10
1. GLOBLAL STIFFNESS MATRIX Elemen Batang 1 L = 500 cm x = 400 cm y = 300 cm E = 2,100 A = 100 I = 5,000 420 0 0 -420 0 0 0 1 252 0 -1 252 0 252 84,000 0 -252 42,000 -420 0 0 420 0 0 0 -1 -252 0 1 -252 0 252 42,000 0 -252 84,000 0.8 0.6 0.0 0.0 0.0 0.0 -0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 0.6 0.0 0.0 0.0 0.0 -0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.8 -0.6 0.0 0.0 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 -0.6 0.0 0.0 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0.0 0.0 1.0 0.8 -0.6 0.0 0.0 0.0 0.0 420 0 0 -420 0 0 0.8 0.6 0.0 0.0 0.6 0.8 0.0 0.0 0.0 0.0 0 1 252 0 -1 252 -0.6 0.8 0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0 0 252 84,000 0 -252 42,000 0.0 0.0 1.0 0.0 0.0 0.0 0.0 0.8 -0.6 0.0 -420 0 0 420 0 0 0.0 0.0 0.0 0.8 0.0 0.0 0.0 0.6 0.8 0.0 0 -1 -252 0 1 -252 0.0 0.0 0.0 -0.6 0.0 0.0 0.0 0.0 0.0 1.0 0 252 42,000 0 -252 84,000 0.0 0.0 0.0 0.0 336.00 -0.60 -151.20 -336.00 0.60 -151.20 0.80 0.60 0.00 0.00 0.00 0.00 252.00 0.81 201.60 -252.00 -0.81 201.60 -0.60 0.80 0.00 0.00 0.00 0.00 0.00 252.00 ### 0.00 -252.00 42,000.00 0.00 0.00 1.00 0.00 0.00 0.00 -336.00 0.60 151.20 336.00 -0.60 151.20 0.00 0.00 0.00 0.80 0.60 0.00 -252.00 -0.81 -201.60 252.00 0.81 -201.60 0.00 0.00 0.00 -0.60 0.80 0.00 0.00 252.00 ### 0.00 -252.00 84,000.00 0.00 0.00 0.00 0.00 0.00 1.00 269.16 201.12 -151.20 -269.16 -201.12 -151.20 201.12 151.85 201.60 -201.12 -151.85 201.60 -151.20 201.60 ### 151.20 -201.60 42,000.00 -269.16 -201.12 151.20 269.16 201.12 151.20 -201.12 -151.85 -201.60 201.12 151.85 -201.60 -151.20 201.60 ### 151.20 -201.60 84,000.00 Elemen Batang 2 L = 300 cm 270 E = 2,100 A = 40 I = 1,000 280.00 0.00 0.00 -280.00 0.00 0.00 0.00 0.93 140.00 0.00 -0.93 140.00 0.00 140.00 ### 0.00 -140.00 14,000.00 -280.00 0.00 0.00 280.00 0.00 0.00 0.00 -0.93 -140.00 0.00 0.93 -140.00 0.00 140.00 ### 0.00 -140.00 28,000.00 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 t/cm 2 cm 2 cm 4 [Kl] = [T] = [T] T = [Kg] 1 = [T] T [Kl] [T] [Kg] 1 = [Kg] 1 = u1 v1 q1 u2 v2 q2 u1 v1 [Kg] 1 = q1 u2 v2 q2 a = o t/cm 2 cm 2 cm 4 [Kl] = [T] = 1 2 3 5 Tm 4 T 5 m 3 m E = 2100 t/cm 2 A = 100 cm 2 I = 5000 cm 4 E = 2100 t/cm 2 A = 40 cm 2 I = 1000 cm 4 4 m a Cos a = 4/5 Sin a = 3/5

Contoh Soal FRAME 2D

Embed Size (px)

