3.1 Derivation of the Stiffness Matrix for a Bar in Local Coorinates. In 3.1 we will perform Steps 1-4 of Logan s FEM. Derive the truss element equations. 1. Set the element type. 2. Select a isplacement functions. 3. Define the stress/strain relationships. 4. Derive the stiffness matrix. Steps 1-4 complete using an x y elemental/local coorinate system. Nee to put element equations in terms of global coorinate system to solve truss loa-eformation problems. x ŷ y x
Step 1: Set the element type. Constant cross section (A). Constant material prop s on length L (E). element affors stretch in only the axial irection,. External loas are applie at the noes. State of the truss/bar element is given by isplacements at noes:, x 2x y Note: at this point we are not concerne with any particular truss geometry.
Step 2: Select the isplacement function (see section 3.2). u(x): polynomial interpolating noal isplacements. Conforming element: isplacement continuous within the element. continuity in the moel DOF s between elements. element shoul be capable of rigi boy motion. Completeness: he necessary level of ifferentiation is possible through u(x) choice (we can have a constant strain in the bar). see step 3 of this proceure an the MWR to come later in the course. y
Step 3: Define the stress-strain relationships. As we are focuse on the loa-isplacement problem, the stress strain relationships are of primary interest. Our conforming element can be applie within this DE. u ε x = an σx = Eεx Set by element type. x 2x ux ( ) = + x L ε x 2x 2x an = σx = E L L Set by isplacement function.
Step 4: Derive the element stiffness matrix,. Relate noal forces to the noal isplacements through efinition of internal tension,. f = an f =+ 2x Aσ = AE 2x = x L k Same concept as with our spring element. he sense of the noal loas is governe by the reference frame s x irection. AE 1 1 f L 1 1 = f 2x 2x Element equations.
3.3 ransformation of Vectors in wo Dimensions. he isplacement,, at a noe point can be communicate using any reference frame. When we assemble truss elements, we will have ifferent elemental x irections for each element. Nee to represent the noal an f vectors in a common frame. A irection, i, can be forme from a combination of basis vectors of another frame. +
3.3 (cont ) C S = 1y S C 1y r C S =, r = S C 1 2 2 2 ( ) rir = C + S = 1 for i= j i j ( ) rir = CS SC = 0 for i j i j R R R 1 = he planar rotation operator is an orthonormal (orthogonal) matrix.
3.4 (Forming the) Global Stiffness Matrix. Up until now we have ignore the iy an jy noal isplacements. Clearly a truss noe must isplace normal to an element s axis. We augment/pa the element equations to account for the transverse egrees of freeom. he truss assembly will have to ensure that all transverse motions are restraine. Whether or not this occurs can be etecte by the eterminant of the stiffness matrix (singular?). x ŷ y x
We alreay efine R as a transformation from the elemental frame to the global frame. 1 0 1 0 f 0 0 0 0 1y f1y 1 0 1 0 = 2x 2x 0 0 0 0 2y f2y AE L f f f 1y 2 2 0 2 2 1y f 1y 2 2 0 f R x x R x 2x2 1y an = 0 = 2 2x2 R2x2 2x 0 f x f2x 2x2 R2x2 2x 2y f 2y 2y f 2y C S 0 0 R 0 0 2 2 0 S C x 2x2 = = 0 0 0 C S 2x2 R2x2 0 0 S C Eq. (3.4.15) of Logan.
Since is block iagonal an is forme by a submatrix that is orthogonal: 1 = Each set of element equations can be written in terms of the global reference frame by using the for each element: { f} = k{ } f = k f = k Element stiffness matrix in terms of global coorinates (no carat ).
he change in coorinate system oes not change the generic sense of the element stiffness matrix. k ( e) = AE CS S CS S 2 2 CS S CS S 2 2 C CS C CS 2 2 2 2 L C CS C CS Assembly K N = k e= 1 ( e) Notes: C an S are particular to each element. he rotation takes the global frame into alignment with the element frame ( + clockwise about z axis protruing from the page).
Steps 5 an 6: Assemble Global Equations, apply bounary conitions, solve. Bounary conitions are likely in terms of global isplacement values at each noe, an. 3.5 Computation of Stress for a Bar in the xy Plane. Steps 7 an 8: Recovery an Interpretation. F ix Any piece of information (stress an strain) can be recovere from the noal isplacements. (1) E σ = KD { 1y 2x Ny} iy σ σ [ 1 1] = L 1 2x E [ 1 1] (2) 2x = L 2 3x E [ 1 1] ( N ) ( N 1) x = L N Nx
Element strains an stresses are in terms of local isplacements (isplacement along the elements axial irection). he noal isplacements in the specific element frame can always be recapture from the orthogonal transformation,. C S 0 0 1y S C 0 0 1y = 0 0 C S 2x 2x 0 0 S C 2 y 2 y C S 0 0 1y = 0 0 C S 2x 2x 2 y C S 0 0 * = 0 0 C S See equations (3.5.6) thru (3.5.8) pg.# 79. A shorthan matrix operator. Not a true linear transformation.
3.6 Solution of a Plane russ. Problem 3.20 Bounary conitions aligne with a global reference frame. Applie loas also easily expresse in global coorinate system y x E = A = 5 in L = 100 in x ŷ 6 1 10 psi 2 k (1) = C C S C C S (1) (1) (1)2 (1) (1) (1)2 AE C S S C S S L 1 C C S C C S (1) (1) (1)2 (1) (1) (1)2 C S S C S S (1)2 (1) (1) (1)2 (1) (1) (1)2 (1) (1)2 (1) (1) k (2) = C C S C C S AE C S S C S S L 2 C C S C C S (2) (2) (2)2 (2) (2) (2)2 C S S C S S (2)2 (2) (2) (2)2 (2) (2) (2) (2) (2)2 (2) (2) (2)2 (2)2 (2) (2)2 (2) (2) ( 2 lbf 5 in ) 1 10 6 2 AE in kip = = 35.36 2L 2 100 in in
Orientation of each element calculate inepenently: C S = cosθ = cos(45 ) = 0.707 = cosθ = sin(45 ) = 0.707 (1) (1) (1) (1) C S Step 5: Assembly: = cosθ = cos(135 ) = 0.707 = cosθ = sin(135 ) = 0.707 (2) (2) (2) (2) (1) f F1 x 0.5 0.5 0.5 0.5 0 0 (1) f F 1 1 0.5 0.5 0.5 0 0 y y (1) (2) f2x + f 2x F2 x kip 1.0 0.0 0.5 0.5 (1) (2) = 35.36 f F = 2y f2y 2 y in 1.0 0.5 0.5 + (2) f F 0.5 0.5 3x 3x (2) f F SYM 0.5 3 y 3 y 1y 2x 2 y 3x 3y Note: K is symmetric an singular.
Step 6: Apply bounary conitions an solve. = 1y = 3x = 3y = 0.0 Reucing the assemble system equations F 2y = 10.0 kip F 2x = 0.0 kip F1 x 0.5 0.5 0.5 0.5 0 0 F 0.5 0.5 0.5 0 0 1y 1y F2 x kip 1.0 0.0 0.5 0.5 2x 35.36 F = 2 y in 1.0 0.5 0.5 2 y F 0.5 0.5 3x 3x F SYM 0.5 3 y 3 y We solve irectly for the remaining state variables: F2x kip 1.0 0.0 2x 35.36 F = in 0.0 1.0 2y 2y 2x 0.0 = 2 y 0.283