Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 Chapter 3a Development of Truss Equations Learning Objectives To derive the stiffness matri for a bar element. To illustrate how to solve a bar assemblage by the direct stiffness method. To introduce guidelines for selecting displacement functions. To describe the concept of transformation of vectors in two different coordinate systems in the plane. To derive the stiffness matri for a bar arbitrarily oriented in the plane. To demonstrate how to compute stress for a bar in the plane. To show how to solve a plane truss problem. To develop the transformation matri in threedimensional space and show how to use it to derive the stiffness matri for a bar arbitrarily oriented in space. To demonstrate the solution of space trusses. 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 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. Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part 3/53 Development of Truss Equations Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part 4/53 Development of Truss Equations Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part 5/53 Development of Truss Equations Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part 6/53 Development of Truss Equations 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part 7/53 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. 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 L, and modulus of elasticity E. The nodal degrees of freedom are the local aial displacements u and u at the ends of the bar.

CIVL 7/87 Chapter 3 - Truss Equations - Part 8/53 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 0 d d The following assumptions are considered in deriving the bar element stiffness matri:. The bar cannot sustain shear force: f f 0. Any effect of transverse displacement is ignored. 3. Hooke s law applies; stress is related to strain: E y y

CIVL 7/87 Chapter 3 - Truss Equations - Part 9/53 Step - Select Element Type We will consider the linear bar element shown below. The bar element has a constant cross-section A, an initial length L, and modulus of elasticity E. 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 L u u u u L L u

CIVL 7/87 Chapter 3 - Truss Equations - Part 0/53 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 L L The linear displacement function u plotted over the length of the bar element is shown here. Step 3 - Define the Strain/Displacement and Stress/Strain Relationships The stress-displacement relationship is: du d u u L 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 L

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 Step 4 - Derive the Element Stiffness Matri and Equations The nodal force sign convention, defined in element figure, is: f T f T therefore, uu u u f AE f AE L L Writing the above equations in matri form gives: f AE u f L u Notice that AE/L for a bar element is analogous to the spring constant k for a spring element. 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 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 L Eample - Bar Problem Consider the following three-bar system shown below. Assume for elements and : A = in and E = 30 (0 6 ) psi and for element 3: A = in and E = 5 (0 6 ) psi. Determine: (a) the global stiffness matri, (b) the displacement of nodes and 3, and (c) the reactions at nodes and 4.

CIVL 7/87 Chapter 3 - Truss Equations - Part 3/53 Eample - Bar Problem For elements and : For element 3: 6 300 () () 6 k lb 0 lb in in k node numbers for element 30 6 50 (3) 6 lb 0 lb in in k 3 node numbers for element 3 4 node numbers for element 3 30 As before, the numbers above the matrices indicate the displacements associated with the matri. Eample - Bar Problem Assembling the global stiffness matri by the direct stiffness methods gives: E E E 3 0 0 6 0 K 0 0 0 0 Relating global nodal forces related to global nodal displacements gives: F 0 0u F 6 0 u 0 F3 0 u3 F 4 0 0 u 4

CIVL 7/87 Chapter 3 - Truss Equations - Part 4/53 Eample - Bar Problem The boundary conditions are: u u4 0 F 0 00 F 6 0 u 0 F3 0 u 3 F 4 0 0 0 Applying the boundary conditions and the known forces (F = 3,000 lb) gives: 3,000 u 0 6 0 u3 Eample - Bar Problem Solving for u and u 3 gives: u u 3 0.00 in 0.00in The global nodal forces are calculated as: F 0 0 0,000 F 6 0 0.00 3,000 0 lb F3 0 0.00 0 F 4 0 0 0,000

CIVL 7/87 Chapter 3 - Truss Equations - Part 5/53 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. 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part 6/53 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. 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part 7/53 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: NN 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. Transformation of Vectors in Two Dimensions In many problems it is convenient to introduce both local and global (or reference) coordinates. Local coordinates are always chosen to conveniently represent the individual element. Global coordinates are chosen to be convenient for the whole structure.

