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		<title>Prediksi kekuatan Balok beton bertulang</title>
		<link>http://syont.wordpress.com/2009/11/09/prediksi-pada-balok-beton/</link>
		<comments>http://syont.wordpress.com/2009/11/09/prediksi-pada-balok-beton/#comments</comments>
		<pubDate>Mon, 09 Nov 2009 09:05:57 +0000</pubDate>
		<dc:creator>syont</dc:creator>
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Menghitung momen tersedia :
Titik berat tulangan tarik,
CG_tul = c_v + d_tul + (D_tul/2)
= 30.00 + 8.00 + (13.00/2)
= 44.00 mm
Tingi effektif balok,
d = h &#8211; CG_tul
= 400.00 &#8211; 44.00
= 356.00 mm
Rasio tulangan
rho = A_st / (b * d)
= 398.20 / (200.00 * 356.00) = 0.56 %
rho_maks = 0.75 * rho_b = (382.5 * beta_1 * [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=894&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p><img class="aligncenter size-full wp-image-895" title="2009-11-04_232749" src="http://syont.files.wordpress.com/2009/11/2009-11-04_232749.jpg?w=450&#038;h=214" alt="2009-11-04_232749" width="450" height="214" /></p>
<p>Menghitung momen tersedia :</p>
<p>Titik berat tulangan tarik,<br />
CG_tul = c_v + d_tul + (D_tul/2)<br />
= 30.00 + 8.00 + (13.00/2)<br />
= 44.00 mm<br />
Tingi effektif balok,</p>
<p>d = h &#8211; CG_tul<br />
= 400.00 &#8211; 44.00<br />
= 356.00 mm</p>
<p>Rasio tulangan</p>
<p>rho = A_st / (b * d)</p>
<p>= 398.20 / (200.00 * 356.00) = 0.56 %</p>
<p>rho_maks = 0.75 * rho_b = (382.5 * beta_1 * f_c) / ((600 + f_y) * f_y)</p>
<p>= (382.5 * 0.85 * 21.15) / ((600 + 400.00) * 400.00) = 1.72 %</p>
<p>Kontrol : rho &lt; rho_maks 0.56 % &lt; 1.72 % &#8230; Okey</p>
<p>Gaya tarik baja tul.,</p>
<p>T_s = A_st * f_y = 398.20 * 400.00 = 159278.75 N</p>
<p>Gaya tekan beton,</p>
<p>C_c = 0.85 * f_c * a_t * b</p>
<p>= 0.85 * 21.15 * a_t * 200.00 = 3595.50*a_t N</p>
<p>Persamaan kondisi kesetimbangan, T_s = C_c</p>
<p>Tinggi blok tekan beton,</p>
<p>a = 159278.75 / 3595.50 = 44.30 mm</p>
<p>Momen nominal,</p>
<p>M_n = T_s * (d &#8211; a/2)<br />
= 159278.75 * (356.00 &#8211; 44.30/2)<br />
= 53175252.00 N.mm = 53.18 kN.m</p>
<p><img class="aligncenter size-full wp-image-896" title="2009-11-04_232728" src="http://syont.files.wordpress.com/2009/11/2009-11-04_232728.jpg?w=450&#038;h=288" alt="2009-11-04_232728" width="450" height="288" /></p>
<p>Perbandingan :</p>
<p>M &amp; V dibuat incremental, Gaya N tidak ada</p>
<p>@ maksimum M = 45.8 kN.m, V = 45.8 kN, P = 0.0 kN</p>
<p>Vmaks &lt; (1/6)*Sqrt(21.15)*200*356 = 54.57 kN</p>
<p>Gaya tarik baja atau tekan beton, R1 = 159/151 = ~5%</p>
<p>Lengan momen, R2 = 333.85 / 303.00 = ~10%</p>
<p>Momen nominal, R3 = 53.18/45.7 = ~16%</p>
<p><img class="aligncenter size-full wp-image-905" title="mphinovn" src="http://syont.files.wordpress.com/2009/11/mphinovn.png?w=450&#038;h=284" alt="mphinovn" width="450" height="284" /></p>
<p>M dibuat incremental, Gaya V &amp; N tidak ada</p>
<p>@ maksimum M = 71.6 kN.m, V = 0.0 kN, P = 0.0 kN</p>
<p>Gaya tarik baja atau tekan beton, R1 = 159/233 = ~32%</p>
<p>Lengan momen, R2 = 333.85 / 308.00 = ~8%</p>
<p>Momen nominal, R3 = 53.18/71.6 = ~26%</p>
<p><img class="aligncenter size-full wp-image-907" title="mphimvn" src="http://syont.files.wordpress.com/2009/11/mphimvn.png?w=450&#038;h=284" alt="mphimvn" width="450" height="284" /></p>
<p>M, V &amp; N dibuat incremental</p>
<p>@ maksimum M = 40.3 kN.m, V = 40.2 kN, P = 40.3 kN</p>
<p>Pmaks &lt; 0.1*21.15*200*400 = 169.20 kN</p>
<p>Momen nominal, R3 = 53.18/40.39 = ~32%</p>
<p><img class="aligncenter size-full wp-image-908" title="mphinov" src="http://syont.files.wordpress.com/2009/11/mphinov.png?w=450&#038;h=284" alt="mphinov" width="450" height="284" /></p>
<p>M &amp; N dibuat incremental, Gaya V tidak ada</p>
<p>@ maksimum M = 61.7 kN.m, V = 0.0 kN, P = 61.7 kN</p>
<p>Pmaks &lt; 169.20 kN</p>
<p>Momen nominal, R3 = 53.18/61.7 = ~14%</p>
<p><img class="aligncenter size-full wp-image-914" title="chart" src="http://syont.files.wordpress.com/2009/11/chart.jpg?w=450&#038;h=405" alt="chart" width="450" height="405" /></p>
<p>Steel rebars and concrete properties</p>
<p><img class="aligncenter size-full wp-image-898" title="2009-11-04_234227" src="http://syont.files.wordpress.com/2009/11/2009-11-04_234227.jpg?w=450&#038;h=247" alt="2009-11-04_234227" width="450" height="247" /></p>
<p>Concrete Materials</p>
<p><img class="aligncenter size-full wp-image-902" title="stress_strain_concrete_compression" src="http://syont.files.wordpress.com/2009/11/stress_strain_concrete_compression.png?w=450&#038;h=310" alt="stress_strain_concrete_compression" width="450" height="310" /></p>
<p>Rebar materials</p>
<p><img class="aligncenter size-full wp-image-910" title="rebar" src="http://syont.files.wordpress.com/2009/11/rebar.png?w=450&#038;h=296" alt="rebar" width="450" height="296" /></p>
<p><img class="aligncenter size-full wp-image-911" title="steelmat" src="http://syont.files.wordpress.com/2009/11/steelmat.png?w=436&#038;h=333" alt="steelmat" width="436" height="333" /></p>
<p><em><strong>&#8230; in draft</strong></em></p>
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		<title>SAP2000 distribusi beban 1way</title>
		<link>http://syont.wordpress.com/2009/10/26/sap2000-distribusi-beban-1way/</link>
		<comments>http://syont.wordpress.com/2009/10/26/sap2000-distribusi-beban-1way/#comments</comments>
		<pubDate>Mon, 26 Oct 2009 09:18:08 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

