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# STRESS ANALYSIS CES 4141

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CES 4141 Stress Analysis Summer 1998 Direct Stiffness Beam Application We ve looked at creating the global stiffness matrix for a structure built entirely of truss elements These elements only have two local degrees of freedom axial motion at each end In this section we ll extend the ideas we used for trusses to create beam elements and then assemble a global stiffness matrix for frames Then we ll look at special cases where we may use only portions of the full element stiffness matrix 1 Overview Just as we did for trusses we will consider each whole beam member as afmite element and each joint becomes a node We ll determine the forcedisplacement relationship for each element separately then combine each individual contribution to the whole structure in a Global Struc tural Stz ness Matrix KG What is di erent from a truss Each node now has axial shear and moment degrees of freedom rather than just axial These notes will present the stiffness for each beam member at the local level then transform them into the global coordinate system then sum the contribution of all elements then discuss solutions for displacements reactions and internal forces 2 Element Stiffness Matrix In Local Coordinates We look at a single beam member in its own local coordinate system Con v6 sider the element to the right The local y global coordinate system x y will align with the member orientation for any arbitrary orientation The local displacement degrees of freedom DOF are labeled as v1 v6 Now we can apply the stiffness by de nition procedure to nd the stiffness matrix for this arbitrary element Holding v11 v2 v60 we get the left column of the 6x6 stiffness matrix by application of the slope de ection equations Hold ing v10 v21 v3 v60 we get the Global d0f r second column etc This is illustrated on r2 r5 page 158 in Hoit However we here include the axial displacements which 3 r1 6 r4 r r v4 local dof v V3 Fig 1 local beam element will give us a 6x6 local element stiffness matrix instead of the 4x4 given on page 158 which ignores axial After the above procedure is completed by holding all six DOF in turn to 1 and all others zero we get the local stiffness matrix k given on the next page 1 of 7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis Summer 1998 AE iAE 0 0 0 0 L L 12EI 6E1 12EI 6E1 3 2 0 3 2 L L L L 0 6EI 4E1 76 131 2E1 2 L 2 L k L L 1 7AE AE 0 0 0 0 L L 712E176EI 12EI 76EI L3 L2 L3 L2 6E1 2E1 6E1 4E1 0 2 T 0 7 L L L Note that this will be the same no matter what the orientation of the element since all quanti ties are with respect to the element local coordinate system which will always be set up to lie along the axis of the member Note also that the order is dependent on what order we chose to label V1 V6 so we ll always label the dof ofan elementjust as we did in Fig 1 Now we can relate the local forces to local displacements by AE iAE 0 0 0 0 L L 12EI 6EI 712EI 6EI Sl L3 L2 L3 L2 V1 32 6E1 4E1 6E1 2E1 V2 0 2 T 0 2 7 S3 L L v3 2 S4 iAE AE v4 0 0 0 0 S5 L L v5 36 12EI 6E1 12EI 6E1 vg 0 7 0 7 7 6E1 2E1 6E1 4E1 0 2 L 0 7 2 L L L 3 Displacement Transformation Matrix Since a frame structure consists of many members of possibly many orientations we ll want to transform all local displacements into a single uniform global coordinate system The question is then if we were to nd the displacement at both ends of an element in terms of the local coordi nates V1 V6 what is that displacement in terms of the global coordinates r1 r6 For the truss case at this point we were transforming 4 global degrees of freedom to 2 local degrees of freedom and thus the transformation matrix a is a 2X4 In this case we are relating 6 global to 6 local 2 of 7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis Summer 1998 degrees offreedom so a will be a 6x6 Refer now to the procedure we used in the truss development packet to nd the transformation matrix We moved an arbitrarily oriented member a unit value in each of the global degrees of freedom in turn holding all other motions to zero and finding what that does to the local compo nent We do the same here now and get the 6 equations below v1 r1cos9x r2quot cos9y v2 7r1cos9yr2cosex V3 r3 I v4 r4 cos 9x r5cos 9y v5 7r4cos9yr5cosex v6 r6 Using again Lx cos 9x and Ly cos 9y we rewrite these equations in matrix form and de ne the transformation matrix a LxLyO 0 00 iLnyO 0 00 0 010 00 0 OOLxLyO 0 007Lny0 0 00 0 01 Now we can relate global to local displacements in matrix form as v 61 r 5 Now we ve found a way to relate local displacements to global displacements We ll also want to do this for local and global forces 4 Force Transformation Matrix We can also identify the transformation matrix for the localglobal force relationship This time we ll get the global on the left hand side and the local on the right hand side by applying a unit force in local coordinates and nding the resultant global components Just as was the case for the truss element the resultant is the transpose of the a transformation matrix we found in Eq 4 We can then relate local to global forces by R aTS 6 5 Element Stiffness Matrix In Global Coordinates Now we ll use the transformation matrices we just derived to nd the stiffness matrix for a single element in terms of global coordinates The local stiffness matrix will remain a 6x6 We ll repeat the process we used for truss elements since the speci c element in question 3 of 7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis Summer 1998 does not matter Start with S k v 7 Transform local displacements V to global displacements r by substituting Eq 5 into Eq 7 to get S kar 8 Premultiply both sides by aT the left hand side becomes Eq 6 and the nal expression is R aTkar 9 We have the displacements and forces in terms of global coordinates now The stiffness matrix is still local and is pre and postmultiplied by aT and a Let s separate this component and rename Ke 7 aTk a 10 This gives us a nal expression of R K e r 11 Let s write out and multiply through Eq 10 Performing the matrix multiplications gives us L2x12EIL2y AE712EIJLxLv 76E1Ly 11sz 12EIL2y leEIjoLv 76E1Ly L L3 L3 L2 L L3 L L3 L2 f A1121331LxLy AELzy121331sz 6E12Lx 71E7121331LxLy AfELz izisaiszj 6E12Lx L L L L L L L L L 76E1Ly 6E1Lx 6E1Ly i EILx E 2 2 L 2 2 L Ke L L L L 12 z LzH 123EIL2y 7 121331LCLy 6E12Ly L2X 121331L2y 7 121321LxLy 6E1Ly L L L L L L L L L 7 121331LXLy 19sz 121IL2XJ 76E12Lx 7 121321LJCLy AELzy121331sz L EiLx L L L L L L L L L L 76E1Ly 6E1Lx E 6E1Ly 76E1Lx L2 L2 L L2 L2 L 6 Assembly of Global Stiffness Matrix Before going on let s review what we have derived Eq 11 relates the displacements to forces all in global coordinates for a single element of arbitrary orientation The orientation of the individual member is accounted for in the Global element sti ness matrix Ke What s left to do We now have the contribution of a single beam element For a structure with multiple members we assemble a Ke for each member in the frame then add them each into a global structural sti ness matrix that covers every degree of freedom in the entire structure We ll call that KG KG will be a square matrix with as many rows and columns as there