DESCRIPTION

Anstruk 2

Citation preview

Contoh1. GLOBLAL STIFFNESS MATRIXElemen Batang 1L =500cmx =400cmy =300cmE =2,100t/cm2A =100cm2I =5,000cm442000-42000012520-1252[Kl] =025284,0000-25242,000-42000420000-1-25201-252025242,0000-25284,0000.80.60.00.00.00.0-0.60.80.00.00.00.0[T] =0.00.01.00.00.00.00.00.00.00.80.60.00.00.00.0-0.60.80.00.00.00.00.00.01.00.8-0.60.00.00.00.00.60.80.00.00.00.0[T]T =0.00.01.00.00.00.00.00.00.00.8-0.60.00.00.00.00.60.80.00.00.00.00.00.01.0[Kg]1 =[T]T [Kl] [T]0.8-0.60.00.00.00.042000-420000.80.60.00.00.00.00.60.80.00.00.00.0012520-1252-0.60.80.00.00.00.0[Kg]1 =0.00.01.00.00.00.0025284,0000-25242,0000.00.01.00.00.00.00.00.00.00.8-0.60.0-42000420000.00.00.00.80.60.00.00.00.00.60.80.00-1-25201-2520.00.00.0-0.60.80.00.00.00.00.00.01.0025242,0000-25284,0000.00.00.00.00.01.0336.00-0.60-151.20-336.000.60-151.200.800.600.000.000.000.00[Kg]1 =252.000.81201.60-252.00-0.81201.60-0.600.800.000.000.000.000.00252.0084,000.000.00-252.0042,000.000.000.001.000.000.000.00-336.000.60151.20336.00-0.60151.200.000.000.000.800.600.00-252.00-0.81-201.60252.000.81-201.600.000.000.00-0.600.800.000.00252.0042,000.000.00-252.0084,000.000.000.000.000.000.001.00u1v1q1u2v2q2269.16201.12-151.20-269.16-201.12-151.20u1201.12151.85201.60-201.12-151.85201.60v1[Kg]1 =-151.20201.6084,000.00151.20-201.6042,000.00q1-269.16-201.12151.20269.16201.12151.20u2-201.12-151.85-201.60201.12151.85-201.60v2-151.20201.6042,000.00151.20-201.6084,000.00q2Elemen Batang 2L =300cma =270oE =2,100t/cm2A =40cm2I =1,000cm4280.000.000.00-280.000.000.000.000.93140.000.00-0.93140.00[Kl] =0.00140.0028,000.000.00-140.0014,000.00-280.000.000.00280.000.000.000.00-0.93-140.000.000.93-140.000.00140.0014,000.000.00-140.0028,000.00-0-100001-00000[T] =001000000-0-100001-00000001-010000-1-00000[T]T =001000000-010000-1-00000001[Kg]2 =[T]T [Kl] [T]-01000028000-28000-0-10000-1-00000011400-11401-00000[Kg]2 =001000014028,0000-14014,000001000000-010-2800028000000-0-10000-1-000-1-14001-1400001-00000001014014,0000-14028,000000001-0.000.93140.000.00-0.93140.00-0-10000[Kg]2 =-280.00-0.00-0.00280.000.00-0.001-000000.00140.0028,000.000.00-140.0014,000.000010000.00-0.93-140.00-0.000.93-140.00000-0-10280.000.000.00-280.00-0.000.000001-000.00140.0014,000.000.00-140.0028,000.00000001u2v2q2u3v3q30.930.00140.00-0.93-0.00140.00u20.00280.00-0.00-0.00-280.00-0.00v2[Kg]2 =140.00-0.0028,000.00-140.000.0014,000.00q2-0.93-0.00-140.000.930.00-140.00u3-0.00-280.000.000.00280.000.00v3140.00-0.0014,000.00-140.000.0028,000.00q32. OVERALL STIFFNESS MATRIXM2 = 5 Tm = 500 TcmF2 = -4 Tu1v1q1u2v2q2u3v3q3F1=269.16201.12-151.20-269.16-201.12-151.200.000.000.000u1G1201.12151.85201.60-201.12-151.85201.600.000.000.000v1M1-151.20201.6084,000.00151.20-201.6042,000.000.000.000.000q1F2 = -4-269.16-201.12151.20270.10201.12291.20-0.93-0.00140.00u2u2G2 = 0-201.12-151.85-201.60201.12431.85-201.60-0.00-280.00-0.00v2v2M2 = 500-151.20201.6042,000.00291.20-201.60112,000.00-140.000.0014,000.00q2q2F30.000.000.00-0.93-0.00-140.000.930.00-140.000u3G30.000.000.00-0.00-280.000.000.00280.000.000v3M30.000.000.00140.00-0.0014,000.00-140.000.0028,000.000q33. REARRAGEMENTu2v2q2u1v1q1u3v3q3-4=270.10201.12291.20-269.16-201.12151.20-0.93-0.00140.00u2u20201.12431.85-201.60-201.12-151.85-201.60-0.00-280.00-0.00v2v2500291.20-201.60112,000.00-151.20201.6042,000.00-140.000.0014,000.00q2q2F1-269.16-201.12-151.20269.16-201.12151.200.000.000.000u1G1-201.12-151.85201.60201.12-151.85-201.600.000.000.000v1M1151.20-201.6042,000.00-151.20201.6042,000.000.000.000.000q1F3-0.93-0.00-140.000.000.000.000.930.00-140.000u3G3-0.00-280.000.000.000.000.000.00280.000.000v3M3140.00-0.0014,000.000.000.000.00-140.000.0028,000.000q34. UNKNOWN DISPLACEMENT AND REACTIONSDisplacement diselesaikan dg Metode Eliminasi Gauss-4=270.10201.12291.20u20201.12431.85-201.60v2500291.20-201.60112,000.00q2270201291-4201432-2020Disusun seperti ini291-202112,00050010.741.08-0.014812.15-1.000.0000Kolom pertama diusahakan menjadi satu1-0.69384.621.7170dapat dilakukan dengan cara membagi dengan kolom pertama10.741.08-0.0148Usahakan kolom pertama menjadi nol01.40-2.080.0148baris ke 2 dikurangi baris ke 10-1.44383.541.7318baris ke 3 dikurangi baris ke 110.741.08-0.0148Kolom kedua diusahakan menjadi satu01-1.480.0106dapat dilakukan dengan cara membagi dengan kolom kedua01-266.92-1.205210.741.08-0.014801-1.480.0106Usahakan kolom kedua baris 3 menjadi nol00265.431.2158baris ke 3 dikurangi baris ke 2q2 =0.00458v2 =0.010560.00679=0.01735u2 =-0.01481-0.00494-0.01292=-0.03267u2=-0.03267v20.01735q20.00458ReaksiF1=-269.16-201.12-151.20-0.03267=4.611tonG1-201.12-151.85201.600.017354.859tonM1151.20-201.6042,000.000.00458183.941ton cmF2-0.93-0.00-140.00-0.611tonG2-0.00-280.000.00-4.859tonM2140.00-0.0014,000.0059.553ton cm5. MEMBER FORCESElemen Batang 1f1[T] [Kg] {u}g1m1f2g2m2f10.80.60.00.00.00.0269.16201.12-151.20-269.16-201.12-151.200g1-0.60.80.00.00.00.0201.12151.85201.60-201.12-151.85201.600m10.00.01.00.00.00.0-151.20201.6084,000.00151.20-201.6042,000.000f20.00.00.00.80.60.0-269.16-201.12151.20269.16201.12151.20-0.03267g20.00.00.0-0.60.80.0-201.12-151.85-201.60201.12151.85-201.600.01735m20.00.00.00.00.01.0-151.20201.6042,000.00151.20-201.6084,000.000.00458f1336.00252.000.00-336.00-252.000.000g1-0.600.81252.000.60-0.81252.000m1-151.20201.6084,000.00151.20-201.6042,000.000f2-336.00-252.00-0.00336.00252.00-0.00-0.03267g20.60-0.81-252.00-0.600.81-252.000.01735m2-151.20201.6042,000.00151.20-201.6084,000.000.00458f16.60tong11.12tonm1183.94ton cmf2-6.60tong2-1.12tonm2376.32ton cmElemen Batang 2f2[T] [Kg] {u}g2m2f3g3m3f2-0-100000.930.00140.00-0.93-0.00140.00-0.0327g21-000000.00280.00-0.00-0.00-280.00-0.000.0174m2001000140.00-0.0028,000.00-140.000.0014,000.000.0046f3000-0-10-0.93-0.00-140.000.930.00-140.000g30001-00-0.00-280.000.000.00280.000.000m3000001140.00-0.0014,000.00-140.000.0028,000.000f2-0.00-280.000.000.00280.000.00-0.03267g20.93-0.00140.00-0.930.00140.000.01735m2140.00-0.0028,000.00-140.000.0014,000.000.00458f30.00280.000.00-0.00-280.000.000g3-0.930.00-140.000.93-0.00-140.000m3140.00-0.0014,000.00-140.000.0028,000.000f2-4.86tong20.61tonm2123.68ton cmf34.86tong3-0.61tonm359.55ton cm

