metode elemen hingga, beam element

Kelompok 2

Diketahui :Beam dengan spesifikasi sebagai berikut :

Profile betonModulus Elastisitas E : 2100000 kg /cm2

Penampang (beton uk. 20 x 60 cm) Iz : 360000 cm4

Beban Merata Q : 100 Kg / cm'Beban Terpusat P : 2000 Kg

Ditanya : a. Perpindahan pada setiap titik simpul ? b. Reaksi Tumpuan ? c. Gaya Batang ?

Tabulasi Panjang & Nama Batang Struktur

Tugas Mata Kuliah Metode Element Hingga

Awal - Akhir1 - 2 a Cm2 - 3 b Cm3 - 4 c Cm

300.00300.00

No. simpul

Tabulasi Panjang & Nama Batang Struktur

Nama Batang Panjang Batang

300.00


Kelompok 2

MATRIKS KEKAKUAN INDIVIDU / LOKAL :

Dimana elemen a,b dan c sama panjang yaitu 300cm, maka

6L = 1800 cm4L2 = 360000 cm2L2 = 180000 cmL3 = 27000000 cm

12 1800 -12 1800[ K ] = 2100000 x 360000 1800 4E+05 -1800 180000

-12 -1800 12 -18001800 2E+05 -1800 360000

28000.000

336000 50400000 -336000 50400000= 50400000 1.008E+10 -50400000 5040000000

-336000 -50400000 336000 -5040000050400000 5040000000 -50400000 1.008E+10

27000000.00


Kx11 Kx12

Kx21 Kx22


Kelompok 2

PERSAMAAN KEKAKUAN STRUKTUR

F1 = ka11.d1 + ka12.d2

1 2 3

d1 d2 d3

Batang a Batang b

TUGAS METODE ELEMEN HINGGA

F2 = ka21.d1 + ka22.d2 + kb11.d2 + kb12.d3

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


F1 - F red = ka11.d1 + ka12.d2F2 - F red = ka21.d1 + ka22.d2 + kb11.d2 + kb12.d3F3 - F red = kb21.d2 + kb22.d3 + kc11.d3 + kc12.d4

F4 = kc21.d3 + kc22.d4


Kelompok 2

MATRIKS KEKAKUAN STRUKTUR SEBELUM DIKURANGIN F red

F = K d

F1 d1

F2 d2

F3 d3

F4 d4

V1 336000 50400000 -336000 50400000 0 0 0 0 V1

M1 50400000 10080000000 -50400000 5040000000 0 0 0 0 χ1

V2 -336000 -50400000 672000 0 -336000 50400000 0 0 V2

M2 50400000 5040000000 0 20160000000 -50400000 5040000000 0 0 χ2

V3 0 0 -336000 -50400000 672000 0 -336000 50400000 V3

M3 0 0 50400000 5040000000 0 20160000000 -50400000 5040000000 χ3

V4 0 0 0 0 -336000 -50400000 336000 -50400000 V4

M4 0 0 0 0 50400000 5040000000 -50400000 10080000000 χ4

SYARAT BATAS

ka11 ka12 0 0

Ka21 ka22 + kb11 kb12 0

0 kb21 kb22 + kc11 kc12

0 0 kc21 kc22

=

=

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


SYARAT BATAS

v1 = 0 v3 = 0

χ1 = Berputar χ3 = Berputar

v2 = 0 v4 = Berpindah

χ2 = Berputar χ4 = Tetap

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

Maka Matriks menjadi

V1 336000 50400000 -336000 50400000 0 0 0 0 0

M1 50400000 10080000000 -50400000 5040000000 0 0 0 0 χ1

V2 -336000 -50400000 672000 0 -336000 50400000 0 0 0

M2 50400000 5040000000 0 20160000000 -50400000 5040000000 0 0 χ2

V3 0 0 -336000 -50400000 672000 0 -336000 50400000 0

M3 0 0 50400000 5040000000 0 20160000000 -50400000 5040000000 χ3

V4 0 0 0 0 -336000 -50400000 336000 -50400000 V4

M4 0 0 0 0 50400000 5040000000 -50400000 10080000000 χ4

M1 10080000000 5040000000 0 0 0 χ1

M2 5040000000 20160000000 5040000000 0 0 χ2

M3 0 5040000000 20160000000 -50400000 5040000000 χ3

V4 0 0 -50400000 336000 -50400000 V4

M4 0 0 5040000000 -50400000 10080000000 χ4

Gaya yang terjadi akibat beban merata ( F red )

