Analisis Struktur II Struktur Rangka ?· Garis putus-putus menunjukkan kondisi terakhir struktur rangka…

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<p>Analisis Struktur IIStruktur Rangka Bidang</p> <p>Diketahui Struktur rangka batang bidang seperti gambar di bawah : Diketahui Struktur rangka batang bidang seperti gambar di bawah :</p> <p>Data Batang : Data Batang :</p> <p>A1 = 5 cm2</p> <p>A1 = 5 cm2</p> <p>A2 = 6 cm2</p> <p>A2 = 6 cm2</p> <p>L1 = 400 cm2 L1 = 400 cm2</p> <p>L2 = 300 cm2 L2 = 300 cm2</p> <p>E1 = E2 = E kg/cm2</p> <p>E1 = E2 = E kg/cm2</p> <p>E = 200000 kg/cm2</p> <p>E = 200000 kg/cm2</p> <p>P1 = 3320 kg</p> <p>P2 = 240 kg</p> <p>Ditanyakan :</p> <p>Ditanyakan :</p> <p>Jawab : Penomoran Global :</p> <p>Catatan :</p> <p>Tabel Data Batang :</p> <p>Sehingga matriks perpindahan yang terjadi di titik 1 : ( i ) ( j ) ( k ) L (cm) A (cm2) E (kg/cm</p> <p>2) ( o ) Cos Sin </p> <p>{DF} = -1 cm 1 2 1 500 5 200000 323.13 0.80 -0.60</p> <p>-1 2 3 1 400 6 200000 0 1.00 0.00</p> <p>Penomoran Global : Matriks Kekakuan Batang terhadap Sistem Koordinat Global [SMS] :</p> <p>Matriks Kekakuan Batang terhadap Sistem Koordinat Lokal [SM] :</p> <p>2. Reaksi di masing-masing tumpuan.</p> <p>3. Gaya ujung masing-masing batang.</p> <p>Garis putus-putus menunjukkan kondisi terakhir struktur rangka bidang akibat</p> <p>adanya beban luar yang bekerja di titik 1.</p> <p>1. Nilai P yang bekerja, jika perpindahan yang diperbolehkan terjadi di titik 1 adalah :</p> <p>D1 = 1 cm dan D2 = 1 cm. </p> <p>2. Reaksi di masing-masing tumpuan.</p> <p>3. Gaya ujung masing-masing batang.</p> <p>Asumsi beban yang bekerja di titik 1 adalah P1 dan P2 dengan arah gaya dan perpindahan</p> <p>yang terjadi seperti gambar berikut ini :</p> <p>1. Perpindahan yang terjadi.</p> <p>1</p> <p>21</p> <p>2</p> <p>3</p> <p>2</p> <p>1</p> <p>4</p> <p>3</p> <p>6</p> <p>5Y</p> <p>X</p> <p>1</p> <p>2</p> <p>L1</p> <p>L2</p> <p>1</p> <p>2</p> <p>3</p> <p>P2</p> <p>P1D2</p> <p>D1</p> <p>1</p> <p>2</p> <p>L1</p> <p>L2</p> <p>1</p> <p>2</p> <p>3</p> <p>P2</p> <p>P1</p> <p>1</p> <p>21</p> <p>2</p> <p>3</p> <p>2</p> <p>1</p> <p>4</p> <p>3</p> <p>6</p> <p>5</p> <p>Y</p> <p>X</p> <p>[ ] [ ] [ ] [ ]TTTS R.SM.RSM