For duralumin (see Section 1.6):E=75,000 MPa.One finds △=16.7mm 2.Denoting by W the elastic energy due to flexure,one has' 高+高r heam with 1 (G两*G.(e.+2e,)×b Using Castigliano theorem,one has aF then: ∫尚=+∫高票 0≤x≤E/2:M=Fx/2;T=-F12 2≤x≤e:M=t-T=F2 r"高受×+nc- …+总-x-+× △'= Fe3 Fe k 8(ED+4(G两 Approximate calculation: (En=B,xe,xbxe+e2+E eb 2 12 then: (EI)=7090 MKS+7.8 MKS with E=60 MPa (cf.1.6) negligible TTo establish this relation,see Chapter 15,Equation 15.17. 2 See calculation of this coefficient in 18.2.1,and more precise calculation in 18.3.5. 2003 by CRC Press LLCFor duralumin (see Section 1.6): E = 75,000 MPa. One finds 2. Denoting by W the elastic energy due to flexure, one has1 with2 : Using Castigliano theorem, one has then: Approximate calculation: then: 1 To establish this relation, see Chapter 15, Equation 15.17. 2 See calculation of this coefficient in 18.2.1, and more precise calculation in 18.3.5. D = 16.7 mm W 1 2 -- M2 · Ò EI ----------- x 1 2 -- k · Ò GS ------------T 2 dx beam Ú d + beam Ú = k · Ò GS ------------ # 1 Gc ec + 2ep ( ) ¥ b --------------------------------------- D¢ ∂W ∂F = -------- D¢ M · Ò EI ----------- dM dF ------- x k · Ò GS ------------T dT dF ------ dx beam Ú d + beam Ú = 0 £ £ x
/2: M = = Fx/2; T F– /2
/2 £ £ x
: M F 2 = = --( )
– x ; T F /2 D¢ 1 · Ò EI ----------- Fx 2 ------ x 2 ¥ -- x F 2 --( )
– x ( )
– x 2 ----------------dxº
/2
Ú d + 0
/2 ÚÓ ˛ Ì ˝ Ï ¸ = º k · Ò GS ------------- F 2 -- dx 2 – ¥ –------ F 2 -- dx 2 ¥ ------
/2
Ú + 0
/2 ÚÓ ˛ Ì ˝ Ï ¸ + D¢ F
3 48 · Ò EI ----------------- F
4 ------ k · Ò GS = + ------------ · Ò EI # Ep ep b ec + ep ( )2 2 ---------------------- Ec ec 3 b 12 ¥ ¥ ¥ + ¥ ------- · Ò EI 7090 MKS 7.8 MKS negligible = = + with Ec 60 MPa ( ) cf. 1.6 TX846_Frame_C18a Page 344 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC