Review of Strain Energy Methods and Introduction to Stiffness Matrix Methods of Structural Analysis


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1 uke University epartment of Civil and Environmental Engineering CEE 42L. Matrix Structural Analysis Henri P. Gavin Fall, 22 Review of Strain Energy Methods and Introduction to Stiffness Matrix Methods of Structural Analysis Strain Energy Strain energy is stored within an elastic solid when the solid is deformed under load. In the absence of energy losses, such as from friction, damping or yielding, the strain energy is equal to the work done on the solid by external loads. Strain energy is a type of potential energy. Consider the work done on an elastic solid by a single point force F. When the elastic solid carries the load, F, it deforms with strains (ɛ and γ) and the material is stressed (σ and τ). is a displacement in the same location and in the same direction as a point force, F. and F are colocated. The work done by the force F on the elastic solid is the area under the force vs. displacement curve. W = F d () This work is stored as strain energy U within the elastic solid. U = 2 V (σ xx ɛ xx + σ yy ɛ yy + σ zz ɛ zz + τ xy γ xy + τ xz γ xz + τ yz γ yz ) dv. (2) This is a very general expression for the strain energy, U, and is not very practical for structural elements like bars, beams, trusses, or frames.
2 2 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin. Bars For a bar in tension or compression, we have internal axial force, N, only, N x N x x dl ε xx dl σ xx so σ yy =, σ zz =, τ xy =, τ xz =, and τ yz =, and U = 2 V σ xx ɛ xx dv, where σ xx = N/A and ɛ xx = N/. Substituting dv = A dx we get U = 2 L N(x) 2 E(x) A(x) dx, (3) and if N, E, and A are constant U = N 2 L 2 E A. Alternatively, we may express the strain as a function of the displacements along the bar u x (x), ɛ xx = u x (x)/ x, and σ xx = E u x (x)/ x. Again substituting dv = A dx, U = 2 L E(x) A(x) and if E, A and u x / x = (u 2 u )/L are constants, ( ) 2 ux (x) dx, (4) x U = 2 L (u 2 u ) 2
3 Strain Energy and Matrix Methods of Structural Analysis 3.2 Beams For a beam in bending we have internal bending moments, M, and internal shear forces, V. For slender beams the effects of shear deformation are usually neglected. M zz y v" dl M zz x σ xx dl As in the axially loaded bar, σ yy =, σ zz =, τ xy =, τ xz =, and τ yz =, and U = 2 V σ xx ɛ xx dv. For bending, σ xx = My/I and ɛ xx = My/EI. Substituting dv = da dx, where A y2 da = I, so U = 2 L U = 2 A L M(x) 2 y 2 da dx, E(x) I(x) 2 M(x) 2 E(x) I(x) dx. (5) Alternatively, we may express the moment in terms of the curvature of the beam, φ 2 u y / x 2, M(x) = E(x) I(x) 2 u y (x) x 2, from which σ xx = E ( 2 u y / x 2 ) y and ɛ xx = ( 2 u y / x 2 ) y, so that U = 2 where, again, A y2 da = I, so U = 2 L L A E(x) E(x) I(x) ( 2 ) 2 u y (x) y 2 da dx x 2 ( 2 ) 2 u y (x) dx. (6) x 2
4 4 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin.3 Summary External work is done by a set of forces, F i, on a linear elastic solid, producing a set of displacements, i, in the same locations and directions. F Fi i j F j n Fn F F i Fi Fi i n j Fj F j j F n F n n The work done by these forces is W = 2 F + 2 F F The external forces are resisted by internal moments, M, and axial forces, N. The total strain energy stored within the solid is U = M 2 2 L E I dx + Nj 2 L j (7) 2 j E j A j where the first term is the integral over all lengths of all the beams and the second term is the sum over all the bars. If torsion and shear are included, then two additional terms are T 2 2 L G J dx and V 2 2 L G A/α dx. Alternatively, we can think of external forces producing curvatures ( 2 u y / x 2 ) by bending, and axial stretches ( u x / x). In this case U = ( 2 ) 2 u y E I dx + E j A j (u 2 L x 2 2j u j ) 2 (8) 2 j L j If torsion and shear are included, then two additional terms are ( ) 2 ( ) 2 uxθ uy G J dx, and G A/α dx, 2 L x 2 L x where u xθ is the torsional rotation about the xaxis, u xθ / x is the torsional shear strain, γ xθ, (on the face perpendicular to the xaxis and in the θdirection) and u y / x is the shear strain, γ xy, (on the face perpendicular to the xaxis and in the ydirection). Analyses using expressions of the form of equations (3), (5), or (7) are called force method or flexibility method analyses. Analyses using expressions of the form of equations (4), (6), or (8) are called displacement method or stiffness method analyses.