are total degrees of freedom frozen and unfrozen Let s do this by example 4 of 7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis r6 20 ft l 5 r2 r1 T r III r3 5 kips CD 20 ft i r8 r9 gr r7 Fig 2 Frame example r6 r2 r5 r 1 r r3 Fig 3 Element 1 r2 r9 r7 Fig 4 Element 2 A10 in2 1 500 in4 E 29000 ksi Summer 1998 Example 1 AssemblV of KG for a truss structure We ll assemble the global structural stiffness matrix for the structure in Fig 2 then solve for displacements reac tions and internal forces Note that we have labeled the unfrozen DOF as r1 r5 then the frozen DOF as r6 r9 R1 R5 will represent the external forces in this case R15 R2R3R4R50 R6 R9 represent the reactions The process is simply to create Ke for each of the two members using Eq 12 then to add them by keeping track of how the local degrees of freedom correspond to the global degrees of freedom Note that this problem is indeterminate but this does not affect our procedure or solution Element 1 the element labeled 1 connects nodes 1 and 2 and global DOF r1 r6 Finding Lx and Ly and applying Eq 12 Lx change in x L 20 020 1 Ly change in y L 0 020 0 r4 r6 r5 r1 r2 r3 12083 0 0 712083 0 0 r4 0 126 15104 0 7126 15104 r6 K61 0 15104 2417e3 0 715104 12083e3 5 13 712083 0 0 12083 0 0 r1 0 7126 715104 0 126 715104 r2 0 1510412083e3 0 715104 2417e3 r3 Element 2 This element connects nodes 2 and 3 and global DOF r1r3 r7r9 Lx change in x L 20 2020 0 Ly change in x L 20 020 1 this gives us Ke2 through Eq 12 as r1 r2 r3 r7 r8 r9 126 0 15104 7126 0 15104 r1 0 12083 0 0 712083 0 r2 K62 15104 0 2417e3 715104 0 12083e3 r3 14 7126 0 715104 126 0 715104 r7 0 712083 0 0 12083 0 r8 15104 0 12083e3 715104 0 2417e3 r9 Note that we have listed above and to the right of each element global stiffness matrix the global displacements that correspond to the six local degrees of freedom Now we have all element stiffness matrices We simply set up the global structural stiffness matrix as a 9x9 with all zeros initially and add in the element matrices KG Ke1Ke2 15 After assembly de ning known forces and displacements and partitioning as we did for the 50f7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis Summer 1998 truss example we have con rm this yourself 5 12209 0 15104 712083 0 0 7126 0 15104 r1 0 0 12209 715104 0 715104I7126 0 712083 0 r2 0 15104 715104 4834e3 0 12083e315104 715104 0 12083e3 r3 0 712083 0 0 12083 0 0 0 0 0 r4 0 0 71510412083e3 0 2417e3 15104 0 0 0 r5 16 39 R6 71 71275 150 0 15104 126 000 0 39 R7 7126 0 715104 0 0 I 0 126 0 715104 0 R8 0 712083 0 0 0 0 0 12083 0 0 R9 15104 0 12083e3 0 0 0 715104 0 2417e3 0 Note that we get all 5 known forces at the top of the force vector and the last 4 components in the displacement vector are known because we labeled the unfrozen degrees of freedom first 7 The Solution Procedure The knowns Rk R1 R5 external forces rk r6 r9 frozen displacements all 0 The unknowns Ru R6 R9 reactions ru r1 r5 displacements rewrite Eq 16 in shorthand as R K11K12 ru 17 RE 1511703 rk now expanding we getmake sure you can follow this step RkK11ruK12rk 18 Ru K21ruK22rk 19 We can solve the system in Eq 18 which is 5 equations and 5 unknowns for ru then solve the system in Eq 19 a 4 equation 4 unknown system The solutions are below rl 0696 in R6 7187k r2 7155e73ln R7 500k r3 72488e73 rad 39 R8 187k r4 0696 1n R9 750 k r5 1234e73 rad m Now force recovery is given by applying Eq 8 S kar for each element recalling that we are using the local element stiffness matrices given in Eqs 13 and 14 and being careful to apply the correct displacements r by referring to Figures 3 and 4 6 of 7 3200 Direct Stiffness Beam Application CES 4141 Stress Analysis Summer 1998 Element 1 r4 r6 r5 r1 r2 r3 12083 0 0 712083 0 0 1 0 0 0 0 0 0696 in r4 0 126 15104 0 7126 15104 01 0 0 0 0 0 r6 0 15104 2417e3 0 71510412083e3 0 0 1 0 0 0 1234e73 rad 5 20 712083 0 0 0 0 0 0 0 0 1 0 0 0696 in r1 0 7126 715104 0 126 715104 0 0 0 0 10 7155e73rad r2 0 1510412083e3 0 715104 2417e3 0 0 0 0 0 1 72488e73rad r3 which gives us S4 0 S6 7187k A 450 k 1n SS 0 which is just 21 S1 0 S2 187k V 187 k S3 7450 k in L87 k Element 2 The above procedure is applied to element 2 and gives con rm this 13987 k 450 kin gt5k SkA w 750 kin 187k What should be obvious now is that although the procedure is quite systematic it gets quite lengthy to solve by hand Not only putting together the stiffness matrix but the solution of Eq 18 using Gaussian Elimination is time consuming However given the systematic nature of the method it is easily programmed This is why matrix methods are universally chosen as the method of choice in engineering analysis software What do we do about hinges mixed beam and trusses and beams with no axial displacement 7 of 7 3200 Direct Stiffness Beam Application Basic Linear Algebra Objectives 1 Write a matrix and know it s size 2 Add matrices 3 Subtract matrices 4 Multiply matrices 5 De ne a matrix transpose 6 De ne an Identity matrix 7 De ne a symmetric matrix Matrix methods are a subset of linear algebra A matrix has a size de ned by its number of rows and columns 12 23 A 54 10 00 02 The size ofmatrix A is 3 rows by 2 columns In common language it is said quotmatrix A is 3 by 2quot Any term in the matrix is referred to by its row and column position The row position is always given rst ie A22 10 The diagonal of a matrix is de ned as all terms where the row number and the column number are the same ie A1l and A22 Matrices can be added Matrices MUST be the same size The resulting values is the sum of the corresponding terms 12 92 B 34 21 31 65 00 Then A B C Gives the result C 11 31 Basic Linear Algebra Page 1 of 4 Subtraction is handled in a similar way 24 D A B gives the result D 31 67 Matrices can be multiplied but not divided Matrix multiplication is NOT commutative it is order dependent The terms of the result consist of multiplying a row of the rst matrix times a column of the second matrix 34 21 GAF l 73 10 The 11 term row one column one of G is G11 12342373 1271 The location where the result goes is dictated by the row and column number being multiplied The result G32 comes from multiplying row 3 of A times column 2 of F The nal matrix G is 1271 482 146 Matrix sizes must match G A F N m N W M w Must be same Multiplication is associative and distributive ABCABCABC and ABCACBC Basic Linear Algebra Page 2 of 4 There is a multiplicative identity matrix 1 De ned as a matrix that has ones on the diagonal and zeros everywhere else off diagonal terms The identity matrix is always square same number of rows as columns For example a 3 by 3 identity matrix looks like 10 00 00 I 00 10 00 00 00 10 The identity matrix has the property A I I A A The inverse of a matrix is the matrix that when multiplied times the original matrix will give the identity matrix The inverse of matrix A is designated as Al AA1A1AI A matrix MUST be square to be inverted Not all matrices can be inverted An invertible matrix is called nonsingular or invertible If it cannot be inverted then it is called singular Matrix inversion is very in the solution of equations 5x1 4952 x3 0 4x16x2 4953 x4 1 x1 4x26x3 4x4 0 x2 4x35x42 Is represented by 5410 X1 0 4641 m 1 A x b 1464 X3 0 0145 X4 2 The set of equations is represented by Axb Using matrix algebra Basic Linear Algebra Page 