PELAT CANGKANG KELOMPOK 1
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PELAT CANGKANG
KELOMPOK 1
Soal1.Turunkan dengan seksama dari mana persamaan pelat
2.Pelat, tebal 12 cm, Fc = 30 Mpa, beban hidup 250 kg/m2. Hitung dan gambarkan Mx(max), My(max) dengan Wmax, dengan teori Hirschfeld, Navier, M. Levy, Stiglat/Wipper dan Pieper/Martins. Berikan masukan dan kesimpulan. Jika diketahui Ly = 4 m, Lx = 5 m dan kondisi perletakan sederhana
3.Soal 2, jika perletakan seluruh sisinya jepit. Hitung dan gambarkan Mx(max), My(max) dengan Wmax, dengan teori Hirschfeld, Stiglat/Wipper, dan Pieper /Martins. Berikan Masukan dan kesimpulan.
Dq
yw
yxw
xw
4
4
22
4
4
4
LyLx
HIRSCHFELD [sederhana]
8.054
LxLy
291,08.018.0
1 4
4
4
4
k
Beban Mati (WD)Berat sendiri pelat = 0,12 x 24 =
2,88 kN/m2
Beban Hidup (WL)WL = 250 Kg/m2 = 2,50 kN/m2
Wu = WD + WL = 2,88 + 2,5 = 5,38 kN/m2 = 538 kg/m2
4633 10144)120)(1000(121
121 mmxbhI
MPafcE 500.23254700'4700
Arah yPy = (1 – K)p = (1 – 0,291) 538 = 381,442 kg/m
My
kgm
xLyPyM y
884,762
)4(442,38181
81 22
Arah XPx = K x p = 0,291 x 538 = 156,558
kg/m
Mx
kgm
xLxPxM x
244,489
)5(558,15681
81 22
Lx = 5 m
Mx = 489,244 kgm
Ly = 4 m 762,884
kgm
TEORI HIRSCHFELD
Lendutan
mmmxx
EIWuL
299,5005299,01044,1)1023500(
)4(53838453845
45
4
4
mmL 33,84804000
480
!!.33,8299,5 KOmmmm
Lendutan
mmmxx
EIWuL
06,100106,01044,1)10500.23(
)4(53838413841
45
4
4
mmL 33,84804000
480
!!.33,806,1 KOmmmm
METODE NAVIER• Elastisitas Beton
• Kekakuan Pelat
292 /1029,3/35,32863306000 mkgmmNE
kgmv
EhD 500.493)2,01(12)12,0(1029,3
)1(12 2
39
2
3
METODE NAVIER• Lendutan maksimum
1,..
sin.sin162
2
2
2
260
nm
bn
amnm
byn
axm
DqWmaks
2
2
2
2
26
41
51.1.1
4)2(1sin.5
)5,2(1sin)500.493(
53816
Wmaks
mmmWmaks 732,1001732,0
METODE NAVIER• Momen Maksimum
Mmax (arah x)
06,873)14,3()538(16
1,)(16
2
2
amn
nmmnfamn o
kgmkgmMxMx
bynaxmamnbnamDMx
6
222
222
102237,099.023.223)4/)2(1sin().5/)5,2(1sin().06,873.()4/1(2,0)5/1()500.493)(14,3(
)/(sin)./(sin.)/()/(
METODE NAVIER• Mmax (arah y)
kgmkgmMyMy
bynaxmamnnmbnDMy
6
222
222
10115.1499.115.115.1)4/)2(1sin().5/)5,2(1sin().06,873.()1/1(2,0)4/1()500.493)(14,3(
)/(sin)./(sin.)/()/(
Ly = 4 m 1.115 x 10^6
kgmLx = 5 m
Mx = 223 x 10^6 kgm
METODE NAVIER
METODE M.LEVY• Elastisitas Beton
• Kekakuan Pelat
292 /1029,3/35,32863306000 mkgmmNE
kgmv
EhD 500.493)2,01(12)12,0(1029,3
)1(12 2
39
2
3
METODE M.LEVY• Lendutan maksimum
1cosh22..)1(4
3845
5
21
5
44
mm
mtghmmD
qaD
qaWmaksm
mmmWmaks
mmtghWmaks
65,11065,110215,710872,8
256,152)4(
256,1cosh22256,1.256,1.1
)1(493500
5)538(4)493500(384
)5)(538(5
333
5
211
5
44
METODE M.LEVY• Momen Maksimum• Mmax (arah x)
axmAmBmmqaxaqxMx
m
sinh)1(22)(
5,3,1
222
0106,0256,1cosh114,3)2256,1tanh256,1(2
cosh_2tanh(2
5555
mm
mmAm
00345,0256,1cosh114,32
cosh2
5555 mm
Bm
METODE M.LEVY• Mmax (arah x)
