𝜼 𝑿𝟏
𝑀
-8
0,911
-4
0,612
-0,896
0
0,342
2,736
0
0,128
1,024
0
0
0
0
-0,028
-0,224
0
0
0
+8
𝜂 𝑀
-0,712
E3: branched holder indeterminate to
the first degree
𝜂 𝐶
1
2
3
4
5
6
1
-6
2
-4
3
-2
5
4
𝜂 𝑀
5
1
2
3
4
𝜂 𝑀
6
7
2
6
+
𝐶 𝜂 𝑋
7
𝜂 𝑋
-8
=𝜂 𝐶
7
𝜂 𝑀𝟔
𝜼 𝑿𝟏
𝑀𝟔
𝜂 𝑀
-8
0,911
+8
-6
0,612
-1,404
-4
0,342
-1,264
-2
0,128
-0,976
0
0
0
2
-0,028
1,776
0
0
0
-0,712
E3: branched holder indeterminate to
the first degree
𝜂 𝐶
1
2
3
4
5
6
3
1
2
2,5
2
𝜂 𝑩
3
4
5
6
2
1,5
1
0,5
3
4
7
𝜂 𝑴𝟑
5
𝜂 𝑴𝟑
6
+
𝐶 𝜂 𝑋
7
𝜂 𝑋
1
=𝜂 𝐶
7
𝜼 𝑿𝟏
𝑀
-3
𝜂 𝐵
3
0,911
0,267
2,5
0,612
0,664
2
0,342
0,974
1,5
0,128
1,116
1
0
1
0,5
-0,028
0,584
0
0
0
Literature
In Hungarian:
◦ Kurutzné Kovács Márta: Tartók statikája, Műegyetemi kiadó 2006.
◦ Pásztor Erzsébet, Tamássy Tamás: Tartók statikája példatár I., Tankönyvkiadó Budapest 1992. J9-1275.
(source of illustrations on a yellow background) offered!
◦ BME, PTE MIK órai jegyzetek
◦ Gáspár Zsolt, Tarnai Tibor: Statika, egyetemi jegyzet, Műegyetemi Kiadó, Budapest 2006.
In English:
R.C. Hibbeler: Structural Analysis, 9th edition, chapter 10 offered!