Question

Problem 3: (25%) The steel frame shown in the Figure is subjected to a force P(t) = 15 sin(ot) kips at the beam level. The damping ratio 5 is 5%, E = 3 x 10 ksi, total weight of the beams = 100 kips, all the columns are 12" x 12" in size and this is the case of resonance. Determine: a) Damped and un-damped natural periods of the structure. b) The maximum dynamic horizontal displacement (u.) at the top of the frame. c) The maximum shearing force at the base of column-1 and column-2 EI = 0 P(t) 12 ft Column-1 Column-2 10 ft 5 ft 8 ft

Answer 1

* Answers EI. PIE 12ft @ 3 Th loft 5ft 8ft Moment of inertia of column 3 12 X 12 1 1798 in 4 Given E= 3x10²ksi Stiffness of the system K=ko + ko tko tko 12 E1 12 EL 12 ET 39 EI ЗЕТ 63 + + 62 b3 b3 39 x 3 x 103 x 1728 (12 x 12)3 61.708 klon = 67.708 Kirsten = 812500 lb/ft 0 Natural period i given Om = 1000 kips :M 3 We have con = m = 1000 X10 32.2 m=31055-9 812500 31055.9 con = 5.114 rad /sec

217 w 217 :: Natural period, The = 1.288 5.114 + To = 1.228 Sec * Caven, damping ratio = E= 5%. COD นอก .. Damped frequency ſing² Vi-(0.05) +Wp = 0.9987 X5.114 = 5.1076 rad/sec = 0.9987 Damped natural period 211 1.230 sec Тр: 5.1076 To = 1.23 sec (10) Marimum dynamic horizontal displacement (): (F%) (sin wt t-0) Geven ſpl)= VCD-v=) + (Pre)? we have, a= I= 15 sinalt kins when , YA re frequency ratio, wn (fo/ka) and maa, displacement Xman - 5.1016 (1-12)+ (2/2) (assume , w=wo , since not given) NOCO V= > 5.114

So, r= 0.9987 and given , Fo=15 kips & wehave , k = 67.708 Kips/ir i manimum displacement (15/64.708) amara VC -0.9987)+ (ax0.9967 x 0.05) → Xmaa 0.8955 in 100 Max. Shearing force at the base of column -1 and column-2i @ Maa Shearing force at column-18 FO Kolmaa = 3E I b3 x amaa 3X3 X 10PX1778 x 0.8955 (12 X 12) 3 3 FO = 4.664 Kirs column - ai 6 Max. Shearing force at FO ka amaa 12 EI h3 x aman 12 X3 X 10x1728 x 0.8955 (12 x 12) 3 ro = 18.656 Kirs