Tham khảo tài liệu 'mechanics analysis composite materials episode 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 266 Mechanics and analysis of composite materials 0 PB22 x 2-ttRB 1 e r sin tx - cos Zr 4 T nD 1 e sin tx - cos Zx 0x 6-3 cos tx - sin tx 2nRB v 7 where r R and z Ấ. As can be seen solution in Eqs. is supplemented with a boundary-layer solution that vanishes with a distance from the cylinder end. To determine transverse shear stress TJZ we integrate the first equation in Eqs. under the condition Tr- z 0 0. As a result the shear stress acting in the angle-ply layer is specified by the following expression 4 1 4 ZKX 4I 4 ZK 0 K Substitution of Eqs. and calculation yield 4-9 e cos tx sin tx z cos tx -I- sin Zx z2 . Transverse normal stress can be found from the following equation similar to Eq. ơz rT zỊ R Aĩì e J22 1 Đ dz D For a thin cylinder we can neglect z R in comparison with unity. Using Eqs. and for the angle-ply layer we get ơỊ1 z e- cos Zx sin Zx z cos zx sin Zx z2 cos lx sin Zx z3 . we get C44 0 and A- 44 0. For the cylinder under study e mm . the reference surface is within the angle-ply layer. Free-body diagram for the cylinder loaded with torque T see Figs. and yields Chapter 5. Mechanics of laminates As can be seen the first equation in Eqs. follows from this solution ifx 00. Distribution of shear stress Tlr9 z h and normal stress ci1 z h acting at the interface between the angle-ply and the hoop layer of the cylinder along its length is shown in Fig. . Consider now the problem of torsion see Fig. . Constitutive equations in Eqs. that we need to use for this problem are Taking coordinate of the reference surface in accordance with Eq. . Fig. . Forces and moments acting on an element of the cylinder under torsion. 1 r y 0 ------------------------- x R 0 NV1. 447 . Mr . Fig. . Distribution of normalized transverse shear stress