5 Strain Energy and Matrix Methods of Structural Analysis 5 2 Castigliano s Theorems 2. Castigliano s Theorem  Part I U = F d... strain energy F j U() j U + j j j F i = U i = U i A force, F i, on an elastic solid is equal to the derivative of the strain energy with respect to the displacement, i, in the direction and location of the force, F i. 2.2 Castigliano s Theorem  Part II U = df... complementary strain energy F+ j F j F j U* U*(F) F j j i = U F i = U F i A displacement, i, on an elastic solid is equal to the derivative of the complementary strain energy with respect to the force, F i, in the direction and location of the displacement, i. If the solid is linear elastic, then U = U.
6 6 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin 3 Superposition Superposition is an extremely powerful idea that helps us solve problems that are statically indeterminate. To use the principle of superposition, the system must behave in a linear elastic fashion. The principle of superposition states: Any response of a system to multiple inputs can be represented as the sum of the responses to the inputs taken individually. By response we can mean a strain, a stress, a deflection, an internal force, a rotation, an internal moment, etc. By input we can mean an externally applied load, a temperature change, a support settlement, etc. 4 etailed Example of Castigliano s Theorem and Superposition An example of a statically indeterminate system with external loads w(x) and three redundant reaction forces, R B, R C, and R, is shown below. y w(x) EI A B L C x H In general, the displacements at the locations of the unknown reaction forces are known, and, in this example these displacements will be taken as zero: B =, C =, =. Invoking the principle of superposition, we may apply the external loads, (w(x)) and the unknown reactions (R B, R C, and R ) individually, and then sumup the responses to each individual load. Further, we may represent the response to a reaction force, (e.g., R B ) as the response to a unit force colocated with the reaction force, times the value of the reaction force. Note that all four systems to the right of the equal sign in the following figure are statically determinate. Expressions for Mo(x), m (x), m 2 (x), m 3 (x), N o (x), n (x), n 2 (x), and n 3 (x) may be found from static equilibrium alone.
7 Strain Energy and Matrix Methods of Structural Analysis 7 w(x) A B C M(x) R B R C N(x) R = + w(x) A B C M (x) o m (x) B N (x) o n (x) * R B + m (x) 2 C n (x) 2 * R C + m (x) 3 n (x) 3 * R In equation form, the principle of superposition says: M(x) = M o (x) + m (x)r B + m 2 (x)r C + m 3 (x)r (9) N = N o + n R B + n 2 R C + n 3 R () (Note that in this particular example, N o (x) =, n =, n 2 =, n 3 =, m (x) = for x > x B, and m 2 (x) = for x > x C.)
8 8 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin The total strain energy, U, in systems with bending strain energy and axial strain energy is, U = L M(x) 2 dx + N 2 H () 2 EI 2 We are told that the displacements at points B, C, and are all zero and we need to assume the structure behaves linear elastically in order to invoke superposition in the first place. Therefore, from Castigliano s Second Theorem, i = U F i = U F i, we obtain three expressions for the facts that B =, C =, and =. B = = U R B C = = U R C = = U R Inserting equation () into the three expressions for zero displacement at the fixed reactions, noting that EI and are constants in this problem, and noting that the strain energy, U, depends on the reactions R, only through the internal forces, M and N, we obtain B = = L M(x) M(x) dx + H EI R B N N R B C = = L M(x) M(x) dx + H EI R C N N R C = = L M(x) M(x) dx + H EI R N N R Now, from the superposition equations (9) and (), M(x)/ R B = m (x), M(x)/ R C = m 2 (x), M(x)/ R = m 3 (x), N(x)/ R B = n, N(x)/ R C = n 2, and N(x)/ R = n 3. Inserting these expressions and the superposition equations (9) and () into the above equations for B, C, and, B = = L [M o + m R B + m 2 R C + m 3 R ] m dx + H EI [N o + n R B + n 2 R C + n 3 R ] n C = = L [M o + m R B + m 2 R C + m 3 R ] m 2 dx + H EI [N o + n R B + n 2 R C + n 3 R ] n 2 = = L [M o + m R B + m 2 R C + m 3 R ] m 3 dx + H EI [N o + n R B + n 2 R C + n 3 R ] n 3 These three expressions contain the three unknown reactions R B, R C, and R. Everything else in these equations (m (x), m 2 (x)... n 3 ) can be found without knowing the unknown
9 Strain Energy and Matrix Methods of Structural Analysis 9 reactions. By taking the unknown reactions out of the integrals (they are constants), we can write these three equations in matrix form. L m m EI L m 2m EI L m 3m EI dx + n n H dx + n 2n H dx + n 3n H L m m 2 EI L m 2m 2 EI L m 3m 2 EI dx + n n 2 H dx + n 2n 2 H dx + n 3n 2 H L m m 3 EI L m 2m 3 EI L m 3m 3 EI dx + n n 3 H dx + n 2n 3 H dx + n 3n 3 H R B R C R = L Mom EI L Mom 2 EI L Mom 3 EI (2) This 3by3 matrix is called a flexibility matrix, F. The values of the terms in the flexibility matrix depend only on the responses of the structure to unit loads placed at various points in the structure. The flexibility matrix is therefore a property of the structure alone, and does not depend upon the loads on the structure. The vector on the righthandside depends on the loads on the structure. Recall that this matrix looks a lot like the matrix from the threemoment equation. All flexibility matrices share several properties: dx + Non H dx + Non 2H dx + Non 3H All flexibility matrices are symmetric. No diagonal terms are negative. Flexibility matrices for structures which can not move or rotate without deforming are positive definite. This means that all of the eigenvalues of a flexibility matrix describing a fixed structure are positive. The unknowns in a flexibility matrix equation are forces (or moments). The number of equations (rows of the flexibility matrix) equals the number of unknown forces (or moments). There are some fascinating cases in which the behavior does depend upon the loads, but that is a story for another day!
10 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin It is instructive to now examine the meaning of the terms in the matrix, F F = L m m EI dx + n n H = δ displacement at due to unit force at F A B C F 2 = L m 2 m EI dx + n 2n H = δ 2 displacement at 2 due to unit force at F 2 A B C F 2 = L m m 2 EI dx + n n 2 H = δ 2 displacement at due to unit force at 2 F 2 A B C F 3 F 3 = L m 3 m EI dx + n 3n H = δ 3 displacement at 3 due to unit force at A B C The fact that F 2 = F 2 is called Maxwell s Reciprocity Theorem.
11 Strain Energy and Matrix Methods of Structural Analysis 5 Introductory Example of the Stiffness Matrix Method In this simple example, elements are springs with stiffness k. A spring with stiffness k > connecting point i to point j, will have a force f = k(d j d i ) where d i is the displacement of point i and d j is the displacement of point j. (Tension is positive so d i points into the spring and d j points away from the spring.) The stiffness matrix for this structure can be found using equilibrium and forcedeflection relationships (f = kd) for the springs. #: F x = : f k d + k 2 (d 2 d ) = #2: F x = : f 2 k 2 (d 2 d ) k 4 d 2 + k 3 (d 3 d 2 ) = #3: F x = : f 3 k 3 (d 3 d 2 ) k 5 d 3 = In matrix form these three equations may be written: k + k 2 k 2 d k 2 k 2 + k 3 + k 4 k 3 d 2 k 3 k 3 + k 5 d 3 = The displacements are found by solving the stiffness matrix equation for d, d = K f. The matrix K is called a stiffness matrix. All stiffness matrices are symmetric. f f 2 f 3
12 2 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin All diagonal terms of all stiffness matrices are positive. Stiffness matrices are diagonally dominant. This means that the diagonal terms are usually larger than the offdiagonal term. If the structure is not free to translate or rotate without deforming, then the stiffness matrix is positive definite. This mathematical property guarantees that the stiffness matrix is invertible, and a unique set of displacements, d, can be found by solving K d = f. The total potential energy, U, in this system of springs is U = 2 k d k 2(d 2 d ) k 3(d 3 d 2 ) k 4d k 5d 2 3. You should be able to confirm that this is equal to U = 2 dt K d Also, note that no matter what the values of the displacements, d, may be, the energy U is always positive. The statement 2 dt Kd > d is another way of saying that K is positivedefinite. The set of forces required to deflect coordinate i by a deflection of unit equals the ith column of the stiffness matrix. For example consider the case in which d =, d 2 =, and d 3 =, k + k 2 k 2 k 2 k 2 + k 3 + k 4 k 3 k 3 k 3 + k 5 = which is equal to the first column of the stiffness matrix. k + k 2 k 2,
