MAE 323: Lecture 1. Review

Transcription

1 This review is divided into two parts. The first part is a mini-review of statics and solid mechanics. The second part is a review of matrix/vector fundamentals. The first part is given as an refresher to aid the student in validating problems later on, because a large part of this course concerns validation of finite element solutions to structural problems. B validation, we mean that the analst should check various solution values against manual calculation or tabular look-up data whenever possible The second part is given because man students will have onl a passing familiarit with matrices and how to manipulate and calculate things with them. As this course contains a component on finite element formulation and numerical implementation, this is essential. The finite element method involves matrices and this is inescapable Alex Grishin MAE 323 Lecture 1 1

2 Statics and Basic Solid Mechanics 2011 Alex Grishin MAE 323 Lecture 1 2

3 Q: What does it mean to perform a structural analsis? A: In the most general terms, it means that we subject a geometric domain (the structure) to an external force (b external, we mean onl that it is prescribed). We then calculate the structural response to this load. The primar variable that we calculate is displacement (including an rotations). After that, we calculate stresses and strains as derived from these displacements. In the simplest possible terms, we might characterize the situation something like: Step 1: F = K u prescribed calculated calculated = C ε Step 2: calculated prescribed calculated When we validate a structural solution, we will tr to estimate some of the calculated values using other methods as a check 2011 Alex Grishin MAE 323 Lecture 1 3

4 The other methods will tpicall involve statics or simple solid mechanics (especiall small-deflection beam theor otherwise known as Euler-Bernoulli beam theor). 6m Example 1: 4m 40 kn 2m 10 kn A x B R A R B Problem: Find the force reactions R A and R B of the pinned-pinned beam 2011 Alex Grishin MAE 323 Lecture 1 4

5 Solution: Use Newton s Third Law to sum the moments AND forces 2m 4m 10 kn 6m 40 kn A x B R A R B Summing Forces: Sum moments about point A: R A F x = 0 F = R + R = 0 A B + R = 50 B (1) Now plug (2) back into (1) to wrap it up Ax Az 6R = 180 R B M M B = 30 = M = 0 A = R 6 + (40 4) + (10 2) = 0 B (2) 2011 Alex Grishin MAE 323 Lecture 1 5

6 Finall: R R A A F x = 0 F = R + R = 0 A B + 30 = 50 = 20 So, the answer is: R R A B = 20 = 30 It doesn t matter which point we use to sum moments. We could just as easil have chosen point B: Bx Bz 6R = 120 R A M M A = 20 = M = 0 B = R 6 + ( 40 2) + ( 10 4) = 0 A 2011 Alex Grishin MAE 323 Lecture 1 6

7 We could have chosen to sum moments about both points A and B. In that case, we wouldn t have needed to sum forces. Summing moments and forces is a ver useful tool in checking that a structure is in equilibrium. If our calculated force and moment reactions agree with those calculated b the software, then we can be confident that the model is properl constrained Summing moments and forces works even in more general cases (not just beams). In those cases, it helps to make use of their vector form: N R 0 N + F = 0 i= 1 M = R + r F = 0 i M 0 i i i= 1 i= 1 i N Force Sum Law Moment Sum Law In this form of the moment sum law, moments are calculated from a particular reference frame and r i is the vector distance from the origin to the force F i 2011 Alex Grishin MAE 323 Lecture 1 7

8 Example 2: Three beams connected endto-end, oriented along each axis z O x 4m 3m 1.5m 4 kn First the forces 3 kn R F = 0 i= 1 i Rx R + 4 kn + 0 kn = 0 0 R z So 0 ˆ ˆ ˆ R0 = 0i + 4j 3k = 4 kn 3 from now on, we ll write vectors in column format 2011 Alex Grishin MAE 323 Lecture 1 8

9 Example 2 (cont.): z O x 4m 3m 1.5m 4 kn N Now, the moments M = R + r F = 0 i M 0 i i i= 1 i= = RM kn N i= 1 M Recall the cross products can be evaluated using determinants i i j k i j k 0 = R M = 0 knm Alex Grishin MAE 323 Lecture 1 9

10 Once we ve determined force and moment reactions of a structure, we can use some simple formulae to estimate the resulting stresses A-A A-A T P Case 1: Axial Load Stress: = Distribution: P A Case 2: Torsion Load Tr Stress: τ =, τ max = J Distribution: Tr J max 2011 Alex Grishin MAE 323 Lecture 1 10

11 A-A A-A k = 4 / 3 P P k = 2 k = 3 / 2 Case 3: Bending Stress: = Distribution: M I Case 4: Transverse Shear Stress: PQ τ =, τ max = k Ib Distribution: P A 2011 Alex Grishin MAE 323 Lecture 1 11

