Stiffness Matrix for a Bar Element - Memphis

1 Chapter 3b Development of Truss Equations Learning Objectives To derive the Stiffness Matrix for a bar Element . To illustrate how to solve a bar assemblage by the directstiffness method. To introduce guidelines for selecting displacementfunctions. To describe the concept of transformation of vectors intwo different coordinate systems in the plane. To derive the Stiffness Matrix for a bar arbitrarily orientedin the plane. To demonstrate how to compute stress for a bar in theplane. To show how to solve a plane truss problem. To develop the transformation Matrix in three-dimensional space and show how to use it to derive thestiffness Matrix for a bar arbitrarily oriented in space. To demonstrate the solution of space Matrix for a Bar ElementInclined, or Skewed SupportsIf a support is inclined, or skewed, at some angle for the global xaxis, as shown below, the boundary conditions on the displacements are not in the global x-ydirections but in the x -y 7/8117 Chapter 3 - Truss Equations - Part 21/44 Stiffness Matrix for a Bar ElementInclined, or Skewed, SupportsWe must transform the local boundary condition of v 3= 0 (in local coordinates) into the global x-ysystem.

2 Stiffness Matrix for a Bar ElementInclined, or Skewed, SupportsTherefore, the relationship between of the components of the displacement in the local and the global coordinate systems at node 3 is:3333'cos sin'sin cosuuvv We can rewrite the above expression as: 333'[]dtd We can apply this sort of transformation to the entire displacement vector as: 11'[][] 'TdTdordTd 3cos sinsin cost CIVL 7/8117 Chapter 3 - Truss Equations - Part 22/44 Stiffness Matrix for a Bar ElementInclined, or Skewed, SupportsWhere the Matrix [T1]Tis:Both the identity Matrix [I] and the Matrix [t3] are 2 x 2 force vector can be transformed by using the same [] [0] [0][] [0] [] [0][0] [0] [ ]ITIt 1'[]fTf In global coordinates, the force-displacement equations are: []fKd Stiffness Matrix for a Bar ElementInclined, or Skewed, SupportsApplying the skewed support transformation to both sides of the equation gives:By using the relationship between the local and the global displacements, the force-displacement equations become:Therefore the global equations become: 11[] [][]Tf TKd 11'[][][] 'TfTKTd 11112211223333[][][]''''xyxTyxyFuFvFuTKT FvFuFv []fKd 1[] 'TdTd CIVL 7/8117 Chapter 3 - Truss Equations - Part 23/44 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemDetermine the Stiffness Matrix for each the plane truss shown below.

3 Assume E= 210 GPa, A= 6 x 10-4m2for Element 1 and 2, and A= (6 x 10-4)m2for Element 3. 222222222 CSCSCCSSCSCSAEkLCSCSCCSSCSCS Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe global elemental Stiffness Matrix for Element 1is:(1)cos0 112262 42(1)00000101(210 10 / )(6 10)000010101uvuvkN mmm k22222222 CSCSCCSSCSCSAEkLCSCSCCSSCSCS (1)sin1 CIVL 7/8117 Chapter 3 - Truss Equations - Part 24/44(2)cos1 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe global elemental Stiffness Matrix for Element 2is:22 3 362 42(2)10 1000 00(210 10 / )(6 10 )10 10100 00uv u vkN mmm k22222222 CSCSCCSSCSCSAEkLCSCSCCSSCSCS (2)sin0 (3)2cos2 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe global elemental Stiffness Matrix for Element 3is.

4 11 3 36242(3)1 1111 111(210 10 / )(6 2 10 )1111221111uv u vkN mmm k22222222 CSCSCCSSCSCSAEkLCSCSCCSSCSCS (3)2sin2 CIVL 7/8117 Chapter 3 - Truss Equations - Part 25/44 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemUsing the direct Stiffness method, the global Stiffness Matrix is: ,15mNKWe must transform the global displacements into local coordinates. Therefore the transformation [T1] is:222222221000 0 00100 0 00010 0 00001 0 000000000 13[] [0] [0][] [0] [] [0][0] [0] [ ]ITIt Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe first step in the Matrix transformation to find the product of [T1][K]. 1TK 0 0 ,260 101 NTKm 2222222251000 0 0 0 00010101,0101000001 0 0 0 10Nm 1T KCIVL 7/8117 Chapter 3 - Truss Equations - Part 26/44 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe next step in the Matrix transformation to find the product of [T1][K][T1]T.