13 Strain Energy and Matrix Methods of Structural Analysis 3 This fact may be used to derive the stiffness matrix: d =, d 2 =, d 3 = f = k + k 2, f 2 = k 2, f 3 =... st column d =, d 2 =, d 3 = f = k 2, f 2 = k 2 + k 3 + k 4, f 3 = k nd column d =, d 2 =, d 3 = f =, f 2 = k 3, f 3 = k 3 + k rd column
14 4 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin 6 Basic Concepts of the Stiffness Matrix Method The previous example illustrates some of the basic concepts needed to apply the stiffness matrix method to structures made out of bars and beams. There are, however, a few additional complications. isplacements in structures can be vertical, horizontal, or rotational, and structural bars and beams have a more complicated forcedisplacement relationships than those of simple springs. In applying the matrix stiffness method of structural analysis, structures are described in terms of elements that connect nodes which can move in certain coordinate directions. 6. Elements In the stiffness matrix method, structures are modeled as assemblies of elements such as bars, beams, cables, shafts, plates, and walls. Elements connect the nodes of the structural model. Like the simple springs in the previous example, structural elements have clearly defined, albeit more complicated, forcedisplacement relationships. The stiffness properties of structural elements can be determined from equilibrium equations, Castigliano s Theorems, the principle of minimum potential energy, and/or the principle of virtual work. Structural elements can be mathematically assembled with one another (like making a structural system using a set of tinkertoys), into a system of equations for the entire structure. 6.2 Nodes The forcedisplacement relationship of a structural element is defined in terms of the forces and displacements at the nodes of the element. Nodes define the points where elements meet. The nodes in the model of a truss are at the joints between the truss bars. The nodes in the model of a beam or a frame are at the reaction locations, at locations at which elements connect to each other, and possibly at other intermediate locations. 6.3 Coordinates Coordinates describe the location and direction at which forces and displacements act on an element or on a structure. Trusses are loaded with vertical and horizontal forces at the joints. The joints of a 2 truss can move vertically and horizontally; so there are two coordinates per node in a 2 truss. Beams and frames carry vertical and horizontal loads as well as bending moments. The nodes of a 2 frame can move vertically, horizontally, and can rotate; so there are three coordinates per node in a 2 frame. Structural coordinates can be classified into two sets. isplacement coordinates have unknown displacements but know forces. Reaction coordinates have unknown forces but known displacements (usually zero).
15 Strain Energy and Matrix Methods of Structural Analysis Elements, Nodes and Coordinates Planar (2) truss bar elements have two nodes and four coordinates, two at each end. Space (3) truss bar elements have two nodes and six coordinates, three at each end. Planar (2) frame elements have two nodes and six coordinates, three at each end. Space (3) frame elements have two nodes and twelve coordinates, six at each end.
16 6 CEE 42L. Matrix Structural Analysis uke University Fall 22 H.P. Gavin 6.5 Structural Nodes and Coordinates Planar (2) truss nodes and coordinates Planar (2) frame nodes and coordinates
17 Strain Energy and Matrix Methods of Structural Analysis 7 7 Relate the Flexibility Matrix to the Stiffness Matrix Column j of the stiffness matrix: The set of forces at all coordinates required to produce a unit displacement at coordinate j Column j of the flexibility matrix: The set of displacements at all coordinates resulting from a unit force at coordinate j Stiffness matrix equation: Flexibility matrix equation: K K 2 K 3 K n K 2 K 22 K 23 K 2n K 3 K 32 K 33 K 3n K = F and K = F K n K n2 K n3 K nn F F 2 F 3 F n F 2 F 22 F 23 F 2n F 3 F 32 F 33 F 3n F n F n2 F n3 F nn d d 2 d 3. d n f f 2 f 3. f n = = f f 2 f 3. f n d d 2 d 3. d n A useful fact for 2by2 matrices... [ ] a b = c d ad bc [ d b c a ]... you should be able to prove this fact to yourself.
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