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		<description><![CDATA[Beberapa waktu yang lalau ada isu dari pembaca di blog ini dan dibeberapa beberapa forum penggunanya , menyatakan bugs SAP2000 pada distribusis beban area ke frame satu arah (one way) disini sya akan coba sampaikan perbedaan dari pemodelan dustribusi beban dari area ke frame baik yng satu arah dan dua arah.

Diberi contoh sederhana balok induk [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=878&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Beberapa waktu yang lalau ada isu dari pembaca di blog ini dan dibeberapa beberapa forum penggunanya , menyatakan bugs SAP2000 pada distribusis beban area ke frame satu arah (one way) disini sya akan coba sampaikan perbedaan dari pemodelan dustribusi beban dari area ke frame baik yng satu arah dan dua arah.</p>
<p style="text-align:center;"><img class="size-medium wp-image-879 aligncenter" title="steel_deck1" src="http://syont.files.wordpress.com/2009/10/steel_deck1.jpg?w=300&#038;h=150" alt="steel_deck1" width="300" height="150" /></p>
<p>Diberi contoh sederhana balok induk dan sekunder/anak struktur baja dengan beban merat 100 kg/m2 dan modul bentang 6&#215;6m, pemodelan ini tergantung sistem pelat yang digunakan jika hanya pelat tanpa bondek dalam artian begesting papan/konvensional maka akan cenderung bekerja secara dua arah ( two ways). Sedangkan model akan cenderung satu arah (one way) jika menggukakan steel deck sesuai alurnya.</p>
<p><img class="aligncenter size-full wp-image-890" title="halfslab" src="http://syont.files.wordpress.com/2009/10/halfslab.jpg?w=450&#038;h=198" alt="halfslab" width="450" height="198" /></p>
<p>Perbandingan bentangnya juga akan berpengaruh terhadap distribusi beban, dapat dilakukan penyederhanaan menjadi satu arah jika Lx/Ly ~ 2,5.</p>
<p><img class="aligncenter size-full wp-image-880" title="2009-10-25_150921" src="http://syont.files.wordpress.com/2009/10/2009-10-25_150921.png?w=450&#038;h=453" alt="2009-10-25_150921" width="450" height="453" /></p>
<p>Asumsi dan pemodelan distribusi beban dua arah.</p>
<p><img class="aligncenter size-full wp-image-881" title="2009-10-25_150926" src="http://syont.files.wordpress.com/2009/10/2009-10-25_150926.png?w=450&#038;h=453" alt="2009-10-25_150926" width="450" height="453" /></p>
<p>Tampilan hasil keluaran gaya lentur balok.</p>
<p><img class="aligncenter size-full wp-image-882" title="2009-10-25_151102" src="http://syont.files.wordpress.com/2009/10/2009-10-25_151102.png?w=450&#038;h=453" alt="2009-10-25_151102" width="450" height="453" /></p>
<p>Asumsi dan pemodelan distribusi beban satu arah, terlihat SAP2000 tidak menggunakan asumsi balok arah tegak lurus sisi pendek sebagai penerima beban malah sebaliknya. Ini mungkin yang dimaksud state pembaca blog ini yg disampaikan beberapa waktu lalu, saat itu saya coba tanyakan dibagian mana namun not responding :) Dengan sumsi beban diatas maka hasilnya jelas akan berbeda karena kenyataannya arah <em>steel deck </em>akan dipasang alur pendek untuk tujuan efektifitas maupun kerja shear connector.</p>
<p><img class="aligncenter size-full wp-image-883" title="2009-10-25_151106" src="http://syont.files.wordpress.com/2009/10/2009-10-25_151106.png?w=450&#038;h=453" alt="2009-10-25_151106" width="450" height="453" /></p>
<p>Seperti terlihat balok sekunder/anak tidak menerima distribusi beban sehingga gaya lentur dll yg terjadi adalah nul. Apakah itu sesuai asumsi pemodelan yang anda harapkan??</p>
<p><img class="aligncenter size-full wp-image-884" title="2009-10-25_151217" src="http://syont.files.wordpress.com/2009/10/2009-10-25_151217.png?w=450&#038;h=453" alt="2009-10-25_151217" width="450" height="453" /></p>
<p>Dengan memutar sumbu lokal element area, maka akan seperti diatas. Asumsi yang biasa diharapkan pada penerapan steel deck dan beton half slab.</p>
<p><img class="aligncenter size-full wp-image-886" title="2009-10-25_153305" src="http://syont.files.wordpress.com/2009/10/2009-10-25_153305.png?w=272&#038;h=189" alt="2009-10-25_153305" width="272" height="189" /></p>
<p>Terlihat distribusi gaya lentur sesuai dengan penerapan beban jika dilakukan secara perhitungan tangan/kalkulator (M3=(1/8)*200*6^2 = 900.00 Kgf.m)</p>
<p><img class="aligncenter size-full wp-image-885" title="2009-10-25_151221" src="http://syont.files.wordpress.com/2009/10/2009-10-25_151221.png?w=450&#038;h=453" alt="2009-10-25_151221" width="450" height="453" /></p>
<p>Seperti terlihat dari beberapa pemodelan  diatas terlihat SAP2000 menggunakan asumsi balok yang menerima beban adalah arah tegak lurus sumbu lokal-1. Seperti juga yg disampaikan sya sebelumnya <em>shortest by default&#8217;s not always be the best</em>.</p>
<p><img class="aligncenter size-full wp-image-892" title="phs" src="http://syont.files.wordpress.com/2009/10/phs.jpg?w=400&#038;h=300" alt="phs" width="400" height="300" /></p>
<p>Pengguna mempunyai keleluasaan penerapan beban sesuai harapan model struktur yg akan dikerjakan, karena adanya berbagai macam jenis slab beton monolit, <em>precast hollow sla</em>b, <em>half slab</em>, <em>steel deck</em> dll yg mempunyai perilaku berbeda-beda.</p>
<p>Tambahan pada balok sekunder/anak sambungan yg digunakan berupa <em>simple shear connection</em>, sehingga ujung dibuat sendi/realease. Hal ini tidak berbeda jauh dengan kenyataan jika <em>single span</em>, namun akan berbeda saat adanya balok lain <em>continous spans</em> karena adanya top <em>wiremesh</em> yang menerus yang akan menahan perputaran ujung balok.</p>
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		<title>rangkuman PPKGURG-1987</title>
		<link>http://syont.wordpress.com/2009/09/01/rangkuman-ppkgurg-1987/</link>
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		<pubDate>Tue, 01 Sep 2009 04:13:28 +0000</pubDate>
		<dc:creator>syont</dc:creator>
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		<description><![CDATA[Berikut adalah rangkuman Peraturan Perencanaan Ketahanan Gempa untuk Rumah dan Gedung PPKGURG-1987, distribusi beban gempa statik eqivalen tsb dipelajari dan digunakan saat kuliah dulu untuk tugas kampus maupun desain beneran bangunan rendah pada saat masih fresh graduates. Dapat digunakan untuk perhitungan beban gempa tiap lantai untuk gedung beraturan dgn prinsip dominan ragam getar adalah mode [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=863&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Berikut adalah rangkuman Peraturan Perencanaan Ketahanan Gempa untuk Rumah dan Gedung PPKGURG-1987, distribusi beban gempa statik eqivalen tsb dipelajari dan digunakan saat kuliah dulu untuk tugas kampus maupun desain beneran bangunan rendah pada saat masih fresh graduates. Dapat digunakan untuk perhitungan beban gempa tiap lantai untuk gedung beraturan dgn prinsip dominan ragam getar adalah mode pertama, ragam getar mode lebih tinggi dan ragam getar puntir tidak banyak berpengaruh.</p>
<p><img class="alignnone size-full wp-image-870" title="inaseismicitymaps" src="http://syont.files.wordpress.com/2009/09/inaseismicitymaps.png?w=450&#038;h=229" alt="inaseismicitymaps" width="450" height="229" /></p>
<p>Saat ini sudah dikeluarkan yang terbaru yaitu SNI 1726-2002, memang terlihat mencolok koefisien gempa <em>C</em> yang digunakan ini kerena perbedaan penerapan faktor daktilitas struktur. Pada PPKGURG-1987 digunakan notasi daktilitas struktur <em>K</em> sebagai faktor pengali (1*K), sedangkan pada SNI 1726-2002 digunakan notasi <em>R</em> sebagai faktor pembagi (1/R).</p>
<p>1. <a href="http://syont.files.wordpress.com/2009/09/rangkuman-ppkgurg-1987-a.pdf">rangkuman PPKGURG 1987 a</a> download as PDF</p>
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<p>2. <a href="http://syont.files.wordpress.com/2009/09/rangkuman-ppkgurg-1987-b-peta-wilayah-gempa.pdf">rangkuman PPKGURG 1987 b (Peta Wilayah Gempa)</a>, download as PDF</p>
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<p>Sedangkan untuk rangkuman SNI 1726-2002 dapat merujuk <a href="http://www.haki-konstruksi.com/images/course_haki/perhitungan_beban_gempa_rev.pdf">kesini</a>.</p>
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		<title>SAP2000 &#8211; Daftar kebutuhan material baja/beton</title>
		<link>http://syont.wordpress.com/2009/08/05/sap2000-daftar-kebutuhan-material-bajabeton/</link>
		<comments>http://syont.wordpress.com/2009/08/05/sap2000-daftar-kebutuhan-material-bajabeton/#comments</comments>
		<pubDate>Wed, 05 Aug 2009 05:21:07 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

