Perhitungan Pelat Bondek m10 [PDF]

Citation preview

PERHITUNGAN PELAT BONDEK Data pelat lantai bondek : 1. Bondek jenis W-1000 (Fy = 500 Mpa) 2. Tebal beton = 140 mm 3. Tebal floor deck = 0,75 mm , mutu baja : fy=500 Mpa 4. Mutu beton = K350 (f’c = 29 Mpa) 5. Tipe Wiremesh : a. Wiremesh M10-150 , mutu U-50



A. Wiremesh M10-150 , mutu U-50 h



= 90 mm



D



= 10 mm



b



= 1000 mm



P



= 25 mm



E



= 200000 MPa



f’c



= 29 MPa



fyD



= 500 Mpa



fyv



= 500 Mpa



d’



= 25mm



d



= h – d’ = 90 – 25 = 65 mm



Menghitung Mpr Pada Pelat Momen Negatif (Tumpuan) Tulangan yang digunakan D10, ntarik = 6D10 (As = 471 mm2) ntekan = (As’ = 750 mm2) Cc+Cs = Ts Asumsi awal tulangan tekan belum leleh dan tulangan tarik sudah leleh sehingga dapat diperoleh persamaan sebagai berikut: 𝜀′𝑠 = ( 𝜀𝑠 = ( 𝜀𝑦 =



𝑐−𝑑′



) 𝑥 0,003 < 0,002



𝑐



𝑑−𝑑′ 𝑐



) 𝑥 0,003 ≥ 0,002



𝑓𝑦 500 = = 0,0025 𝐸 200000



Kontrol Kekuatan Gaya tekan baja tulangan: 𝑐 − 𝑑′ 𝐶𝑠 = 𝐴𝑠′ 𝐸𝑠 ( ) 𝑥 0,003 𝑐 𝐶𝑠 = 750 𝑥 200000 𝑥 ( 𝐶𝑠 = 450.000 (



𝑐 − 𝑑′ ) 𝑥 0,003 𝑐



𝑐−25 𝑐



)N



Gaya tekan beton : 𝐶𝑐 = 0,85 𝑓 ′ 𝑐 𝑎 𝑏 𝐶𝑐 = 0,85 𝑓 ′ 𝑐 β c 𝑏 𝐶𝑐 = 0,85 𝑥 29 𝑥 0,843 𝑥 𝑐 𝑥 1000 𝐶𝑐 = 20779,95 𝑐 N



Gaya tarik baja tulangan: 𝑇𝑠 = 𝐴𝑠 𝑓𝑦 𝑇𝑠 = 471 𝑥 500 𝑥 1,25 𝑇𝑠 = 294375 N



Keseimbangan gaya 𝐶𝑐 + 𝐶𝑠 = 𝑇𝑠 20779,95 𝑐 + 450.000 (



𝑐 − 25 ) = 294375 𝑐



20779,95 𝑐 2 + 450000 𝑐 − 11250000 = 294375 𝑐 20779,95 𝑐 2 − 155625 𝑐 −11250000 = 0 𝑐1 = 27,312 𝑚𝑚 𝑐2 = −19,823 𝑚𝑚



Kontrol asumsi awal 𝐸𝑠 =



𝑓𝑦 𝐸



𝜀′𝑠 = ( 𝜀𝑠 = (



500



= 200000 = 0,0025



𝑐−𝑑′



27,312−25



𝑐



27,312



) 𝑥 0,003 = (



𝑑−𝑑′



65−25



𝑐



27,312



) 𝑥 0,003 = (



) 𝑥 0,003 = 0,0003 < 0,002 (belum leleh)



) 𝑥 0,003 = 0,0044 > 0,002 (sudah leleh)



Momen nominal terpasang 𝐶𝑐 = 20779,95 𝑐 N 𝐶𝑐 = 20779,95 𝑥 27,312 = 567541,994 N 𝐶𝑠 = 450.000 (



