Unit 21 Influence Coefficients

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1 Unit 21 Influence Coefficients Readings: Rivello 6.6, 6.13 (again), 10.5 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems

2 Have considered the vibrational behavior of a discrete system. How does one use this for a continuous structure? First need the concept of.. Influence Coefficients which tell how a force/displacement at a particular point influences a displacement/force at another point --> useful in matrix methods finite element method lumped mass model (will use this in next unit) --> consider an arbitrary elastic body and define: Figure 21.1 Representation of general forces on an arbitrary elastic body Unit 21-2

3 q i = generalized displacement (linear or rotation) Q i = generalized force (force or moment/torque) Note that Q i and q i are: at the same point have the same sense (i.e. direction) of the same type (force displacement) (moment rotation) For a linear, elastic body, superposition applies, so can write: q 1 = C 11 Q 1 + C 12 Q 2 + C 13 Q 3 q i = C ij Q j q 2 = C 21 Q 1 + C 22 Q 2 + C 23 Q 3 or in Matrix Notation: q 3 = C 31 Q 1 + C 32 Q 2 + C 33 Q 3 q 1 C 11 C 12 C 13 Q 1 q 2 = C 21 C 22 C 23 Q 2 q 3 C 31 C 32 C 33 Q 3 Unit 21-3

4 Note: --> row { } --> column [ ] --> full matrix or { q }= [ ] i C ij { Q j } or q = C Q ~~ ~ C ij = Flexibility Influence Coefficient and it gives the deflection at i due to a unit load at j C 12 = is deflection at 1 due to force at 2 Figure 21.2 Representation of deflection point 1 due to load at point 2 (Note: C ij can mix types) Unit 21-4

5 Very important theorem: Maxwell s Theorem of Reciprocal Deflection (Maxwell s Reciprocity Theorem) Figure 21.3 Representation of loads and deflections at two points on an elastic body q 1 due to unit load at 2 is equal to q 2 due to unit load at 1 i.e. C 12 = C 21 Generally: C ij = C ji symmetric Unit 21-5

6 This can be proven by energy considering (path independency of work) --> Application of Flexibility Influence Coefficients Look at a beam and consider 5 points Figure 21.4 Representation of beam with loads at five points q 1 C 11 C 12 C 13 C 14 C 15 Q 1 q 2 C 21 C 22 L L M Q 2 q 3 = M O M Q 3 q 4 M O M Q4 q 5 C 51 L L L C 55 Q 5 Beam The deflections q 1 q 5 can be characterized by: Unit 21-6

7 Since C ij = C ji, the [C ij ] matrix is symmetric Thus, although there are 25 elements to the C matrix in this case, only 15 need to be computed. So, for the different loads Q 1.Q 5, one can easily compute the q 1.q 5 from previous work Example: C ij for a Cantilevered Beam Figure 21.5 Representation of cantilevered beam under load find: C ij --> deflection at i due to unit load at j Most efficient way to do this is via Principle of Virtual Work (energy technique) Resort here to using simple beam theory: 2 EI dw = Mx dx 2 ( ) Unit 21-7

8 What is M(x)? --> First find reactions: Figure 21.6 Free body diagram to determine reactions in cantilevered beam 1 V = 1 M = -1x j --> Now find M(x). Cut beam short of x j : Figure 21.7 Free body diagram to determine moment along cantilevered beam Unit 21-8

9 M x = x 1 x + M x j ( ) = 0 Mx ( ) = 1 (x j x) Plugging into deflection equation: 2 EI dw = 1 (x j x) dx 2 for EI constant: dw 1 x 2 = x x + C 1 dx E I j x 2 x w = EI x j 2 Boundary Conditions: + Cx 1 + C 2 x = 0 w = 0 C 2 = 0 x = 0 = 0 C 1 = 0 dx Unit 21-9

10 So: 3 1 x 2 x w = EI x j 2 evaluate at x i : 6 w = 3 1 x i x x j 2 EI i 3 2 One important note: w is defined as positive up, have defined q i as positive down. So: q i = w = x 2 xj 3 1 x i 2 EI i 3 C ij xi x j = EI 2 xi 6 for x i x j Deflection, q i, at x i due to unit force, Q j, at x j Unit 21-10

11 --> What about for x i x j? Does one need to go through this whole procedure again? No! Can use the same formulation due to the symmetry of C ij (C ij = C ji ) --> Thus far have looked at the influence of a force on a displacement. May want to look at the opposite : the influence of a displacement on a force. Do this via Stiffness Influence Coefficients k ij where can write: Q 1 = k 11 q 1 + k 12 q 2 + k 13 q 3 Q 2 = k 21 q 1 + k 22 q 2 + k 23 q 3 Q 3 = k 31 q 1 + k 32 q 2 + k 33 q 3 Unit 21-11

12 MIT or write: { Q }= [ ] i k ij { q j } or: Q = k q ~ ~ ~ If compare this with: q = C Q ~~ ~ k = C -1 ~ ~ [ k ] [ ] 1 ij = C ij inverse matrix Note: Had a similar situation in the continuum case: E = S -1 ~ ~ elasticity (stiffness) compliance (flexibility) Fall, > Look at the Physical Interpretations: Unit 21-12

13 Flexibility Influence Coefficients Figure 21.8 Physical representation of flexibility influence coefficients for cantilevered beam 1 unit level C ij = displacement at i due to unit load at j Note: This is only defined for sufficiently constrained structure Unit 21-13

14 Stiffness Influence Coefficients Figure 21.9 Physical representation of stiffness influence coefficients for cantilevered beam 1 unit displacement k ij = forces at i s to give a unit displacement at j and zero displacement everywhere else (at nodes) (much harder to think of than C ij ) Note: This can be defined for unconstrained structures Unit 21-14

15 --> Can find k ij by: calculating [C ij ] first, then inverting k = ij + ( 1) i j minor of C ij determinant of [C ] ij Note: k -1 may be singular (indicates rigid body modes) --> rotation --> translation --> etc. calculating [k ij ] directly from individual local [k ij ] elements and adding up for the total system Most convenient way Note: This latter method is the basis for finite element methods Unit 21-15

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