CIVL 7/8117 Chapter 3a - Development of Truss Equations 1/80

Transcription

1 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Development of Truss Equations Having set forth the foundation on which the direct stiffness method is based, we will now derive the stiffness matri for a linear-elastic bar (or truss) element using the general steps outlined in Chapter. We will include the introduction of both a local coordinate system, chosen with the element in mind, and a global or reference coordinate system, chosen to be convenient (for numerical purposes) with respect to the overall structure. We will also discuss the transformation of a vector from the local coordinate system to the global coordinate system, using the concept of transformation matrices to epress the stiffness matri of an arbitrarily oriented bar element in terms of the global system. Development of Truss Equations Net we will describe how to handle inclined, or skewed, supports. We will then etend the stiffness method to include space trusses. We will develop the transformation matri in three-dimensional space and analyze a space truss. We will then use the principle of minimum potential energy and apply it to the bar element equations. Finally, we will apply Galerkin s residual method to derive the bar element equations.

2 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Development of Truss Equations Development of Truss Equations

3 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Development of Truss Equations Development of Truss Equations

4 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Development of Truss Equations Development of Truss Equations

5 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Consider the derivation of the stiffness matri for the linearelastic, constant cross-sectional area (prismatic) bar element show below. This application is directly applicable to the solution of pinconnected truss problems. Consider the derivation of the stiffness matri for the linearelastic, constant cross-sectional area (prismatic) bar element show below. where T is the tensile force directed along the ais at nodes and, is the local coordinate system directed along the length of the bar.

6 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Consider the derivation of the stiffness matri for the linearelastic, constant cross-sectional area (prismatic) bar element show below. The bar element has a constant cross-section A, an initial length, and modulus of elasticity E. The nodal degrees of freedom are the local aial displacements u and u at the ends of the bar. The strain-displacement relationship is: du E d From equilibrium of forces, assuming no distributed loads acting on the bar, we get: A T constant Combining the above equations gives: du AE T d constant Taking the derivative of the above equation with respect to the local coordinate gives: d du AE d d

7 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 The following assumptions are considered in deriving the bar element stiffness matri:. The bar cannot sustain shear force: f f. Any effect of transverse displacement is ignored. 3. Hooke s law applies; stress is related to strain: E y y Step - Select Element Type We will consider the linear bar element shown below.

8 CIV 7/87 Chapter 3a - Development of Truss Equations 8/8 Step - Select a Displacement Function A linear displacement function u is assumed: u a a The number of coefficients in the displacement function, a i, is equal to the total number of degrees of freedom associated with the element. Applying the boundary conditions and solving for the unknown coefficients gives: u u u u u u u Step - Select a Displacement Function Or in another form: u u N N u where N and N are the interpolation functions gives as: N N The linear displacement function û plotted over the length of the bar element is shown below.

9 CIV 7/87 Chapter 3a - Development of Truss Equations 9/8 Step 3 - Define the Strain/Displacement and Stress/Strain Relationships The stress-displacement relationship is: du d u u Step 4 - Derive the Element Stiffness Matri and Equations We can now derive the element stiffness matri as follows: T A Substituting the stress-displacement relationship into the above equation gives: T u AE u Step 4 - Derive the Element Stiffness Matri and Equations The nodal force sign convention, defined in element figure, is: f T f T therefore, u u u u f AE f AE Writing the above equations in matri form gives: f AE u f u Notice that AE/ for a bar element is analogous to the spring constant k for a spring element.

10 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Step 5 - Assemble the Element Equations and Introduce Boundary Conditions The global stiffness matri and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations. n n ( ) ( e) e K F K k F f e e Where k and f are the element stiffness and force matrices epressed in global coordinates. Step 6 - Solve for the Nodal Displacements Solve the displacements by imposing the boundary conditions and solving the following set of equations: F Ku Step 7 - Solve for the Element Forces Once the displacements are found, the stress and strain in each element may be calculated from: du u u E d

11 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Eample - Bar Problem Consider the following three-bar system shown below. Assume for elements and : A = in and E = 3 ( 6 ) psi and for element 3: A = in and E = 5 ( 6 ) psi. Determine: (a) the global stiffness matri, (b) the displacement of nodes and 3, and (c) the reactions at nodes and 4. Eample - Bar Problem For elements and : For element 3: 6 3 () () 6 k lb lb in in k node numbers for element (3) 6 lb lb in in k 3 node numbers for element 3 4 node numbers for element 3 3 As before, the numbers above the matrices indicate the displacements associated with the matri.