• Mmax (arah y)
kgmMx
Mx
15,38469,216425,168155,21sinh0106,0)2,01(00345,0)2,0(2)1(14,3)5(5382
)5,25)(5,2)(538( 222
axmAmBmmqaxaqxMy
m
sinh)1(22)(
5,3,1
222
kgmMy
My
1,27863,30012,215421sinh0106,0)2,01()00345,0(2)1(14,3)4)(538(2
)24(2)538(2,0 222
METODE M.LEVY
Lx = 5 m
Mx = 3846,15kgm
Ly = 4 m -
2786,1kgm
METODE STIGLAT/WIPPEL • Momen Maksimum (perletakan sederhana)2,0
25,145
lylx
erpolasiyfxf int55,37
35,22
yfxflylxpMx
1..max
METODE STIGLAT/WIPPEL • Momen Maksimum (arah x)
• Momen Maksimum (arah y)
kgmMx 74,53855,372,0
35,221)4).(5(538max
kgmMy 84,38235,222,0
55,371)4).(5(538max
METODE STIGLAT/WIPPEL
Ly = 4 m
382,84 kgm
Lx = 5 m
Mx = 538,74 kgm
METODE PIEPER/MARTINMomen Lapanganberdasarkan tabel dan hasil interpolasi untuk Ly/Lx = 1,25 di peroleh fx = 17,95dan fy = 30.Maka, Kgm
fLqMx
xxf 303,74995,17
)5(.538. 22
KgmfLq
My
yyf 933,28630
)4(.538. 22
METODE PIEPER/MARTIN
Lx = 5 m
Mx = 749,303kgm
Ly = 4 m 286,933
kgm
HIRSCHFELD JEPIT - JEPIT
8.054
LxLy
291,08.018.0
1 4
4
4
4
k
Beban Mati (WD)Berat sendiri pelat = 0,12 x 24 =
2,88 kN/m2
Beban Hidup (WL)WL = 250 Kg/m2 = 2,50 kN/m2
Wu = WD + WL = 2,88 + 2,5 = 5,38 kN/m2 = 538 kg/m2
4633 10144)120)(1000(121
121 mmxbhI
MPafcE 500.23254700'4700
Arah yPy = (1 – K)p = (1 – 0,291) 538 = 381,442 kg/m
My
kgm
xLyPyM jepit
589,508
)4(442,381121
121 22
kgm
xLyPyM lapangan
295,254
)4(442,381241
241 22
Ly = 4 m My = 254,295 kgm
Mx = 508,589kgm
TEORI HIRSCHFELD
Arah XPx = K x p = 0,291 x 538 = 156,558
kg/m
Mx
kgm
xLxPxM jepit
162,326
)5(558,156121
121 22
kgm
xLxPxM lapangan
081,163
)5(558,156241
241 22
Lx = 5m
Mx = 163,081 kgm
Mx = 326,162 kgm
TEORI HIRSCHFELD
METODE STIGLAT/WIPPEL• Momen Lapangan Arah x• Dari tabel diperoleh αxf =46,3; αyf =87,3
• Momen Lapangan Arah y
2,0
kgmMxlapangan 05,2573,872,0
3,461)4).(5(538max
kgmMylapangan 73,1693,462,0
3,871)4).(5(538max
METODE STIGLAT/WIPPEL• Momen Tumpuan Arah y• Dari tabel diperoleh αxs =-18,7; αys =-22,2
• Momen Tumpuan Arah y
kgmMxtumpuan 48,1542,222,0
7,181)4).(5(538max
kgmMytumpuan 76,5997,182,0
2,221)4).(5(538max
METODE STIGLAT/WIPPEL
Ly = 4 m My = 169,73kgm
My =599,76 kgm
METODE STIGLAT/WIPPEL
Lx = 5m
Mx = 257,05 kgm
Mx = 154,48 kgm
METODE PIEPER/MARTINMomen Lapanganberdasarkan tabel dan hasil interpolasi untuk Ly/Lx = 1,25 di peroleh fx = 24,2 dan fy = 41,95
Maka,Kgm
fLqMx
xxf 785,5552,24
)5(.538. 22
KgmfLq
My
yyf 197,20595,41
)4(.538. 22
METODE PIEPER/MARTINMomen Tumpuanberdasarkan tabel dan hasil interpolasi untuk Ly/Lx = 1,25 di peroleh sx = 15 dan sy = 17,75Maka,
KgmsLqMx
xxs 667,89615
)5(.538. 22
KgmsLq
My
yys 958,48475,17
)4(.538. 22
METODE PIEPER/MARTIN
Ly = 4 m My = 205,197kgm
My =484,958 kgm
METODE PIEPER/MARTIN
Lx = 5m
Mx = 555,785 kgm
Mx = 896,667 kgm
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