T=</p> <p>=</p> <p>0000</p> <p>0101</p> <p>0000</p> <p>0101</p> <p>L</p> <p>EA]SM[</p> <p>zacoeb@ub.ac.id 1</p> <p>Analisis Struktur IIStruktur Rangka Bidang</p> <p>Tabel Data Batang :</p> <p>( i ) ( j ) ( k ) L (cm) A (cm2) E (kg/cm</p> <p>2) ( o ) Cx Cy Cx</p> <p>2Cy</p> <p>2CxCy Matriks Rotasi Transformasi [RT] :</p> <p>1 2 1 500 5 200000 323.13 0.80 -0.60 0.64 0.36 -0.48</p> <p>2 3 1 400 6 200000 0 1.00 0.00 1.00 0.00 0.00</p> <p>Matriks Kekakuan Batang [SMS]i :</p> <p>Matriks Rotasi Transformasi Transpose [RT]T :</p> <p>Batang 1 :</p> <p>1280.00 -960.00 1280.00 960.00 3</p> <p>[SMS]1 = -960.00 720.00 960.00 -720.00 4</p> <p>-1280.00 960.00 1280.00 -960.00 1</p> <p>960.00 -720.00 -960.00 720.00 2</p> <p>Batang 2 : Batang 1 :</p> <p>3000.00 0.00 -3000.00 0.00 5 Matriks Kekakuan Lokal [SM] Matriks Transformasi [RT] Matriks Transformasi Transpose [RT]T</p> <p>[SMS]2 = 0.00 0.00 0.00 0.00 6 2000.00 0.00 2000.00 0.00 0.80 -0.60 0.00 0.00 0.8 0.6 0 0</p> <p>-3000.00 0.00 3000.00 0.00 1 0.00 0.00 0.00 0.00 0.60 0.80 0.00 0.00 -0.6 0.8 0 0</p> <p>0.00 0.00 0.00 0.00 2 -2000.00 0.00 2000.00 0.00 0.00 0.00 0.80 -0.60 0 0 0.8 0.6</p> <p>0.00 0.00 0.00 0.00 0.00 0.00 0.60 0.80 0 0 -0.6 0.8</p> <p>Matriks Kekakuan Struktur [SJ] : Matriks Kekakuan Global [SMS] [RT]T[SM][RT] [RT]</p> <p>T[SM]</p> <p>(catatan : sekaligus dilakukan penataan) 1280 -960 1280 -960 3 1280 -960 1280 -960 1600 0 1600 0</p> <p>-960 720 -960 720 4 -960 720 -960 720 -1200 0 -1200 0</p> <p>4280.00 -960.00 1 -1280 960 1280 -960 1 -1280 960 1280 -960 -1600 0 1600 0</p> <p>-960.00 720.00 2 960 -720 -960 720 2 960 -720 -960 720 1200 0 -1200 0</p> <p>[SJ] = 1280.00 960.00 3</p> <p>960.00 -720.00 4 Batang 2 :</p> <p>-3000.00 0.00 5 Matriks Kekakuan Lokal [SM] Matriks Transformasi [RT] Matriks Transformasi Transpose [RT]T</p> <p>0.00 0.00 6 3000.00 0.00 -3000.00 0.00 1.00 0.00 0.00 0.00 0.8 0.6 0 0</p> <p>1 2 3 4 5 6 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 -0.6 0.8 0 0</p> <p>-3000.00 0.00 3000.00 0.00 0.00 0.00 1.00 0.00 0 0 0.8 0.6</p> <p>Analisis Beban Join : 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0 0 -0.6 0.8</p> <p>{AC} = {AJ} + {AE} Matriks Kekakuan Global [SMS] [RT]T[SM][RT] [RT]</p> <p>T[SM]</p> <p>karena beban yang bekerja berupa