27,312 −25 ) 27,312



= 38093,146 N



a = β x c = 0,843 x 27,312



= 23,024 m



𝑎



𝑀𝑝𝑟1 = 𝐶𝑐 (𝑑 − ) + 𝐶𝑠 (𝑑 − 𝑑′ ) 2



𝑀𝑝𝑟1 = 567541,994 (65 −



23,024 2



) + 38093,146 (65 − 25)



Mpr1 = 31880412,02 Nmm = 31,880 kNm ϕMpr1 = 0,9 x 31,880 = 28,692 kNm 𝜙𝑀𝑝𝑟1 > 𝑀𝑢 𝟐𝟖, 𝟔𝟗𝟐 kNm > 6,308 kNm (OK)



Menghitung Mpr Pada Balok Momen Positif Tulangan yang digunakan D10, ntarik = (As = 750mm2) ntekan = 6D10 (As’ = 471 mm2) Cc+Cs = Ts Asumsi awal tulangan tekan belum leleh dan tulangan tarik sudah leleh sehingga dapat diperoleh persamaan sebagai berikut: 𝑐 − 𝑑′ 𝜀′𝑠 = ( ) 𝑥 0,003 𝑐



𝑑 − 𝑑′ 𝜀𝑠 = ( ) 𝑥 0,003 𝑐 𝜀𝑦 =



𝑓𝑦 500 = = 0,0025 𝐸 200000



Kontrol Kekuatan 𝑐 − 𝑑′ 𝐶𝑠 = 𝐴𝑠′ 𝐸𝑠 ( ) 𝑥 0,003 𝑐 𝐶𝑠 = 471 𝑥 200000 𝑥 ( 𝐶𝑠 = 282600 (



𝑐 − 𝑑′ ) 𝑥 0,003 𝑐



𝑐−25 𝑐



)N



𝐶𝑐 = 0,85 𝑓 ′ 𝑐 β c 𝑏 𝐶𝑐 = 0,85 𝑥 29 𝑥 0,843 𝑥 𝑐 𝑥 1000 𝐶𝑐 = 20779,95 𝑐 N 𝑇𝑠 = 𝐴𝑠 𝑓𝑦 𝑇𝑠 = 1,25 𝑥 750 𝑥 500 𝑇𝑠 = 468750 N



Keseimbangan gaya 𝐶𝑐 + 𝐶𝑠 = 𝑇𝑠 20779,95 𝑐 + 282600 (



𝑐−25 𝑐



) = 468750



20779,95 𝑐 2 + 282600 𝑐 − 7065000 = 468750 𝑐 20779,95 𝑐 2 − 186150 𝑐 − 7065000 = 0 𝑐1 = 25,454 𝑚𝑚



𝑐2 = −14,496 𝑚𝑚 Kontrol asumsi awal 𝐸𝑠 =



𝑓𝑦 𝐸



500



= 200000 = 0,0025



𝑐−𝑑′



𝜀 ′𝑠 = ( 𝜀𝑠 = (



𝑐



25,454 −25



) 𝑥 0,003 = (



𝑑−𝑑′ 𝑐



25,454



) 𝑥 0,003 = 0,001 < 0,002 (belum leleh)



65−25



) 𝑥 0,003 = (23,454 ) 𝑥 0,003 = 0,0051 > 0,002 (sudah leleh)



Momen nominal terpasang 𝐶𝑐 = 20779,95 𝑐 N 𝐶𝑐 = 20779,95 𝑥 25,454 = 528932,847 N 𝐶𝑠 = 282600 (



25,454−25 ) 25,454



= 5040,481 N



a = β x c = 0,843 x 25,454 = 21,458 m 𝑎



𝑀𝑝𝑟2 = 𝐶𝑐 (𝑑 − ) + 𝐶𝑠 (𝑑 − 𝑑′ ) 2



𝑀𝑝𝑟2 = 528932,847 (65 −



21,458 2



) + 5040,481 (65 − 25)



Mpr2 = 28907333,78 Nmm = 28,907 kNm ϕMpr2 = 0,9 x 28,907 = 26,016 kNm 𝜙𝑀𝑝𝑟2 > 𝑀𝑢 𝟐𝟔, 𝟎𝟏𝟔 kNm > 3,934 kNm (OK)