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By William R. Spillers

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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. ^-iN^F^ = Rt-AN^N^Rt-A Λ,-ιίΝ^-^ΓΛ + ^ ^ ^ or simply Q = Rt-ANU)-1 [KT^ + N^Ri-^Ri-iN^] (N^)"1*,-,. 10) This is the desired result.

2. The /th member. 46 z Global coord. system The Node Method for Space Frames The member description of a beam in space is considerably more complicated than that of a plane beam and for that reason it is shown rather schematically in Fig. 2. In its own local coordinate system, each member is again oriented along the x axis. In space, a beam is acted upon by a force vector and a moment vector at each end, a total of twelve quantities. But since these twelve quantities are related by the six equilibrium equations of a rigid body in space, there are six quantities which may be specified arbitrarily.

42 The Node Method for Plane Frames IR2N2+K2N2+R2 I R2N2+K2N2-R2 + . 6 0 0 1 , 0 -1 0 (δ,)χΊ 8= (βι)» (δ 2 ) χ (δ2)„ _Θ2 \ EXERCISES 1. e. structures whose elements not only lie in a plane but also have negligible length changes). 2. g. from Handbook of Frame Con­ stants, Portland Cement Assoc, Chicago). 43 Automated Structural Analysis 3. g. when Ft =fi+). 4. 3 to perform plastic analysis. Answer: In this exercise, the simplest plastic analysis problem is con­ sidered in which: 1. Loads are only allowed to act at nodes.

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