3 of 4 A A x A b I x A 1 b x A 1 b In practice the inverse is never calculated a solution process is used There are two types of solution methods used in practice 1 Direct and 2 Iterative Direct solutions can be thought of as variations of the Gauss Elimination method The iterative schemes can be thought of as variations of the GaussSidel solution technique The transpose is de ned as switching the rows with the columns A superscript T designates the transpose For example the transpose of matrix H is HT 12 43 10 12 54 33 00 54 11 00 H 43 11 31 78 HT 10 00 82 19 00 78 19 3by4 4by3 The transpose can also be taken of a set of multiplied matrices by reversing the order of the multiplications and individually transposing the matrices HADTDTATHT A matrix is symmetric when the rows are equal to the corresponding column In other words the matrix is equal to its transpose 60 20 10 00 10 00 20 50 10 20 00 00 10 10 60 10 20 10 00 20 10 40 10 20 10 00 20 10 50 10 00 00 10 20 10 50 Basic Linear Algebra Page 4 of 4 Modeling Spring Supports Objectives 1 Use springs to model xed boundary conditions 2 Model slanted Boundary Conditions with rotated springs 3 Choose the correct spring value for modeling xed supports So far we have handled xed support conditions by eliminating a DOF at the location of xity This has the effect of specifying the displacement at that DOF as exactly zero In addition this method reduces the total number of unknowns to only those that are not speci ed Another method for specifying xity is to allow the DOF to exist and then place a very stiff spring at that DOF to restrain movement For example a simple single bay frame can be modeled as having fixed supports with either of the following methods feds PM 3 s s Actual ModeI No DOF at Fixity DOF with stiff springs Methods for Modeling a Fixed Support If a roller support were needed this could be handled by placing a very stiff spring at the roller location offering vertical support only i Q j 3 Model with roller support Model with Stiff spring Methods for Modeling a Roller Support Modeling 7 Spring Supports Page 1 of1 Inclined Supports The use of springs becomes necessary when the support is needed in a direction other than one of the global directions For example if an inclined roller were needed this could be handled by placing a spring at the roller location However the spring must be rotated to offer support perpendicular to the incline Since the rotation of a stiffness matrix is a standard transformation this is easily accomplished 6 Inclined roller support Stiff spring representing roller Model for II IGIIIIOG Support The use of rotated spring supports is a common method of specifying nonglobal boundary conditions Ifthe values of the springs are very large when rotated the off diagonal values also become very large These off diagonal values can cause numerical instability during the solution of the equilibrium stiffness equations The solution to this problem is to use spring stiffness values appropriate to the problem The rule of thumb for choosing support stiffness values is that the value should be approximately 1000 times greater than the structure they are connected to For example when using the structure below that requires an inclined roller we need to check the stiffness of the structure that the spring is to support As can be seen the support is to restrain displacements of the type drawn dlsplacement Unlt dlsplacement In the dlrectlon of the restralnt Modeling 7 Spring Supports Page 2 of2 Since stiffness is de ned as force divided by displacement we need to nd the approximate stiffness of the structure for the above displacement This is approximately the stiffness of a straight cantilever subjected to a tip displacement The stiffness in this case is KZ3E3 L Where L is the total length of the cantilever Using this as a guide the approximate stiffness of the spring should be 3 000 E L3 K Spring 5 Lets look at another example of trying to find the approximate structural stiffness Using the frame structure below lets find what the support spring values should be gt i ml dlmn Frame Example for Sprlng Support As before the first step is to draw the approximate de ected shape for a displacement in the direction of the required support Now looking at this deflected shape we can see that the member that absorbs most of the vertical displacement is the beam Also this beam is primarily deformed in a perpendicular displacement mode Modeling 7 Spring Supports Page 3 of3 K 39 a K E 3 L 1 l l A 1 Vartlcal unlt dlsplacament Horlzontal unlt dlsplacemant Approxlmlh Stl nm Value for Spran Support This means that the approximate vertical structural stiffness to use as a fixed spring support would be K H 1 2000 E Spring L3 Where the length and properties used are those for the beam The approximate horizontal displacement and resulting stiffness is also shown It is important to note that these methods are approximate As such it is not crucial that the structure s stiffness be evaluated exactly The usual method for checking if the choice was correct is to look at the results and see if the displacement in the direction of the support is sufficiently small as to be negligible Ifit is not then increase the magnitude of the spring stiffness and rerun the problem It is important to remember that modeling is not an exact process and often requires a trial and error process to develop a satisfactory model Often times programs do not have a rotated spring boundary condition option If this is the case there is a simple alternative Translational springs can be modeled using truss or beam members This process is exact since a truss or beam can act as an axial member Therefore a truss bar can be placed so that the axis is parallel to the line of support The values for the area Young s Modulus and length need to be chosen so the truss has an equivalent stiffness of the required spring For example if a truss bar is to be used for the inclined cantilever we would model it as shown below Modeling 7 Spring Supports Page 4 of4 K Length Truu bar laud u oqulvulont auppon oprlng Note that a length of one and a Young s modulus of one could be used This is an arbitrary choice that causes the area to be equal to the desired stiffness of the spring since the stiffness is Modeling 7 Spring Supports Page 5 of5 Virtual Work Derivation Objectives Calculate the internal Virtual work for a truss 2 Calculate the internal Virtual work for a beam 3 Calculate the external Virtual work Derivation Assumptions Virtual Work will handle all of the requirements needed for this course Assumes that the structure is at a stable equilibrium position If a small perturbation movement from this position is introduced the structure will return to its equilibrium position In addition the work done to move the structure in an energy sense is zero The work of the applied loads moving through the perturbation is stored in the structure as elastic energy Two types of work External work is the work caused by the loads acting on the structure Internal work is a potential energy type work that is stored in beams columns and trusses as a result of stretching and bending The equation for the Virtual work balance can be written as 6 W2 6 W Definitions R is a vector of externally