beban di titik (joint load ), maka {AE} = 0 1280 -960 1280 -960 5 2400 0 -2400 0 2400 0 -2400 0</p> <p>{AC} = {AJ} -960 720 -960 720 6 -1800 0 1800 0 -1800 0 1800 0</p> <p> P1 1 -1280 960 1280 -960 1 -2400 0 2400 0 -2400 0 2400 0</p> <p> P2 2 AFC 960 -720 -960 720 2 1800 0 -1800 0 1800 0 -1800 0</p> <p> = 0 3</p> <p>0 4 ARC Matriks Kekakuan Struktur [SJ] :</p> <p>0 5 (catatan : sekaligus dilakukan penataan)</p> <p>0 6</p> <p>4280.00 -960.00 1</p> <p>1. Mencari Nilai P : -960.00 720.00 2</p> <p>{DF} = [SFF]-1</p> <p> {AFC} [SJ] = 1280.00 960.00 3</p> <p>dengan operasi matriks, maka : 960.00 -720.00 4</p> <p>{DF} = [SFF]-1</p> <p> {AFC} -3000.00 0.00 5</p> <p>[SFF]{DF} =[SFF] [SFF]-1</p> <p> {AFC} 0.00 0.00 6</p> <p>[SFF]{DF} =[ I ] {AFC} 4280 -960 -1 -3320 1 2 3 4 5 6</p> <p>{AFC} = [SFF]{DF} -960 720 -1 240</p> <p>SRR</p> <p>SFR</p> <p>SRR</p> <p>SFR</p> <p>[ ]</p> <p>=</p> <p>22</p> <p>22</p> <p>22</p> <p>22</p> <p>yyxyyx</p> <p>yxxyxx</p> <p>yyxyyx</p> <p>yxxyxx</p> <p>iS</p> <p>CCCCCC</p> <p>CCCCCC</p> <p>CCCCCC</p> <p>CCCCCC</p> <p>L</p> <p>EASM</p> <p>=</p> <p>cossin</p> <p>sincos</p> <p>cossin</p> <p>sincos</p> <p>]R[ T</p> <p>00</p> <p>00</p> <p>00</p> <p>00</p> <p>=</p> <p>cossin</p> <p>sincos</p> <p>cossin</p> <p>sincos</p> <p>]R[ TT</p> <p>00</p> <p>00</p> <p>00</p> <p>00</p> <p>zacoeb@ub.ac.id 2</p> <p>Analisis Struktur IIStruktur Rangka Bidang</p> <p>Analisis Beban Join :</p> <p> P1 = -3320 kg {AC} = {AJ} + {AE}</p> <p> P2 240 karena beban yang bekerja berupa beban di titik (joint load ), maka {AE} = 0</p> <p>{AC} = {AJ}</p> <p>2. Mencari Nilai Reaksi Tumpuan : -3320 1</p> <p>{AR} = [SRF] {DF} 240 2 AFC</p> <p>320 3 1280.00 960.00 -1 320 = 0 3</p> <p> = -240 4 kg 960.00 -720.00 -1 -240 0 4 ARC</p> <p>3000 5 -3000.00 0.00 3000 0 5</p> <p>0 6 0.00 0.00 0 0 6</p> <p>Kontrol : 1. Perpindahan yang terjadi :</p> <p>H = 0 {DF} = [SFF]-1</p> <p> {AFC}</p> <p>H3 + H5 + P1 = 0 0.000333 0.000444 -3320</p> <p>320 + 3000 + (-3320) = 0 0.000444 0.001981 240</p> <p>0 = 0 ..Ok.