=

=

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Batang a Batang b

Sy₁' = - qL/2 = -15000 Kg Sy2' = - qL/2 = -15000 Kg

Sy2' = - qL/2 = -15000 Kg Sy3' = - qL/2 = -15000 KgMz₁' = -qL2/12 = -750000 Kg.cm Mz2' = -qL2/12 = -750000 Kg.cmMz₂' = qL2/12 = 750000 Kg.cm Mz3' = qL2/12 = 750000 Kg.cm

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

Matriks setelah di kurangin F red

-750000 10080000000 5040000000 0 0 0 χ1

0 5040000000 20160000000 5040000000 0 0 χ2

750000 0 5040000000 20160000000 -50400000 5040000000 χ3

-2000 0 0 -50400000 336000 -50400000 V4

0 0 0 5040000000 -50400000 10080000000 χ4

PERPINDAHAN TITIK

d = K-1 F red

χ1 1.15741E-10 -3.30688E-11 1.65344E-11 4.96032E-09 1.65344E-11 -750000χ2 -3.30688E-11 6.61376E-11 -3.30688E-11 -9.92063E-09 -3.30688E-11 0χ3 1.65344E-11 -3.30688E-11 1.15741E-10 3.47222E-08 1.15741E-10 750000V4 4.96032E-09 -9.92063E-09 3.47222E-08 2.23214E-05 9.4246E-08 -2000χ4 1.65344E-11 -3.30688E-11 1.15741E-10 9.4246E-08 5.12566E-10 0

=

=

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


χ1 = -0.000084 radχ2 = 0.000020 radχ3 = 0.000005 radV4 = -0.022321 cmχ4 = -0.000114 rad

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

MATRIKS REAKSI TITIK SIMPUL SEBELUM DIKURANGIN F Red

V1 = 336000 50400000 -336000 50400000 0 0 0 0 0

M1 = 50400000 10080000000 -50400000 5040000000 0 0 0 0 -0.000084

V2 = -336000 -50400000 672000 0 -336000 50400000 0 0 0

M2 = 50400000 5040000000 0 20160000000 -50400000 5040000000 0 0 0.000020

V3 = 0 0 -336000 -50400000 672000 0 -336000 50400000 0

M3 = 0 0 50400000 5040000000 0 20160000000 -50400000 5040000000 0.000005

V4 = 0 0 0 0 -336000 -50400000 336000 -50400000 -0.022321

M4 = 0 0 0 0 50400000 5040000000 -50400000 10080000000 -0.000114

V1 = -3250M1 = -750000V2 = 4500M2 = 0V3 = 750M3 = 750000V4 = -2000M4 = 0

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


M4 = 0

MATRIKS REAKSI TITIK SIMPUL DIKURANGIN F Red

V1 = -3250 -15000 11750 KgM1 = -750000 -750000 0 KgcmV2 = 4500 -30000 34500 KgM2 = 0 - 0 = 0 KgcmV3 = 750 -15000 15750 KgM3 = 750000 750000 0 KgcmV4 = -2000 -2000 0 KgM4 = 0 0 0 Kgcm

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

KONTROL REAKSI PADA SETIAP TITIK SIMPUL ( AKSI = REAKSI )

∑V = 0AKSI = REAKSI

= Q.L + P= 62000 ……………………………… ( OK )

Tinjau moment di titik 1∑M = 0

AKSI = REAKSI=== …………………… ( OK )

V1+V2+V362000

(V2*L1)+(V3*(L1+L2)(34500*300)+(15750*(300+300))

19800000

(Q*(L1+L2)*(0.5*(L1+L2))+(P*(L1+L2+L3))(100*(300+300)*(0.5*(300+300))+(2000*(300+300+300))

19800000

TUGAS METODE ELEMEN HINGGATUGAS METODE ELEMEN HINGGA

Kelompok 2

MENCARI GAYA BATANG

Batang AF = K d

Gaya dalam sebelum dikurangi fred

Sy₁ 336000 50400000 -336000 50400000 V1Mz₁ = 50400000 10080000000 -50400000 5040000000 * χ1Sy₂ -336000 -50400000 336000 -50400000 V2Mz₂ 50400000 5040000000 -50400000 10080000000 χ2

Sy₁ 336000 50400000 -336000 50400000 0Mz₁ = 50400000 10080000000 -50400000 5040000000 * -0.000084Sy₂ -336000 -50400000 336000 -50400000 0Mz₂ 50400000 5040000000 -50400000 10080000000 0.000020