applied concentrated loads Individual terms in the vector R can be identi ed as R r is a vector of global external displacements for a structure The external Virtual work done on a structure can be calculated by 6W2ZRJrJZRJ r RTrrTR J J where the bar over the vector indicates that the quantity is a Virtual quantity Virtual Work Derivation Page 1 of 5 Or iRx rx dx where RX and rX are continuous functions of load and displacement S is the vector of internal forces They are concentrated values at particular points on a member For example the end moments on a beam are internal forces S 2 S5 8 3 Definition of Internal Forces V is the vector of internal displacements They are concentrated values corresponding to the internal loads For example the end rotations of a beam which correspond to the end moments v 2 V5 v 3 Definition of Internal Displacements The internal virtual work is 5W12S jvZSJVJ TgtXltVVTgtXltS Virtual Work Derivation Page 2 of 5 Internal Work for Trusses Consider a bar subjected to an axial load Sl It has a displacement V1 at the end of the bar S 1 g l 5 3 1 l L 1 v1 A Deformed Truss Bar Clearly the work force times displacement is W S1 V1 From mechanics we know that the displacement of an aXial member is related to the applied force by Solving for V1 we get Substituting into the work equation we have W1 S1 S1 L A E For the Virtual work either 81 or V1 need to be made Virtual EaL 6W1V1 1V 1S1 AE Internal Work for Beam Elements The stored energy in a beam can only come from bending ignoring aXial and shear work Virtual Work Derivation Page 3 of 5 The internal work for any continuum can be measured as strain energy Strain energy is the product of stress times strain X 9 Baa Deformed Beam Differential Element Pure Bending of Beam Element The internal work equal to the strain energy of the beam is W I IasdAdxI j jasdydzdx lzngth area lzngih mam hzxghi Again from mechanics we know the following where M is the moment on the face y is the distance from the centroid I is the moment of inertia and E is Youngs modulus Substituting we get dmx W 7 I E lzngth area The moment M moment of inertia I and Youngs Modulus E are all constant for a cross section and can be removed from the integration on the area dy and dz This leaves Virtual Work Derivation Page 4 of 5 MM 2 W1 d dz dx E12 y But the integral of y2 is the de nition of the moment of inertia This nal equation is of the form W Mde E lzngth For virtual work one of the moments needs to be made a virtual quantity MM 6W1 dx E length M Note that 1s de ned as the curvature w Therefore beam work can be thought of as moment times curvature integrated over the length of the beam 6W1 Myax lzngth This is equivalent to moment times rotation Virtual Work Derivation Page 5 of 5 Direct Stiffness Frame Example Objectives 1 Apply the direct stiffness procedure to a structure 2 Form the required a matrices 3 Solve for structural displacements and forces Solve the following structure for displacements and forces using the direct stiffness method W forceunit length zoj l Vl f L Direct Stiffness Example The rst step is to choose the global DOF DOF for Direct Stiffness Example Direct Stiffness Frame Example Page 1 of 5 The second step is to form the a matrices for each element Unit Displacement at DOF 2 Direct Stiffness Frame Example Page 2 of 5 Unit Displacement at DOF 3 The element coordinate systems are chosen and shown on the rst displaced shape Multiply and sum the results for each element to get K KalKga1a Kza2K1K2 Where the subscript again represents which element produced the contribution to the global stiffness 4E 0 6E L L2 K17 0 0 0 6E 0 12E L2 L3 Direct Stiffness Frame Example Page 3 of 5 L L K2EI 4E10 2 L L 0 00 8E 2E 6E L L L2 6E 0 12E L2 L3 A diagonal term in a stiffness matrix says there is an element connected to that DOF In order for a structure to be stable there should always be a nonzero diagonal in the global stiffness The element stiffness may however have zero diagonal stiffness terms An off diagonal term in a stiffness matrix means that an element is linking the two DOF where the term is located For example the K12 has a nonzero value This means that some element or elements connects DOF l and DOF 2 Next we need the load vector R First consider the concentrated loads acting directly at the DOF Rmmentmted 0 20 Next include loads within the span of an element Direct Stiffness Frame Example Page 4 of 5 Using the conversion equation we get the global load vector contribution R2 a S Ifwe multiply out the matrices we get R2 Note that the element contributes directly to the first two DOF The nal form of the load vector is R Rmnczntmtzd R1 R2 wL2 12 wL2 12 20 L J Using MathCad 7 you can easily perform the required multiplications Direct Stiffness Frame Example Page 5 of 5 Consistent Deformations Derivation Objectives 1 De ne consistent deformation variables 2 State consistent deformation procedure steps Consistent deformations is based on compatibility Compatibility states that a structure can deform in any shape provided the displacements conform to the boundary conditions and the material restrictions ie the material can not tear apart Consistent deformations process Remove all the redundant reactions and forces ie add hinges until it becomes determinate Then apply forces at the removed redundants to put the structure back into a compatible displacement Example take the uniformly loaded propped cantilever beam One redundant or extra support Remove the prop or end support we have a cantilever beam 2 Apply a force at the point where the support was removed 3 Need to calculate the force needed to push the tip back up to the support Original Structure r I 10 Need r I r reaction Two Equivalent Structures Reaction 1o rreanlicin Using Superposition Derivation of Consistent Deformation Method The tip de ection under uniform load is Consistent Deformation Derivation Page 1 of 3 wL4 8E r10 The tip de ection due to a unit force at the end of the beam quot If 1 unit load ALA Displacement Caused by Unit Load L3 fu 3E Apply a tip force equal to the ratio of the de ection without the prop to that of the unit force wL4 E Rpmp 7 L3 3E De nitions r is the nal displacement at removed redundant i This is usually zero but can also be a nal support displacement such as a support settlement ris is the displacement at redundant i on the determinate base structure due to settlement of the supports This is only a rigid body displacement and only requires geometry to calculate rit is the displacement at redundant i on the determinate base structure due to temperature rie is the displacement at redundant i on the determinate base structure due to initial imperfections no is the displacement at redundant i on the determinate base structure due to the applied loading Consistent Deformation Derivation Page 2 of 3 j is the displacement at redundant i on the determinate base structure due to a unit force at redundant j Xj is the unknown redundant force at redundant j This can be a support reaction or an internal force like a moment or an axial force The second index refers to the load applied that caused the de ection Consistent Deformation Solution Procedure 1 Remove all redundant forces to form a determinate base structure This can involve removing