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By Academician Leonid Brekhovskikh, Dr. Valery Goncharov (auth.)

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I) Absolutely rigid boundary. Particle displacement is forbidden: u(xo,t) = o. 6) In practice, this case can be realized if an end face of the rod is stuck to a massive wall of a material with very large Young's modulus. ii) Absolutely soft (free) boundary, say, a rod in a rather rarefied medium or vacuum. The forces (stresses) vanish at such a boundary: F/sIXQ = Eou/oxlxQ = 0, or ou/oxlxQ = o. 7) An interface with air is a good approximation to an absolutely soft boundary. iii) Contact between two rods with equal cross sections glued together with different material constants Pl' El and P2' E 2, respectively.

29) O. 27, 28) by use of the well-known vector identity Au = grad div u curl curl u. 29) becomes (A + 2fl) grad div u - fl curl curl u + Ib = O. 1. 0) (" a 0 0 000 b) a;, ~ 0 0 0 -a 0 ~ (: "a a ") a a D Solution a: Stresses are the same along every axis. Hence, isotropic compression (expansion) takes place. Solution b: Refer the tensor to the principal axes. ll), (a - 0)3 + 2a 3 - 3a Z(a - 0) = 0 has the roots 01 = 3a, Oz = 03 = O. Hence, in this coordinate system the tensor has the form all = 3a, a ik = 0, i, k i= 1, which corresponds to the elongation (contraction) of a bar.

Determine the reflection and transmlSSlOn coefficients of a bending wave at the rod's section x = 0 where the point mass m is placed. 7 Exercises 37 Solution: We have for x < 0 and x > 0, respectively: '1 = '2 = exp[i(kx - wt)] + Vexp[ -i(kx + wt)] + V1 exp(kx - iwt), Wexp[i(kx - wt)] + W1exp( - kx - iwt). At x = 0 the displacement, slope, and moment must be continuous: 'l(O,t) = '2(0,t), (o'dox)x=o = (O'2/0X)x=O, (o2'dox 2)x=o = (02'2/8x 2)x=o· The difference between shear forces at x = 0 causes the mass motion with acceleration o2'dot 2: EI(03'2/0X3 - o3'dox 3)x=o = m(02'dot 2)x=o.

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