5 22222222T51000 0 0 0 00010100101000001 0 0 0 , 2 6 0 1 011 NTKmT 0 1 0 ,260 1011 TKNTm T1T 1 TKStiffness Matrix for a Bar ElementExample 9 Space Truss ProblemThe displacement boundary conditions are:112 3'0uvv v 51122331122331, 2 6 0 1 0 1 0 ''''xyxyxyNmFFFFFF uvuvuv CIVL 7/8117 Chapter 3 - Truss Equations - Part 27/44 Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemBy applying the boundary conditions the global force-displacement equations ,260 10' Solving the equation 231,000'0xxFkNF 3' Stiffness Matrix for a Bar ElementExample 9 Space Truss ProblemTherefore:The global nodal forces are calculated 0 1 0001 0 01, 2 6 0 1 0''xyxNmmyxyFFFFFF 11500500xyFkNFkN 230'707yyFFkN 11112211223333[][][]''''xyxTyxyFuFvFuTKT FvFuFv CIVL 7/8117 Chapter 3 - Truss Equations - Part 28/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsThe differential internal work (strain energy) dUin a one-dimensional bar Element is:Let s derive the equations for a bar Element using the principle of minimum potential energy.

6 The total potential energy, p, is defined as the sum of the internal strain energy Uand the potential energy of the external forces :pU ()()()xxdUyzx d Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsSumming the differential energy over the whole bar gives:If we let the volume of the Element approach zero, then:xxdUd dV 0xxxVUddV For a linear-elastic material (Hooke s law) as shown below:xxE 0xxxVEd dV 212xVEdV 12xxVUdV CIVL 7/8117 Chapter 3 - Truss Equations - Part 29/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsThe potential energy of the external forces is:The internal strain energy statement becomes12xxVUdV sVSXudVTu dSfu Mbx ixii1where Xbis the body force (force per unit volume), Txis the traction (force per unit area), and fixis the nodal concentrated force.

7 All of these forces are considered to act in the local Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element Equations1. Formulate an expression for the total potential Assume a displacement Obtain a set of simultaneous equations minimizing the total potential energy with respect to the displacement the following steps when using the principle of minimum potential energy to derive the finite Element 7/8117 Chapter 3 - Truss Equations - Part 210/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsWe can approximate the axial displacement as:Consider the following bar Element , as shown below:220 Lpxx1x1xsVSAdx f uf u2X u dVT u dS bx 1122uuNNu 11xNL 2xNL Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element Equationswhere N1and N2are the interpolation functions gives as.

8 Using the stress-strain relationships, the axial strain is:xdudx 1211xuuLL 11 BLL The axial stress-strain relationship is: []xxD []{}xBd 1122udNdNudxdx CIVL 7/8117 Chapter 3 - Truss Equations - Part 211/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsThe total potential energy expressed in Matrix form is:For the one-dimensional stress-strain relationship [D] = [E] where Eis the modulus of elasticity. Therefore, stress can be related to nodal displacements as:where {P} represented the concentrated nodal loads. [][]xDB d 0 LTTTTpxxVSAdxdPuXdVuT dS2 bxStiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsIf we substitute the relationship between and into the energy equations we get: u d 0 LTTTTpTTTTsVSAdBDBddxdP2dN XdV dN TdS bxIn the above expression for potential energy pis a function of the d, that is: p= p( ).

9 1,u2uHowever, [B] and [D] and the nodal displacements uare not a function of x. x x CIVL 7/8117 Chapter 3 - Truss Equations - Part 212/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationswhereIntegration the energy expression with respect to xgives: [][][]2 TTTTpALdBDBd d f [][]VSfP NXdVNXdS TTbbWe can define the surface tractions and body-force matrices as: []xSfNTdS Ts []VfNXdV TbbStiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsMinimization of pwith respect to each nodal displacement requires that:For convenience, let s define the following1200ppuu *[][][]TTTUdBDBd 1*122111[]1uLUuuEuLLL CIVL 7/8117 Chapter 3 - Truss Equations - Part 213/44 Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsSimplifying the above expression gives:The loading on a bar Element is given as: *22112222 EUuuuuL 112 2 Txxdf uf uf Therefore, the minimum potential energy is: 12112202pxAEuufuL 12222202pxAEuufuL Stiffness Matrix for a Bar ElementPotential Energy Approach to Derive Bar Element EquationsThe above equations can be written in Matrix form as:The Stiffness Matrix for a bar Element is.

10 This form of the Stiffness Matrix obtained from the principle of minimum potential energy is identical to the Stiffness Matrix derived from the equilibrium equations. 112211011pxxufAEufdL 1111 AEkL CIVL 7/8117 Chapter 3 - Truss Equations - Part 214/44 Stiffness Matrix for a Bar ElementExample 10 - Bar ProblemConsider the bar shown below:The energy equivalent nodal forces due to the distributed load are: 0[]xSfNTdS T 00Lx1fLfCxdxfxL 1x2xStiffness Matrix for a Bar ElementExample 10 - Bar ProblemThe total load is the area under the distributed load curve, or:The equivalent nodal forces for a linearly varying load are:ff 1x2x21()( )22 CLFLCL 11of the total load33xFf 222of the total load33xFf 0Lx1 LCx dxxL 00L23L3 CxCx23 LCx3L 22CL6CL3 CIVL 7/8117 Chapter 3 - Truss Equations - Part 215/44 Stiffness Matrix for a Bar ElementExample 11 - Bar ProblemConsider the axially loaded bar shown below.