supports adding hinges or cutting members The choice of which forces to remove is arbitrary however clever choices can make the calculations simpler 2 Calculate the displacements at each of the removed redundants due to all the applied loading This should include loads temperature and settlement Each type of loading is considered to be its own load case Therefore applied loads are considered as load case 0 Temperature is considered as load case t etc 3 Apply a unit force at each redundant one at a time and find the displacements at all of the other removed redundants Each redundant creates a separate load case designated as load case j where j corresponds to the redundant at which the load is applied This will develop the terms fij The applied unit load is at redundant j The calculated displacements are at all other redundants i Each load case will fill out a column of the F matrix Applying the unit load at another redundant will develop another column of the F matrix 4 Apply the compatibility equation all redundants rrxsmrmr10 ijXj There is one compatibility equation for each removed redundant This will create a set of simultaneous equations with Xj s as the unknowns Solving the set of equations gives the values for the removed redundant forces Consistent Deformation Derivation Page 3 of 3 Structural Analysis Program Steps Objectives 1 Analyze a general structure using a nite element analysis program General structural analysis programs perform exactly the same steps as outlined before Below are the step required in all Finite Element analysis programs De ne the Nodes 1 De ne the geometric location 2 De ne the DOF for each node Coordinates Nodes are the analysis points in a structure connection and load points The have XY and Z locations in space This is called the nodal coordinates Degrees of Freedom In 2D a node is moves in the X and Y directions and rotates about the Zaxis In 3D a node is allowed to move in the X Y and Z directions and rotate about all three axes All nodes are generally assumed to have the six DOF shown below e Six possible DOF for a node For a 2D problem in the XY plane you only need to consider the following DOF Structural Analysis Program Steps Page 1 of 1 Three possible DOF in 2D for a nude bi ActuI DOF lar gonlrll 2D frlml Illmant You MUST fix restrain all DOF that do not have stiffness provided by a support or a connected element De ne all elements Elements are de ned by the nodes they connect to and their properties EI A This allows the formation of all element matrices stiffness in global directions ny the force transformation matrix FTCl K821 the LV vector the pointer form of the aL matrix and the member loads in global and local directions Sr and a S Define all loads Loads consist of two types Concentrated and Distributed Concentrated are applied directly to the nodes Distributed are applied on an element Structural Analysis Program Steps Page 2 of 2 Load Cases In design the building codes require you different combinations of loadings For example you might need to check dead plus live load 34dead 12 live wind and so on You de ne the base load cases applied load like dead weight wind etc For example three load cases are de ned below 10KlFt 8KFt 23KlFt lllllllll lllmlw 39 13 Kft 11KFt lllllllll lllmlw Dead Load Live Load Wind Load Load Combinations You could combine any number of load cases to form a combination For example load cases 1 and 2 Dead Live to form a combination needed to perform a code design SSTAN The beam member in SSTAN is a three dimensional bending member Therefore it has six force results at each end The coordinate system for these results is dictated by the K node location The following figures show the two possible results Note that SSTAN gives the results in terms of local coordinate planes Structural Analysis Program Steps Page 3 of 3 Knoll M a 14 14quotquotqu Pooltlvo Beam Forces K Node nght Structural Analysis Program Steps Page 4 of 4 SSTAN Capabilities Objectives 1 Use the SSTAN analysis program The general analysis program SSTAN is available for download from wwwceu edusoftwarehtm It is a 3D nite element program that has a graphics post processor It is a batch program that assumes the structural information is contained in an input le named nameinp The results are put into a le named nameout The users manual is available in the help le from STANPLOT The input is prepared and stored in the input le The general rules for the input le are The rst line in the le is an analysis title IT MUST BE THE FIRST LINE It is also important that it does not begin with any of the header names like COORDINATE FRAME etc The second line is the analysis control information line It is of the form N1N2N3N4 Where N1 is the number of nodes in the structure N2 is the number of different element types in the structure Again this means if you are using both truss and beams there are two regardless of the number of member properties N3 is the number of load cases N4 is the number of load combinations The rest of the INPUT is on a free formatted header basis All data is alranged by groups and is signi ed by a header For example nodal coordinates are signi ed by the header COORDINATE This data is order independent That is these data blocks can be placed in the INPUT le in any order All data groups must end with a blank line The following headers are available COORDINATE This speci es the nodal coordinates of the structure to be analyzed Note that the program assumes the structure is 3D BOUNDARY This speci es the boundary conditions or nodal DOF Nodes can be either released or xed and only in the global XYZ coordinate system SSTAN Capabilities Page 1 ofl TRUSS This specifies the truss element data Trusses can have initial tension rigid end offsets and the ability to not take either compression or tension not both The zero compression option is useful for slender bracing members The zero tension option is useful for gap elements or uplift problems BEAM This specifies the bending member data These elements can be used for beams as well as columns Beams can have uniform loads applied to the member rigid end offsets and can include PA effects LOADS This specifies the concentrated loads applied to the structure They can be concentrated loads or moments COMBINATIONS This section allows linear combinations of the basic load cases This option is not valid when using noncompressiontension trusses or PA effects PLATE MEMBRANE SHELL BRICK and AXISYMMETRIC These are other types of members available These constitute what are usually called finite elements and will not be discussed here but covered in later chapters Several examples will be given to demonstrate the use of SSTAN The coordinate system assumed is SSTAN is a right hand rule system All displacements forces and moments are given in this righthanded system The results for element forces are given at the nodes of a member in the local coordinate system The local coordinate system for the truss and beam members are given below SSTAN Capabilities Page 2 of2 Axlll ram anur MamM K node TenSIon J nude Positive Axial Force 3 Onl y I node Mumml Truss Forces Mm Frame Forces In 13 Plane Force Output for Truss and Beam Use of STANPLOT STANPLOT is a graphics post processor for the SSTAN program STANPLOT