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. By validation, we mean that the analyst should check various solution values against manual calculation or tabular look-up data whenever possible The second part is given because many students will have only a passing familiarity 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 is inescapably a matrix method. 2011 Alex Grishin MAE 323 Lecture 1 1
Statics and Basic Solid Mechanics 2011 Alex Grishin MAE 323 Lecture 1 2
Q: What does it mean to perform a structural analysis? A: In the most general terms, it means that we subject a geometric domain (the structure) to an external force (by external, we mean only that it is prescribed). We then calculate the structural response to this load. The primary variable that we calculate is displacement (including any 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 try to estimate some of the calculated values using other methods as a check 2011 Alex Grishin MAE 323 Lecture 1 3
The other methods will typically involve statics or simple solid mechanics (especially small-deflection beam theory otherwise known as Euler-Bernoulli beam theory). 6m Example 1: 4m 40 kn y 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
Solution: Use Newton s Third Law to sum the moments AND forces y 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 10 40 = 0 y 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 Ay = R 6 + (40 4) + (10 2) = 0 B (2) 2011 Alex Grishin MAE 323 Lecture 1 5
Finally: R R A A F x = 0 F = R + R 10 40 = 0 y 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 easily have chosen point B: Bx Bz 6R = 120 R A M M A = 20 = M = 0 By = R 6 + ( 40 2) + ( 10 4) = 0 A 2011 Alex Grishin MAE 323 Lecture 1 6
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 very useful tool in checking that a structure is in equilibrium. If our calculated force and moment reactions agree with those calculated by the software, then we can be confident that the model is properly 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
Example 2: Three beams connected endto-end, oriented along each axis z O y x 4m 3m 1.5m 4 kn First the forces 3 kn R 0 2 + F = 0 i= 1 i Rx0 0 0 0 R + 4 kn + 0 kn = 0 y0 R z0 0 3 0 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
Example 2 (cont.): z O y x 4m 3m 1.5m 4 kn N Now, the moments M = R + r F = 0 i M 0 i i i= 1 i= 1 2 4 0 4 0 = M 0 + 0 4 + 1.5 0 0 0 3 3 3 kn R i M 0 i= 1 N M Recall the cross products can be evaluated using determinants i j k i j k 0 = R + 4 0 0 + 4 1.5 3 = 0 knm 0 4 0 0 0 3 0 2011 Alex Grishin MAE 323 Lecture 1 9
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
A-A A-A k = 4 / 3 P P k = 2 k = 3 / 2 Case 3: Bending Stress: = Distribution: My I Case 4: Transverse Shear Stress: PQ τ =, τ max = k Ib Distribution: P A 2011 Alex Grishin MAE 323 Lecture 1 11
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 simply 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 y Tensile stress Note the section coordinate system has nothing to do with any global system. It is simply a convenience for defining distance from the neutral axis *See J. Gere and S. Timoshenko, Mechanics of Materials, 3rd ed. Boston: PWS-KENT Publishing Co., 1984. 2011 Alex Grishin MAE 323 Lecture 1 12
For pure (torsional) shear, Both positive and negative shears exist for a given external torque, but they will have the same maximum absolute values -τ +τ -τ +τ +T -τ -τ +τ +τ And, for 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
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 y 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 everything we need from the moment vector we calculated earlier 2011 Alex Grishin MAE 323 Lecture 1 14
The applied moment at O was: M 0 4.5 = 12 16 knm The corresponding moment reaction is: 4.5 R M 0 = 12 knm 16 z O y 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 15
We use the applied moment loads to estimate stresses: y τ Max. torsional stress at O: Tr 4.5 0.1 = = = 2864.8kPa J π (0.1) / 2 max max 4 Max. bending stress at O (xz plane): 1 y ( 12 0.1*sin(tan (3 / 4))) max 4 I yy π (0.1) / 4 z O x 4m 3m 1.5m 4 kn 3 kn M z = = = 9167.3kPa Max. bending stress at O (xy plane): 1 M z y 16 0.1*cos(tan (3 / 4)) max 4 I zz π (0.1) / 4 = = = 16297.5kPa 2011 Alex Grishin MAE 323 Lecture 1 16
But we haven t finished. The bending stresses in the XZ and XY planes are in the same direction and orientation meaning they should be added if we want the total overall maximum bending stress for the structure. Thus, max( ) = max( _ ) + max( _ ) = (9167.3 + 25464.8) bending bending xy bending yz kpa Notice we re using the formula*: max( bending ) = 25464.8 kpa max M yz = + I y M z y I z *see http://en.wikipedia.org/wiki/second_moment_of_area 2011 Alex Grishin 17 MAE 323 Lecture 1
But this formula only works if the product of inertia I yz is negligible or zero so that should be checked before applying this formula. Also, where did the negative sign in the first term come from? To understand that, we have to go back to the moment-stress sign convention we spoke of earlier (remember that this coordinate system is not necessarily the same as the global origin. It is purely conventional) max Compressive stress Neutral axis M z M y y z z +M y Tensile stress When we use two coordinate axes with this convention, one of them must always receive a negative sign to remain consistent 2011 Alex Grishin MAE 323 Lecture 1 18
