Perhitungan Beban 1.14

TUGAS BESAR BAJA 2009

Perhitungan Beban Dengan jarak gording 1.14 m

1. Beban Mati

Gording kuda-kuda bentang 20,00 m

Jarak antar gording ( l )

Jarak antar gording: 1.14 m.Berat penutup atap (Seng Gelombang): 10 kg/m2. (tebal : 20 mm, panjang:180 cm)

Berat gording diperkirakan 9.20 kg/m.

Berat atap, (1.14).(10) = 11.4 kg/m

Berat gording, = 9.20 kg/m +

qD = 20.6 kg/m

= 20.6 kg/m

qD = 20.6 kg/m=> RDx = = = 25.75 kg

RDy = = = 51.5 kg

RDx = (25.75 ).cos 30° = 22.30 kg

RDy = (51.5 ).sin 30° = 25.75 kg

qD = 20.6 kg/m => MDx = = =32.19

kgm

MDy = = = 64.38 kgm

MDx = (32.19).cos 30° = 27.88 kgm

MDy = (64.38).sin 30° =32.19 kgm

2. Beban Hidup

Berat pekerja = 100 kg

PL = 100 kg => RL = = = 50 kg

RLx = (50).cos 30° = 43,30 kg

RLy = (50).sin 30° = 25.00 kg

PL = 100 kg => MLx = = = 62.5 kgm

MLy = = = 125 kgm

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MLx = (62.5).cos 30° = 54.12 kgm

MLy = (125).sin 30° = 62,5 kgm

3. Beban Angin

Angin tekan, q = {(0,02).α – (0,4)}.p

= {(0,02).(30) – (0,4)}.(35).(1.14)

= 7.98 kg/m

q = 7.98 kg/m => Rw = = = 19.95 kg

Mw = = = 24.94 kgm

Angin hisap, q = (- 0,4).p

= (- 0,4).(35).(1.14)

= -15.96 kg/m

q = -15.96 kg/m=> Rw = = = -39.9 kg

Mw = = = - 49.87 kgm

Kombinasi Pembebanan

A. Pembebanan Sementara

1. Arah Tegak Lurus Bidang Atap

RUx1 = (1,2).(RDx) + (0,5).(RLx) + (1,3).(Rw)

= (1,2).(22.30) + (0,5).(43,30) + (1,3).(19.95)

= 74.34 kg

RUx2 = (1,2).(RDx) + (0,5).(RLx) + (1,3).(Rw)

= (1,2).( 22.30) + (0,5).(43.30) + (1,3).(-39.9)

= -3.46 kg

2. Arah Sejajar Bidang Atap

RUy = (1,2).(RDy) + (0,5).(RLy)

= (1,2).(25.75) + (0,5).(25.00)

= 51.12 kg

B. Pembebanan Tetap

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RUx = (1,2).(RDx) + (0,5).(RLx)

= (1,2).(22.30) + (0,5).(43.30)

= 48.41 kg


RUy = (1,2).(RDy) + (0,5).(RLy)

= (1,2).(25.75) + (0,5).(25.00)

= 43.4kg

Jadi tiap gording menerima beban sebesar,

arah tegak lurus bidang atap = 1.14.( 48.41) = 55.19 kg

arah sejajar bidang atap = 1.14.( 43.4) = 49.48 kg

Kombinasi Momen

A. Momen Akibat Beban Sementara


MUx1 = (1,2).(MDx) + (0,5).(MLx) + (1,3).(Mw)

= (1,2).(27.88) + (0,5).(54.12) + (1,3).(24.94)

= 92.94 kgm

MUx2 = (1,2).(MDx) + (0,5).(MLx) + (1,3).(Mw)

= (1,2).( 27.88) + (0,5).( 54.12) + (1,3).(-49.87)

= -4.3 kgm


MUy= (1,2).(MDy) + (0,5).(MLy)

= (1,2).( 32.19) + (0,5).(62.5)

= 69.9 kgm

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B. Momen Akibat Beban Tetap


MUx= (1,2).(MDx) + (0,5).(MLx)

= (1,2).( 27.88) + (0,5).(54.12)

= 60.52 kgm


MUy= (1,2).(MDy) + (0,5).(MLy)

= (1,2).( 32.19) + (0,5).(62.5)

= 69.9 kgm

Dipasang sagrod di tengah bentang sehingga,

MUy= .(69.9)

= 34.95 kgm

Tabel hasil perhitungan kombinasi momen dan reaksi.

Keterangan Beban TetapBeban Sementara

Angin Tekan Angin Hisap

Momen

(Kgm)

Mx

My

60.52

69.9

92.94 -4.3

69.9

Reaksi

(Kg)

Rx

Ry

48.41

43.4

74.34 -3.46

43.3

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Tegangan Pada Profil C

Dari tabel profil kanal 150 x 50 x 20 x 3,2Ix = 280 cm4

Iy = 28 cm3

Zx = 37,4 cm3

Zy = 8,2 cm3

Lendutan Pada Profil C

∆max =

=

= 2,08 cm

qD = 20.6 kg/m=> qDx = (20.6).cos 30° = 17.84 kg/m

qDy = (20.6).sin 30° = 10.3 kg/m

PD = 100 kg => PDx = (100).cos 30° = 86.60 kg

PDy = (100).sin 30° = 50kg

Lendutan terhadap sumbu x,

qx = (1,2).qDx = (1,2).( 17.84) = 21.41 kg/m

Px = (1,2).PDx = (1,2).( 86.60) = 103.92 kg

∆x = +

=

= 0,06 + 0.46

= 0.52 cm

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Lendutan terhadap sumbu y,

qy = (1,2).qDy = (1,2).(10.30) = 12.36 kg/m

Py = (1,2).PDy = (1,2).(50.00) = 60.00 kg

∆y = +

=

= 0,021 + 0,33

= 0,35cm

∆ =

=

= 0.62 cm < ∆max = 2,08 cm

Jadi gording dengan profil C 150x50x20, dengan tebal 3,2 mm dapat digunakan karena telah

memenuhi persyaratan.

q = 6.76 kg/m

t = 3,2 mm Cy = 1,54 cm

A = 8.61 cm2 Xo = 3.77 cm

Ix = 280 cm4 J = 2938 cm4

Iy = 28 cm4 Cw = 1398m6

Zx = 37,4 cm3

Zy = 8.2 cm3

ix = 5,71 cm

iy = 1,81 cm

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Perhitungan Beban 1.14