can plot any of the SSTAN element types displaced shapes axial and moment results for beams and truss members and stress contours for membranes plates shells and solid elements The program allows windowing and changing of the structures view as well as many other useful options STANPLOT has many additional capabilities It is an essential tool in checking the structure to see that its input and analysis are correct Using the plotting program things like connectivity nodal coordinates symmetry in displacements and many other things can be checked Verifying the structure and results is the most important part of a structural analysis SSTAN Solution Errors If a structure is unstable the stiffness matrix generated is singular As a result during the equation solving process an error will be generated specifying a singular matrix or that there is a Negative on the diagonal Some typical causes of a singular matrix are l The structure is unstable The cause of this is generally improper specification of boundary conditions As in the above example all out of plane DOF were fixed if this were not the case an error would have occurred giving a singular matrix SSTAN Capabilities Page 3 of3 Another possible cause is if the structure were to be placed on rollers At least ONE horizontal fiXity must be supplied for stability In the direct stiffness method all active DOF must have some stiffness associated with them Even if no load is applied to the horizontal direction the structure must be stabilized 2 The elements are not connected correctly causing an unstable structure Usually the real structure is stable but the model is unstable due to a typographical error in element connectivity Check to be sure the elements are connected to the proper nodes STANPLOT can be very helpful for this type of error 3 The element properties are specified incorrectly If an element property is left out or read in as zero this can cause the structure to be unstable This is common in 3d structures where the torsional property is left out One final point to note is that when distributed loads are used the maximum moment in the span is given along with the beam member forces This is useful for design where the maximum moment is required A good way to check the structure is through a graphical display of the structure and its results SSTAN has a graphical post processor that allows display of the structure displaced shapes and stress and displacement contours for finite elements To use the graphics post processor just run the program STANPLOT in the directory where the result data is contained SSTAN Capabilities Page 4 of4 Drawing Displaced Shapes Objectives 1 Draw basic displaced shapes 2 Draw displaced shape for beam structures 3 Draw displaced shape for frame structures Use only independent unknowns Slanted and axial members link DOF Displaced shapes are important for veri cation ALWAYS drawn very exaggerated First set of basic shapes are for cantilevers 90 Rotational Spring Free Three Basic Cantilever Shapes Second set of basic shapes are for propped cantilevers Rotational Spring amp ADA Q 61 e2 91 gt 92 Propped Cantilever Basic Shapes Drawing Displaced Shapes Page 1 of 5 Propped Can Hver amp as W Sp ng Beam Acting as Spring for Propped Cantiliver Assumes NO axial deformations Only displacements perpendicular to a member are allowed Circular Arc Small Deformation Theory Perpendicular Displacement General procedure is 1 To draw displaced shapes we start at a support 2 Let the opposite ends of the members that are connected to that support displace This means displacing and rotating the joint 3 Allow any other portion of the structure to follow as a rigid body 4 Move to the next joint and continue with other members 5 Continue until you reach the end Drawing Displaced Shapes Page 2 of 5 Vi WT Drawing Deflected Shapes Simple Frame The next shape to try is a portal frame Drawing De ected Shapes Portal Frame Drawing Displaced Shapes Page 3 of 5 The nal structure is one with a slanted member 1 ca 10 co Ice Additional examples of displaced shapes Drawing Displaced Shapes Page 4 of 5 Examples of Deformed Shapes Hint If you think a displacement is independent then try applying its displacement Holding others zero and seeing if the axial or perpendicular assumptions are Violated Axial Deformation Attempted Shape Axial Deformation Violation Drawing Displaced Shapes Page 5 of 5 CES 4141 Stress Analysis Summer 1998 Direct Stiffness Truss Application The following notes present the direct stiffness method for matrix truss analysis We will general ize to include beam elements later Symbols We ll stay consistent with the notation in Hoit r displacements in terms of global coordinate system global displacements V displacements in terms of element coordinate system local displacements a transformation matrix from global to local coordinates for displacements V ar R forces in terms of global coordinate system global forces S forces in terms of local coordinate system local forces b transformation from global to local coordinates for forces S bR k local element stiffness matrix local coordinates Ke element stiffness matrix in global coordinates KG Global structural stiffness matrix 1 Overview Application of the stiffness method of structural analysis requires subdividing the structure into a set of nite elements where the endpoints are called nodes For the case of trusses we will consider each whole truss member as a nite element and each joint becomes a node We ll deter mine the forcedisplacement relationship for each element separately then combine each individ ual contribution to the whole structure in a Global Structural Stiffness Matrix KG Now for any given loading scenario we can use Gaussian elimination to solve for all unknown displacements Once these displacements are known we can solve for all unknown forces which are the reac tions These notes will present the stiffness for each member at the local level then transform them into the global coordinate system then sum the contribution of all elements then discuss solu tions for displacements reactions and internal forces 2 Element Stiffness Matrix In Local Coordinates First we look at a single truss member lobal in its own local coordinate system and y g V2 32 define its stiffness Consider the element to LX the right The local coordinate system x y will align with the member orientation for any arbitrary orientation On this same gure the local displacements DOF are labeled with V1 at the end with the local axis origin and V2 at the far end 10 al NOW we can apply the Stiffness by de F1g 1 local truss member st1ffness nition procedure to find the stiffness matrix for this arbitrary element Holding Vll V20 we get the left column of the 2x2 stiffness matrix Then holding Vl0 V2l we get the right column Removing the AEL as a constant the result is given in Fig 1 Note that the stiffness is denoted by a lower case k The local forces S and the local displacements V can be related by 1 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 S kv expanded for the single truss element as S1 A E 1 1 V1 1 32 L 71 1 v2 Note that this will be the same no matter what the orientation of the element since al quanti ties are with respect to the element local coordinate system which