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Solutions to problems Structural Stability Theory and Practice by Sukhvarsh Jerath

Solutions to exercises for Structural Stability Theory and Practice by Sukhvarsh Jerath

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Solutions to problems Structural Stability Theory and Practice by Sukhvarsh Jerath

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  1. You can access complete document on following URL. Contact me if site not loaded https://solumanu.com/product/structural-stability-theory-and-practice-jerath/ smtb98@gmail.com Solution Manual Structural Stability Theory and Practice Buckling of Columns, Beams, Plates, and Shells 1 Contact me in order to access the whole complete document - Email: smtb98@gmail.com WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 - Telegram: https://t.me/solutionmanual

  2. PROBLEMS 1.1 Find the critical load Pcr for the rigid bar column in Fig. P1.1 by using the equilibrium method. The column is restricted by a rotational spring of stiffness c at the support. c P L A B Fig. P1.1 M  0  PL sin θ – cθ = 0 or c P L For small deformation, sin θ ≈ θ c P L A   sin    cr 2

  3. 1.2 Solve problem 1.1 by the energy method. 1 2 2 U c  V PL 1 cos      1 2 2 U V c PL 1 cos          d d  0 For equilibrium, or c For small deformations, sin θ ≈ θ or c P L PL sin 0      cr 3

  4. 1.3 Determine the critical load Pcr for the rigid bar column in Figs. P 1.3 a and b by using the equilibrium method. P P k2 L L k k1 45o (a) (b) Fig. P 1.3 (a) M  0  PL A sin k AD 0               BC BC 2 L cos L 2       4 2 δ = Extension of the spring          AD L sin  4 2                       PL sin k 2 cos L L 2 L sin 0       4 2 4 2 4

  5. or                  2 PL sin 2 kL cos cos sin sin sin cos cos sin     4 2 4 2 4 2 4 2           2 2 kL sin cos cos sin 0 +   4 2 4 2 or          2 2 PL sin kL cos2 kL cos sin 0       2 2  cos2 cos sin    2 2 P kL  sin  If θ is small, cos θ ≈ 1, sin θ ≈ θ kL P   cr 2 (b) M  sin F k   For small deformations, sin θ ≈ θ, cos θ ≈ 1 2( ) PL k L L       2( ) 0 k k L       or 2 2 2 ( k L k k    0  PL  k A k ( L sin ) cos  L 0  (1)    0 L   2  horizontal 2( sin ) 0  (2)    1 0 1 0 0 k L PL k L                    )  2 1 2 2 k L PL k L  2 2  0  k L ( k k ) 2 1 2 5

  6. or ( or 2 2 2 2 k k )( k L PL ) k L 0     1 2 2 1 2 k k  P L  cr k k 1 2 6

  7. 1.4 Solve Problem 1.3 by energy method. (a) 1 2 1 2 2 U k                        2 2 2 2 U k 4 L cos 2 L 4 L 2cos      4 2 4 2 V=-PL(1-cos θ) U   or 1 2 V 2                         2 2 2 k 4 L cos 2 L 4 L 2cos PL (1 cos )         4 2 4 2 d d  0 For equilibrium, d d or d d or 1 2 1 2 1 2                                      2 2 k 4 L (2)cos sin 4 L 2sin PL sin 0         4 2 4 2 4 2  1 2                             2 2 k 2 L cos sin cos sin 2 L cos sin PL sin 0         2 2 2 2 2 2          kL cos2 cos sin    2 2 P  sin  For small θ, cos θ ≈ 1, sin θ ≈ θ kL P   cr 2 d d d d At θ = 0 2 d d    2cos2 kL          1 2 cos sin PL sin      2 1 2 2  2         2 kL 2sin2 sin cos PL cos       2 2 2  1 2 kL        2 kL PL   2 2 d d d  P P , 0 For    cr 2 2 2 kL  P P , 0 When    cr 2 2 d Hence, the system is stable for P < Pcr and is unstable for P >Pcr in the initial position. 7

  8. (b) 1 2 1 2 2 2 U k k ( L sin )       1 2 V = -PL(1-cosθ) 1 2 For equilibrium     For small deformations, sin θ ≈ θ and cos θ ≈ 1 1 2 2 2 k k ( L sin ) PL (1 cos )          1 2 k 2( L sin )( cos ) L  PL sin       k k 2( L sin )      1  2 0 0 k L PL k L                      2 2  k L k k  2 1 2 These are same equations as in Problem 1.3(b). 1 2 k k  P L  Hence, cr k k 1 2 2       2     For initial position, θ = δ = 0 2 K   2 K   2 k L sin k L cos2 PL cos        2 2 2  2 k k  1 2 2  2 cos k L      2 k L PL    11 2 2   k k    22 1 2 2 2     K K k L    12 21 2   2 0 Two degrees of freedom system is in stable equilibrium if or D1 = K11 >0, or k2L2 – PL > 0, or P < k2L K K k L D K K or 2 2 2 2 1 2 2 ( )( ) k L PL k k k L    k k P L k k    2 PL k L  11 12 2 2  0 or 0    2 k L k k 21 22 2 1 2 0  1 2  or 1 2 This is a sample only. Lat solved problem in chapter1 is #6. 8

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