		<guid isPermaLink="false">http://syont.wordpress.com/?p=859</guid>
		<description><![CDATA[Perbandingan beberapa jenis sistem struktur misal pada struktur baja, misal perbandingan struktur atap balok biasa, castellated, dan truss perlu mengetahui berat struktur tersebut keseluruhan untuk masing2 sistem. Biasanya struktur yang teringan adalah struktur yang baik terhadap konsumsi material, namun bukan berati yg terbaik karena perlu tinjauan lain seperti sistem sambungan, kekakuan, fix arsitek, dll.
Untuk mengetahui [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=859&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Perbandingan beberapa jenis sistem struktur misal pada struktur baja, misal perbandingan struktur atap balok biasa, castellated, dan truss perlu mengetahui berat struktur tersebut keseluruhan untuk masing2 sistem. Biasanya struktur yang teringan adalah struktur yang baik terhadap konsumsi material, namun bukan berati yg terbaik karena perlu tinjauan lain seperti sistem sambungan, kekakuan, fix arsitek, dll.</p>
<p>Untuk mengetahui secara cepat pada program bantu SAP2000 V11 sudah dapat menampilkan tabulasi kebutuhan material melalui menu <em><strong>Display &#8211;&gt; Show Tables </strong></em>lalu pilih <em>Material List by Section Property</em>.</p>
<p><a href="http://syont.files.wordpress.com/2009/08/2009-08-04_202436.jpg"><img class="alignnone size-full wp-image-860" title="2009-08-04_202436" src="http://syont.files.wordpress.com/2009/08/2009-08-04_202436.jpg?w=450&#038;h=355" alt="2009-08-04_202436" width="450" height="355" /></a></p>
<p>Daftar tabel material yang ditampikan adalah jenis penampang yg digunakan, panjang total dan beratnya tiap jenis penampang tsb, serta jumlah batangnya.</p>
<p><a href="http://syont.files.wordpress.com/2009/08/2009-08-04_202501.jpg"><img class="alignnone size-full wp-image-861" title="2009-08-04_202501" src="http://syont.files.wordpress.com/2009/08/2009-08-04_202501.jpg?w=450&#038;h=291" alt="2009-08-04_202501" width="450" height="291" /></a></p>
<p>Untuk jenis struktur baja ini sudah cukup representatif walau masih cukup konservatif karena adanya perhitungan yg dimasukkan pada overlap balok-kolom namun cukup beruna. Sedangkan untuk jenis struktur beton bertulang hanya menampilkan estimasi kubikasi beton yg diperlukan, program SAP2000 tidak <em>capable</em> menampilkan daftar kilogram besi tulangan yang perlu, modul tersebut adalah CSI Detailer yg ada pada program lain yaitu CSi SAFE.</p>
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			<media:title type="html">2009-08-04_202436</media:title>
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		<title>hXc dead fans !!</title>
		<link>http://syont.wordpress.com/2009/06/18/hxc-dead-fans/</link>
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		<pubDate>Thu, 18 Jun 2009 11:36:54 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Uncategorized]]></category>