will always be set up to lie along the axis of the member 3 Displacement Transformation Matrix Since a truss structure consists of many members of possibly many orientations we ll want to transform all local displacements into a single uniform global coordinate system This global coordinate system is shown in Fig 1 and is labeled xy The question is then if we were to nd the displacement at both ends of an element in terms of the local coordinates V1 V2 in terms of x what is that displacement in terms of the global coordinates r1 r2 r3 r4 in terms of x y Consider Fig 2 below The top left picture shows an individual element with the local dis placements V1 V2 the global displacements r1 r2 r3 r4 that we wish to conVert to and the local coordinate system with the origin at the lower end of the member x y Let s displace the ele ment at node 1 a unit Value in the r1 direction r11 r2r3r40 We can see this in the top right picture A displacement in global coordinate r1 is equal to a displacement in the local coordinate of VI r1cos 9x where ex is the angle of the element with the horizontal axis So cos 9x transforms r1 to V1 We can repeat this procedure for each of the global coordinates in turn be set ting each to 1 while holding all others to 0 r10 r21 r3r40 r1r20 r31 r40 and r1r2r30 r41 Note this is the same procedure we used to nd the stiffness matrix by de ni tion Let s do one more Set r41 r1r2r30 seen in the bottom picture in Fig 2 The transfor mation from r4 to V2 is v2 r4 cos 9y r4 V2 y 2 r3 X7 y r node2 r1cos9x r1 A x r1 node 1 X 7 Fig 2 global to local displacements r4 gtxlt r4 cos 9y 2 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 Although we didn t do the other two cases you should be able to Visualize that a displacement ofVl will be caused by both a displacement r1 and r2 and V2 caused by r3 and r4 If we complete the procedure for all four we can express V1 V2 in terms ofrl r2 r3 r4 as follows v1 r1quot cos9xr2cos9y 2 v2 r3quot cos9xr4cos9y 3 Let s represent Lx cos 9X and Ly cos 9y and rewrite these equations in matrix form as V Lx Ly 0 0 r2 Try expanding this and getting Eqs 2 and 3 4 v2 0 0 Lx Ly r3 r Let s de ne the transformation matrix and call it a a LxLy 0 0 5 0 0 LxLy Now we can relate global to local displacements in matrix form as v 61 r 6 Now we Ve found a way to relate local displacements to global displacements We ll also want to do this for local and global forces 4 Force Transformation Matrix We can also identify the transformation matrix for the localglobal force relationship Refer ring to Fig 2 replace local displacements with local forces V1 V2 with 81 S2 and replace glo bal displacements with global forces r1 r2 r3 r4 with R1 R2 R3 R4 This time we ll get the global on the left hand side and the local on the right hand side by applying a unit force in local coordinates and nding the resultant global components The resultant is R1 Lx 0 R2 Ly 0 S1 That is R aTS where aT indicates transpose of a 7 R3 0 Lx S2 R4 0 Ly 5 Element Stiffness Matrix In Global Coordinates Now we ll use the transformation matrices we just deriVed to nd the stiffness matrix for a single element in terms of global coordinates This will expand our element stiffness matrix from a 2x2 to a 4x4 Why We only haVe 2 degrees of freedom per truss element in local coordinates but we haVe 4 DOF per element in global degrees offreedom See Fig 2 3 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 Let s rewrite Eq 1 which is the force displacement relationship all in local coordinates S k v 8 Now we ll transform local displacements V to global displacements r by substituting Eq 6 into Eq 8 to get S kar 9 Finally if we premultiply both sides by transa the left hand side becomes Eq 7 and the nal expression is R aTkar 10 We have the displacements and forces in terms of global coordinates now The stiffness matrix is still local and is pre and postmultiplied by transa and a Let s separate this component and rename it as T K e a ka 11 where Ke now represents the local stiffness matrix in terms of global coordinates This gives us a final expression of R K e r 12 Let s write out and multiply through Eq 11 LxO Kez4E Ly 0 171 LxLy 0 0 13 L 0Lx7110 OLxLy OLy Performing the matrix multiplications gives us Lx2 LxL y 7Lx2 iLxLy 2 2 Kg ATE LxLy Ly iLxLy iLy 14 7Lx2 iLxL y Lx2 LxL y iLxLy 7Ly2 LxLy Ly2 And finally Eq 12 can be written fully as 2 2 R1 Lx LxLy iLx iLxLy r1 R2 AE LxLy Ly2 iLxLy 7Ly2 r2 15 L R3 7Lx2 iLxLy Lx2 LxLy r3 R4 r4 iLxLy 7Ly2 LxL y L y2 4 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 6 Assembly of Global Stiffness Matrix Before going on let s review what we have derived Eq 15 relates the displacements to forces all in global coordinates for a single element of arbitrary orientation The orientation of the individual member is accounted for in the Global element sti ness matrix Ke What s left to do We now have the contribution of a single element For a structure with mul tiple members we assemble a Ke for each member in the truss structure then add them each into a global structural sti ness matrix that covers every degree of freedom in the entire structure We ll call that KG KG will be a square matrix with as many rows and columns as there are total degrees of freedom frozen and unfrozen Let s do this by example i node r3 D element Fig 3 Example 1 a y r2 r4l x r1 r3 Fig 4 Element 1 r5 y Fig 5 Element 2 an r1 5 of 10 Example 1 Assemblv of KG for a truss structure We ll assemble the global structural stiffness matrix for the structure in Fig 3 If we solved this via chapter 4 Hoit by developing stiffness by de nition we d create a 2x2 stiffness matrix representing only the unfrozen DOF r1 r2 In the problem at the left note that we have labeled the unfrozen DOF as r1 and r2 then the frozen DOF as r3 r4 r5 r6 we know that r3r4r5r60 this will help later For each DOF there is an associated R representing a force R1 R2 will be the external forces applied to the structure and R3 through R6 represent the reaction at the pins The process is simply to create Ke for each of the two members using Eq 14 then to add them by keeping track of how the local degrees of freedom correspond to the glo bal degrees of freedom Element 1 the element labeled 1 connects nodes 2 and 3 and global DOF r1 r2 r3 r4 We need to nd Lx and Ly so we can apply Eq 14 If we de ne the local origin x y for element 1 at node 2 we have LxchangeinxL 3031 LychangeinyL003 0 this gives us Eq 14 r1 r2 r3 r4 0333 0 70333 0 r1 Kel AE 0 0 0 0 a 16 70333 0 0333 0 r3 0 0 0 0 r4 Element 2 This element connects nodes 2 and 3 and global DOF r1 r2 r5 r6 We pick the local origin x y at node 2 again giving us Lx change in x L 3 05 06 Ly change in x L 4 05 08 this gives us Ke2 through Eq 14 as 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 r1 r2 r5 r6 0072 0096 70072 70096 K82 AE 0096 0128 70096 70128 70072 70096 0072 0096 70096 70128 0096 0128 r1 r2 r5 r6 17 Note that we have listed above and to the right of each element global stiffness matrix the global displacements that correspond to the four local degrees of freedom Now we have all element stiffness matrices We simply set up the global structural stiffness matrix as a 6x6 with all zeros initially and add in the element matrices KG Kel Ke2 18 where for example we