By William R. Spillers

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**Example text**

Since it is assumed that the members are rigidly connected, all member ends which meet at a joint have the same x and y displacement components and the same rotation. Proceeding more formally, associated with joint i are an applied force vector Ph applied moment Mh a displacement vector dt and a rota tion et. These are again described for the entire structure by the matrices and δ = AL PJ Displaced position Initial position Fig. 3. The ith point. 37 Automated Structural Analysis but here the elements are δ,= and (Vox Pi = (Pi)y Mi j J again is the number of movable joints.

The stiffness matrix. For a uniform straight beam in bending the values of Ft are shown in Fig. 3. 3, it is again possible to identify the contribution of each member to the system matrix. Let ■Ni N2 and N= NKN = 2 Ν,Κ{Ν„ LNBA 41 Automated Structural Analysis the contribution of the /th member to the system matrix, NKN, is Ν,Κ,Ν, = RtNi+ IMC! col A I — Λ,Ν,+Κ,Ν,+Λ, col. 3 included at the end of this book illustrates the work dis cussed in this chapter and provides a numerical example. 6 shows an example for which the matrices just presented will be described in detail.

For example, (R^f^) is the end of member force on the positive end of member /—1 in the global coordinate system. In Eqs. 7) the parentheses have been added for clarity. To find Cu it is convenient to use the equilibrium equation of joint / — 1, Eqs. - = f}rFi, and Eq. 2), ^ = Ni+Ri8A^Ni-RA -> ec = Ä i (Nr)- 1 (A < -N i + Ä i 6 i l ). 9) Starting from Eq. 6) and using Eq. 8) it follows that 6i-1=-Ct-i(Ri-if;_l), but using Eqs. 1) leads to Applying Hooke's law, Ft = XA, and then the first of Eqs.