UNIVERSIDAD DE LAS FUERZAS ARMADAS ESPE ELEMENTOS FINITOS APLICADOS PROYECTO DEL 3ER PARCIALDeterminar las frecuencias naturales y modos de vibracin de una armadura plana con 3 elementos, utilizando MathCAD y la metodologa vista en vigas. La topologa de las vigas es la siguienteDatos de elementos, nodos y conectividadORIGIN 1 :=elementos 3 :=nodos 3 :=conectividad112233:=Grados16650901802.8970.8731.571= :=Matriz de rigidezTS -000C00000S -000C0:=CkA E LC2C S C2-C - S C S S2C - S S2-C2-C - S C2C S C - S S2-C S S2 :=EK TT k T := TKLA E simplifyC2S2C3S -C2S2 -C3S C3S -C4C3S C4-C2S2 -C3S C2S2C3S -C3S C4-C3S -C4PARA EL ELEMENTO 1: BARRA 1C cos 1( ):= S sin 1( ):=K1C2S2C3S -C2S2 -C3S C3S -C4C3S C4-C2S2 -C3S C2S2C3S -C3S C4-C3S -C4:=zeros00000000:= z1 augment zerosT zerosT, ( ):= z2 augment z1 z1 ,( ) :=K1augment K1 z1 , ( ):= K1stack K1 z2 , ( ):=K10.0550.2210.055 -0.221 -00000.2210.8860.221 -0.886 -00000.055 -0.221 -0.0550.22100000.221 -0.886 -0.2210.886000000000000000000000000000000000000=PARA EL ELEMENTO 2: BARRA 2C cos 2( ):= S sin 2( ):=K2C2S2C3S -C2S2 -C3S C3S -C4C3S C4-C2S2 -C3S C2S2C3S -C3S C4-C3S -C4:=z1 augment zeros zeros ,( ) :=zeros00000000:=K2augment zerosT K2,zerosT, := K2stack z1 K2,z1 , ( ):=K20000000000000000000.2420.203 -0.242 -0.20300000.203 -0.1710.2030.171 -00000.242 -0.2030.2420.203 -00000.2030.171 -0.203 -0.171000000000000000000=PARA EL ELEMENTO 3: BARRA 3( ) ( )C cos 3( ):= S sin 3( ):=K3C2S2C3S -C2S2 -C3S C3S -C4C3S C4-C2S2 -C3S C2S2C3S -C3S C4-C3S -C4:=zeros00000000:= z1 augment zerosT zerosT, ( ):= z2 augment z1 z1 ,( ) :=K3augment z1 K3, ( ):=K3stack z2 K3, ( ):=K30000000000000000000000000000000000000000000000000000000000000000=K K1K2+ K3+ :=K0.0550.2210.055 -0.221 -00000.2210.8860.221 -0.886 -00000.055 -0.221 -0.2980.0180.242 -0.203000.221 -0.886 -0.0181.0570.2030.171 -00000.242 -0.2030.2420.203 -00000.2030.171 -0.203 -0.171000000000000000000=Matriz de Masa ConcentradaTCS -00SC0000CS -00SC:=ma A L 21000010000100001 :=M TT ma T := maM2 A L simplify1.000001.000001.000001.0PARA EL ELEMNETO 1: BARRA 1C cos 1( ):= S sin 1( ):=M1C2S2+0000C2S2+0000C2S2+0000C2S2+:=zeros00000000:= z1 augment zerosT zerosT, ( ):= z2 augment z1 z1 ,( ) :=M1 augment M1 z1 ,( ) := M1 stack M1 z2 ,( ) :=M11000000001000000001000000001000000000000000000000000000000000000=PARA EL ELEMENTO 2: BARRA 2C cos 2( ):= S sin 2( ):=M2C2S2+0000C2S2+0000C2S2+0000C2S2+:=z1 augment zeros zeros ,( ) :=zeros00000000:=M2 augment zerosT M2 ,zerosT, ( ):= M2 stack z1 M2 ,z1 ,( ) :=M20000000000000000001000000001000000001000000001000000000000000000=PARA EL ELEMENTO 3: BARRA 3C cos 3( ):= S sin 3( ):=M3C2S2+0000C2S2+0000C2S2+0000C2S2+:=zeros00000000:= z1 augment zerosT zerosT, ( ):= z2 augment z1 z1 ,( ) :=M3 augment z1 M3 ,( ) :=M3 stack z2 M3 ,( ) :=M30000000000000000000000000000000000001000000001000000001000000001=M M1 M2 + M3 + :=M1000000001000000002000000002000000002000000002000000001000000001=Resolucion A L 210000000010000000020000000020000000020000000020000000010000000012t d1xdd22t d1ydd22t d2xdd22t d2ydd22t d3xdd22t d3xdd22t d1xdd22t d1ydd2A E L0.10.20.1 -0.2 -00000.20.90.2 -0.9 -00000.1 -0.2 -0.300.2 -0.2000.2 -0.9 -0.010.20.2 -00000.2 -0.20.20.2 -00000.20.2 -0.2 -0.2000000000000000000d1xd1yd2xd2yd3xd3xd1xd1y +El nodo 1 y 2 estan empotrados, y el nodo 3 es de libre movimiento.d3y t ( ) d3y ei t =2t d3ydd22- d3y ei t ( ) =d3x t ( ) d3x ei t =2t d3xdd22- d3x ei t ( ) = A L 22002d3x - 2d3y - 2E I L30.2420.203 -0.203 -0.171d3xd3y + 0 = A L 22E I L3= A L4 22 E I = -20020.2420.203 -0.203 -0.171+ simplify0.242 2 -0.203 -0.203 -0.171 2 -M0.242 2 -0.203 -0.203 -0.171 2 -:=eigenvals M ( )2.0 - 0.41258068808114941851 +2.0 - 0.00041931191885058149307 +Given2.0 - 0.41258068808114941851 + 0 =Find () 0.20629034404057470925 Given2.0 - 0.00041931191885058149307 + 0 =Find () 0.00020965595942529074653 1 0.20629034404057470925 := 2 0.00020965595942529074653 :=Frecuencias Naturales de la ArmaduraFRECUENCIA 1:12 1 L2E I A 12 = 10.642324441447738635326E I A L2= 2 1 0.642 =FRECUENCIA 2:2 2 0.02 =22 2 L2E I A 12 = 20.0204771071895075426706246E I A L2= Modos de Vibracin 1 :=0.242 2 -0.203 -0.203 -0.171 2 -0.171 -0.203 -0.203 -0.242 -=d3xd3y0.171 -0.203 -=d3xd3y0.8423645320197044335 = d3xd3y0.203 -0.242 -=d3xd3y0.83884297520661157025 = 2 :=0.242 2 -0.203 -0.203 -0.171 2 -0.2420.203 -0.203 -0.171=d3xd3y0.2420.203 -=d3xd3y1.1921182266009852217 - = d3xd3y0.203 -0.171=d3xd3y1.1871345029239766082 - = 0 =