SolSec8.3

# SolSec8.3 - Problems and Solutions Section 8.3(8.21 through...

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Unformatted text preview: Problems and Solutions Section 8.3 (8.21 through 8.33) 8.21 Use equations (8.47) and (8.46) to derive equation (8.48) and hence make sure that the author and reviewer have not cheated you. Solution: u ( x , t ) = Χ 1 ( τ 29 ξ 3 + Χ 2 ( τ 29 ξ 2 + Χ 3 ( τ 29 ξ + Χ 4 ( τ 29 (8.46) u (0, t ) = υ 1 ( τ 29 υ ξ (0, τ 29 = υ 2 ( τ 29 υ ( λ , τ 29 = υ 3 ( τ 29 υ ξ ( λ , τ 29 = υ 4 ( τ 29 (8.47) Substituting(8.46) into (8.47) u (0, t ) = Χ 4 ( τ 29 = υ 1 ( τ 29 υ ξ (0, τ 29 = Χ 3 ( τ 29 = υ 2 ( τ 29 υ ( λ , τ 29 = Χ 1 ( τ 29 λ 3 + Χ 2 ( τ 29 λ 2 + Χ 3 ( τ 29 λ + Χ 4 ( τ 29 = υ 3 ( τ 29 υ ξ ( λ , τ 29 = 3 Χ 1 ( τ 29 λ + 2 Χ 2 ( τ 29 λ + Χ 3 ( τ 29 = υ 4 ( τ 29 Τηισγιωεσ Χ 1 = 1 λ 3 2( υ 1- υ 3 29 + λ ( υ 2 + υ 4 29 ( 29 Χ 2 = 1 λ 2 3( υ 3- υ 1 29- λ ( υ 4 + 2 υ 2 29 ( 29 Χ 3 = υ 2 Χ 4 = υ 1 8.22 It is instructive, though tedious, to derive the beam element deflection given by equation (8.49). Hence derive the beam shape functions. Solution: Substituting (8.48) into (8.46) gives u ( x , t ) = 1 -3 ξ 2 λ 2 + 2 ξ 3 λ 3 υ 1 ( τ 29 + λ ξ λ- 2 ξ 2 λ 2 + ξ 3 λ 3 υ 2 ( τ 29 + 3 ξ 2 λ 2- 2 ξ 3 λ 3 υ 3 ( τ 29 + λ- ξ 2 λ 2 + ξ 3 λ 3 υ 4 ( τ 29 8.23 Using the shape functions of Problem 8.22, calculate the mass and stiffness matrices given by equations (8.53) and (8.56). Although tedious, this involves only simple integration of polynomials in x . Solution: T ( t ) = 1 2 ρΑ υ τ ( ξ , τ 29 ( 29 2 λ ∫ δξ = 1 2 υ Τ Μ υ ϖηερε υ = υ 1 ( τ 29 υ 2 ( τ 29 υ 3 ( τ 29 υ 4 ( τ 29 [ ] Τ And M is given by equation (8.35). Similarly V ( t ) = 1 2 ΕΙ υ ξξ ( ξ , τ 29 [ ] 2 λ ∫ δξ = 1 2 υ Τ Κ υ where K is given by (8.56) 8.24 Calculate the natural frequencies of the cantilevered beam given in equation (8.69) using l = 1 m and compare your results with those listed in Table 6.1. Solution: M = ρΑ 840 312 54-6.5 2 6.5-.75 54 6.5 156-11-6.5-.75-11 1 Κ = 8 ΕΙ 24-12 3 2-3 1 2-12-3 12-3 3 1 2-3 1 Following the procedures of section 4.2 ϖ 1 = 3.5177 EI r A , w 2 = 22.2215 EI r A w 3 = 75.1571 EI rA , w 4 = 218.138 EI r A From continuous theory, the natural frequencies of a cantilevered beam are ϖ i = b i EI rA where b 1 = 3.51601, b 2 = 22.0345, b 3 = 61.6972, b 4 =120.9019. The predictions of the first two natural frequencies are quite accurate while the predictions of the third and fourth natural frequencies are terrible. 8.25 Calculate the finite element model of a cantilevered beam one meter in length using three elements. Calculate the natural frequencies and compare them to those obtained in Problem 8.23 and with the exact values listed in Table 6.4....
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SolSec8.3 - Problems and Solutions Section 8.3(8.21 through...

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