12 We should also note common conventions regarding the sign of these stresses. In general, the sign of a stress or strain component does not depend simpl on a coordinate reference, but the behavior of the material over which it acts* So, for a beam in bending we have: Compressive stress Neutral axis x +M Tensile stress *See J. Gere and S. Timoshenko, Mechanics of Materials, 3rd ed. Boston: PWS-KENT Publishing Co., Alex Grishin MAE 323 Lecture 1 12

13 For pure (constant) torsion, Both positive and negative shears exist for a given constant external torque, but the will have the same maximum absolute values -τ +τ -τ +τ +T -τ -τ +τ +τ And, for pure tension/compression, we have: + -P +P + Reversing the sign of the load reverses the sign of the stress IN ALL CASES 2011 Alex Grishin MAE 323 Lecture 1 13

14 Coordinate Conventions for Bending Stress The following common coordinate convention ma be used to determine the signs of bending stresses + M XY plane + M xmax z x = x M zz I zz z xmin 2011 Alex Grishin MAE 323 Lecture 1 14

15 -M XY plane -M xmin z x = x M zz I zz z xmax 2011 Alex Grishin MAE 323 Lecture 1 15

16 To see how these formulae can be used to estimate stresses, consider the previous example. Suppose that this structure has a constant circular cross-section of radius 0.1m z O A-A x 4m R=0.1m 3m 1.5m 4 kn A-A 3 kn Let s use the formulae to calculate torsional shear and bending stresses at O It just so happens that in this case, we can calculate everthing we need from the moment vector we calculated earlier 2011 Alex Grishin MAE 323 Lecture 1 16

17 The applied moment at O was: M = knm The corresponding moment reaction is: 4.5 R M 0 = 12 knm 16 z O x 4m 3m 1.5m 4 kn Since the moment component in the x direction acts along the structure length at O, this moment results in a torsional shear stress. The other two moments result in bending: 4.5 Torsion M0 = 12 Bending 16 Bending 3 kn 2011 Alex Grishin MAE 323 Lecture 1 17

18 We use the applied moment loads to estimate stresses: τ Max. torsional stress at O: Tr = = = kPa J π (0.1) / 2 max max 4 Max. bending stress at O (xz plane): M z sin(tan Max. bending stress at O (x plane): (3/ 4)) 1 max = = = 4 I π ( 0.1) / 4 M cos(tan (3/ 4)) = 4 π / 4 z max = = I zz ( 0.1) z O x 4m 3m 1.5m 4 kn 3 kn kpa kpa 2011 Alex Grishin MAE 323 Lecture 1 18

19 But we haven t finished. The bending stresses in the XZ and XY planes are in the same direction and orientation meaning the should be added if we want the total overall maximum bending stress for the structure. Thus, max( ) = max( _ ) + max( _ ) = ( ) bending bending x bending z kpa Notice we re using the formula*: max( bending ) = kpa max = M z + I M I z z *see Alex Grishin 19 MAE 323 Lecture 1

20 But this formula onl works if the product of inertia I z is negligible or zero (as happens when the cross section is smmetric) so that should be checked before appling this formula. Also, how did we know what signs to use? Simpl use the moment sign convention mentioned on slides 14 and 15. This sign convention is revealed in the expression: M = EI 2 d v 2 dx This form of the differential equation of a beam can be read as stating: The moment per unit length is negativel proportional to the curvature Thus, in the previous problem, the bending moment in XY plane is positive (even though the sign of the applied moment vector is negative), and we expect the maximum stress to occur at =+r. In the XZ plane, the bending moment is negative (agreeing with the moment vector) and we expect the maximum stress to occur at z=-r 2011 Alex Grishin MAE 323 Lecture 1 20

21 Of course, because of the radial smmetr of this cross-section, we could have used a shortcut: * M * z max( bending ) = I Neutral axis * z max z * Where M * is the vector resultant moment, z* is the distance to the max. stress location (which is simpl rin our case), and I * is the corresponding second moment of inertia (which won t change because of smmetr). Notice that this is effectivel a rotational transformation * M * * * max 4 I * π Compressive stress Tensile stress 2011 Alex Grishin MAE 323 Lecture M M z = = = kpa (0.1) / 4

22 Questions for Statics and Basic Solid Mechanics: In example 2, we estimated the maximum bending and pure (torsional) shear stress. Are these the onl stresses that exist for that problem? How would ou calculate the maximum transverse shear stress in example 2? 2011 Alex Grishin MAE 323 Lecture 1 22

23 Questions for Statics and Basic Solid Mechanics (cont.): What is the sign of the stress/strain deformation implied below? -P +P τ τ +T τ τ 2011 Alex Grishin MAE 323 Lecture 1 23