12 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Eample - Bar Problem Assembling the global stiffness matri by the direct stiffness methods gives: E E E 3 6 K Relating global nodal forces related to global nodal displacements gives: F u F u 6 F3 u3 F 4 u 4 Eample Bar Problem The boundary conditions are: u u4 F F 6 u F3 u 3 F 4 Applying the boundary conditions and the known forces (F = 3 lb) gives: 3 u 6 u3

13 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Eample Bar Problem Solving for u and u 3 gives: u u 3. in.in The global nodal forces are calculated as: F F 6. 3 lb F3. F 4 Selecting Approimation Functions for Displacements Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement function.. Common approimation functions are usually polynomials.. The approimation function should be continuous within the bar element.

14 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Selecting Approimation Functions for Displacements Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement function. 3. The approimating function should provide interelement continuity for all degrees of freedom at each node for discrete line elements, and along common boundary lines and surfaces for two- and three-dimensional elements. Selecting Approimation Functions for Displacements Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement function. For the bar element, we must ensure that nodes common to two or more elements remain common to these elements upon deformation and thus prevent overlaps or voids between elements. The linear function is then called a conforming (or compatible) function for the bar element because it ensures both the satisfaction of continuity between adjacent elements and of continuity within the element.

15 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Selecting Approimation Functions for Displacements Consider the following guidelines, as they relate to the onedimensional bar element, when selecting a displacement function. 4. The approimation function should allow for rigid-body displacement and for a state of constant strain within the element. Completeness of a function is necessary for convergence to the eact answer, for instance, for displacements and stresses. Selecting Approimation Functions for Displacements The interpolation function must allow for a rigid-body displacement, that means the function must be capable of yielding a constant value. Consider the follow situation: u a a u u Therefore: Since u = a then: u Nu N u N N a u a N N a This means that: N N The displacement interpolation function must add to unity at every point within the element so the it will yield a constant value when a rigid-body displacement occurs.

16 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Transformation of Vectors in Two Dimensions In many problems it is convenient to introduce both local and global (or reference) coordinates. ocal coordinates are always chosen to conveniently represent the individual element. Global coordinates are chosen to be convenient for the whole structure. Transformation of Vectors in Two Dimensions Given the nodal displacement of an element, represented by the vector d in the figure below, we want to relate the components of this vector in one coordinate system to components in another.

17 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 Transformation of Vectors in Two Dimensions et s consider that d does not coincident with either the local or global aes. In this case, we want to relate global displacement components to local ones. In so doing, we will develop a transformation matri that will subsequently be used to develop the global stiffness matri for a bar element. Transformation of Vectors in Two Dimensions We define the angle to be positive when measured counterclockwise from to. We can epress vector displacement d in both global and local coordinates by: d uiv juivj

18 CIV 7/87 Chapter 3a - Development of Truss Equations 8/8 Transformation of Vectors in Two Dimensions Consider the following diagram: Using vector addition: a b i Using the law of cosines, we get: a i cos a cos Similarly: b i sin b sin Transformation of Vectors in Two Dimensions Consider the following diagram: The vector a is in the i direction and b is in the jdirection, therefore: a a i cos i b b j sin j

19 CIV 7/87 Chapter 3a - Development of Truss Equations 9/8 Transformation of Vectors in Two Dimensions Consider the following diagram: The vector i can be rewritten as: i cos i sin j The vector j can be rewritten as: j sin i cos j Therefore, the displacement vector is: cos sin sin cos uiv ju i j v i j u i vj Transformation of Vectors in Two Dimensions Consider the following diagram: Combining like coefficients of the local unit vectors gives: u cos v sin u u sin v cos v u C Su C cos v S C v S sin

20 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Transformation of Vectors in Two Dimensions The previous equation relates the global displacement d to the d local displacements C S The matri is called the transformation matri. S C The figure below shows u epressed in terms of the global coordinates and y. u Cu Sv Eample - Bar Element Problem The global nodal displacement at node is u =. in and v =. in for the bar element shown below. Determine the local displacement. Using the following epression we just derived, we get: u CuSv o o u cos6 (.) sin6 (.).3 in

21 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Global Stiffness Matri We will now use the transformation relationship developed above to obtain the global stiffness matri for a bar element. Global Stiffness Matri We known that for a bar element in local coordinates we have: f AE u f u f kd We want to relate the global element forces f to the global displacements d for a bar element with an arbitrary orientation. f u f y v k f=kd f u f y v

22 CIV 7/87 Chapter 3a - Development of Truss Equations /8 Global Stiffness Matri Using the relationship between local and global components, we can develop the global stiffness matri. We already know the transformation relationships: u u cos v sin u u cos v sin Combining both epressions for the two local degrees-offreedom, in matri form, we get: u u C S v u C S u v * d=td * C S T C S Global Stiffness Matri A similar epression for the force transformation can be developed. f f f C S y * f C S f T f f f y Substituting the global force epression into element force equation gives: * f =kd Tfkd Substituting the transformation between local and global displacements gives: * * * d=td TfkTd

23 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Global Stiffness Matri The matri T* is not a square matri so we cannot invert it. et s epand the relationship between local and global displacement. u C S u v S C v d=td u C Su v S C v where T is: C S S C T C S S C Global Stiffness Matri We can write a similar epression for the relationship between local and global forces. f C S f f y S C f y f =Tf f C S f f y S C f y Therefore our original local coordinate force-displacement epression f AE u f u f =kd

24 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Global Stiffness Matri May be epanded: f u f y v AE f u f y v The global force-displacement equations are: fkd Tf ktd Multiply both side by T - we get: f T ktd - where T - is the inverse of T. It can be shown that: T T T Global Stiffness Matri The global force-displacement equations become: Where the global stiffness matri k is: k T T kt T f = T ktd Epanding the above transformation gives: AE CS k CS C CS C CS S CS C CS C CS S CS S S We can assemble the total stiffness matri by using the above element stiffness matri and the direct stiffness method. n n ( ) ( e) e K F K k F f F Kd e e

25 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Global Stiffness Matri ocal forces can be computed as: f u f y v AE f u f y v C S u S C v C Su S C v AE f CuSvCu Sv f y AE f CuSvCu Sv f y Eample 3 - Bar Element Problem For the bar element shown below, evaluate the global stiffness matri. Assume the cross-sectional area is in, the length is 6 in, and the E is 3 6 psi. C AE CS k C CS CS CS S C CS CS CS S C CS S S Therefore: cos3 3 sin3 o o C S

26 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Eample 3 - Bar Element Problem The global elemental stiffness matri is: k in ( in ) 3 psi Simplifying the global elemental stiffness matri is: k lb in Computation of Stress for a Bar in the -y Plane For a bar element the local forces are related to the local displacements by: f AE u f u The force-displacement equation for f is: AE u f u The stress in terms of global displacement is: E C S u u C S v v E Cu Sv Cu Sv

27 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 Eample 4 - Bar Element Problem For the bar element shown below, determine the aial stress. Assume the cross-sectional area is 4-4 m, the length is m, and the E is GPa. The global displacements are known as u =.5 mm, v =, u =.5 mm, and v =.75 mm. E Cu Sv Cu Sv (.5) () (.5) (.75) kn 8.3 m MPa KN m Solution of a Plane Truss We will now illustrate the use of equations developed above along with the direct stiffness method to solve the following plane truss eample problems. A plane truss is a structure composed of bar elements all lying in a common plane that are connected together by frictionless pins. The plane truss also must have loads acting only in the common plane.

28 CIV 7/87 Chapter 3a - Development of Truss Equations 8/8 Eample 5 - Plane Truss Problem The plane truss shown below is composed of three bars subjected to a downward force of kips at node. Assume the cross-sectional area A = in and E is 3 6 psi for all elements. Determine the and y displacement at node and stresses in each element. Eample 5 - Plane Truss Problem Element Node Node C S 9 o 3 45 o o

29 CIV 7/87 Chapter 3a - Development of Truss Equations 9/8 Eample 5 - Plane Truss Problem The global elemental stiffness matri are: element : C S k () in in 6 ( )(3 ) CS CS C AE CS S k C CS S u v u v C CS CS CS C CS psi lb in S S element : C S () k in 6 ( )(3 ) 4 in u v u v 3 3 psi lb in element 3: C S k (3) in in 6 ( )(3 ) u v u v 4 4 psi lb in Eample 5 - Plane Truss Problem The total global stiffness matri is: K lb in The total global force-displacement equations are: element element element u, v F F y 5 5 F F 3 y F 4 F 4 y

30 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Eample 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: u, v F F y 5 5 F F 3 y F 4 F 4 y Eample 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: u 5, v Solving the equations gives: u.44 in v.59 The stress in an element is: E Cu Sv Cu Sv where i is the local node number in

31 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Eample 5 - Plane Truss Problem Element Node Node C S 9 o 3 45 o o E Cu Sv Cu Sv 6 3 3,965 psi () element v 6 3 (.77) (.77),47 () element u v psi Eample 5 - Plane Truss Problem Element Node Node C S 9 o 3 45 o o E Cu Sv Cu Sv 6 3, 35 psi (3) element 3 u

32 CIV 7/87 Chapter 3a - Development of Truss Equations 3/8 Eample 5 - Plane Truss Problem et s check equilibrium at node : F f () cos(45 ) (3) f F y f () sin(45 ) () f,lb Eample 5 - Plane Truss Problem et s check equilibrium at node : F psi in psi in (,47 )( )(.77) (,35 )( ) F psi in psi in y (3,965 )( ) (,47 )( )(.77),

