Theory and Problems of Strength of Materials Part 11

Tham khảo tài liệu 'theory and problems of strength of materials part 11', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | CHAP. 11 STATICALLY INDETERMINATE ELASTIC BEAMS 295 y Fig. 11-10 and we must supplement these equations with additional relations stemming from beam deformations. The bending moment along the length ABC is conveniently written in terms of singularity functions dJT 2 Integrating R 2 where C is a constant of integration. As the first boundary condition we have when X 0 the slope dyldx 0. Substituting in Eq. 2 . we have 0--0 0-0 C for G o As the second boundary condition when X L dyldx 0. Substituting in Eq. 2 we find 0- z o 3 Next integrating Eq. 2 W C find Ma 2 2 3 Ely w 6 4 7 4 The third boundary condition is when X 0. y 0 so from Eq. 4 we have c2 0. The fourth boundary condition is when X L y 0 so from Eq. 4 we have MAL2 RAL . 0 - 4 5 2 o 24 The expressions for MA given in Eqs. 3 and 5 may now be equated to obtain a single equation containing RA as an unknown. Solving this equation we find wl 3- Ị Substituting this value in Eq. 3 we find MA 2. From statics we have s F - w Re 0 .-. Rc and t 5 2 Ma 2 - Me - h 0 .MC . The beam in Fig. 11-11 of flexural rigidity EỈ is clamped at A supported between knife edges at s and loaded by a vertical force p at the unsupported tip c. Determine the deflection at c. 296 STATICALLY INDETERMINATE ELASTIC BEAMS CHAP 11 Fig. 11-11 The reactions at A are the moment MA and shear force RA as shown in Fig. 11-11. From statics we have 3 ĨMA MAi - RB L -0 2 . P Rb ữ 2 ihese two equations contain the three unknowns MA RA and RB. Thus we must supplement these two statics equations with another equation arising from deformation of the beam. Using the x-y coordinate system shown the differential equation of the deformed beam in terms of singularity functions is -MaM - Rfí x J The first integration yields G 4 where C is a constant of integration. The first boundary condition is that when X 0. dy dx 0 hence from 4 C 0. The next integration yields .