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Balok BETON

By Ayub Pujiyanto | Like Facebook AURELIA SANJAYA

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<h2 class="title">
Desain Balok Kantilever</h2>
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 </div>
Salah satu follower <a href="https://www.twitter.com/juragan_sipil" target="_blank">@juragan_sipil</a>
 di twitter menawarkan sebuah permasalahan sederhana dalam desain balok 
kantilever beton bertulang. Balok kantilever adalah balok yang salah 
satu tumpuannya adalah jepit, sementara ujung yang lain bebas. Kira-kira
 soalnya seperti ini: 
<img alt="soal balok kantilever" border="0" height="226" src="http://duniatekniksipil.web.id/wp-content/uploads/2012/11/soalbalokkantilever_thumb.jpg" style="border: 0px none; display: block; float: none; margin-left: auto; margin-right: auto;" title="soal balok kantilever" width="484" /> 
Nah, di sini saya punya 2 poin sekaligus asumsi: 
<ol>
<li>Apakah <img alt="Q_{DL} " class="latex" src="http://s.wordpress.com/latex.php?latex=Q_%7BDL%7D%20%20&bg=ffffff&fg=000000&s=-2" title="Q_{DL} " /> sudah termasuk berat sendiri? Kita asumsikan SUDAH. Jadi kita ngga usah hitung lagi berat sendirinya.</li>
<li>Beban hidup dan mati di atas, apakah diaplikasikan di sepanjang balok? Kita asumsikan IYA. Beban <img alt="Q_{DL} " class="latex" src="http://s.wordpress.com/latex.php?latex=Q_%7BDL%7D%20%20&bg=ffffff&fg=000000&s=-2" title="Q_{DL} " /> dan <img alt="Q_{LL} " class="latex" src="http://s.wordpress.com/latex.php?latex=Q_%7BLL%7D%20%20&bg=ffffff&fg=000000&s=-2" title="Q_{LL} " /> bekerja di sepanjang balok.</li>
<li>Kombinasi pembebanan tidak ditentukan. Jadi, kita asumsikan menggunakan kombinasi pembebanan sesuai ACI/SNI.</li>
</ol>
Tahap I. Analisis Struktur 
Karena baloknya kantilever, maka momen lenturnya negatif, artinya 
serat atas mengalami tarik, serat bawah mengalami tekan. Jadi, yang kita
 desain kali ini adalah tulangan tarik atau tulangan atas. 
<ol>
<li>Momen di ujung batang (tumpuan) akibat beban mati dan beban hidup. 
<img alt="M_{DL} = \dfrac12\times Q_{DL} \times L^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7BDL%7D%20%3D%20%5Cdfrac12%5Ctimes%20Q_%7BDL%7D%20%5Ctimes%20L%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="M_{DL} = \dfrac12\times Q_{DL} \times L^2 " /> 
<img alt="M_{DL} = \dfrac12 \times 7.5 \times 3^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7BDL%7D%20%3D%20%5Cdfrac12%20%5Ctimes%207.5%20%5Ctimes%203%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="M_{DL} = \dfrac12 \times 7.5 \times 3^2 " /> 
<img alt="M_{DL} = 33.75 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7BDL%7D%20%3D%2033.75%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_{DL} = 33.75 \text{kNm} " /> 
Dengan cara yang sama, 
<img alt="M_{LL} = 18 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7BLL%7D%20%3D%2018%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_{LL} = 18 \text{kNm} " /></li>
<li>Menghitung momen ultimit dengan Kombinasi Pembebanan, 
<img alt="M_{u_1} = 1.4\times M_{DL} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7Bu_1%7D%20%3D%201.4%5Ctimes%20M_%7BDL%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_{u_1} = 1.4\times M_{DL} " /> 
<img alt="M_u{_1} = 47.25 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_u%7B_1%7D%20%3D%2047.25%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_u{_1} = 47.25 \text{kNm} " /> 
— 
<img alt="M_{u_2} = 1.2\times M_{DL} + 1.6 \times M_{LL} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7Bu_2%7D%20%3D%201.2%5Ctimes%20M_%7BDL%7D%20%2B%201.6%20%5Ctimes%20M_%7BLL%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_{u_2} = 1.2\times M_{DL} + 1.6 \times M_{LL} " /> 
<img alt="M_{u_2} = 69.3 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_%7Bu_2%7D%20%3D%2069.3%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_{u_2} = 69.3 \text{kNm} " /> <—menentukan 
— 
Jadi, <img alt="M_u = 69.3 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_u%20%3D%2069.3%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_u = 69.3 \text{kNm} " /></li>
</ol>
 