33 CIV 7/87 Chapter 3a - Development of Truss Equations 33/8 Eample 6 - Plane Truss Problem Develop the element stiffness matrices and system equations for the plane truss below. Assume the stiffness of each element is constant. Use the numbering scheme indicated. Solve the equations for the displacements and compute the member forces. All elements have a constant value of AE/ Eample 6 - Plane Truss Problem Develop the element stiffness matrices and system equations for the plane truss below. Member Node Node Elemental Stiffness k 3 k 3/4 3 3 k /

34 CIV 7/87 Chapter 3a - Development of Truss Equations 34/8 Eample 6 - Plane Truss Problem Compute the elemental stiffness matri for each element. The general form of the matri is: C CS C CS AE CS S CS S k C CS C CS CS S CS S Member Node Node Elemental Stiffness k 3 k 3/4 3 3 k / Eample 6 - Plane Truss Problem For element : k () u v u v k u v u v Member Node Node Elemental Stiffness k 3 k 3/4 3 3 k /

35 CIV 7/87 Chapter 3a - Development of Truss Equations 35/8 Eample 6 - Plane Truss Problem For element : k () u v u v 3 3 k u v u v 3 3 Member Node Node Elemental Stiffness k 3 k 3/4 3 3 k / Eample 6 - Plane Truss Problem For element 3: k (3) u v u v 3 3 k u v u v 3 3 Member Node Node Elemental Stiffness k 3 k 3/4 3 3 k /

36 CIV 7/87 Chapter 3a - Development of Truss Equations 36/8 Eample 6 - Plane Truss Problem Assemble the global stiffness matri by superimposing the elemental global matrices. K u v u v u v 3 3 k 3 3 u v u v u v 3 3 element element element 3 Eample 6 - Plane Truss Problem The unconstrained (no boundary conditions satisfied) equations are: u F F v y u P v P u F 3 3 v F 3 3y k 3 3 The displacement at nodes and 3 are zero in both directions. Applying these conditions to the system equations gives: F F y k 3 u P v P F3 3 F3 y

37 CIV 7/87 Chapter 3a - Development of Truss Equations 37/8 Eample 6 - Plane Truss Problem Applying the boundary conditions to the system equations gives: k 3 u P v P Solving this set of equations is fairly easy. The solution is: u P P P 3P v k k Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) : C S f Cu Sv Cu Sv f y k f Cu Sv Cu Sv f y f k Cu P P k k k Cu k f P P k P P P P fy fy

38 CIV 7/87 Chapter 3a - Development of Truss Equations 38/8 Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) : C S f Cu Sv Cu Sv 3 3 f y k f3 Cu Sv Cu3 Sv3 f 3y f k Cu Sv PP P3 P k k k P f3 kcu Sv PP P3 P k k k P Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) 3: f f y f f 3 3y The solution to this simple problem can be readily checked by using simple static equilibrium equations.

39 CIV 7/87 Chapter 3a - Development of Truss Equations 39/8 Eample 7 - Plane Truss Problem Consider the two bar truss shown below. Determine the displacement in the y direction of node and the aial force in each element. Assume E = GPa and A = 6-4 m Eample 7 - Plane Truss Problem The global elemental stiffness matri for element is: cos k () () sin () 6 4 (6 ) Simplifying the above epression gives: k () u v u v ,

40 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Eample 7 - Plane Truss Problem The global elemental stiffness matri for element is: cos () sin () k () ( )(6 ) Simplifying the above epression gives: k () u v u v ,.5.5 Eample 7 - Plane Truss Problem The total global equations are: F u F y v F u 5, F y v F 3 u 3 F 3y.5.5 v3 element element The displacement boundary conditions are: u u v u v 3 3

41 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Eample 7 - Plane Truss Problem The total global equations are: F u FP y v F , F y F 3 F 3 y.5.5 By applying the boundary conditions the force-displacement equations reduce to: P 5,( v ) Eample 7 - Plane Truss Problem Solving the equation gives: v P 5 (. ).5 By substituting P =, kn and = -.5 m in the above equation gives: v.337m The local element forces for element are: u.5 f.6.8 v.337 5, f.6.8 u v The element forces are: f 76.6kN f 76.7kN

42 CIV 7/87 Chapter 3a - Development of Truss Equations 4/8 Eample 7 - Plane Truss Problem The local element forces for element are: u.5 f v.337 3,5 f 3 u3 v 3 The element forces are: f,6kn f,6kn 3 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space et s derive the transformation matri for the stiffness matri for a bar element in three-dimensional space as shown below:

43 CIV 7/87 Chapter 3a - Development of Truss Equations 43/8 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space The coordinates at node are, y, and z, and the coordinates of node are, y, and z. Also, let, y, and z be the angles measured from the global, y, and z aes, respectively, to the local ais. Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space The three-dimensional vector representing the bar element is gives as: duivjwk uivjwk

44 CIV 7/87 Chapter 3a - Development of Truss Equations 44/8 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space Taking the dot product of the above equation with u( ii ) v( ji ) w( ki ) u i gives: By the definition of the dot product we get: y y z z ii C ji C ki C y z where ( ) ( y y ) ( z z ) C cos C cos C cos y y z z where C, C y, and C z are projections of i on to i, j, and k, respectively. Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space Therefore: u CuCyv Czw The transformation between local and global displacements is: u v u CCyCz w u C C y Czu v w d T * d * T C CyCz C C y Cz

45 CIV 7/87 Chapter 3a - Development of Truss Equations 45/8 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space The transformation from the local to the global stiffness matri is: k T T kt C C y C z AE CCyC z k C C C y Cz C y Cz C CC y CC z C CC y CC z C y y C CC CC y z CC y y CC y z AE CC CC z y z Cz CC CC z y z C z k C CC y CC z C CC y CC z CC C y y CC C y z CC y y CC y z CC CC z y z C CC z CC z y z Cz Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space The global stiffness matri can be written in a more convenient form as: C CCy CC z CC y Cy CC y z CC z CC y z C z AE k

46 CIV 7/87 Chapter 3a - Development of Truss Equations 46/8 Eample 8 Space Truss Problem Consider the space truss shown below. The modulus of elasticity, E =. 6 psi for all elements. Node is constrained from movement in the y direction. To simplify the stiffness matrices for the three elements, we will epress each element in the following form: AE k Eample 8 Space Truss Problem Consider element : ( ) ( y y ) ( z z ) () ( 7) (36) 8.5in () C C C y z

47 CIV 7/87 Chapter 3a - Development of Truss Equations 47/8 Eample 8 Space Truss Problem Consider element : u v w u v w 6 (.3 in )(. psi ) k 8.5in lb in Eample 8 Space Truss Problem Consider element : ( ) ( y y ) ( z z ) () ( 7) (36) (7) 8in () C C C y z

48 CIV 7/87 Chapter 3a - Development of Truss Equations 48/8 Eample 8 Space Truss Problem Consider element : u v w u v w (.79 in )(. psi) k 8in lb in Eample 8 Space Truss Problem Consider element 3: ( ) ( y y ) ( z z ) (3) ( 7) ( 48) 86.5in (3) C C C y z

49 CIV 7/87 Chapter 3a - Development of Truss Equations 49/8 Eample 8 Space Truss Problem Consider element 3: u v w u v w (.87 in )(. psi ) k 86.5in lb in The boundary conditions are: u v w u3 v3 w3 u v w v Eample 8 Space Truss Problem Canceling the rows and the columns associated with the boundary conditions reduces the global stiffness matri to: u w 9,,45 K,45 4,45 The global force-displacement equations are: 9,,45u,45 4,45 w, Solving the equation gives: u.7 in w.64 in

50 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Eample 8 Space Truss Problem It can be shown, that the local forces in an element are: ui v i f i AE C Cy Cz C Cy Cz wi f j C Cy Cz C Cy C z uj v j w j The stress in an element is: ui v i E wi C C y C z C C y C z u j v j w j Eample 8 Space Truss Problem The stress in element is: () () 955 psi The stress in element is:.7 6 () (), 43 psi

51 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Eample 8 Space Truss Problem The stress in element 3 is:.7 6 (3) (3),843 psi Inclined, or Skewed Supports If a support is inclined, or skewed, at some angle for the global ais, as shown below, the boundary conditions on the displacements are not in the global -y directions but in the -y directions.

52 CIV 7/87 Chapter 3a - Development of Truss Equations 5/8 Inclined, or Skewed, Supports We must transform the local boundary condition of v 3 = (in local coordinates) into the global -y system. Inclined, or Skewed, Supports Therefore, the relationship between of the components of the displacement in the local and the global coordinate systems at node 3 is: u' 3 cos sin u3 v' 3 sin cos v3 We can rewrite the above epression as: cos sin sin cos d' [ t ] d t We can apply this sort of transformation to the entire displacement vector as: T d' [ T ] d or d [ T ] d'