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		<description><![CDATA[EarthCrisis sebuah bands pengusung philosophy straightXedge dari N.Y.H.C sudah membuat seseorang fans nya proud to be tatoos di wajahnya. Saya ngga tahu persis siapa, hanya dari blog komunitas. The bands banyak membicarakan tentang animal liberation, veganism, drugs free, pemahaman poverty&#38;ignorance, buruh dan industri, polution, law &#38; the rich, criminals etc.

Tatoos sendiri biasanya merupakan suatu gambaran [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=842&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>EarthCrisis sebuah bands pengusung philosophy straightXedge dari N.Y.H.C sudah membuat seseorang fans nya proud to be tatoos di wajahnya. Saya ngga tahu persis siapa, hanya dari blog komunitas. The bands banyak membicarakan tentang animal liberation, veganism, drugs free, pemahaman poverty&amp;ignorance, buruh dan industri, polution, law &amp; the rich, criminals etc.</p>
<p><a href="http://syont.files.wordpress.com/2009/06/earth-crisis-lover.jpg"><img class="alignnone size-full wp-image-843" title="earth crisis lover" src="http://syont.files.wordpress.com/2009/06/earth-crisis-lover.jpg?w=450&#038;h=276" alt="earth crisis lover" width="450" height="276" /></a></p>
<p>Tatoos sendiri biasanya merupakan suatu gambaran kisah seseorang yg ingin membekas dan dikenangnya, mungkin itu sebuah &#8220;<em>history of my life</em>&#8221; yg berupa symbol pengalaman kehidupan pribadi seseorang terkadang ada unsur promise. Menurut saya sesuatu yg hanya perlu diketahui sendiri sehingga letaknya seharusnya bukan di bagian tubuh yg terlihat exposed. Apalagi membentuk sebuah corak tulisan nama sebuah bands, menurut pandangan saya tidak perlu sejauh itu. <em>&#8230; but it&#8217;s in your hands, .. your ways</em>.</p>
<p><a href="http://syont.files.wordpress.com/2009/06/crazyface.jpg"><img class="alignnone size-full wp-image-848" title="CrazyFace" src="http://syont.files.wordpress.com/2009/06/crazyface.jpg?w=450&#038;h=337" alt="CrazyFace" width="450" height="337" /></a></p>
<p>ya c&#8217;mon dudes i know what you&#8217;re thinking about.</p>
<blockquote><p>in my minds, &#8230; in my heart, and</p>
<p>&#8230; in my eyes.</p></blockquote>
<p>i never heard &#8220;<em>in my skins</em>&#8221; &#8230; right??? This pictures are look more essences i think.</p>
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<p style="text-align:center;"><em><strong>&#8220;times goes by &#8230; but thes spirit still remains.</strong></em>&#8220;</p>
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		<title>Inersia effektif kolom tinjauan gempa</title>
		<link>http://syont.wordpress.com/2009/05/22/inersia-effektif-kolom-tinjauan-gempa/</link>
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		<pubDate>Fri, 22 May 2009 12:20:53 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

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		<description><![CDATA[Tulisan ini melanjutkan postingan saya sebelumnya mengenai pengaruh retak terhadapt distribusi gaya pada portal akibat beban gempa.
Grafik dibawah adalah hasil perbandingan data eksperimen dengan rekomendasi nilai kekakuan kolom efektif untuk bentuk pesegi dan jenis beton normal (Elwood., K.J., et al 2006)

Rekomendasi nilai yg digunakan adalah :