Tahan II. Desain Tulangan Balok 
Teori dan prosedurnya bisa dibaca di <a href="http://duniatekniksipil.web.id/244/desain-balok-beton-bag-1/" target="_blank">sini</a> 
<ol>
<li><img alt="f`_c = 32 \text{MPa} " class="latex" src="http://s.wordpress.com/latex.php?latex=f%60_c%20%3D%2032%20%5Ctext%7BMPa%7D%20%20&bg=ffffff&fg=000000&s=-2" title="f`_c = 32 \text{MPa} " />, dan <img alt="f_y = 320 \text{MPa} " class="latex" src="http://s.wordpress.com/latex.php?latex=f_y%20%3D%20320%20%5Ctext%7BMPa%7D%20%20&bg=ffffff&fg=000000&s=-2" title="f_y = 320 \text{MPa} " /></li>
<li>Untuk <img alt="f`_c = 32 \text{MPa} " class="latex" src="http://s.wordpress.com/latex.php?latex=f%60_c%20%3D%2032%20%5Ctext%7BMPa%7D%20%20&bg=ffffff&fg=000000&s=-2" title="f`_c = 32 \text{MPa} " />, maka 
<img alt="\beta_1 = 0.85 - 0.05 \big(\dfrac{f_c-30}{7}\big) " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbeta_1%20%3D%200.85%20-%200.05%20%5Cbig%28%5Cdfrac%7Bf_c-30%7D%7B7%7D%5Cbig%29%20%20&bg=ffffff&fg=000000&s=-2" title="\beta_1 = 0.85 - 0.05 \big(\dfrac{f_c-30}{7}\big) " /> 
<img alt="\beta_1 = 0.836 " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cbeta_1%20%3D%200.836%20%20&bg=ffffff&fg=000000&s=-2" title="\beta_1 = 0.836 " /></li>
<li>Asumsi, jarak antara serat terluar sampai tulangan tarik adalah mmmm… 50 mm. Jadi, 
<img alt="d = 400-50 = 350 \text{mm} " class="latex" src="http://s.wordpress.com/latex.php?latex=d%20%3D%20400-50%20%3D%20350%20%5Ctext%7Bmm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="d = 400-50 = 350 \text{mm} " /></li>
<li><img alt="j_d \approx 0.875d = 306.25 \text{mm} " class="latex" src="http://s.wordpress.com/latex.php?latex=j_d%20%5Capprox%200.875d%20%3D%20306.25%20%5Ctext%7Bmm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="j_d \approx 0.875d = 306.25 \text{mm} " /></li>
<li>Kebutuhan tulangan, 
<img alt="A_s = \dfrac{M_u}{\phi f_y j_d} " class="latex" src="http://s.wordpress.com/latex.php?latex=A_s%20%3D%20%5Cdfrac%7BM_u%7D%7B%5Cphi%20f_y%20j_d%7D%20%20&bg=ffffff&fg=000000&s=-2" title="A_s = \dfrac{M_u}{\phi f_y j_d} " /> 
<img alt="A_s = \dfrac{69.3 \times 1000000}{0.8 \times 320 \times 306.25} " class="latex" src="http://s.wordpress.com/latex.php?latex=A_s%20%3D%20%5Cdfrac%7B69.3%20%5Ctimes%201000000%7D%7B0.8%20%5Ctimes%20320%20%5Ctimes%20306.25%7D%20%20&bg=ffffff&fg=000000&s=-2" title="A_s = \dfrac{69.3 \times 1000000}{0.8 \times 320 \times 306.25} " /> 
<img alt="A_s = 883.93 \text{mm}^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=A_s%20%3D%20883.93%20%5Ctext%7Bmm%7D%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="A_s = 883.93 \text{mm}^2 " /></li>
<li>Tulangan minimum, 
<img alt="A_{smin} = \dfrac{1.4}{f_y}bd " class="latex" src="http://s.wordpress.com/latex.php?latex=A_%7Bsmin%7D%20%3D%20%5Cdfrac%7B1.4%7D%7Bf_y%7Dbd%20%20&bg=ffffff&fg=000000&s=-2" title="A_{smin} = \dfrac{1.4}{f_y}bd " /> 