CIVL 7/87 Chapter 3 - Truss Equations - Part 8/53 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. Transformation of Vectors in Two Dimensions Let s consider that d does not coincide 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part 9/53 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: duiv ju i v j Transformation of Vectors in Two Dimensions Consider the following diagram: Using vector addition: ab i Using the law of cosines, we get: a i cos a cos Similarly: b i sin b sin

CIVL 7/87 Chapter 3 - Truss Equations - Part 0/53 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 Transformation of Vectors in Two Dimensions Consider the following diagram: The vector i can be rewritten as: icos isin j The vector j can be rewritten as: j sin icos j Therefore, the displacement vector is: cos sin sin cos uiv ju i j v i j u i vj

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 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 C cos S sin u C Su v S C v 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 CuSv

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 Eample - Bar Element Problem The global nodal displacement at node is u = 0. in and v = 0. 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 cos60 (0.) sin60 (0.) 0.3 in Global Stiffness Matri We will now use the transformation relationship developed above to obtain the global stiffness matri for a bar element.

CIVL 7/87 Chapter 3 - Truss Equations - Part 3/53 Global Stiffness Matri We known that for a bar element in local coordinates we have: f AE u f L 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 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 0 0 v u 0 0 C S u v * d=td * C S 0 0 T 0 0 C S

CIVL 7/87 Chapter 3 - Truss Equations - Part 4/53 Global Stiffness Matri A similar epression for the force transformation can be developed. f 0 0 f f C S y * f 0 0 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 Global Stiffness Matri The matri T* is not a square matri so we cannot invert it. Let s epand the relationship between local and global displacement. u C S 0 0u v S C 0 0 v u 0 0 C S u v 0 0 S C v d=td where T is: C S S C T 0 0 0 0 0 0 0 0 C S S C

CIVL 7/87 Chapter 3 - Truss Equations - Part 5/53 Global Stiffness Matri We can write a similar epression for the relationship between local and global forces. f C S 0 0f f S C 0 0 f f 0 0 C S f y y f y 0 0 S C f y f =Tf Therefore our original local coordinate force-displacement epression f AE u f L u f =kd Global Stiffness Matri May be epanded: f 0 0u f y 0 0 0 0 v AE f L 0 0u f y 0 0 0 0 v The global force-displacement equations are: f kd 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

CIVL 7/87 Chapter 3 - Truss Equations - Part 6/53 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 CS S CS S S C C CS L C CS C CS S CS 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 Global Stiffness Matri Local forces can be computed as: f 0 0u f y 0 0 0 0 v AE f L 0 0u f y 0 0 0 0 v 0 0 C S 0 0 u 0 0 0 0 S C 0 0 v 0 0 0 0 C Su 0 0 0 0 0 0 S C v AE L f CuSvCu Sv f y AE 0 f L CuSvCu Sv f y 0

CIVL 7/87 Chapter 3 - Truss Equations - Part 7/53 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 60 in, and the E is 30 0 6 psi. C AE CS k C CS CS CS S S S C CS L CS CS S C CS Therefore: C cos30 o 3 S sin30 o Eample 3 - Bar Element Problem The global elemental stiffness matri is: k 3 3 3 3 4 4 4 4 3 3 60in 3 3 3 3 4 4 4 4 3 3 4 4 4 4 6 ( in ) 300 psi 4 4 4 4 Simplifying the global elemental stiffness matri is: 0.750 0.433 0.750 0.433 6 0.433 0.50 0.433 0.50 k 0 0.750 0.433 0.750 0.433 0.433 0.50 0.433 0.50 lb in

CIVL 7/87 Chapter 3 - Truss Equations - Part 8/53 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 L u The force-displacement equation for f is: AE L u f u The stress in terms of global displacement is: E L 0 0 C S u u C S 0 0 v v E Cu Sv Cu Sv L Eample 4 - Bar Element Problem For the bar element shown below, determine the aial stress. Assume the cross-sectional area is 4 0-4 m, the length is m, and the E is 0 GPa. The global displacements are known as u = 0.5 mm, v = 0, u = 0.5 mm, and v = 0.75 mm. E Cu Sv Cu Sv L 6 0 0 3 3 (0.5) (0) (0.5) (0.75) 4 4 3 8.3 MPa 8.3 0 kn m KN m

CIVL 7/87 Chapter 3 - Truss Equations - Part 9/53 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. Eample 5 - Plane Truss Problem The plane truss shown below is composed of three bars subjected to a downward force of 0 kips at node. Assume the cross-sectional area A = in and E is 30 0 6 psi for all elements. Determine the and y displacement at node and stresses in each element.