Untuk nilai  P/(A_g*f&#8217;c) &#60;= 0.2 maka faktor EI_eff/EI_g = 0.2
Untuk nilai  0.2 &#60; P/(A_g*f&#8217;c) &#60; 0.5 [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=700&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p align="left">Tulisan ini melanjutkan postingan saya sebelumnya mengenai pengaruh retak terhadapt distribusi gaya pada portal akibat beban gempa.</p>
<p align="left">Grafik dibawah adalah hasil perbandingan data eksperimen dengan rekomendasi nilai kekakuan kolom efektif untuk bentuk pesegi dan jenis beton normal (Elwood., K.J., et al 2006)</p>
<p align="left"><a href="http://syont.files.wordpress.com/2007/06/2009-05-22_153508.png"><img class="alignnone size-full wp-image-696" style="border:0 initial initial;" title="2009-05-22_153508" src="http://syont.files.wordpress.com/2007/06/2009-05-22_153508.png?w=450&#038;h=276" alt="2009-05-22_153508" width="450" height="276" /></a></p>
<p align="left">Rekomendasi nilai yg digunakan adalah :</p>
<ul>
<li>Untuk nilai  <em>P/(A_g*f&#8217;c) </em>&lt;= 0.2 maka faktor <em>EI_eff/EI_g</em> = 0.2</li>
<li>Untuk nilai  0.2 &lt; <em>P/(A_g*f&#8217;c) </em>&lt; 0.5 maka faktor <em>EI_eff/EI_g</em> = (5/3) - <em>(P/(A_g*f&#8217;c))</em> - (4/30)</li>
<li>Untuk nilai  <em>P/(A_g*f&#8217;c) &gt;</em> 0.5 maka faktor <em>EI_eff/EI_g</em> = 0.7</li>
</ul>
<p>Agak berbeda dgn rekomendasi peraturan ACI/SNI yang menentukan nilai reduksi sebesar 0.7, jika dibandingkan dengan rekomendasi diatas maka besarnya berkisar <em>P/(A_g*f&#8217;c)</em> &gt; 0.5 dan ini biasanya jarang ditemui kisaran nilai pada desain sesungguhnya, seharusnya tidak berlaku generalize dalam arti perlu penyesuaian.</p>
<p>Sedangkan FEMA-356 merekomendasikan nilai yg digunakan adalah :</p>
<ul>
<li>Untuk kolom dengan beban aksial gravitasi <em>P &lt; 0.3*A_g*f&#8217;c</em> maka kekakuan effektif adalah <em>0.5*EI_g</em></li>
<li>Untuk nilai <em>P &gt; 0.5*A_g*f&#8217;c</em> maka kekakuan effektif adalah <em>0.7*EI_g</em></li>
</ul>
<p>Dinding geser sebagai berikut :</p>
<ul>
<li>Tanpa terlihat retak, maka kekakuan effektif adalah <em>0.8*EI_g</em></li>
<li>Terlihat retak, maka kekakuan effektif adalah <em>0.5*EI_g</em></li>
</ul>
<p>Sedangkan peraturan beton New Zealand Standard (1995) merekomendasikan :</p>
<p>Untuk balok,</p>
<ul>
<li>pesegi, nilai <em>I_e = 0.4*I_g</em></li>
<li>balok T, L nilai <em>I_e = 0.35*I_g</em></li>
</ul>
<p>Untuk kolom,</p>
<ul>
<li>Nilai <em>I_e</em> = 0.80*<em>I_g</em>  jika perbandingan  <em>N / (fc * A_g)</em> &gt; 0.5</li>
<li>Nilai <em>I_e</em> = 0.60*<em>I_g</em>  jika perbandingan  <em>N / (fc * A_g) </em>= 0.2</li>
<li>Nilai <em>I_e </em>= 0.40*<em>I_g</em>  jika perbandingan  <em>N / (fc * A_g)</em> = −0.05</li>
</ul>
<p>Untuk dinding geser,</p>
<ul>
<li>Nilai <em>I_e</em> = 0.45* <em>I_g </em>  jika perbandingan   <em>N / (fc *A_g )</em>= 0.2</li>
<li>Nilai <em>I_e</em> = 0.25* <em>I_g<span style="font-style:normal;">   jika perbandingan  <em>N / (fc *A_g )</em> = 0</span></em></li>
<li>Nilai <em>I_e</em> = 0.50* <em>I_g<span style="font-style:normal;">   jika perbandingan   <em>N / (fc *A_g )</em> = −0.1</span></em></li>
</ul>
<p> </p>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:918px;width:1px;height:1px;">For Columns 0.80 Ig when N* / fc′ Ag &gt; 0.5</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:918px;width:1px;height:1px;">0.60 Ig N* / fc′ Ag = 0.2</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:918px;width:1px;height:1px;">0.40 Ig N* / fc′ Ag = −0.05     </p>
<div>For Columns 0.80 Ig when N* / fc′ Ag &gt; 0.5</div>
<div>0.60 Ig N* / fc′ Ag = 0.2</div>
<div>0.40 Ig N* / fc′ Ag = −0.05</div>
</div>
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		<slash:comments>4</slash:comments>
	
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		<title>pengingat itu perlu</title>
		<link>http://syont.wordpress.com/2009/05/18/pengingat-itu-perlu/</link>
		<comments>http://syont.wordpress.com/2009/05/18/pengingat-itu-perlu/#comments</comments>
		<pubDate>Mon, 18 May 2009 13:40:09 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

		<guid isPermaLink="false">http://syont.wordpress.com/?p=682</guid>
		<description><![CDATA[waktu jaman kul dulu sya mempunyai keinginan suatu saat dapat membuat program implementasi desain struktur beton, baja, pondasi dll. Program tersebut diharapkan dapat menampilkan detail langkah perhitungan, sudah beberapa modul desain telah jadi dengan OpenOffice.org CALC terlihat bagus namun rumit, terutama pada pencabangan dan iterasi. Ada juga yang lebih mendekati sempurna seperti gabungan Maxima (computing) [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=682&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>waktu jaman kul dulu sya mempunyai keinginan suatu saat dapat membuat program implementasi desain struktur beton, baja, pondasi dll. Program tersebut diharapkan dapat menampilkan detail langkah perhitungan, sudah beberapa modul desain telah jadi dengan OpenOffice.org CALC terlihat bagus namun rumit, terutama pada pencabangan dan iterasi. Ada juga yang lebih mendekati sempurna seperti gabungan Maxima (computing) dan Emacs (TEX formated), namun hanya bisa berjalan di Linux.  