Question

Section 4.4 Finite Element Formulation of Frames 235 256 of 929 where the transformation matrix is sine cose 0 0 0 0 0 sine 0 0 0 -sine cose 0 0 In the previous section, we developed the stines matributed to bending for a beancement. This matracounts for lateral deplacements and rotaties teach mode andis TO 0 0 0 60-126 0 0 0 0 0 LO 621041 To represent the contribution of each term to nodal degrees of freedom, the degrees of freedom are shown above and alongside the stiffness matrisin EQ (4:53). In Section 4.1 we derived the stiffness matrix for members under axial loading as AE. AE O o (454) 0 0 0 AE 2 00 AL 0 0 Chapter 4 All Members, Beams, and Frames Adding Eqs (4.33) and (454) results in the stiffness matrix for a frame element with respect to local coordinate system AE 0 0 AE 0 0 EX 0 0 4E 6E ZE L 0 AE 2 0 12E 6E 0 0 L Note that we need to represent 14 (455) with respect to the global coordinate sys sem. To perform this task, we must substitute for the local displacements in term of the global displacements in the strain energy equation using the transformation matrix and performing the minimization. (See Problem 413.) These steps result in the relationship IKT-' where" is the stiffness matris for a frame element expressed in the global code dinate system.XYNext, we will demonstrate finite element modeling of frames with another example

Answer 1

obtain the general stiffness matrix of a bar element arbitrarily oriented in three-dimensional space as shown in Figure . Let the coordinates of node 1 be taken as Xı, Yı, and 21, and let those of node 2 be taken as X2, yz, and 22. Also, let 0,0y, and 0. be the angles measured from the global x,y, and 2 axes, respectively, to the local x' axis. Here x' is directed along the element from node 1 to node 2. We must now determine (T ) such that

y, v x', W2 vi- fix, X, U z, W Figure Bar in three-dimensional space along with local nodal displacements {d'} = [1 ]{d}. We begin the derivation of (T ) by considering the vector d = d expressed in three dimensions as u'i' + d'j' + w'k' = ui + vj + wk (1) where i', i', and k' are unit vectors associated with the local x', y, and z' axes, respec- tively, and i, j, and k are unit vectors associated with the global x, y, and z axes. Also w and w' now denote the displacements in the z and 2 directions, respectively. Taking the dot product of Eq. (1) with i', we have

(2) w +0+0 = u(i' - i) + v(i' . j) + w(i'k) and, by definition of the dot product, X2 = Сх L i'.j= y2y1 = C, ( 3) i.kr 22 L 21 L = C, z1) 9j1/2 where L = [(x2x1)? +(y2 yu)? + (22 and Ct = cos 0x Cy = cos Oy C = cos 0 (4) Here Cx, Cy, and C are the projections of i' on i, j, and k, respectively. Therefore, using Eqs. ( 3) in Eq. ( 2), we have u = Cxu + Cyv + C-W (5)

For a vector in space directed along the x' axis, Eq. (5) gives the components of that vector in the global x, y, and z directions. Now, using Eq. (5), we can write the local axial displacement at node 1 and 2 in explicit form as U1 Vi s! W1 ui un Cx CyC: 0 0 0 0 0 Cx C C (6) 12 U2 W2 111 V1 W1 Now {d} = { }} {d} Cx 0 0 0 Cx Cy C. U12 , and define (T )= [ U2 W2 (7) Using Eq. (7), we write Eq. (6) in matrix form as {d'} = (T ]{d} (a) Here (T ) is the transformation matrix, which enables the local displacement matrix {d'} to be expressed in terms of the displacement matrix {d} components in the global coordinate system. Based on Eq. (a), it will be convenient to express the global force matrix in terms of the local force matrix using (T ) as

{f} = [1 ]"lt} (b) Now in local coordinates, the local forces are related to the local displacements by {f'} = [K]{d} (0) Upon substituting for {d'} from Eq. (a) into Eq. (c) and premultiplying both sides by {T }', we have [T]"{f'} = [T]"[k'][T }{d} (d) Now using Eq. (b) in the left side of Eq. (c), we obtain {f} = {T ]"[k'][T }{d} (e) The global forces are related to the global displacements by {f} = [k]{d}

Comparing the right sides of Eqs. (e) and (f), we then observe that the global stiffness matrix for a bar arbitrarily oriented in space is [k] = [T]"K'][T] Using Eq. (7) for (T ) Eq. for [k'] we obtain (k) as follows: [C, o Cy0 C 0 AE Cx Cy C0 0 0 (k)= (8) 0 0 0 0 CC, C. 1 EL CX 1 0 Cy 0 Simplifying Eq. ( 8), we obtain the explicit form of (k) as [c CC, CC CC Cc CC,C. GC C C,C. AE c? C C CC: [k] = 14 C? (9) C C CC: C Cc: Symmetry c? Equation (9) is the basic form of the stiffness matrix for a bar element arbi- trarily oriented in three-dimensional space.