53 CIV 7/87 Chapter 3a - Development of Truss Equations 53/8 Inclined, or Skewed, Supports Where the matri [T ] T is: [] I [] [] T [ T] [] [ I] [] [] [] [ t3] Both the identity matri [I] and the matri [t 3 ] are matrices. The force vector can be transformed by using the same transformation. f ' [ T ] f In global coordinates, the force-displacement equations are: [ ] f K d Inclined, or Skewed, Supports Applying the skewed support transformation to both sides of the equation gives: [ T ] f [ T ][ K] d By using the relationship between the local and the global displacements, the force-displacement equations become: T f ' [ T ][ K][ T ] d' Therefore the global equations become: F u F y v F u T [ T][ K][ T] F y v F ' 3 u ' 3 F ' 3y v ' 3

54 CIV 7/87 Chapter 3a - Development of Truss Equations 54/8 Eample 9 Space Truss Problem Consider the plane truss shown below. Assume E = GPa, A = 6-4 m for element and, and A = (6-4 )m for element 3. Determine the stiffness matri for each element. k C AE CS C CS CS C S CS S S CS CS CS S C CS Eample 9 Space Truss Problem The global elemental stiffness matri for element is: C CS AE CS S k C CS S C CS CS S C CS S CS CS () () cos sin k () kn m m m 6 4 ( / )(6 ) u v u v

55 CIV 7/87 Chapter 3a - Development of Truss Equations 55/8 Eample 9 Space Truss Problem The global elemental stiffness matri for element is: C AE CS k C CS CS S C CS CS S C CS S CS CS S () () cos sin k () 6 4 ( / )(6 ) kn m m m u v u v 3 3 Eample 9 Space Truss Problem The global elemental stiffness matri for element 3 is: C CS AE CS S k C CS S C CS CS S C CS S CS CS k (3) cos sin (3) (3) kn m m 6 4 ( / )(6 ) m u v u3 v3

56 CIV 7/87 Chapter 3a - Development of Truss Equations 56/8 Eample 9 Space Truss Problem Using the direct stiffness method, the global stiffness matri is: K,6 N m We must transform the global displacements into local coordinates. Therefore the transformation [T ] is: [] I [] [] [ T] [] [ I] [] [] [] [ t3] Eample 9 Space Truss Problem The first step in the matri transformation to find the product of [T ][K]. K T K 5 T K, , 5 6 N m N m T

57 CIV 7/87 Chapter 3a - Development of Truss Equations 57/8 Eample 9 Space Truss Problem The net step in the matri transformation to find the product of [T ][K][T ] T. T K T KT T 5, N m T T T T 5 T K,6 N m Eample 9 Space Truss Problem The displacement boundary conditions are: u v v v' 3 F F v y F u 5, 6 N F m y v F ' u ' 3 F ' 3y v ' 3 u

58 CIV 7/87 Chapter 3a - Development of Truss Equations 58/8 Eample 9 Space Truss Problem By applying the boundary conditions the global forcedisplacement equations are: 5.77,,6 N u F kn m.77.5 u' F' 3 3 Solving the equation gives: u.9 mm u' 3 5.6mm Eample 9 Space Truss Problem The global nodal forces are calculated as: F Therefore: F y F , 6 N mm F y F ' F ' 3y F 5 kn F 5 kn y F F' 77 kn y 3y F u F y v F u T [ T][ K][ T] F y v F ' ' u 3 F ' 3y v ' 3 3

59 CIV 7/87 Chapter 3a - Development of Truss Equations 59/8 Potential Energy Approach to Derive Bar Element Equations et s derive the equations for a bar element using the principle of minimum potential energy. The total potential energy, p, is defined as the sum of the internal strain energy U and the potential energy of the eternal forces : U p The differential internal work (strain energy) du in a onedimensional bar element is: du ( y)( z)( ) d Potential Energy Approach to Derive Bar Element Equations If we let the volume of the element approach zero, then: du d dv Summing the differential energy over the whole bar gives: U ddv EddV V V E dv For a linear-elastic material (Hooke s law) as shown below: V E U dv V

60 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Potential Energy Approach to Derive Bar Element Equations The internal strain energy statement becomes U dv V The potential energy of the eternal forces is: XudV Tus ds M b fu i i V S i where X b is the body force (force per unit volume), T is the traction (force per unit area), and f i is the nodal concentrated force. All of these forces are considered to act in the local direction. Potential Energy Approach to Derive Bar Element Equations Apply the following steps when using the principle of minimum potential energy to derive the finite element equations.. Formulate an epression for the total potential energy.. Assume a displacement pattern. 3. Obtain a set of simultaneous equations minimizing the total potential energy with respect to the displacement parameters.