24 Vectors and Arras 2011 Alex Grishin MAE 323 Lecture 1 24

25 Students in this course will alread have some familiarit with vectors and arras, but we re going to review some things AND add some nuances with which man students are probabl unfamiliar In this course, unless I specif otherwise, vectors AND arras will be represented as bold smbols to refer to all their components at once, while individual components will not. And (unless I sa otherwise) vectors will be represented in equations as single-column arras When I write them down in long-hand, vectors will have a bar beneath them while matrices will have a double-bar beneath them. Thus the reaction force vector of example 2 can be written in R ˆi ˆj kˆ 0 0 = R0 = = Alex Grishin MAE 323 Lecture 1 25

26 And the individual components can be written: R R R 0x 0 0z = 0 = 4 = 3 We have alread been careful to follow this convention when writing the stress equations. Consider, for example, the equation for bending stress: = M I Stress is a tensor (an arra. And so the 2 nd moment of inertia) and moment is a vector, so wh aren t the bold in this equation? 2011 Alex Grishin MAE 323 Lecture 1 26

27 The reason is that the equations of slides 9 and 10 are written on a component basis! This is exactl wh we had to find the root mean square (the vector magnitude) for the maximum bending stress due to the compound loading of example 2 (it s wh we needed an additional step that wasn t quite obvious) The Transpose We will find the notion of a vector and matrix transpose useful in writing vector equations. A transpose is simpl an interchange of rows and columns, turning column arras into row arras and vice-versa. We will denote the transpose of an arra with a capital T superscript: ax T a = a, a = a a a a z { } x z 2011 Alex Grishin MAE 323 Lecture 1 27

28 The Dot (or Inner) Product vs The Outer Product Continuing on: Because it is understood that vectors are represented as single-column (as opposed to single-row) arras, an expression like: T a b Will be understood to mean the inner product: Instead of: bx ax a az b = axbx + ab + azbz = a b b { } z ax axbx axb azbz a b b b = a b a b a b = = { } x z x z a z azbx azb azb z Inner Product T a b ab Outer Product Which represents an outer product (both are useful constructs) 2011 Alex Grishin MAE 323 Lecture 1 28

29 Dot Products of Arras Some ambiguit ma still arise, however, when we take dot products of multi-column arras. For example, suppose I have a stress tensor (represented b an arra) xx x xz = x z zx z zz and I want to multipl it b another arra (mabe to transform it): R R R R xx x xz = Rx R R z R R R zx z zz 2011 Alex Grishin MAE 323 Lecture 1 29

30 Dot Products of Arras This results in the familiar matrix product: R + R + R R + R + R R + R + R xx xx x x xz zx xx x x xz z xx xz x z xz zz = Rx xx + R x + Rz zx Rx x + R + Rz z Rx xz + R z + Rz zz R Rzx xx Rz x Rzz zx Rzx x Rz Rzz z Rzx xz Rz z Rzz zz However, this should not be confused with the matrix double-dot product: R = R + R + R : xx xx x x x x + R + R + R xz xz zx xz + R + R + R z z z z zz zz Note that this product produces a scalar value 2011 Alex Grishin MAE 323 Lecture 1 30

31 Index Notation One wa to avoid the ambiguit between different kinds of matrix products is to use index (or indicial) notation. With this notation, row and column indices are explicitl numbered (as are the coordinate axes to which the ma refer),2 ax a1 a = a = a2 a z a x,1 3 z,3 xx x xz = x z = xz z zz Alex Grishin MAE 323 Lecture 1 31

32 And so, if we wanted to represent the product R as before, we could write: R This is commonl interpreted as: n j= 1 Index Notation ij ij jk R jk This is because it is conventional to sum an repeated indices*. For students new to indicial notation, it ma be helpful to write this in pseudo-code: *This is known as the Einstein summation convention. See for more details 2011 Alex Grishin MAE 323 Lecture 1 32

33 First initialize the product: for(i,1,n) P ik = 0 then add entries as follows Index Notation for(k,1,o) P(i,k)=0.0 end(k) end(i) P = R ik ij jk for(i,1,n) for(j,1,m) for(k,1,o) P(i,k)=P(i,k)+R(i,j)*(j,k) end(k) end(j) end(i) 2011 Alex Grishin MAE 323 Lecture 1 33

34 With this notation in hand, we can easil distinguish between an kind of matrix product whatsoever. As an added bonus, it immediatel tells us if a particular kind of product is even possible Examples: Vector Dot Product Vector Outer Product Index Notation a b a b ab A B = A B A B = A B T = = aibi T = aib j Arra Dot Product ij jk Arra Double-Dot Product : ij ji The expression for arra Dot Product tells us that the arras A and B need not be square so long as the share an index. This is the same as saing the number of columns in A has to equal the number of rows in B 2011 Alex Grishin MAE 323 Lecture 1 34

35 Questions for Vectors and Arras: Do matrices have to be square (equal number of rows and columns) in order to find their dot product? What about their double-dot product? Does the dot product of these two matrices exist? a11 a12 a13 b11 b12 b13 a a a b b b What about these two? a11 a12 a13 b11 b12 b13 a a a b b b T 2011 Alex Grishin MAE 323 Lecture 1 35