280 Mechanics of Materials Now the mass falls through a distance h+δa+2 where og/2 is the effect of the rod extension on the mid-point of the beam.(This assumes that the beam remains straight and rotates about the fixed support position.) work done by faling mass If P=reaction at one end of beam then p=2 w(++)- wL+we⊙g 8AE (1) WL3 For a centrally loaded beam δ=48E1 WE×33 We 8B= 48×200×109×23×10-6=8.18×106 (2) WL For an axially loaded rod δR=AE WE×2.3 We =7×202×10-0×200×109=27.3×10 (3) Substituting (2)and (3)in (1), T2 W径×2.3 2.25×103 WE WE 0+818×10+54.6×10= 8(径×202×10-6)×200×109 W至 +2×8.18×106 +220+204-3xw22w W2×2.3 45 W是 +16.36×100 45+275×10-6WE+41.2×106WE=4.58×10-9W径+61.1×10-9W2 45+316.2WE×10-6=65.68×10-9W2 316.2×10-6, 45 Then 65.68×10-We-65.68×109=0 W2-4.8×103WE-685×106=0280 Mechanics of Materials Now the mass falls through a distance L where 6R/2 is the effect of the rod extension on the mid-poin the beam remains straight and rotates about the fixed support position.) f the beam. (This .. work done by falling mass = W If P = reaction at one end of beam then WE p=- 2 ssumes that WL3 6=- 48 EI For a centrally loaded beam (2) WEX 33 - WE 48 x 200 x lo9 x 23 x - 8.18 x lo6 6.q = WL AE For an axially loaded rod 6R = - .. Substituting (2) and (3) in (l), WE Wix2.3 2.25 x 103 -+ [ :09 8.18 x lo6 + 8 (4 x 202 x x 200 x lo9 w’, 2 x 8.18 x lo6 + W2 x 2.3 8 x 314 x loW6 x 200 x lo9 + - 2.25 x 103 wE 2.25 x 103 wE 45+ 8.18 x lo6 + 54.6 x lo6 w’, 16.36 x lo6 45+275 x WE+41.2 x WE = 4.58 x W’,+61.1 x W’, 45 + 316.2 WE x = 65.68 x low9 W’, Then 316.2 x 45 65.68 x 65.68 x low9 = W’,- WE - .. W’,-4.8 x lo3 WE-685 x lo6 = 0