T EGA EGAN NGA GAN N KRI RIT T IS & BAT BA T ASA SAN N BE BER RLA LAKU KU FO RM RMU ULA EULER Oleh : CHO CH O ER ERUR UR RO ROB BACH JURUS JUR USA AN T EK EKNI NIK K SI SIP IL UNIVERSITAS BRAWIJAYA
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BEBAN TEKUKKRIT ITIS IS
Beban tekuk kritis adalah beban maksimal yang dapat menyebabkan suatu kolom menekuk. Beban aksial kritis yang menyebabkan tekuk berkaitan dengan bentuk terdefleksi yang terjadi (atau ditentukan dari momen lentur PxΔ) Formula Euler untuk tekuk elastis: Pcr : Pcritical : Beban tekuk kritis
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Beban tekuk kritis tergantung pada: - panjang elemen struktur (L) - tahanan kedua ujung elemen struktur panjang efektif elemen struktur (Le = kL) - kekakuan elemen struktur (jenis material modulus elastisitas material/E) - bentuk serta ukuran penampang (momen inersia)
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TE T EGANGAN TE T EKU K KRIT KRITIS IS
Beban tekuk kritis untuk kolom dapat dinyatakan dalam tegangan tekuk kritis ( σcr) dengan cara membagi beban dengan luas penampang. Tegangan tekuk kritis ( jika lebih kecil dari tegangan aksial/normal ) :
P cr cr σ cr
=
A
π =
2
EI 2
AL
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BEBAN TEKUK(TEGAN ANG GAN TEKUK) vs PANJANG KOLO LOM M
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TEGAN ANG GAN LELEH & TEGAN ANG GAN TE TEKUK
Dua syarat desain kolom: Kolom tidak mengalami tekuk Kekuatan batas tidak dilampaui
Tergantung pada rasio kelangsingan kolom, keduanya saling mengontrol. Dalam kenyataannya kolom tidak
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BATAS BA TAS BER BER L AK AKU U FOR FORM M U L A EU EU LE LER R
“Formula Euler digunakan jika tegangan kritis kurang dari
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FORMULA EULER 2
P cr
π EImin =
2
AL
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K ONDI DISI SI TA TAHA HANA NAN N UJ UJUNG UNG vs v s PANJA PANJAN NG EFE EFEK K T I F d a n BE BEBA BAN N T EK UK K RITI RITIS S
Le = k xL Le : panjang efe efekti ktif f
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CONTOH SOAL Platform pemandnagan di sebuah taman satwa liar dipikul oleh sederetan kolom pipa aluminium yang mempunyai panjang L = 3,25 m dan diameter luas d = 100 mm.
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Solusi
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Dengan mensubstitusikan harga Pcr ini ke
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Sehingga kita dapat menghitung momen inersia, luas penampang dan radius girasi kolom tersebut
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Rasio kelangsingan L/r kolom ini kira-kira
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