Sekarang sya rebuild module desain agar dapat menampilkan hasil perhitungan sesuai yg sya inginkan dulu, tujuan dari itu adalah &#8220;<em>&#8230; untuk mengingat</em>&#8221;  karena lupa tak dapat dihindari menjadikan setidaknya ada kontrol standar laporan perhitungan desain dan yang terpenting adalah kreativitas dan enjoy permainan logika pikiran.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-18_173328.png"><img class="alignnone size-full wp-image-683" title="2009-05-18_173328" src="http://syont.files.wordpress.com/2009/05/2009-05-18_173328.png?w=450&#038;h=342" alt="2009-05-18_173328" width="450" height="342" /></a></p>
<p>Program sya menggunakan Python(TM) yang sudah di compile menjadi <em>exe</em>, dapat berjalan di semua sistem operasi Win32, Linux, Mac, Solaris,BSD,  bahkan PDA atau Handphone,  jadi agak nyesel kenapa sya dulu belajarnya F77 ribet ngga produktif (<em>my opinion</em>).  Pada windows dapat dijalankan melalui DOS bawaannya, pilih menu <strong><em>Start </em></strong>&#8211;&gt; <strong><em>Programs </em></strong>&#8211;&gt; <strong><em>Acessories </em></strong>&#8211;&gt; <em><strong>Comand Prompt </strong></em>mungkin pengguna yang dulu sudah pernah memakai SAP90 atau Microfeap sudah familiar dgn command prompt.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-18_174659.png"><img class="alignnone size-full wp-image-684" title="2009-05-18_174659" src="http://syont.files.wordpress.com/2009/05/2009-05-18_174659.png?w=450&#038;h=342" alt="2009-05-18_174659" width="450" height="342" /></a></p>
<p>diatas adalah modul desain balok beton bertulang mengacu SKSNI-T15-1991-03 merujuk pada buku catatan2 kuliah sya dulu. Data yang diperlukan adalah dimensi penampang balok, material dan gaya terfaktor.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-18_174736.png"><img class="alignnone size-full wp-image-685" title="2009-05-18_174736" src="http://syont.files.wordpress.com/2009/05/2009-05-18_174736.png?w=450&#038;h=342" alt="2009-05-18_174736" width="450" height="342" /></a></p>
<p>keluaran, <em>scroll down</em> untuk melihat hasil laporan keseluruhan. Dapat dilakukan fasilitas<em> copy / paste</em> pada word processor untuk dilakukan editing jenis huruf dan page layout. Jenis font yang baik ditampilkan adalah Courier New 9pt.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-18_174745.png"><img class="alignnone size-full wp-image-686" title="2009-05-18_174745" src="http://syont.files.wordpress.com/2009/05/2009-05-18_174745.png?w=450&#038;h=342" alt="2009-05-18_174745" width="450" height="342" /></a></p>
<p>..</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-18_174801.png"><img class="alignnone size-full wp-image-687" title="2009-05-18_174801" src="http://syont.files.wordpress.com/2009/05/2009-05-18_174801.png?w=450&#038;h=342" alt="2009-05-18_174801" width="450" height="342" /></a></p>
<p>Hasil keluaran yang telah dilakukan sedikit editing dan dibuat dalam format PDF, dapat dilihat dibawah atau download <a href="http://syont.files.wordpress.com/2009/05/glassboxdi_v02_print.pdf">disini</a>.</p>
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<p>Sedangkan programnya sendiri dapat didownload dari <a href="http://www.4shared.com/file/106313292/35477b92/glassboxDI_v02.html">4shared</a> simpenan sya. Masih perlu banyak perbaikan diantaranya dalam penentuan tinggi effektif <em>d, </em>sketsa balok dan penulangannya. Pengembangan lainnya desain beton : Pelat, Balok-T, Kolom( penulangan, diagram interaksi, check biaksial), desain gempa, pelat pondasi, pilecap. Desain struktur baja : batang tekan, tarik, balok, kolom, dll, tancepin ke program analisa sruktur portal/rangka batang biar manage gaya internal lebih mudah, dll namanya juga ngembangin keinginannya banyak :) cuman perlu <em>side-by-side, it&#8217;s not an easy tasks but possible</em>.</p>
<p>Program yang saya buat tersebut masil dalam alpha version, diberi calon  nama <em>glassboxDI </em>masih ada kemungkinan ganti namanya mungkin lebih endo jadi <em>kota<strong>k</strong></em><em>acaID</em>, maksudnya hasil program desain outputnya itu transparan ngga <strong>in&gt;&gt;out </strong>aja, ngga tau tuh program running apa aja. Rencananya sedang dalam proses juga tapi masih permulaan untuk dibuat GUI fronted biar lebih mudah, pake yg cross platform juga.</p>
<p><em>funny computing again</em> :)</p>
<p>* updates v0.2 (alpha realese) &#8211; 19 May 2009</p>
<p><em> </em></p>
<p><em> </em></p>
<p><em> </em></p>
<p><em> </em></p>
<p><em> </em></p>
<p><em> </em></p>
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		<title>Tekuk lateral pada rangka batang yg menerus</title>
		<link>http://syont.wordpress.com/2009/05/15/tekuk-lateral-pada-rangka-batang-yg-menerus/</link>
		<comments>http://syont.wordpress.com/2009/05/15/tekuk-lateral-pada-rangka-batang-yg-menerus/#comments</comments>
		<pubDate>Fri, 15 May 2009 12:09:30 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