1235 Tm4 T5 m3 mE = 2100 t/cm2A = 100 cm2I = 5000 cm4E = 2100 t/cm2A = 40 cm2I = 1000 cm44 maCos a = 4/5Sin a = 3/5123,68 tcm59,55 tcm183,94 tcm376,32 tcmBMD

CONTOH SOLUSI PERSAMAAN4X + 3Y + Z = 13X + 2Y + 3Z = 143X + 2Y + 5Z = 224311312314Dapat disusun seperti ini325221.000.750.253.251.002.003.0014.00Kolom pertama diusahakan menjadi satu1.000.671.677.33dapat dilakukan dengan cara membagi dengan kolom pertama1.000.750.253.25Usahakan kolom pertama menjadi nol0.001.252.7510.75baris ke 2 dikurangi baris ke 10.00-0.081.424.08baris ke 3 dikurangi baris ke 11.000.750.253.25Kolom kedua diusahakan menjadi satu0.001.002.208.60dapat dilakukan dengan cara membagi dengan kolom kedua0.001.00-17.00-49.001.000.750.253.250.001.002.208.60Usahakan kolom kedua baris 3 menjadi nol0.000.0019.2057.60baris ke 3 dikurangi baris ke 2Z =3.00Y =8.60-6.60=2.00X =3.25-0.75-1.50=1.00


Contoh-contoh Soal UKDI

Contoh-contoh Soal UKDI

Documents
Contoh Soal Dan Latihan Soal Fluida

Contoh Soal Dan Latihan Soal Fluida

Documents
contoh soal statistika

contoh soal statistika

Documents
Referensi Contoh Soal

Referensi Contoh Soal

Documents
CONTOH - CONTOH SOAL BEBRAS

CONTOH - CONTOH SOAL BEBRAS

Documents
Contoh Soal USM

Contoh Soal USM

Education
Contoh Soal Pmd

Contoh Soal Pmd

Documents
Contoh Soal & Pembahasan

Contoh Soal & Pembahasan

Documents
Kumpulan Contoh Soal

Kumpulan Contoh Soal

Documents
Contoh Soal Kelarutan

Contoh Soal Kelarutan

Documents
Contoh Soal PDP

Contoh Soal PDP

Documents
dieng.blob.core.windows.net · Contoh soal tes FDB Contoh soal tes : VCD Contoh soal tes : INA Contoh soal tes : MED - FTO Contoh soal tes : Tes Potensi Akademik ( TPA ) Pergantian

dieng.blob.core.windows.net · Contoh soal tes FDB Contoh soal tes : VCD Contoh soal tes : INA Contoh soal tes : MED - FTO Contoh soal tes : Tes Potensi Akademik ( TPA ) Pergantian

Documents
contoh soal termodinamika

contoh soal termodinamika

Documents
Contoh Soal Mek.ban

Contoh Soal Mek.ban

Documents
Contoh Soal

Contoh Soal

Documents
contoh soal psikotest

contoh soal psikotest

Documents
Contoh Soal Ips

Contoh Soal Ips

Documents
Contoh Soal Fisika

Contoh Soal Fisika

Documents
Contoh Soal Otk

Contoh Soal Otk

Documents
CONTOH-CONTOH SOAL

CONTOH-CONTOH SOAL

Documents
HIDROSTATIKA HIDRODINAMIKA · CONTOH SOAL 1 CONTOH SOAL 2 CONTOH SOAL 3 CONTOH SOAL 4 CONTOH SOAL 1 CONTOH SOAL 2 CONTOH SOAL 3 CONTOH SOAL 4 CONTOH SOAL 5 CONTOH SOAL 6 CONTOH SOAL

HIDROSTATIKA HIDRODINAMIKA · CONTOH SOAL 1 CONTOH SOAL 2 CONTOH SOAL 3 CONTOH SOAL 4 CONTOH SOAL 1 CONTOH SOAL 2 CONTOH SOAL 3 CONTOH SOAL 4 CONTOH SOAL 5 CONTOH SOAL 6 CONTOH SOAL

Documents
Contoh soal kolokium

Contoh soal kolokium

Education
Contoh soal trafo

Contoh soal trafo

Documents
125492121 Contoh Contoh Soal UKDI

125492121 Contoh Contoh Soal UKDI

Documents
Contoh Soal & Pembahasan

Contoh Soal & Pembahasan

Documents
Contoh Contoh Soal Thermodinamika.ppt

Contoh Contoh Soal Thermodinamika.ppt

Documents
Contoh Soal OSCE

Contoh Soal OSCE

Documents
Contoh Soal Pd

Contoh Soal Pd

Documents
Contoh soal cahaya

Contoh soal cahaya

Documents
contoh soal icas

contoh soal icas

Documents