Hoja1RESOLVER:SOLUCION:MATRIZ DE RIGIDEZ DE LA ESTRUCTURA:NUDO1231[K11]1[K12]102[K21]1[K22]1+[K11]2[K12]230[K21]2[K22]2MATRIZ DE TRANSFORMACION

C-S0[R] =SC0001

CS0[R]T =-SC0001MATRIZ DE RIGIDEZ UNA BARRA EN LOCAL:[K11][K22]BARRA 11000000[K22] =0120-6000-6004000

[K22]g =[R]*[K22]*[R]T

C-S0[R] =SC00018000-72360600096-480CS00-6004000[R]T =-SC0001

0.8-0.606443.24742.4360[R] =0.60.80[K22]g =4742.43676.8-480001360-4804000

0.80.60[R]T =-0.60.80001BARRA 2

11785.1130.0000.000[K11] =0.000196.418833.3330.000833.3334714.045

[K11]g =[R]*[K11]*[R]T

C-S0[R] =SC00018333.3333193497138.8884997471589.2554152865-8333.3333193497138.8884997471589.2554152865CS00833.33334714.045[R]T =-SC0001

5990.7655-5794.3475589.2554152865[K11]g =-5794.34755990.7655589.25541528650.70710678120.70710678120589.2556274186589.25562741864714.045[R] =-0.70710678120.70710678120001

0.7071067812-0.70710678120[R]T =0.70710678120.70710678120001MATRIZ DE RIGIDEZ DE LA ESTRUCTURA:[K] = [K22]1+[K11]26443.24742.43605990.7655-5794.3475589.255415286512433.9655-1051.9475949.2554152865[K] =4742.43676.8-480+-5794.34755990.7655589.2554152865=-1051.94759667.5655109.2554152865360-4804000589.2556274186589.25562741864714.045949.2556274186109.25562741868714.045ENSAMBLAJE DE CARGAS000{F} =10000{R} = ?0000CALCULO DE {R}010606.601120001500039166.667

39166.6672100010606.60115000

120002100010606.60110606.60112000-10606.6012100010606.601{Y'}1 =39166.6671500012000{Y'}2 =-10606.6012100010606.601-39166.667-150000.8-0.600.70710678120.70710678120[R] =0.60.80[R] =-0.70710678120.70710678120001001TRANSFORMANDO AL SISTEMA GLOBAL0.8-0.6012000-30000.60.80*2100024000nudo1{Y}1 =00139166.667=39166.6670.8-0.6012000-30000.60.80*2100024000nudo2001-39166.667-39166.6670.70710678120.70710678120-10606.6010-0.70710678120.70710678120*10606.60114999.99898488nudo2{Y}2=00115000=150000.70710678120.70710678120-10606.6010-0.70710678120.70710678120*10606.60114999.99898488nudo3001-15000-1500011-30000-30002400014999.9989848838999.99898488{R} = {Y}1 +{Y}2 =-39166.667+15000=-24166.667PROBLEMA COMPLEMENTARIO10000-300013000{Q} = {F}2 -{R}2 =0-38999.99898488=-38999.998984880-24166.66724166.667HALLAMOS LOS DESPLAZAMIENTOS{Q} = {K} *{D} =12433.9655-1051.9475949.2554152865-1051.94759667.5655109.2554152865949.2556274186109.25562741868714.045{D} = {K}-1 *{Q} =0.00008187680.0000090113-0.0000090321130000.00000901130.0001044451-0.0000022911x-38999.99898488-0.0000090321-0.00000229120.000115769924166.667nudo 20.4946825054X{D} = -4.0115812755Y2.7697096672mDESPLAZAMIENTOS EN SISTEMA LOCALbarra1{d2}1 = 0.80.600.4946825054-2.011202761-0.60.80x-4.0115812755=-3.50607452370012.76970966722.7697096672barra20.7071067812-0.707106781200.49468250543.1864096773{d1}2 = 0.70710678120.70710678120x-4.0115812755=-2.48682296910012.76970966722.7697096672FUERZAS EN EXTREMOS DE BARRASbarra1-1000000-2.01120276120112.0276099871{q'1}1 = 0-120600x-3.5060745237=2082.55474317860-60020002.76970966727643.0640486658

1000000-2.011202761-20112.0276099871{q'2}1 = 0120-600x-3.5060745237=-2082.55474317860-60040002.769709667213182.48338312

barra211785.1130.0000.0003.186409677337552.1981116576{q'1}2 = 0.000196.418833.333x-2.4868229691=1819.63367217290.000833.3334714.0452.769709667210984.18361688

