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SolSec 6.8

SolSec 6.8 - 6 64 Problems and Solutions Section 6.8(6.64...

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6- Problems and Solutions Section 6.8 (6.64 through 6.68) 6.64 Calculate the response of the damped string of Example 6.8.1 to a disturbance force of f ( x,t ) = (sin π x / l ) sin10 t . Solution: f ( x , t ) = σιν πξ λ σιν10 τ . Assume a solution of the form: w n ( x , t ) = Τ ν ( τ 29 Ξ ν ( ξ 29 where X n ( x ) = σιν νπξ λ Substitute into (6.158) ρ && Τ ν + γ & Τ ν - τ - νπ λ 2 Τ ν σιν νπξ λ = σιν πξ λ σιν10 τ Multiply by sin n πξ λ and integrate over the length of the string: ρ && Τ ν + γ & Τ ν + τ νπ λ 2 Τ ν λ 2 = 0 φορ ν =1 σιν10 τ φορ ν 1 Only the particular solution is of interest since we are looking for the response to the disturbance force. Therefore, dropping the subscripts: ρ && Τ + γ & Τ + τ π λ 2 Τ = σιν10 τ && Τ + γ ρ & Τ + χπ λ 2 Τ = σιν10 τ ρ ϖηερε χ = τ ρ Solution is T = Α σιν(20 τ - φ 29 64

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6- where A = 1 ρ χ 2 π 2 λ 2 -100 2 + 100 γ 2 ρ 2 = λ 2 ρ 2 χ 2 π 2 -100 λ 2 ( 29 2 + 100 γ 2 λ 4 φ = ταν -1 10 γ ρ χ 2 π 2 λ 2 -100 = ταν -1 10 γ λ 2 ρχ 2 π 2 -100 ρλ 2 ϖ ( ξ , τ 29 = Α σιν(10 τ - φ 29 σιν πξ λ where A and φ are given above. 65
6- 6.65 Consider the clamped-free bar of Example 6.3.2. The bar can be used to model a truck bed frame. If the truck hits an object (at the free end) causing an impulsive force of 100 N, calculate the resulting vibration of the frame. Note here that the truck cab is so massive compared to the bed frame that the end with the cab is modeled as clamped. This is illustrated in Figure P6.65. Solution: Assume constant area and constant material properties. Equation of motion: ρΑϖ ττ - ΕΑϖ ξξ = φ ( ξ , τ 29 = -100 δ ( ξ - λ 29 δ ( τ 29 Mode shapes (eigenvalues) of a fixed-free bar are (Table 6.1) X n ( x ) = σιν (2 ν -1 29 πξ 2 λ Assume a solution of the form: w n ( x , t ) = Ξ ν ( ξ 29 Τ ν ( τ 29 . Substitute into the equation of motion: && T n - - (2 ν -1 29 π 2 λ 2 χ 2 Τ ν σιν (2 ν -1 29 πξ 2 λ = - 100 ρΑ δ ( ξ - λ 29 δ ( τ 29 δξ && Τ ν + ϖ ν 2 Τ ν { } σιν (2 ν -1 29 πξ 2 λ = - 100 ρΑ δ ( ξ - λ 29 δ ( τ 29 where c 2 = Ε ρ ανδ ϖ ν = (2 ν -1 29 πχ 2 λ . Multiply by sin (2 n -1 29 πξ 2 λ and integrate over the length of the rod: && T n + ϖ ν 2 Τ ν = - 2 λ 100 ρΑ σιν (2 ν -1 29 πξ 2 λ δ ( ξ - λ 0 λ 29 δ ( τ 29 = - 200 ρΑλ σιν (2 ν -1 29 π 2 δ ( τ 29 which has the solution: T n ( t ) = - 200 ρΑλϖ ν σιν (2 ν -1 29 π 2 σιν ϖ ν τ The total solution is: w n ( x , t ) = - 400 ρΑ (2 ν -1 29 πχ σιν (2 ν -1 29 π 2 ν =1 sin (2 n -1 29 πχτ 2 λ σιν (2

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SolSec 6.8 - 6 64 Problems and Solutions Section 6.8(6.64...

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