MT-1620 al.2002 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 Paul A Lagace @2001
MIT - 16.20 Fall, 2002 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 Paul A. Lagace © 2001
MT-1620 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 马 刁 Paul A Lagace @2001 Unit 21-2
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
MT-1620 al.2002 q= generalized displacement (linear or rotation Q generalized force(force or moment/torque Note that Q; and g;are at the same point have the same sense (i.e. direction of the same type force← displacement) ( moment e> rotation) For a linear, elastic body, superposition applies, so can write q1=C12+ C2 22+ C123 q= ci e Q C22 22+23 03 十C23 q C1Q1+(3 C32Q2+C3Q3 or in matrix notation q2 32 Paul A Lagace @2001 Unit 21-3
MIT - 16.20 Fall, 2002 qi = generalized displacement (linear or rotation) Qi = generalized force (force or moment/torque) Note that Qi and qi 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 = C11 Q1 + C12 Q2 + C13 Q3 qi = Cij Qj q 2 = C21 Q1 + C22 Q2 + C23 Q3 q 3 = C31 Q1 + C32 Q2 + C33 Q3 or in Matrix Notation: q 1 C11 C12 C13 Q1 q 2 = C21 C22 C23 Q2 q 3 C31 C32 C33 Q3 Paul A. Lagace © 2001 Unit 21 - 3
MT-1620 al.2002 Note:l」->roW 3-> column []--> full matrix q g=cQ Ci= Flexibility Influence Coefficient and it gives the deflection at i due to a unit load at M12=is deflection at 1 due to force at 2 Figure 21.2 Representation of deflection point 1 due to load at point 2 苏(Noe: Ci can mix types) Paul A Lagace @2001 Unit 21-4
MIT - 16.20 Fall, 2002 Note: | | --> row { } --> column [ ] --> full matrix or { } q i = [ ] Cij { } Qj or q = C Q ~ ~ ~ Cij = Flexibility Influence Coefficient and it gives the deflection at i due to a unit load at j C12 = is deflection at 1 due to force at 2 Figure 21.2 Representation of deflection point 1 due to load at point 2 (Note: Cij can mix types) Paul A. Lagace © 2001 Unit 21 - 4
MT-1620 al.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 q1 due to unit load at 2 is equal to q 2 due to unit load at 1 Generally C:=C symmetric Paul A Lagace @2001 Unit 21-5
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 q1 due to unit load at 2 is equal to q2 due to unit load at 1 i.e. C12 = C21 Generally: Cij = Cji symmetric Paul A. Lagace © 2001 Unit 21 - 5
MT-1620 al.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 QQQ 2 Beam The deflections qr.. 5 can be characterized by q「CnC12CBC4Csg C22 3 Paul A Lagace @2001 Unit 21-6
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 q1…q5 can be characterized by: q 1 C11 C12 C13 C14 C15 Q1 q 2 C21 C22 LL M Q2 q 3 = M O M Q3 q 4 M O M Q4 q 5 C51 L L L C55 Q5 Paul A. Lagace © 2001 Unit 21 - 6
MT-1620 al.2002 Since ci cin the [cirl 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 Q1.Q5, one can easily compute the q1..5 from previous work Example: Ci for a Cantilevered Beam igure 21.5 Representation of cantilevered beam under load 己 1unit X find: C:--> deflection at i due to unit load at Most efficient way to do this is via Principle of Virtual Work (energy technique Resort here to using simple beam theory E M dx Paul A Lagace @2001 Unit 21-7
MIT - 16.20 Fall, 2002 Since Cij = Cji, the [Cij] 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 Q1….Q5, one can easily compute the q1….q5 from previous work… Example: Cij for a Cantilevered Beam Figure 21.5 Representation of cantilevered beam under load find: Cij --> 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 E I dw = Mx dx2 ( ) Paul A. Lagace © 2001 Unit 21 - 7
MT-1620 al.2002 What is M(x)? > First find reactions Figure 21.6 Free body diagram to determine reactions in cantilevered beam V=1 M=-1X: X Now find M(x). Cut beam short of Figure 21.7 Free body diagram to determine moment along cantilevered beam 1 S(x) Paul A Lagace @2001 Unit 21-8
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 = -1xj V = 1 --> Now find M(x). Cut beam short of xj: Figure 21.7 Free body diagram to determine moment along cantilevered beam Paul A. Lagace © 2001 Unit 21 - 8
MT-1620 al.2002 ∑M=0+ 1·x1-1x+M(x)=0 →M Plugging into deflection equation E d x for ei constant dw E E Cx+ C2 Boundary Conditions @ x=0w=0→C1=0 d w @x=0 0→C1=0 Paul A Lagace @2001 Unit 21-9
MIT - 16.20 Fall, 2002 ∑ Mx = 0 + ⇒ 1⋅ x − 1⋅ x + M x j ( ) = 0 ⇒ M x( ) = −1 ( xj − x ) Plugging into deflection equation: 2 E I dw = −1 ( xj − x ) dx2 for E I constant: dw 1 x 2 = − xx − + C1 dx E I j 2 3 1 x 2 x w = − xj 2 − + Cx + C2 EI 6 1 Boundary Conditions: @ x = 0 w = 0 ⇒ C2 = 0 d w @ x = 0 = 0 ⇒ C1 = 0 dx Paul A. Lagace © 2001 Unit 21 - 9
MT-1620 al.2002 So El evaluate at x W 2EⅠ3 x One important note w is defined as positive up, have defined q; as positive down So q =-W XX 2EⅠ X: x forx≤ EⅠ2 Deflection, q at x due to unit force, Q atxi Paul A Lagace @2001 Unit 21-10
MIT - 16.20 Fall, 2002 So: 3 1 x2 x w = − EI xj 2 − 6 evaluate at xi: 1 xi 3 2 w = − x xj 2 EI 3 i One important note: w is defined as positive up, have defined qi as positive down. So: 3 2 qi = − w = 1 x xj − xi 2 EI i 3 ⇒ C EI xx x ij i i = − 1 2 2 3 j 6 for xi ≤ xj Deflection, qi, at xi due to unit force, Qj, at xj Paul A. Lagace © 2001 Unit 21 - 10