<img alt="A_{smin} = 459.375 \text{mm}^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=A_%7Bsmin%7D%20%3D%20459.375%20%5Ctext%7Bmm%7D%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="A_{smin} = 459.375 \text{mm}^2 " /> –> OK kebutuhan tulangan masih lebih besar, gunakan  <img alt="A_s = 883.93 \text{mm}^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=A_s%20%3D%20883.93%20%5Ctext%7Bmm%7D%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="A_s = 883.93 \text{mm}^2 " /></li>
<li>Tulangan maksimum, 
<img alt="\rho_{max} = 0.75 \rho_b " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Crho_%7Bmax%7D%20%3D%200.75%20%5Crho_b%20%20&bg=ffffff&fg=000000&s=-2" title="\rho_{max} = 0.75 \rho_b " /> 
<img alt="\rho_{max} = 0.75 \big( \dfrac{0.85f`_c \beta_1}{f_y} \big( \dfrac{600}{f_y+600} \big) \big) " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Crho_%7Bmax%7D%20%3D%200.75%20%5Cbig%28%20%5Cdfrac%7B0.85f%60_c%20%5Cbeta_1%7D%7Bf_y%7D%20%5Cbig%28%20%5Cdfrac%7B600%7D%7Bf_y%2B600%7D%20%5Cbig%29%20%5Cbig%29%20%20&bg=ffffff&fg=000000&s=-2" title="\rho_{max} = 0.75 \big( \dfrac{0.85f`_c \beta_1}{f_y} \big( \dfrac{600}{f_y+600} \big) \big) " /> 
<img alt="\rho_{max} = 0.035 " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Crho_%7Bmax%7D%20%3D%200.035%20%20&bg=ffffff&fg=000000&s=-2" title="\rho_{max} = 0.035 " /> 
Atau, 
<img alt="A_{smax} = \rho_{max} b d = 2844.8 \text{mm}^2 " class="latex" src="http://s.wordpress.com/latex.php?latex=A_%7Bsmax%7D%20%3D%20%5Crho_%7Bmax%7D%20b%20d%20%3D%202844.8%20%5Ctext%7Bmm%7D%5E2%20%20&bg=ffffff&fg=000000&s=-2" title="A_{smax} = \rho_{max} b d = 2844.8 \text{mm}^2 " /> –> OK, masih lebih kecil dari tulangan maksimum.</li>
<li>Tinggal pilih, mau pake diameter berapa. 
D16, As = 201 mm2 –> jumlah yang dibutuhkan 4.39 dibulatkan ke atas menjadi 5.  (5D16, As = 1005 mm2) 
D19, As = 283 mm2 –> jumlah yang dibutuhkan 3.12 dibulatkan ke atas menjadi 4.  (4D19, As = 1132 mm2) 
D22, As = 380 mm2 –> jumlah yang dibutuhkan 2.33 dibulatkan ke atas menjadi 3.  (3D22, As = 1140 mm2)</li>
<li>Kita akan memilih 5D16, karena itu yang paling mendekati luas tulangan yang diperlukan (883.93 mm2). 
<img alt="image" border="0" height="222" src="http://duniatekniksipil.web.id/wp-content/uploads/2012/11/image_thumb.png" style="border-width: 0px; display: inline;" title="image" width="203" /></li>
<li>Cek ulang kapasitas penampang. Hitung dulu tinggi blok tekan, a 
<img alt="a = \dfrac{\phi f_y A_s}{0.85f`_cb} " class="latex" src="http://s.wordpress.com/latex.php?latex=a%20%3D%20%5Cdfrac%7B%5Cphi%20f_y%20A_s%7D%7B0.85f%60_cb%7D%20%20&bg=ffffff&fg=000000&s=-2" title="a = \dfrac{\phi f_y A_s}{0.85f`_cb} " /> 
<img alt="a = 31.53 \text{mm} " class="latex" src="http://s.wordpress.com/latex.php?latex=a%20%3D%2031.53%20%5Ctext%7Bmm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="a = 31.53 \text{mm} " /></li>