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		<description><![CDATA[Tinjau truss gantung yang dominan akibat beban tetap. Batang bawah dgn panjang L = 1.5*4 = 6.0m dapat dibuat batang menerus. Batang atas juga dapat dibuat menerus tanpa masalah karena batang tarik dan ada sokongan gording, berbeda perilakunya pada batang bawah karena mengalami tekan dan akan menekuk jika tidak diberikan sokongan samping.

Beban terpusat tetap

Gaya aksial batang [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=672&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Tinjau <em>truss </em>gantung yang dominan akibat beban tetap. Batang bawah dgn panjang <em>L </em>= 1.5*4 = 6.0m dapat dibuat batang menerus. Batang atas juga dapat dibuat menerus tanpa masalah karena batang tarik dan ada sokongan gording, berbeda perilakunya pada batang bawah karena mengalami tekan dan akan menekuk jika tidak diberikan sokongan samping.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_175117.png"><img class="alignnone size-full wp-image-673" title="2009-05-15_175117" src="http://syont.files.wordpress.com/2009/05/2009-05-15_175117.png?w=450&#038;h=279" alt="2009-05-15_175117" width="450" height="279" /></a></p>
<p>Beban terpusat tetap</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_175128.png"><img class="alignnone size-full wp-image-674" title="2009-05-15_175128" src="http://syont.files.wordpress.com/2009/05/2009-05-15_175128.png?w=450&#038;h=279" alt="2009-05-15_175128" width="450" height="279" /></a></p>
<p>Gaya aksial batang &#8211; beban tetap</p>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Pmaks=3776</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Pmaks = 3776.00</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">N1=4231</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>N1 = 4231.00</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">N2=587.5</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>N2 = 587.50</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lbt=1.5</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lbt = 1.50</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lky=4*Lbt*(0.75+0.25*(N2/N1))</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lky = 4.71</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">A2L45=4.3*2</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>A2L45 = 8.60</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">iy2L45=1.874</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>iy2L45 = 1.87</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; faktor tekuk</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_y=(Lky*100)/iy2L45</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_y = 251.24</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; &gt; 200 .. Not Okey</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">E_s = 2100000</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>E_s = 2100000.00</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">F_y=2400</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>F_y = 2400.00</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_g=Pi*Sqrt(E_s/(0.7*F_y))</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_g = 111.07</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_sy = Lmbd_y/Lmbd_g</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_sy = 2.26</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; untuk  Lmbd_sy &gt; 1</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">omega_y = 2.381*Lmbd_sy^2</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>omega_y = 12.18</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; Tegangan Normal Kritis</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">F_crit = omega_y * ( Pmaks / A2L45)</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>F_crit = 5348.96</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; &gt; F_ijin = 1600 kg/cm^2 &#8230; Not Okey</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">A2L70=9.4*2</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>A2L70 = 18.80</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">iy2L70=2.91</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>iy2L70 = 2.91</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; faktor tekuk</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_y=(Lky*100)/iy2L70</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_y = 161.80</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; &lt; 200 .. Okey</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_g=Pi*Sqrt(E_s/(0.7*F_y))</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_g = 111.07</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">Lmbd_sy = Lmbd_y/Lmbd_g</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>Lmbd_sy = 1.46</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; untuk  Lmbd_sy &gt; 1</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">omega_y = 2.381*Lmbd_sy^2</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>omega_y = 5.05</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; Tegangan Normal Kritis</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">F_crit = omega_y * ( Pmaks / A2L70)</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;"><span style="white-space:pre;"> </span>F_crit = 1014.76</div>
<div id="_mcePaste" style="position:absolute;left:-10000px;top:708px;width:1px;height:1px;">&#8216; &lt; F_ijin = 1600 kg/cm^2 &#8230; Okey</div>
<p><em>Pmaks </em>= 3776 Kgf</p>
<p><em>N1</em> = 4231 Kgf (from reff.)</p>
<p><em>N2</em>=587.5 Kgf (from reff.)</p>
<p><em>L</em> = 1.5 m</p>
<p><em>Lky</em> = 4*Lbt*(0.75+0.25*(N2/N1)) = 4.71 m</p>
<p><em>A2L45</em> = 4.3*2 = 8.60 cm^2</p>
<p><em>iy2L45 </em>= 1.874 cm</p>
<p><span style="text-decoration:underline;">Faktor tekuk</span></p>
<p><em>Lmbd_y</em> = (<em>Lky</em>*100)/<em>iy2L45</em>  = 251.24  &gt; 200 .. Not Okey</p>
<p><em>E_s</em> = 2100000 kg/cm^2</p>
<p><em>F_y</em>=2400 kg/cm^2</p>
<p><em>Lmbd_g</em> = Pi * Sqrt (<em>E_s</em>/(0.7*<em>F_y</em>))  = 111.07</p>
<p><em>Lmbd_sy</em> = <em>Lmbd_y </em>/ <em>Lmbd_g</em>  = 2.26</p>
<p>untuk  <em>Lmbd_sy </em>&gt; 1</p>
<p><em>omega_y</em> = 2.381*<em>Lmbd_sy</em>^2  = 12.18</p>
<p> Tegangan Normal Kritis</p>
<p><em>F_crit</em> = <em>omega_y</em> * ( <em>Pmaks</em> / <em>A2L45</em>)  = 5348.96 kg/cm^2  &gt; <em>F_ijin</em> (1600 kg/cm^2)  &#8230;  Not Okey</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_172723.png"><img class="alignnone size-full wp-image-675" title="2009-05-15_172723" src="http://syont.files.wordpress.com/2009/05/2009-05-15_172723.png?w=450&#038;h=279" alt="2009-05-15_172723" width="450" height="279" /></a></p>
<p>Mode pertama tekuk lateral, <em>Faktor</em> = 195.5, diterapkan kembali menjadi beban titik dan menghasilkan gaya batang berikut.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_175429.png"><img class="alignnone size-full wp-image-676" title="2009-05-15_175429" src="http://syont.files.wordpress.com/2009/05/2009-05-15_175429.png?w=450&#038;h=279" alt="2009-05-15_175429" width="450" height="279" /></a></p>
<p>Menghasilkan <em>Pmaks</em> = 1955 Kgf, yaitu hanya sekitar ~1/2*Pmaks yg terjadi akibat beban tetap.</p>
<p><em>A2L70</em> = 9.4*2 = 18.80 cm^2</p>
<p><em>iy2L70 </em>= 2.91 cm</p>
<p><span style="text-decoration:underline;">Faktor tekuk</span></p>
<p><em>Lmbd_y</em> = (<em>Lky</em>*100)/<em>iy2L70</em>  = 161.80  &lt; 200 .. Okey</p>
<p><em>Lmbd_sy</em> = <em>Lmbd_y</em>/<em>Lmbd_g</em>  = 1.46</p>
<p>untuk  <em>Lmbd_sy</em> &gt; 1</p>
<p><em>omega_y</em> = 2.381*<em>Lmbd_sy</em>^2  = 5.05</p>
<p> Tegangan Normal Kritis</p>
<p><em>F_crit</em> = <em>omega_y</em> * ( <em>Pmaks</em> / <em>A2L70</em>) <span style="white-space:pre;"> </span> = 1014.76  kg/cm^2  &lt; <em>F_ijin</em> (1600 kg/cm^2)  &#8230;  Okey</p>
<p>Batang bawah diganti dengan 2L70.70.7</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_180028.png"><img class="alignnone size-full wp-image-677" title="2009-05-15_180028" src="http://syont.files.wordpress.com/2009/05/2009-05-15_180028.png?w=450&#038;h=279" alt="2009-05-15_180028" width="450" height="279" /></a></p>
<p>Mode pertama tekuk lateral, <em>Faktor</em> = 1021.8, diterapkan kembali menjadi beban titik dan menghasilkan gaya batang berikut.</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_180422.png"><img class="alignnone size-full wp-image-678" title="2009-05-15_180422" src="http://syont.files.wordpress.com/2009/05/2009-05-15_180422.png?w=450&#038;h=279" alt="2009-05-15_180422" width="450" height="279" /></a></p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_180422.png"></a>Menghasilkan <em>Pmaks</em> = 10218 Kgf, yaitu 2.7*<em>Pmaks</em> yg terjadi akibat beban tetap. Agak berbeda faktor tersebut dibandingkan tinjauan perhitungan tangan diatas yang besarnya hanya  ~1.6</p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_180422.png"></a></p>
<p><a href="http://syont.files.wordpress.com/2009/05/2009-05-15_180422.png"></a></p>
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		<title>Lentur tidak sentris pd profil unsymmetri</title>
		<link>http://syont.wordpress.com/2009/04/24/lentur-tidak-sentris-pd-profil-unsymmetri/</link>
		<comments>http://syont.wordpress.com/2009/04/24/lentur-tidak-sentris-pd-profil-unsymmetri/#comments</comments>
		<pubDate>Fri, 24 Apr 2009 07:50:47 +0000</pubDate>
		<dc:creator>syont</dc:creator>
				<category><![CDATA[Sipil/Struktural]]></category>

		<guid isPermaLink="false">http://syont.wordpress.com/?p=634</guid>
		<description><![CDATA[Profil unsymmetri pada salah sumbunya akan mengalami puntir jika beban yang bekerja berada pada titik berat penampang (Center of Gravity), bukan pada titik pusat geser penampang (Shear Centre). Ditinjau balok kantilever dari profil UNP dengan bentang 3.00m dan beban terpusat 1000 Kgf.

