# Analisa Pelat Dan Cangkang

• January 2021
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Analisa Pelat dan Cangkang Perbandingan Metode Navier dan Levy Nama Kelompok : 1. Rahmat Sidik Irjali (13.11.1001.7311.229) 2. Abdul Gapur (13.11.1001.7311.357) 3. M.Wahyudi (13.11.1001.7311.049)

Data Konstruksi : • Pelat, tebal 12 cm, Fc = 30 Mpa, beban hidup 250 kg/m2. Hitung dan gambarkan Mx(max), My(max) dengan Wmax, dengan teori Navier, M. Levy. Jika diketahui Ly = 4 m, Lx = 5,6 m dan kondisi perletakan sederhana

METODE NAVIER • Elastisitas Beton E  6000 30  32863,35 N / mm 2  3,29  10 9 kg / m 2

• Kekakuan Pelat Eh 3 3,29 109  (0,12) 3 D   493.500 kgm 2 2 12(1  v ) 12(1  0,2 )

METODE NAVIER • Lendutan maksimum

mx ny sin . sin 16q0 a b  m, n  1 Wmaks  6 2 2 2  D m n  m.n.  2  2  b  a Wmaks 

Wmaks = 0,0074682 m = 7,468 m

16  538  6 (493.500)

1 (2,8) 1 (2) sin . sin 5,6 4  1 1   1.1.  2  2  4   5,6 2

2

2

METODE NAVIER • Momen Maksimum 16( f o ) amn  2  m, n  1,4  mn 16(538) amn   445 2 (3,14) .1,4.1,4

METODE NAVIER • Mmax (arah x)

Mx   2 D (m / a ) 2   (n / b) 2 . amn sin (mx / a ). sin (ny / b)

Mx  (3,14 2 )(493.500) (1,4 / 5,6) 2  0,2 (1,4 / 4) 2 .(445). sin(1,4 (2,8) / 5,6). sin(1,4 (2) / 4) Mx  88379,92kgm

• Mmax (arah y)

My   2 D (n / b) 2   (m / n) 2 . amn sin (mx / a ). sin (ny / b)

My  (3,14 2 )(493.500) (1,4 / 4) 2  0,2 (1,4 / 1,4) 2 .(445). sin(1,4 (2,8) / 5,6). sin(1 (2) / 4) My  32761,52kgm

METODE NAVIER

Ly = 4 m

My = 32761,52 kgm

Lx = 5,6 m

Mx = 88379,92 kgm

METODE M.LEVY • Elastisitas Beton E  6000 30  32863,35 N / mm 2  3,29  10 9 kg / m 2

• Kekakuan Pelat Eh 3 3,29  10 9  (0,12) 3 D   493.500 kgm 2 2 12(1  v ) 12(1  0,2 )

METODE M.LEVY • Lendutan maksimum

m 1 2

5qa 4 4qa 4 (1) Wmaks   5 384 D  D m 5 Wmaks 

m.tghm  2 .  m  1,4 2 cosh m 1, 4 1 2

5(538)(5) 4(538)5 (1)  5 384(493500)  493500 1,45 4

4

.

1,222.tgh 1,222  2 m (4)  m   1,222 2 cosh 1,222 2  5,6

Wmaks  8,472 10 3  6,922 10 3  1,55 10 3 m  1,55 mm

METODE M.LEVY • Momen Maksimum • Mmax (arah x)

qx(a  x) mx  qa 2 2  m 2 2Bm  (1   ) Amsinh 2 a m 1, 3, 5 2(m tanh m  2 _ 2(1,222 tanh 1,222  2) Am    0,00315 5 5 5 5  m cosh m 3,14 1,4 cosh 1,222 2 2 Bm  5 5   0,00122 5 5  m cosh m 3,14 1 cosh 1,222

Mx 

METODE M.LEVY • Mmax (arah x) (538)(2,8)(5,6  2,8) 1,4 2,8  538(5,6) 2 3,14 2 (1,4 2 )2(0,2)0,00121  (1  0,2)0,00315sinh 2 5,6 Mx  2108,96  2153,172  4262,132 kgm

Mx 

• Mmax (arah y) (538)2(4  2) 1,4 2  (538)(4) 2 3,14 2 (1,4 2 )2(0,00121)  (1  0,2)0,00315sinh 2 4 My  421,73  3017,412  2595,62 kgm My  0,2

METODE M.LEVY

Lx = 5,6 m

Mx = 4262,132 kgm

Ly = 4 m

-2595,62 kgm

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