Sy₁ -3250Mz₁ = -750000Sy₂ 3250

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Sy₂ 3250Mz₂ -225000

Gaya dalam dikurangi fred

Sy₁ -3250 15000 -18250 KgMz₁ -750000 - 750000 = -1500000 KgcmSy₂ -3250 -15000 11750 KgMz₂ -225000 -750000 525000 Kgcm

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

Batang BF = K d


Sy2 336000 50400000 -336000 50400000 V2Mz2 = 50400000 10080000000 -50400000 5040000000 * χ2Sy3 -336000 -50400000 336000 -50400000 V3Mz3 50400000 5040000000 -50400000 10080000000 χ3

Sy2 336000 50400000 -336000 50400000 0Mz2 = 50400000 10080000000 -50400000 5040000000 * 0.000020Sy3 -336000 -50400000 336000 -50400000 0Mz3 50400000 5040000000 -50400000 10080000000 0.000005

Sy2 1250Mz2 = 225000Sy3 -1250

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Sy3 -1250Mz3 150000

Gaya dalam dikurangi fred

Sy2 1250 15000 -13750 KgMz2 225000 - 750000 = -525000 KgcmSy3 1250 -15000 16250 KgMz3 150000 -750000 900000 Kgcm

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

Batang CF = K d


Sy3 336000 50400000 -336000 50400000 V3Mz3 = 50400000 10080000000 -50400000 5040000000 * χ3Sy4 -336000 -50400000 336000 -50400000 V4Mz4 50400000 5040000000 -50400000 10080000000 χ4

Sy3 336000 50400000 -336000 50400000 0Mz3 = 50400000 10080000000 -50400000 5040000000 * 0.000005Sy4 -336000 -50400000 336000 -50400000 -0.022321Mz4 50400000 5040000000 -50400000 10080000000 -0.000114

Sy3 2000 KgMz3 = 600000 KgcmSy4 -2000 Kg

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Sy4 -2000 KgMz4 0 Kgcm

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

HASIL PERHITUNGAN METOEDE ELEMEN HINGGA KOMBINASI DENGAN SAP

GAMBAR PEMBEBANAN

GAMBAR REAKSI PERLETAKAN

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

DIAGRAM BIDANG LINTANG

DIAGRAM BIDANG MOMENT

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b


Kelompok 2

GAMBAR DEFORMASI BATANG

1 2 3

d1 d2 d3

Batang a Batang b

1 2 3

d1 d2 d3

Batang a Batang b

TUGAS METODE ELEMEN HINGGATUGAS METODE ELEMEN HINGGA

Kelompok 2

HASIL PERHITUNGAN SAP 2000

Joint OutputCase CaseType V RText Text Text cm Radians

1 DEAD LinStatic 0.000000 0.0000842 DEAD LinStatic 0.000000 -0.0000203 DEAD LinStatic 0.000000 -0.0000054 DEAD LinStatic -0.022321 0.000114

Joint OutputCase CaseType F MText Text Text Kgf Kgf-cm

1 DEAD LinStatic 11750 02 DEAD LinStatic 34500 03 DEAD LinStatic 15750 0

Frame Station OutputCase CaseType V2 M3Text cm Text Text Kgf Kgf-cm

0 DEAD LinStatic -11750 050 DEAD LinStatic -6750 462500

100 DEAD LinStatic -1750 675000150 DEAD LinStatic 3250 637500200 DEAD LinStatic 8250 350000250 DEAD LinStatic 13250 -187500300 DEAD LinStatic 18250 -975000

a

b

TABLE: Joint Displacements

TABLE: Joint Reactions

TABLE: Element Forces - Frames


300 DEAD LinStatic 18250 -9750000 DEAD LinStatic -16250 -975000

50 DEAD LinStatic -11250 -287500100 DEAD LinStatic -6250 150000150 DEAD LinStatic -1250 337500200 DEAD LinStatic 3750 275000250 DEAD LinStatic 8750 -37500300 DEAD LinStatic 13750 -600000

0 DEAD LinStatic -2000 -60000050 DEAD LinStatic -2000 -500000

100 DEAD LinStatic -2000 -400000150 DEAD LinStatic -2000 -300000200 DEAD LinStatic -2000 -200000250 DEAD LinStatic -2000 -100000300 DEAD LinStatic -2000 0

a

b

c


metode elemen hingga, beam element

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