ASME Steam Tables: Compact Edition (Crtd) by ASME Research and Technology Committee on Water and Steam in

By ASME Research and Technology Committee on Water and Steam in Thermal Systems Subcommittee on Properties of Steam

This up-to-date and concise publication contains the next: U.S. widely used devices - desk 1. Saturated Water and Steam (Temperature Table), desk 2. Saturated Water and Steam (Pressure Table), and, desk three. Superheated Steam (1 to 15,000 psia); SI devices - desk four Saturated Water and Steam (Temperature Table), desk five. Saturated Water and Steam (Pressure Table), and, desk 6. Superheated Steam (0.005 to a hundred MPa); and, Unit Conversion components - Mollier Diagrams (U.S. and SI units).

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A - Fs . r = 0 (i) stehen 4 Gleichungen fiir die 4 unbekannten Krafte F, F l , F2, Fs zur Verfiigung. Das Ergebnis fUr die Kraft F lautet br F=la 8 e/loO! e/loO! - 1 FQ. Schwach gekriimmte Seilkurven Die Gleichgewichtslage des schwach gekrfunmten Tragwerkes ist durch eine konstante horizontale Komponente der Seilkraft Fs(x) und eine Differentialbeziehung bestimmt: Hyl/(x) = -Py(x) Fsz = H = konst. 1: Fur das in Abb. 1 dargestellte schwach gekrummte Seil ist die Funktion der Seilkurve zu berechnen.

3). Das sind am Tragwerkteil I die Resultierende ~qol der Dreieckbelastung und am Tragwerkteil II eine Resultierende ~qol der Rechteckbelastung und eine Resultierende ~qol der Dreieckbelastung. ==------qo I -qol----i -t.. 3: Freigeschnittenes Tragwerk Tragwerkteil I: cl> L:MiG = 0 (i) Fav Kontrolle am Tragwerkteil I: L:Fw (i) ? == 0 Tragwerkteil II: CI>L:MiC (i) 2 2 2 2 4 FB . 1 - -q 1 . 1 - -q 1 . e . l+FB·1--3qol·-3l =0 FaH =0 =0 --+ FaH = O. Kontrollen am Tragwerkteil II und am Gesamttragwerk: ?

Weiterhin fafit man die Linienkrafte zu ihrer Resultierenden zusammen (Abb. 2). Aus sechs passend gewahlten Gleichgewichtsbedingungen erbalt man die gesuchten Gro:Ben. Schlie:Blich wird man noch Kontrollen durchfiihren, da es immer Gleichgewichtsbedingungen gibt, die zur Losung nicht herangezogen worden sind. Gesamttragwerk: cE> 2:MiB = 0 FAV· 3,5 +4500,0 ·1,5 -1000,0·5,0 -1400,0 ·1,75 (i) -t FEV . FAV = =0 200, OkN, 3,5 - 1400,0 . 1, 75 + 1000,0 . 1,5 - 4500,0 . 1,5 = 0 -t FEV = 2200,0 kN ; -t FAH = 3000,OkN, Tragwerkteil I: F AH .

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