</p> <p>-1</p> <p>V = 0 -1V4 + V6 + P2 = 0</p> <p>240 + 0 + (-240) = 0 2. Nilai Reaksi Tumpuan :</p> <p>0 = 0 ..Ok. {AR} = [SRF] {DF}</p> <p>1280.00 960.00 -1</p> <p>M1 = 0 960.00 -720.00 -1</p> <p>H3.3 + V4.4 + V6.4 = 0 -3000.00 0.00</p> <p>(-320).3 + 240.4 + 0.4 = 0 0.00 0.00</p> <p>(-960) + 960 + 0 = 0</p> <p>0 = 0 ..Ok. 320 3</p> <p>-240 4 kg</p> <p>3. Mencari Gaya Ujung Batang : 3000 5</p> <p>{AM}i = {AML}i + [SM]i {DM}i 0 6</p> <p>{AM}i = {AML}i + [SM]i .[RT].{DJ}i</p> <p>Kontrol :</p> <p>Batang 1 : H = 00 2000 0 -2000 0 H3 + H5 + P1 = 0</p> <p>{AM}i = 0 + 0 0 0 0 320 + 3000 + (-3320) = 0</p> <p>0 -2000 0 2000 0 0.00 = 0 ..Ok.</p> <p>0 0 0 0 0</p> <p>V = 00.80 -0.60 0 0 0 3 V4 + V6 + P2 = 0</p> <p> x 0.60 0.80 0 0 0 4 240 + 0 + (-240) = 0</p> <p>0 0 0.80 -0.60 -1 1 0.00 = 0 ..Ok.</p> <p>0 0 0.60 0.80 -1 2</p> <p>=</p> <p>=</p> <p>=</p> <p>=</p> <p>zacoeb@ub.ac.id 3</p> <p>Analisis Struktur IIStruktur Rangka Bidang</p> <p>M1 = 0</p> <p>0 2000 0 -2000 0 0 3 H3.3 + V4.4 + V6.4 = 0</p> <p>= 0 + 0 0 0 0 x 0 4 (-320).3 + 240.4 + 0.4 = 0</p> <p>0 -2000 0 2000 0 -0.2 1 (-960) + 960 + 0 = 0</p> <p>0 0 0 0 0 -1.4 2 0.00 = 0 ..Ok.</p> <p>400 3. Mencari Gaya Ujung Batang :</p> <p>= 0 {AM}i = {AML}i + [SM]i {DM}i</p> <p>-400 {AM}i = {AML}i + [SM]i .[RT].{DJ}i</p> <p>0</p> <p>Batang 1 :</p> <p>0 2000 0 -2000 0</p> <p>{AM}i = 0 + 0 0 0 0</p> <p>0 -2000 0 2000 0</p> <p>0 0 0 0 0</p> <p>Batang 2 : 0.80 -0.60 0 0 0 3</p> <p>0 3000 0 -3000 0 x 0.60 0.80 0 0 0 4</p> <p>{AM}i = 0 + 0 0 0 0 0 0 0.80 -0.60 -1 1</p> <p>0 -3000 0 3000 0 0 0 0.60 0.80 -1 2</p> <p>0 0 0 0 0</p> <p>0 2000 0 -2000 0 0 3</p> <p>1.00 0.00 0 0 0 5 = 0 + 0 0 0 0 x 0 4</p> <p> x 0.00 1.00 0 0 0 6 0 -2000 0 2000 0 -0.2 1</p> <p>0 0 1.00 0.00 -1 1 0 0 0 0 0 -1.4 2</p> <p>0 0 0.00 1.00 -1 2</p> <p>400</p> <p>0 3000 0 -3000 0 0 5 = 0</p> <p>= 0 + 0 0 0 0 x 0 6 -400</p> <p>0 -3000 0 3000 0 -1 1 0</p> <p>0 0 0 0 0 -1 2</p> <p>3000</p> <p>= 0</p> <p>-3000</p> <p>0</p> <p>Batang 2 :</p> <p>0 3000 0 -3000 0</p> <p>{AM}i = 0 + 0 0 0 0</p> <p>0 -3000 0 3000 0</p> <p>0 0 0 0 0</p> <p>1.00 0.00 0 0 0 5</p> <p> x 0.00 1.00 0 0 0 6</p> <p>0 0 1.00 0.00 -1 1</p> <p>0 0 0.00 1.00 -1 2</p> <p>0 3000 0 -3000 0 0 5</p> <p>= 0 + 0 0 0 0 x 0 6</p> <p>0 -3000 0 3000 0 -1 1</p> <p>0 0 0 0 0 -1 2</p> <p>3000</p> <p>= 0</p> <p>-3000</p> <p>0</p> <p>400 kg</p> <p>400 kg</p> <p>3000 kg</p> <p>3000 kg</p> <p>400 kg</p> <p>400 kg</p> <p>3000 kg</p> <p>3000 kg</p> <p>zacoeb@ub.ac.id 4</p>