-11785.1130.0000.0003.1864096773-37552.1981116576{q'2}2 = 0.000-196.418-833.333x-2.4868229691=-1819.63367217290.000833.3332357.0232.76970966724455.9156127581FUERZAS TOTALES EN BARRAS {P}BARRA 1{P'1}={q'}1 + {Y'}1 20112.02760998711200032112.02760998712082.55474317862100023082.55474317867643.0640486658+39166.667=46809.7310486658{P'1}=-20112.027609987112000-8112.0276099871-2082.55474317862100018917.445256821413182.48338312-39166.667-25984.18361688BARRA 2{P'}={q'}2 + {Y'}2 37552.1981116576-10606.60126945.59711165761819.633672172910606.60112426.234672172910984.18361688+15000=25984.18361688{P'1}=-37552.1981116576-10606.601-48158.7991116576-1819.633672172910606.6018786.96732782714455.9156127581-15000-10544.0843872419

Hoja1 (2)RESOLVER:SOLUCION:MATRIZ DE RIGIDEZ DE LA ESTRUCTURA:NUDO1231[K11]1[K12]102[K21]1[K22]1+[K11]2[K12]230[K21]2[K22]2MATRIZ DE TRANSFORMACION23C-S0[R] =SC0001

CS0[R]T =-SC01001MATRIZ DE RIGIDEZ UNA BARRA EN LOCAL:[K11][K22]EA=100000BARRA 1EI=1000021000000[K22] =0120-6000-6004000

[K22]g =[R]*[K22]*[R]TYC-S0X[R] =SC010010-1206001000000CS00-6004000[R]T =-SC0001

0-101200600[R] =100[K22]g =010000000160004000

010[R]T =-100001BARRA 2

1000000Y[K11] =012060006004000

12X[K11]g =[R]*[K11]*[R]T

C-S0[R] =SC000110000000120600CS006004000[R]T =-SC0001

1000000[K11]g =012060010006004000[R] =010001

100[R]T =010001MATRIZ DE RIGIDEZ DE LA ESTRUCTURA:[K] = [K22]1+[K11]212006001000000101200600[K] =0100000+0120600=010120600600040000600400060060080001000ENSAMBLAJE DE CARGAS12001000000{F} =0{R} = ?-1000000CALCULO DE {R}012001000005006000{Y'}1 =1250100000{Y'}2 =05006000-1250-100000-10100[R] =100[R] =010001001TRANSFORMANDO AL SISTEMA GLOBAL0-100-500100*5000nudo1{Y}1 =0011250=12500-100-500100*5000nudo2001-1250-125010000010*6000.0006000nudo2{Y}2=00110000=1000010000010*60006000nudo3001-10000-1000011-5000-500060006000{R} = {Y}1 +{Y}2 =-1250+10000=8750PROBLEMA COMPLEMENTARIO0-500500{Q} = {F}2 -{R}2 =-1000-6000=-700008750-8750HALLAMOS LOS DESPLAZAMIENTOS{Q} = {K} *{D} =1012006000101206006006008000{D} = {K}-1 *{Q} =0.00009925760.0000004433-0.00000747765000.00000044330.0000992576-0.0000074776x-7000-0.0000074776-0.00000747760.0001261216-8750nudo 20.1119541573X{D} = -0.6291525621Y-1.0549601196mDESPLAZAMIENTOS EN SISTEMA LOCALbarra1{d2}1 = 0-100.11195415730.6291525621100x-0.6291525621=0.1119541573001-1.0549601196-1.0549601196barra21000.11195415730.1119541573{d1}2 = 010x-0.6291525621=-0.6291525621001-1.0549601196-1.0549601196FUERZAS EN EXTREMOS DE BARRASbarra1-10000000.6291525621-6291.5256207662{q'1}1 = 0-120600x0.1119541573=-646.41057065960-6002000-1.0549601196-2177.0927336567

10000000.62915256216291.5256207662{q'2}1 = 0120-600x0.1119541573=646.41057065960-6004000-1.0549601196-4287.0129729389

barra210000.0000.0000.0000.11195415731119.5415729097{q'1}2 = 0.000120.000600.000x-0.6291525621=-708.47437923380.000600.0004000.000-1.0549601196-4597.3320158103

-10000.0000.0000.0000.1119541573-1119.5415729097{q'2}2 = 0.000-120.000-600.000x-0.6291525621=708.47437923380.000600.0002000.000-1.0549601196-2487.4117765281FUERZAS TOTALES EN BARRAS {P}BARRA 1{P'1}={q'}1 + {Y'}1 -6291.52562076620-6291.5256207662-646.4105706596500-146.4105706596-2177.0927336567+1250=-927.0927336567{P'1}=6291.525620766206291.5256207662646.41057065965001146.4105706596-4287.0129729389-1250-5537.0129729389BARRA 2{P'}={q'}2 + {Y'}2 1119.541572909701119.5415729097-708.474379233860005291.5256207662-4597.3320158103+10000=5402.6679841897{P'1}=-1119.54157290970-1119.5415729097708.474379233860006708.4743792338-2487.4117765281-10000-12487.4117765281

Hoja2

Hoja3