Of course, because of the radial symmetry of this cross-section, we could have used a shortcut: * M y* z max( bending ) = I Neutral axis y z * max z y* Where M y* is the vector resultant moment, z* is the distance to the max. stress location (which is simply rin our case), and I y* is the corresponding second moment of inertia (which won t change because of symmetry). Notice that this is effectively a rotational transformation M y* * y * y* max 4 I y* π Compressive stress Tensile stress 2011 Alex Grishin MAE 323 Lecture 1 19 +M M z 20 0.1 = = = 25464.8 kpa (0.1) / 4
Questions for Statics and Basic Solid Mechanics: In example 2, we estimated the maximum bending and pure shear stress. Are these the only stresses that exist for that problem? How would one calculate the maximum transverse shear stress in example 2? Do any compressive or tensile stresses and strains exist in example 2? 2011 Alex Grishin MAE 323 Lecture 1 20
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 21
Vectors and Arrays 2011 Alex Grishin MAE 323 Lecture 1 22
Students in this course will already have some familiarity with vectors and arrays, but we re going to review some things AND add some nuances with which many students are probably unfamiliar In this course, unless I specify otherwise, vectors AND arrays will be represented as bold symbols to refer to all their components at once, while individual components will not. And (unless I say otherwise) vectors will be represented in equations as single-column arrays 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 0 R 0ˆ 4ˆ 3ˆ R i j k 4 3 0 = 0 = + = 2011 Alex Grishin MAE 323 Lecture 1 23
And the individual components can be written: R R R 0x 0 y 0z = 0 = 4 = 3 We have already been careful to follow this convention when writing the stress equations. Consider, for example, the equation for bending stress: = My I Stress is a tensor (an array. And so the 2 nd moment of inertia) and moment is a vector, so why aren t they bold in this equation? 2011 Alex Grishin MAE 323 Lecture 1 24
The reason is that the equations of slides 9 and 10 are written on a component basis! This is exactly why 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 why 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 simply an interchange of rows and columns, turning column arrays into row arrays and vice-versa. We will denote the transpose of an array with a capital T superscript: ax T a = a, a = a a a a z { } y x y z 2011 Alex Grishin MAE 323 Lecture 1 25
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) arrays, an expression like: T a b Will be understood to mean the inner product: Instead of: bx ax ay az by = axbx + ayby + azbz = a b b { } z ax axbx axby azbz a b b b = a b a b a b = = { } y x y z y x y y y z a z azbx azby 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 26
Dot Products of Arrays Some ambiguity may still arise, however, when we take dot products of multi-column arrays. For example, suppose I have a stress tensor (represented by an array) xx xy xz = yx yy yz zx zy zz and I want to multiply it by another array (maybe to transform it): R R R R xx xy xz = Ryx Ryy R yz R R R zx zy zz 2011 Alex Grishin MAE 323 Lecture 1 27
Dot Products of Arrays This results in the familiar matrix product: R + R + R R + R + R R + R + R xx xx xy yx xz zx xx xy xy yy xz zy xx xz xy yz xz zz = Ryx xx + Ryy yx + Ryz zx Ryx xy + Ryy yy + Ryz zy Ryx xz + Ryy yz + Ryz zz R Rzx xx Rzy yx Rzz zx Rzx xy Rzy yy Rzz zy Rzx xz Rzy yz Rzz + + + + + + zz However, this should not be confused with the matrix double-dot product: R = R + R + R : xx xx xy xy yx xy + R + R + R yy yy xz xz zx xz + R + R + R yz yz zy yz zz zz Note that this product produces a scalar value 2011 Alex Grishin MAE 323 Lecture 1 28
Index Notation One way to avoid the ambiguity between different kinds of matrix products is to use index (or indicial) notation. With this notation, row and column indices are explicitly numbered (as are the coordinate axes to which they may refer) y,2 ax a1 a = ay = a2 a z a x,1 3 z,3 xx xy xz 11 12 12 = xy yy yz = 21 22 23 xz yz zz 31 32 33 2011 Alex Grishin MAE 323 Lecture 1 29
And so, if we wanted to represent the product R as before, we could write: R This is commonly interpreted as: n j= 1 Index Notation ij ij jk R jk This is because it is conventional to sum any repeated indices*. For students new to indicial notation, it may be helpful to write this in pseudo-code: *This is known as the Einstein summation convention. See http://en.wikipedia.org/wiki/einstein_summation_convention for more details 2011 Alex Grishin MAE 323 Lecture 1 30
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 31
With this notation in hand, we can easily distinguish between any kind of matrix product whatsoever. As an added bonus, it immediately tells us if a particular kind of product is even possible Examples: Vector Dot Product Vector Outer Product Index Notation a b ab A B = A B A B = A B T = a b = aibi T = aib j Array Dot Product ij jk Array Double-Dot Product : ij ji The expression for array Dot Product tells us that the arrays A and B need not be square so long as they share an index. This is the same as saying the number of columns in A has to equal the number of rows in B 2011 Alex Grishin MAE 323 Lecture 1 32
Questions for Vectors and Arrays: 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 21 22 23 21 22 23 What about these two? a11 a12 a13 b11 b12 b13 a a a b b b 21 22 23 21 22 23 T 2011 Alex Grishin MAE 323 Lecture 1 33