CIVL 7/87 Chapter 3 - Truss Equations - Part 30/53 Eample 5 - Plane Truss Problem Element Node Node C S 90 o 0 3 45 o 0.707 0.707 3 4 0 o 0 Eample 5 - Plane Truss Problem The global elemental stiffness matri are: element : C 0 S k () in C AE CS k C CS 0in 6 ( )(30 0 ) CS CS S C CS S S L CS CS S C CS u v u v 0 0 0 0 psi 0 0 0 0 0 0 0 0 lb in element : C S k () in 6 ( )(30 0 ) 40 in u v u v 3 3 psi lb in element 3: C S 0 k (3) in 0in 6 ( )(30 0 ) u v u v 4 4 0 0 0 0 0 0 psi 0 0 0 0 0 0 lb in

CIVL 7/87 Chapter 3 - Truss Equations - Part 3/53 Eample 5 - Plane Truss Problem The total global stiffness matri is: K.354 0.354 0 0 0.354 0.354 0 0.354.354 0 0.354 0.354 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 50 lb 0.354 0.354 0 0 0.354 0.354 0 0 in 0.354 0.354 0 0 0.354 0.354 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The total global force-displacement equations are: element element element 3 0.354 0.354 0 0 0.354 0.354 0u 0, 000 0.354.354 0 0.354 0.354 0 0 v F 0 0 0 0 0 0 0 0 F 0 y 5 0 0 0 0 0 0 5 0 F 3 0 0.354 0.354 0 0 0.354 0.354 0 0 0 F 3 y 0.354 0.354 0 0 0.354 0.354 0 0 0 F 0 0 0 0 0 0 0 4 F 0 0 0 0 0 0 0 0 4 y 0 Eample 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: 0.354 0.354 0 0 0.354 0.354 0u 0, 000 0.354.354 0 0.354 0.354 0 0 v F 0 0 0 0 0 0 0 0 F 0 y 5 0 0 0 0 0 0 5 0 F 3 0 0.354 0.354 0 0 0.354 0.354 0 0 0 F 3 y 0.354 0.354 0 0 0.354 0.354 0 0 0 F 0 0 0 0 0 0 0 4 F 0 0 0 0 0 0 0 0 4 y 0

CIVL 7/87 Chapter 3 - Truss Equations - Part 3/53 Eample 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: 0 5.354 0.354 u 50 0,000 0.354.354 v Solving the equations gives: u 0.44 0 in v.59 0 The stress in an element is: E Cu Sv Cu Sv L where i is the local node number in Eample 5 - Plane Truss Problem Element Node Node C S 90 o 0 3 45 o 0.707 0.707 3 4 0 o 0 E Cu Sv Cu Sv L 6 300 3,965 psi 0 () element v 6 300 (0.707) (0.707),47 () element u v psi 0

CIVL 7/87 Chapter 3 - Truss Equations - Part 33/53 Eample 5 - Plane Truss Problem Element Node Node C S 90 o 0 3 45 o 0.707 0.707 3 4 0 o 0 E Cu Sv Cu Sv L 6 300, 035 psi 0 (3) element 3 u Eample 5 - Plane Truss Problem Let s check equilibrium at node : F f () cos(45 ) (3) f F y f () sin(45 ) () f 0,000lb

CIVL 7/87 Chapter 3 - Truss Equations - Part 34/53 Eample 5 - Plane Truss Problem Let s check equilibrium at node : F psi in psi in (,47 )( )(0.707) (,035 )( ) 0 F psi in psi in y (3,965 )( ) (,47 )( )(0.707) 0,000 0 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/L

CIVL 7/87 Chapter 3 - Truss Equations - Part 35/53 Eample 6 - Plane Truss Problem Develop the element stiffness matrices and system equations for the plane truss below. Member Node Node Elemental Stiffness k 0 3 k 3/4 3 3 k / 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 L C CS C CS CS S CS S Member Node Node Elemental Stiffness k 0 3 k 3/4 3 3 k /

CIVL 7/87 Chapter 3 - Truss Equations - Part 36/53 Eample 6 - Plane Truss Problem For element : k () u v u v 0 0 0 0 0 0 k 0 0 0 0 0 0 u v u v Member Node Node Elemental Stiffness k 0 3 k 3/4 3 3 k / 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 0 3 k 3/4 3 3 k /

