Chapter 4 Response To Periodic Loading 4-1 Fourier series Expressions of Periodic Loading Trigonometric Form ∞∞ F a0 F()tatbt=+∑∑nn cosθ+ nn sinθ t 2 nn==11 Tp 2π 2π θ1 = θθn ==nn1 Tp Tp 2 T p aFttt= ()cosθ d n = 0,1,2, ,∞ nn∫0 K T p 2 T p bFttt= ()sinθ d n =1, 2, , ∞ nn∫0 K T p 1 For arbitrary periodic form, the integrals is evaluated numerically N −1 N −1 tmt= ∆ Tp qq0 N m qt()d t=+()qtm +∆=∆tfm ∫0 ∑& ∑ 22m=1 m=1 qqtmm= () The beginning and end of the time period usually can be set so that the ordinates qo and qN are equal to zero ∞∞ 2 Tp a0 aFtttnn= ()cosθ d F()tatbt=+ cosθ+ sinθ T ∫0 ∑∑nn nn p 2 nn==11 2 Tp bFttt= ()sinθ d nn∫0 Tp q(t) qm q 2∆t N −1 3 q aFttnmnm= ∑()cosθ 2 T m=1 q p 1 L N −1 q 2∆t q0 N bFttnmnm= ∑()sinθ∆t t Tp m=1 2 Exponential Form Substituting 1 cosθθθnnntitit=+−[exp( ) exp( )] 2 2π θθn ==nn1 (4-7) i Tp sinnnntititθθθ=−[exp( ) − exp( −)] 2 Into Equation (4-1) a ∞∞ F()tatbt=+0 cosθ+ sinθ ∑∑nn nn (4-1) 2 nn==11 ∞∞ a0 1 i F(taititbitit )=+∑∑nn [exp(θθ) + exp( − n )] −θθ nn [exp( ) − exp( − n )] 22nn==11 2 ∞∞ aaib0 nn−+ aib nn =+∑∑exp(itθnn ) + exp( − itθ) 22nn==11 2 3 Exponential Form aaib∞∞−+ aib 2π 0 nn nn θθ==nn Ft( )=+∑∑ exp( itθnn ) + exp( − itθ) n 1 22nn==11 2 Tp 2 Tp aFttt= ()cosθ d nn∫0 n = 0,1,2, Tp