<li>Hitung ulang <img alt="j_d " class="latex" src="http://s.wordpress.com/latex.php?latex=j_d%20%20&bg=ffffff&fg=000000&s=-2" title="j_d " /> 
<img alt="j_d = d-0.5a =334.23 \text{ mm} " class="latex" src="http://s.wordpress.com/latex.php?latex=j_d%20%3D%20d-0.5a%20%3D334.23%20%5Ctext%7B%20mm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="j_d = d-0.5a =334.23 \text{ mm} " /></li>
<li>Hitung momen nominal penampang, <img alt="\phi M_n " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n " /> 
<img alt="\phi M_n = \phi f_y A_s j_d " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%3D%20%5Cphi%20f_y%20A_s%20j_d%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n = \phi f_y A_s j_d " /> 
<img alt="\phi M_n = 0.8 \times 320 \times 1005 \times 334.23 " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%3D%200.8%20%5Ctimes%20320%20%5Ctimes%201005%20%5Ctimes%20334.23%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n = 0.8 \times 320 \times 1005 \times 334.23 " /> 
<img alt="\phi M_n = 85.99E6 \text{ Nmm} " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%3D%2085.99E6%20%5Ctext%7B%20Nmm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n = 85.99E6 \text{ Nmm} " /> 
<img alt="\phi M_n = 85.99 \text{ kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%3D%2085.99%20%5Ctext%7B%20kNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n = 85.99 \text{ kNm} " /></li>
<li>Bandingkan momen nominal penampang, harus lebih besar daripada momen ultimit. 
<img alt="\phi M_n = 85.99 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=%5Cphi%20M_n%20%3D%2085.99%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="\phi M_n = 85.99 \text{kNm} " /> 
<img alt="M_u = 69.3 \text{kNm} " class="latex" src="http://s.wordpress.com/latex.php?latex=M_u%20%3D%2069.3%20%5Ctext%7BkNm%7D%20%20&bg=ffffff&fg=000000&s=-2" title="M_u = 69.3 \text{kNm} " /> 
OK kan?</li>
</ol>
 
<blockquote>
Rahasia: pada step #8 di atas, kalo mau pake D19, 
sebenarnya kita bisa pake 3 buah D19, hasil baginya cukup dekat ke 3… 
yaitu 3.12. Walaupun As-nya nanti menjadi 3×283 = 849 mm2 (lebih kecil 
dari yg diperlukan, 883.93 mm2), tapi kalau kita teruskan ke langkah #9 
sampai selesai dengan menggunakan 3D19, akan diperoleh momen nominal 
73.17 kNm (masih lebih besar daripada momen ultimit) 
Bukankah yang lebih penting dalam desain LRFD, tahanan nominal harus > beban/gaya ultimit.</blockquote>
 
Update: 
Metode di atas adalah salah satu contoh. Masih ada metode-metode lain
 yang dapat digunakan, misalnya dengan persamaan kuadratik, dengan 
grafik, atau dengan hitungan cepat ala konsultan. 
mohon dikoreksi, 
semoga bermanfaat[]</div>

Desain Balok Kantilever Salah satu follower @juragan_sipil di twitter menawarkan sebuah permasalahan sederhana dalam desain balok k...

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