61 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Potential Energy Approach to Derive Bar Element Equations Consider the following bar element, as shown below: A d f u f u p V X u dv b S T u ds s We can approimate the aial displacement as: u u N N u N N Potential Energy Approach to Derive Bar Element Equations Using the stress-strain relationships, the aial strain is: du dn dn u d d d u where N and N are the interpolation functions gives as: u u [ B]{ d} B The aial stress-strain relationship is: [ D]

62 CIV 7/87 Chapter 3a - Development of Truss Equations 6/8 Potential Energy Approach to Derive Bar Element Equations Where [D] = [E] for the one-dimensional stress-strain relationship and E is the modulus of elasticity. Therefore, stress can be related to nodal displacements as: [ ][ ] D B d The total potential energy epressed in matri form is: A d d P u X dv u T ds T T T T b p where {P} represented the concentrated nodal loads. V S Potential Energy Approach to Derive Bar Element Equations If we substitute the relationship between û and ˆd into the energy equations we get: A p d B D B d d d P T T T T T T T T b s V d N X dv d N T ds In the above epression for potential energy p is a function of the d, that is: p = p ( u ). u, However, [B] and [D] and the nodal displacements u are not a function of. S

63 CIV 7/87 Chapter 3a - Development of Truss Equations 63/8 Potential Energy Approach to Derive Bar Element Equations Integration the energy epression with respect to gives: where A p d B D B d d f T T T T [ ] [ ] [ ] T T [ ] b [ ] b f P N X dv N X ds V S We can define the surface tractions and body-force matrices as: T T fs [ N] T ds fb [ N] Xb dv S V Potential Energy Approach to Derive Bar Element Equations Minimization of p with respect to each nodal displacement requires that: p p u u For convenience, let s define the following * U d T [ B] T [ D] T [ B] d u U u u [ E] u *

64 CIV 7/87 Chapter 3a - Development of Truss Equations 64/8 Potential Energy Approach to Derive Bar Element Equations Simplifying the above epression gives: E U u uu u * The loading on a bar element is given as: d T f uf uf Therefore, the minimum potential energy is: p AE u u f u p AE uuf u Potential Energy Approach to Derive Bar Element Equations The above equations can be written in matri form as: p AE u f d u f The stiffness matri for a bar element is: This form of the stiffness matri obtained from the principle of minimum potential energy is identical to the stiffness matri derived from the equilibrium equations. k AE

65 CIV 7/87 Chapter 3a - Development of Truss Equations 65/8 Eample - Bar Problem Consider the bar shown below: The energy equivalent nodal forces due to the distributed load are: T f f [ N] T ds f Cd f S Eample - Bar Problem 3 C C C f 3 6 C d 3 f C C 3 3 The total load is the area under the distributed load curve, or: C F ( )( C ) The equivalent nodal forces for a linearly varying load are: F F f of the total load f of the total load

66 CIV 7/87 Chapter 3a - Development of Truss Equations 66/8 Eample - Bar Problem Consider the aially loaded bar shown below. Determine the aial displacement and aial stress. et E = 3 6 psi, A = in, and = 6 in. Use (a) one and (b) two elements in the finite element solutions. Eample - Bar Problem The one-element solution: The distributed load can be converted into equivalent nodal forces using: T F N T ds F [ ] S F F d

67 CIV 7/87 Chapter 3a - Development of Truss Equations 67/8 Eample - Bar Problem The one-element solution: F 3 6 d F F 6, lb F, lb Eample - Bar Problem The one-element solution: () 6 k The element equations are: u 6, 6 R, The second equation gives: u.6 in R 8,lb 6 ( u) R,

68 CIV 7/87 Chapter 3a - Development of Truss Equations 68/8 Eample - Bar Problem The one-element solution: The aial stress-strain relationship is: { } [ D]{ } { } [ ] EB d u u u E E u , psi ( T ) 6 Eample - Bar Problem The two-element solution: The distributed load can be converted into equivalent nodal forces. For element, the total force of the triangular-shaped distributed load is: (3 in.)(3 lb in) 4,5 lb

69 CIV 7/87 Chapter 3a - Development of Truss Equations 69/8 Eample - Bar Problem The two-element solution: Based on equations developed for the equivalent nodal force of a triangular distributed load, develop in the one-element problem, the nodal forces are: () (4,5) f, 5 3 lb () f 3,lb (4,5) 3 Eample - Bar Problem The two-element solution: For element, the applied force is in two parts: a triangularshaped distributed load and a uniform load. The uniform load is: (3 in)(3 lb / in) 9, lb The nodal forces for element are: (9,) (4,5) () f 3 6, lb () f 7,5 3 lb (9,) (4,5) 3

70 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 Eample - Bar Problem The two-element solution: The final nodal force vector is: () F f, 5 () () F f f 9, () F 3 f 3 R3 7,5 The element stiffness matrices are: 3 () () AE k k Eample - Bar Problem The two-element solution: The assembled global stiffness matri is: element 6 K element The assembled global force-displacement equations are: u,5 R 7,5 6 u 9, 3