CIVL 7/87 Chapter 3 - Truss Equations - Part 37/53 Eample 6 - Plane Truss Problem For element 3: k (3) u v u v 3 3 0 0 0 0 0 0 k 0 0 0 0 0 0 u v u v 3 3 Member Node Node Elemental Stiffness k 0 3 k 3/4 3 3 k / 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 0 0 0 0 0 0 0 0 k 0 3 0 0 0 0 0 3 u v u v u v 3 3 element element element 3

CIVL 7/87 Chapter 3 - Truss Equations - Part 38/53 Eample 6 - Plane Truss Problem The unconstrained (no boundary conditions satisfied) equations are: 0 0 0 0 u F 0 0 0 0 v F y 0 3 u P v P 0 0 u F 3 3 0 3 v F 3 3y k 0 0 The displacement at nodes and 3 are zero in both directions. Applying these conditions to the system equations gives: 0 0 0 0 0 F 0 0 0 0 0 F y k 0 3 u P 0 0 v P 0 0 0 F3 0 3 0 F 3y 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 k P 3P k v

CIVL 7/87 Chapter 3 - Truss Equations - Part 39/53 Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) : C S 0 f Cu Sv Cu Sv f 0 y k f Cu Sv Cu Sv f y 0 f k Cu P P k k k Cu k f P P k P P P P fy 0 fy 0 Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) : f Cu Sv Cu3 Sv3 C S f 0 y k f3 Cu Sv Cu3 Sv3 f 3y 0 f k Cu Sv PP P3 P k k k P f3 kcu Sv PP P3 P k k k P

CIVL 7/87 Chapter 3 - Truss Equations - Part 40/53 Eample 6 - Plane Truss Problem Using the force-displacement relationship the force in each member may be computed. Member (element) 3: C 0 S f f 0 f 0 0 f 0 y 3 3y f Cu Sv Cu Sv 3 3 f 0 y k f3 CuSvCu3 Sv3 f 3y 0 The solution to this simple problem can be readily checked by using simple static equilibrium equations. 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 = 0 GPa and A = 6 0-4 m

CIVL 7/87 Chapter 3 - Truss Equations - Part 4/53 Eample 7 - Plane Truss Problem The global elemental stiffness matri for element is: cos 0.6 5 () 3 sin 0.8 5 () 4 k () 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 5 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 6 4 0 0 (6 0 ) Simplifying the above epression gives: k () u v u v 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 5,00 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 Eample 7 - Plane Truss Problem The global elemental stiffness matri for element is: () cos 0 () sin k () 0 0 0 0 0 0 4 0 0 0 0 0 0 6 4 (0 0 )(6 0 ) Simplifying the above epression gives: k () u v u v 3 3 0 0 0 0 0.5 0.5 5,00 0 0 0 0 0.5 0.5

CIVL 7/87 Chapter 3 - Truss Equations - Part 4/53 Eample 7 - Plane Truss Problem The total global equations are: F 0.36 0.48 0.36 0.48 0 0u F 0.48.89 0.48 0.64 0.5 y v F 0.36 0.48 0.36 0.48 0 0 u 5,00 F y 0.48 0.64 0.48 0.64 0 0v F 3 0 0 0 0 0 0u 3 F 3y 0.5 0 0 0.5 v3 element element The displacement boundary conditions are: u u v u v 3 3 0 Eample 7 - Plane Truss Problem The total global equations are: F 0.36 0.48 0.36 0.48 0 0u FP y 0.48.89 0.48 0.64 0.5 v F 0.36 0.48 0.36 0.48 0 00 5,00 F y 0.48 0.64 0.48 0.64 0 00 F 3 0 0 0 0 0 0 0 F 3 y 0.5 0 0 0.50 By applying the boundary conditions the force-displacement equations reduce to: P 5,00(0.48.89 v )

CIVL 7/87 Chapter 3 - Truss Equations - Part 43/53 Eample 7 - Plane Truss Problem Solving the equation gives: v P 5 (.0 ) 0.5 By substituting P =,000 kn and = -0.05 m in the above equation gives: v 0.0337m The local element forces for element are: u 0.05 f 0.6 0.8 0 0 v 0.0337 5,00 f 0 0 0.6 0.8 u v The element forces are: f 76.6kN f 76.7kN Tension Eample 7 - Plane Truss Problem The local element forces for element are: u 0.05 f 0 0 0 v 0.0337 3,500 f 3 0 0 0 u3 v 3 The element forces are: f,06kn f,06kn 3 Compression

CIVL 7/87 Chapter 3 - Truss Equations - Part 44/53 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space Let s derive the transformation matri for the stiffness matri for a bar element in three-dimensional space as shown below: 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.