Beban terpusat
P = 1000 Kgf
l=3.0 m
Momen lentur
Mp = P * l =  3000.00 Kgf.m

Karakteristik [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=syont.wordpress.com&blog=619354&post=634&subd=syont&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Profil <em>unsymmetri </em>pada salah sumbunya akan mengalami puntir jika beban yang bekerja berada pada titik berat penampang (<em>Center of Gravity</em>), bukan pada titik pusat geser penampang (<em>Shear Centre</em>). Ditinjau balok kantilever dari profil UNP dengan bentang 3.00m dan beban terpusat 1000 Kgf.</p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_103112.png"><img class="alignnone size-full wp-image-635" title="2009-03-24_103112" src="http://syont.files.wordpress.com/2009/04/2009-03-24_103112.png?w=301&#038;h=241" alt="2009-03-24_103112" width="301" height="241" /></a></p>
<p>Beban terpusat<br />
<em>P</em> = 1000 Kgf<br />
<em>l</em>=3.0 m<br />
Momen lentur<br />
<em>Mp = P * l</em> =  3000.00 Kgf.m</p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_103145.png"><img class="alignnone size-full wp-image-636" title="2009-03-24_103145" src="http://syont.files.wordpress.com/2009/04/2009-03-24_103145.png?w=324&#038;h=444" alt="2009-03-24_103145" width="324" height="444" /></a></p>
<p>Karakteristik penampang (cm)<br />
UNP 200&#215;80x7.5&#215;11<br />
d=20;<br />
bf=8;<br />
tf=0.75;<br />
tw=1.1;<br />
w=24.6;<br />
Ix=1950;<br />
Zx=195;<br />
Abaikan <em>fillet radius</em> dalam pemodelan element <em>shell</em></p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_113043.png"><img class="alignnone size-full wp-image-637" title="2009-03-24_113043" src="http://syont.files.wordpress.com/2009/04/2009-03-24_113043.png?w=450&#038;h=279" alt="2009-03-24_113043" width="450" height="279" /></a></p>
<p>Momen akibat berat sendiri profil<br />
<em>Mq </em>= <em>(1/2)*w*l^2 </em>= 110.70 Kgf.m<br />
Momen total akibat beban terpusat dan berat sendiri<br />
<em>Mt = Mp + Mq </em>= 3110.70 Kgf.m</p>
<p>Tegangan yang terjadi akibat beban terpusat <em>P</em> dan berat sendiri <em>q</em><br />
<em>Teg=(Mt*100)/Zx</em> = 1595.23 Kgf/cm^2</p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_134402.png"><img class="alignnone size-full wp-image-639" title="2009-03-24_134402" src="http://syont.files.wordpress.com/2009/04/2009-03-24_134402.png?w=450&#038;h=559" alt="2009-03-24_134402" width="450" height="559" /></a></p>
<p>Ini hanya benar jika beban terpusat bekerja pada titik pusat geser (<em>O</em>) bukan pusat berat penampang (<em>C</em>)</p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_134759.png"><img class="alignnone size-full wp-image-638" title="2009-03-24_134759" src="http://syont.files.wordpress.com/2009/04/2009-03-24_134759.png?w=450&#038;h=558" alt="2009-03-24_134759" width="450" height="558" /></a><br />
Besarnya nilai eksentrisitas jika ingin mengasumsikan tidak ada puntir maka beban terpusat <em>&#8216;P</em>&#8216; bekerja pada jarak &#8216;<em>e</em>&#8216; dari web</p>
<p>Luas web, <em>Aw = tw*d</em> = 22.0 cm^2<br />
Luas flens, <em>Af = tf*bf </em>= 6.0 cm^2</p>
<p>Eksentrisitas, <em>e = (bf/2)/(1+(1/6)*(Aw/Af))</em> = 2.4828 cm</p>
<p><a href="http://syont.files.wordpress.com/2009/04/unpdefshp5.png"><img class="alignnone size-full wp-image-640" title="unpdefshp5" src="http://syont.files.wordpress.com/2009/04/unpdefshp5.png?w=450&#038;h=314" alt="unpdefshp5" width="450" height="314" /></a></p>
<p>Diatas gambar kontur defleksi resultan akibat beban terpusat dan berat sendiri.</p>
<p>Jika beban bekerja pada titik pusat geser dan defleksi akibat berat sendiri diabaikan terlebih dahulu, maka defleksi akibat beban terpusat besarnya adalah sbb:</p>
<p>Modulus Elastisitas Baja, <em>Es</em> = 2038901.9 Kgf/cm^2<em> (default SAP2000 for steel materials</em>)<br />
Betang balok, <em>l</em> = 3.0*100 = 300.0 cm<br />
Lendutan-z, <em>delta = (P*l^3)/(3*Es*Ix) = </em>2.2637 cm</p>
<p>Keadaan berbeda jika beban yang bekerja pada titik berat penampang (C.G.), akan mengalami deformasi puntir akibat eksentrisitas titik berat dan titik geser penampang. Hasil ditunjukkan berikut :</p>
<p><a href="http://syont.files.wordpress.com/2009/04/unpdefshp4.png"><img class="alignnone size-full wp-image-641" title="unpdefshp4" src="http://syont.files.wordpress.com/2009/04/unpdefshp4.png?w=449&#038;h=311" alt="unpdefshp4" width="449" height="311" /></a></p>
<p>Deformasi puntir akibat eksentrisitas, sedangkan distribusi tegangan arah normal atau sb-1 lokal elemen <em>shell </em>diperlihatkan berikut.</p>
<p><a href="http://syont.files.wordpress.com/2009/04/2009-03-24_134511.png"><img class="alignnone size-full wp-image-642" title="2009-03-24_134511" src="http://syont.files.wordpress.com/2009/04/2009-03-24_134511.png?w=450&#038;h=559" alt="2009-03-24_134511" width="450" height="559" /></a></p>
<p>Perbedaan mencapai ~35% lebih besar pada titik <em>corner</em> penampang namun pada titik <em>edge</em> nilainya lebih kecil, dalam artian tegangan pada tumpuan tidak merata. Selanjutnya perlu peninjauan pengaruhnya terhadap rasio tekuk torsi lateral &#8230;</p>
<p>Catatan :</p>
<p>* Ada yg terlewat, nilai yg seharusnya digunakan <em>Af = tf*(bf-tw)</em> = 5.1750 cm^2, dan menghasilkan nilai <em>e</em> = 2.3412 cm (selisih ~6% dari awal) jadi perlu <em>re-run analysis</em> :) wahduh DoItYourself aja ya .</p>
<p>** Lentur tidak sentris dapat diperhitungkan dengan faktor koreksi atau modifikasi dari Bleich.</p>
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