71 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 Eample - Bar Problem The two-element solution: After the eliminating the row and column associated with u 3, we get: 6 u,5 u 9, Solving the equation gives: Solving the last equation gives: 6 u R3 7,5 u u.6 in.55 in R3 8, Eample - Bar Problem The two-element solution: The aial stress-strain relationship is: () d E d.6 E 75 psi ( T)

72 CIV 7/87 Chapter 3a - Development of Truss Equations 7/8 Eample - Bar Problem The two-element solution: The aial stress-strain relationship is: () d E d 3.55 E 5,5 psi ( T) 3 3 Comparison of Finite Element Solution to Eact Solution In order to be able to judge the accuracy of our finite element models, we will develop an eact solution for the bar element problem. The eact solution for the displacement may be obtained by: u AE P( ) d where the force P is shown on the following free-body diagram. P ( ) ( ) 5

73 CIV 7/87 Chapter 3a - Development of Truss Equations 73/8 Comparison of Finite Element Solution to Eact Solution Therefore: u AE P( ) d 3 5 u 5 d C AE 3AE o Applying the boundary conditions: u ( ) C C 3AE 3AE The eact solution for aial displacement is: 5 u ( ) 3AE 3 3 P ( ) 5 ( ) A A Comparison of Finite Element Solution to Eact Solution A plot of the eact solution for displacement as compared to several different finite element solutions is shown below.

74 CIV 7/87 Chapter 3a - Development of Truss Equations 74/8 Comparison of Finite Element Solution to Eact Solution A plot of the eact solution for aial stress as compared to several different finite element solutions is shown below. Comparison of Finite Element Solution to Eact Solution A plot of the eact solution for aial stress at the fied end ( = ) as compared to several different finite element solutions is shown below.

75 CIV 7/87 Chapter 3a - Development of Truss Equations 75/8 Galerkin s Residual Method and Its Application to a One-Dimensional Bar There are a number of weighted residual methods. However, the Galerkin s method is more well-known and will be the only weighted residual method discussed in this course. In weighted residual methods, a trial or approimate function is chosen to approimate the independent variable (in our case, displacement) in a problem defined by a differential equation. The trial function will not, in general, satisfy the governing differential equation. Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem. Galerkin s Residual Method and Its Application to a One-Dimensional Bar Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem. RdV minimum V In the residual methods, we require that a weighted value of the residual be a minimum over the entire domain of the problem. The weighting function allows the weighted integral of the residuals to go to zero. RW dv V

76 CIV 7/87 Chapter 3a - Development of Truss Equations 76/8 Galerkin s Residual Method and Its Application to a One-Dimensional Bar Using Galerkin s weighted residual method, we require the weighting functions to be the interpolation, N i. Therefore: V RN dv i,,, n i Eample - Bar Element Formulation et s derive the bar element formulation using Galerkin s method. The governing differential equation is: d du AE d d Applying Galerkin s method we get: d du AE Ni d i,,, n d d We now apply integration by parts using the following general formula: udv uv vdu

77 CIV 7/87 Chapter 3a - Development of Truss Equations 77/8 Eample - Bar Element Formulation If we assume the following: dn d i u Ni du d d du du dv AE d v AE d d d then integration by parts gives: du du dni NAE i AE d d d d Eample - Bar Element Formulation Recall that: du dn dn u u d d d du u d u Our original weighted residual epression, with the approimation for u becomes: dni du AE d Ni AE d u d u

78 CIV 7/87 Chapter 3a - Development of Truss Equations 78/8 Eample - Bar Element Formulation Substituting N for the weighting function N i gives: dn du AE d NAE d u d u AE d u u AE u u du du AE d NAE d AE A f AE u u f Eample - Bar Element Formulation Substituting N for the weighting function N i gives: dn du AE d NAE d u d u u AE d u AE u u du NAE d du AE d AE A f AE u u f

79 CIV 7/87 Chapter 3a - Development of Truss Equations 79/8 Eample - Bar Element Formulation Writing the last two equations in matri form gives: AE u f u f This element formulation is identical to that developed from equilibrium and the minimum potential energy approach. Problems:. Verify the global stiffness matri for a three-dimensional bar. Hint: First, epand T* to a 6 6 square matri, then epand k to 6 6 square matri by adding the appropriate rows and columns of zeros, and finally, perform the matri triple product k = T T k T. 3. Do problems 3.4, 3., 3., 3.5a,b, 3.8, 3.3, 3.37, 3.43, 3.48, 3.5, and 3.55 on pages in your tetbook A First Course in the Finite Element Method by D. ogan. 4. Use SAP and solve problems 3.63 and 3.64.

80 CIV 7/87 Chapter 3a - Development of Truss Equations 8/8 End of Chapter 3a