MIT - 16.20 Fall, 2002 Unit 21 Influence Coefficients Readings : Rivell o 6.6, 6.13 (again), 10.5 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 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 Paul A. Lagace ? 2001 Unit 21 - 2 ? ? MIT - 16.20 Fall, 2002 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 1 1 Q 1 + C 1 2 Q 2 + C 1 3 Q 3 q i = C i j Q j q 2 = C 2 1 Q 1 + C 2 2 Q 2 + C 2 3 Q 3 q 3 = C 3 1 Q 1 + C 3 2 Q 2 + C 3 3 Q 3 or in Matrix Notation: ? 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 ? Paul A. Lagace ? 2001 Unit 21 - 3 MIT - 16.20 Fall, 2002 Note : | | --> row { } --> column [ ] --> full matrix or {} = [] {} q i C i j 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) Paul A. Lagace ? 2001 Unit 21 - 4 MIT - 16.20 Fall, 2002 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 Paul A. Lagace ? 2001 Unit 21 - 5 ? ? ? ? MIT - 16.20 Fall, 2002 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 Beam The deflections q 1 …q 5 can be characterized by: ? q 1 ? ? C 11 C 12 C 13 C 14 C 15 ? ? Q 1 ? ? ? ? q 2 ? ? ? C 21 C 22 LL M ? ? Q 2 ? ? ? ? ? ? ? ? ? q 3 ? = ? M O M ? ? Q 3 ? ? q 4 ? ?? M O M ? ? Q 4 ? ? ? ? ? ? ? q 5 ? ?? C 51 L L L C 55 ? ? ? Q 5 ? Paul A. Lagace ? 2001 Unit 21 - 6 MIT - 16.20 Fall, 2002 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 () Paul A. Lagace ? 2001 Unit 21 - 7 MIT - 16.20 Fall, 2002 What is M(x)? --> First find reactions: Figure 21.6 Free body diagram to determine reactions in cantilevered beam 1 ? M = -1x j V = 1 --> Now find M(x). Cut beam short of x j : Figure 21.7 Free body diagram to determine moment along cantilevered beam Paul A. Lagace ? 2001 Unit 21 - 8 MIT - 16.20 Fall, 2002 ∑ M x = 0 + ? 1 ? 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 E I constant: dw 1 ? x 2 ? = ? ? xx ? ? + C 1 dx E I ? j 2 ? 3 1 ? x 2 x ? w = ? ? x j 2 ? ? + Cx + C 2 EI ? 6 ? 1 Boundary Conditions: @ x = 0 w = 0 ? C 2 = 0 dw @ x = 0 = 0 ? C 1 = 0 dx Paul A. Lagace ? 2001 Unit 21 - 9 MIT - 16.20 Fall, 2002 So: 3 1 ? x 2 x ? w = ? EI ?? x j 2 ? 6 ? ? evaluate at x i : 1 ? x i 3 2 ? w = ? xx j ?? 2 EI ? ? 3 i One important note: w is defined as positive up, have defined q i as positive down. So: 3 2 q i = ? w = 1 ? ? xx j ? x i ?? 2 EI ? i 3 ? ? C EI xx x ij i i = ? ? ? ? ? ? ? 1 2 2 3 j 6 for x i ≤ x j Deflection, q i , at x i due to unit force, Q j , at x j Paul A. Lagace ? 2001 Unit 21 - 1 0 MIT - 16.20 Fall, 2002 --> 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 vi a … Stiffness Influence Coefficients ≡ k ij where can write: Q 1 = k 1 1 q 1 + k 1 2 q 2 + k 1 3 q 3 Q 2 = k 2 1 q 1 + k 2 2 q 2 + k 2 3 q 3 Q 3 = k 3 1 q 1 + k 3 2 q 2 + k 3 3 q 3 Paul A. Lagace ? 2001 Unit 21 - 1 1 MIT - 16.20 or write: {} = [] {} Q i k i j q j or : Q = k q ~~ ~ If compare this with: q = C Q ~~ ~ ? k = C -1 ~~ Fall, 2002 [] ? 1 k ij [] = C ij inverse matrix Note : Had a similar situation in the continuum case: E = S -1 ~~ elasticity compliance (stiffness) (flexibility) --> Look at the Physical Interpretations: Paul A. Lagace ? 2001 Unit 21 - 1 2 MIT - 16.20 Fall, 2002 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 Paul A. Lagace ? 2001 Unit 21 - 1 3 MIT - 16.20 Fall, 2002 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 Paul A. Lagace ? 2001 Unit 21 - 1 4 MIT - 16.20 Fall, 2002 --> Can find k ij by: ? calculating [ C ij ] first, then inverting + () ij minor of ? 1 k = ij ? 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 Paul A. Lagace ? 2001 Unit 21 - 1 5