By William R. Spillers

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Extra info for Automated Structural Analysis. An Introduction

Example text

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.