add the r5 row and r2 column of Ke2 into the r5 row and r2 column of KG r1 r2 r3 r4 0333 0 70333 0 r1 KG AE 0 0 0 70333 0 0333 0 r3 0 0 0 0 r4 which gives us con rm this yourself r1 r2 r3 r4 r5 0405 0096 70333 0 70072 70096 0 70096 70128 0096 0128 0 70072 70096 0072 0096 r 70096 70128 0096 0128 r6 KG AE 70333 0 0333 0 0 0 0 0 0 0 0 0 70072 70096 0 0 0072 0096 70096 70128 0 0 0096 0128 r1 0072 0096 70072 70096 0 r2AE 0096 0128 70096 70128 19 r2 r5 r6 r1 r2 UI r1 r r3 r4 r5 r N 20 ON If A and E are different for each member we expand them into Kel and K632 before adding 7 The Solution Procedure Now we can write the full system as R1 0405 0096 70333 0 70072 70096 R2 0096 0128 0 0 70096 70128 R3 AE 70333 0 0333 0 0 0 R4 0 0 0 0 0 0 R5 70072 70096 0 0 0072 0096 R6 70096 70128 0 0 0096 0128 6 oflO 62998 21 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 Since we initially labeled the unfrozen DOF first R1 and R2 are the applied external forces r1 and r2 are the sought displacements in global coordinates The knowns Rk R1 R2 external forces rk r3 r4 r5 r6 frozen displacements all 0 The unknowns Ru R3 R4 R5 R6 reactions ru r1 r2 displacements Now we can partition Eq 21 and separate knowns and unknowns R1 0405 0096 70333 0 70072 70096 r 1 R2 0096 0128I 0 0 70096 70128 r2 R3 AE 70333 0 70333 0 0 0 V3 22 R4 0 0 I 0 0 0 0 r4 R5 70072 70096I 0 0 0072 0096 r5 R6 70096 70128l 0 0 0096 0128 r6 and rewrite as 5k AJE 5131312 L 23 Ru K21K22 rk now expanding we getmake sure you can follow this step Rk AEK11ruK12rk 24 Ru AEK21ruK22rk 25 We can solve the system in Eq 24 which is two equations and 2 unknowns for ru then solve the system in Eq 25 a 4 equation 4 unknown system Once we have solved Eqs 24 and 25 we have displacements and reactions Let s note this We Now Have 1 displacements at the nodes 2 reaction forces What we do not have yet is a description of the internal forces in each member Let s note this and get back to it after an example We Still Need 3 member forces internal axial forces and shear and moments when we go to beams First let s complete an example using the assembled system in Eq 22 7 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 Example with loading If we put the 10 kip load on the structure from Fig 3 as shown to the left the solution set up is exactly Eq 22 where we now know that R10 R2 10 We also know there is zero displacement at all supports so we can solve the portion of Eq 22 we pulled off into Eq 24 When we do this we get 0 0 AE 0405 0096 r1 K12 0 26 710 0096 0128 r2 0 0 N 10 kips Fig 6 Example 1 with forces Note I didn t bother to ll in K12 although we have it since it is multiplied by a zero vector anyway Solving Eq 26 for r1 r2 gives us make sure you can do this by hand by Gaussian elimination r1 2252AE r2 9502AE We now have displacements at the nodal DOF free to move We can now turn to Eq 25 to solve for reactions Ru The set of equations to solve are now R3 70333 0 2252 0 R4 AE 0 0 AE 102 0 27 R5 70072 70096 79502 0 R6 70096 70128 AE 0 The reactions R1 R2 R3 R4 come out directly from matrix multiplication We don t even need Gaussian elimination the result is R3 7 5 R4 0 R5 75 R6 100 all in Kips lt 7 5 Let s redraw the problem with the reactions to see if they make sense A quick check on equilibrium con rms this solu y 100 kips tion in Fig 7 Note that displacements r1 r2 are in terms of the unknown values of AE and that we got numbers for the reactions Equi librium is not dependent on material properties while dis placements are so this all makes sense Let s now move onto the topic of force recovery or nd ing internal forces in the members making up the structure Fig 7 Resulting reactions 8 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 8 Internal member force recovery We have already solved for the displacements at all the nodes We also have stiffness matrices that relate forces to displacements and a description of the geometry of each element in the trans formation matrices It certainly seems that we have enough information to nd internal forces since this is analogous to having displacement and stiffness for a spring where we want the force Go back and take a look at Eq 8 which relates forces to displacements in local coordinates through the local element stiffness matrix S kv What do we have now We have the local element stiffness matrix k which is always just AE 171 ke 28 L 1 1 lt gt We don t really have v since this is displacement in local coordinates We solved for r which is displacement in global coordinates But we did nd the relationship between v and r which is what we called the transformation matrix a Now go look at Eq 6 We see that v ar This is how we got Eq 9 which is repeated here S kar 29 This is exactly what we want The left hand side S is in local coordinates so it will give us exactly axial force no matter what the orientation of the member the right hand side is in terms of the global displacements local stiffness and transformation matrix all of which we have for each element Let s write it out as r1 S1AE171LxLy0 0 V2 30 32 L711 OOLxLy r3 r4 For a truss it is always the case that S1 7S2 So we need only solve for either S1 or S2 Expanding just the bottom row in Eq 30 we can express the axial force in any member by r1 7 AE r2 S2 7 31 L iLx iLy Lx Ly r3 r4 Where we choose S2 since will be tension and will be compression as is our convention An important notation comment is necessary We use r1 r2 r3 r4 above to represent the local labeling on the element In practice they may not be 14 but whatever the global notation hap pens to be We will see this when we complete the example 9 of 10 62998 Direct Stiffness Truss Application CES 4141 Stress Analysis Summer 1998 Example completed force recoveg We repeat the figures that de ne the problem so we can see the degrees of free dom and individual elements The displacements we found were r1 2252 AE r2 9502 AE 41 r3r4r5r60 Now we solve for element internal axial forces Element 1 I L Recall that we found the following node LxchangeinxL 3031 D element Ly change in y L 0 03 0 which gives 2252 r1 AE 12 r4 S quot r2775 39 l X 3 71 0 1 0 AE compress1on r1 r3 0 r3 0 r4 jgX Element 2 Lx change in x L 3 05 06 Ly change in x L 4 05 08 which gives g 2252 r1 AE y S A E r2 125 tension 2 5 706 708 06 08 AE r L 0 r5 r1 0 r6 9 Final Comments For truss structures with more members the solution procedure from displacements to reac tions to internal forces stays the same The KG global stiffness matrix may be larger and be made up from more element contributions but the procedure is identical Also we ll see that the structure can be either determinate or indeterminate and the same pro cedure applies The partitioning that is done in Eq 22 will change since the number of known displacements will change but the total number of unknows remains the same For example if we put a roller at node 2 in our example that xes dof 2 r2 then the initial system we solve for displacements Eq 24 will be a single equation and unknown instead of the 2 we had for the given example and the next system to get reactions Eq 25 will be a 5 equation system Next up beams at the element level then building frame KG matrices 10 of 10 62998 Direct Stiffness Truss Application

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