CIVL 7/87 Chapter 3 - Truss Equations - Part 45/53 Transformation Matri and Stiffness Matri for a Bar in Three-Dimensional Space The three-dimensional vector representing the bar element is gives as: duivjwk u i v j wk 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: ii L C ji Cy where L ( ) ( y y ) ( z z ) y y L z z ki C L z C cos C y cos y C z cos z where C, C y, and C z are projections of i on to i, j, and k, respectively.

CIVL 7/87 Chapter 3 - Truss Equations - Part 46/53 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 0 CCyCz 0 0 w u 0 0 0 C C y Czu v w d T * d * T C 0 0 CyCz 0 0 0 0 C C y Cz 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 0 C 0 y C 0 0 0 0 z AE CCyC z k 0 C L 0 0 0 C C y Cz 0 C y 0 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 L 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

CIVL 7/87 Chapter 3 - Truss Equations - Part 47/53 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: AE k L C CCy CC z CC y Cy CC y z CC z CC y z C z Eample 8 Space Truss Problem Consider the space truss shown below. The modulus of elasticity, E =. 0 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 L

CIVL 7/87 Chapter 3 - Truss Equations - Part 48/53 Eample 8 Space Truss Problem Consider element : L ( ) ( y y ) ( z z ) () L ( 7) (36) 80.5in () C C C y z 7 0.894 80.5 36 0.447 80.5 0 0.80 0.40 0 0.40 0.0 0 0 0 0 Eample 8 Space Truss Problem Consider element : u v w u v w 6 (0.30 in )(.0 psi ) k 80.5in lb in

CIVL 7/87 Chapter 3 - Truss Equations - Part 49/53 Eample 8 Space Truss Problem Consider element : L ( ) ( y y ) ( z z ) () 3 3 3 L ( 7) (36) (7) 08in () C C C y z 7 0.667 08 36 0.333 08 7 0.667 08 0.444 0. 0.444 0. 0. 0.444 0.444 0.444 0.444 Eample 8 Space Truss Problem Consider element : u v w u v w 3 3 3 6 (0.79 in )(.0 psi) k 08in lb in

CIVL 7/87 Chapter 3 - Truss Equations - Part 50/53 Eample 8 Space Truss Problem Consider element 3: L ( ) ( y y ) ( z z ) (3) 4 4 4 L ( 7) ( 48) 86.5in (3) C C C y z 7 0.83 86.5 0 48 0.555 86.5 0.69 0 0.46 0 0 0 0.46 0 0.308 Eample 8 Space Truss Problem Consider element 3: u v w u v w 4 4 4 6 (0.87 in )(. 0 psi ) k 86.5in lb in The boundary conditions are: u v w u3 v3 w3 0 0 u v w 0 4 4 4 v 0

CIVL 7/87 Chapter 3 - Truss Equations - Part 5/53 Eample 8 Space Truss Problem Canceling the rows and the columns associated with the boundary conditions reduces the global stiffness matri to: u w 8,997,403 K,403 4,398 The global force-displacement equations are: 8,997,403u 0,403 4,398 w,000 Solving the equation gives: u 0.07in w 0.66 in 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 L 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 L u j v j w j

CIVL 7/87 Chapter 3 - Truss Equations - Part 5/53 Eample 8 Space Truss Problem The stress in element is: 0.07 0 6 ().0 0.66 0.894 0.447 0 0.894 0.447 0. 80.5 0 0 () 948 psi 0 The stress in element is: 0.07 0 6 ().0 0.66 0.667 0.333 0.667 0.667 0.333 0.667 08 0 0 0 (), 445 psi Eample 8 Space Truss Problem The stress in element 3 is: 0.07 0 6 (3).0 0.66 0.83 0 0.555 0.83 0 0.555 86.5 0 0 0 (3),843 psi

CIVL 7/87 Chapter 3 - Truss Equations - Part 53/53 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. 3a. Do problems 3.4, 3.0, 3., 3.5a,b, 3.8, 3.3, 3.37, 3.43, and 3.48 on pages 46-65 in your tetbook A First Course in the Finite Element Method by D. Logan. End of Chapter 3a