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Nuriaman 060404108
Tugas NIM : 060404108
L
L
L
Dimana :
L = 3,75 mP = 12,5 KN
Profil WF 600 X 300 dimana h = 600 mm, t = 30 mm, b = 300 mm, s=300mm.
G = 81.000N/mm2 , E = 210.000N/mm2
Ditannya :
a. Gambarkan diagram Momen Sekunder Ms dan Momen Primer Mp pada balok A, B , dan C
b. Gambarkan diagram tegangan :- Akibat Torsi pada A, B, C- Akibat lintang pada A, B, C- Akibat Momoen Lentur pada A, B, C
Penyelesaian :
Tugas analisa struktur lanjutan
h
b
t
s
1
PA
D C
B
Nuriaman 060404108
a. Diagram Ms dan Mp
Mt = 46,875 Tm
Mt A = = 23,4375 Tm
Mt B = 23,4375 Tm
Momen Torsi Mt = 23,4375 tm = 2.343.750 kgcm L = 7,5 m = 750 cm
Profil yang digunakan 600 X 300 dimana h = 600 mm, t = 30 mm, b = 300 mm
MT = Ms+ Mp
Tugas analisa struktur lanjutan
h
b
t
s
2
Ms
Mp
MT
z
+
_-
Gambar 1: Bidang Torsi (Momen Sekunder dan Momen Primer)
Ms : Momen Sekunder
Mp : Momen Primer
A
C
MT
L LZ
Y
X
Nuriaman 060404108
PERSAMAAN TORSI PADA TAMPANG I
ν ″′ - λ² . ν ′ = - dimana λ² =
Cw = Iy . = Konstanta warping
G = 81.000N/mm2 = 810.000 Kg/cm2
E = 210.000N/mm2 = 2.100.000 Kg/cm2
Menghitung Cw
Iy = 13.872 cm4
h = 57 cm
Cw = 11.267.496 cm6
Menghitung Jt = 3 cmb = 30 cmh-b = 54 cms = 3 cm
J = 1026 cm4
λ = 0.00600085
Kondisi perletakan :
Penyelesaian umum :
ν = A . Sinh λZ + B . Cosh λZ + C + . Z
ν = λZ -
Tugas analisa struktur lanjutan 3
MT
E . CwGJ
E.Cw
h2
4
MT
GJ
MT
GJλ
Sinh λZ
Cosh λL2
Nuriaman 060404108
maka
ν = λZ -
Akibat M1 : (Tegangan Warping)
M1 = - E y . . ν ″′ P = -EIy . . ν ″′.
= - E Cw . ν ″′
Q = P = - E Iy . . ν ″′
P = lintang
τ w = Q = -E Iy . . ν ″′
S = A . X
= .
=
I = ½ Iy
. ν ″′maka : τ w =
= -E . . ν ″′
dimana ν ″′ = .
Tugas analisa struktur lanjutan 4
h2
4h2
41h
b
tf
b4
x =
M1
h
h
h4
Q . Sb . I
h4
b . tf
2b4
b2 tf
8
- E Iy . h4
b2 tf
8tf . ½ Iy
b2 . h16
MT . λ2
GJ- Cosh λx
Cosh λL2
2343750GJλ
Sinh λZ
CoshλL 2
Nuriaman 060404108
Z λZ Cosh λ² Mt Cosh λL/20 0 1 2.14E-05 2343750 4.798175 0.45006375 1.103 2.14E-05 2343750 4.7981150 0.9001275 1.433 2.14E-05 2343750 4.7981225 1.35019125 2.059 2.14E-05 2343750 4.7981300 1.800255 3.108 2.14E-05 2343750 4.7981375 2.25031875 4.7981 2.14E-05 2343750 4.7981300 1.800255 3.108 2.14E-05 2343750 4.7981225 1.35019125 2.059 2.14E-05 2343750 4.7981150 0.9001275 1.433 2.14E-05 2343750 4.798175 0.45006375 1.103 2.14E-05 2343750 4.79810 0 1 2.14E-05 2343750 4.7981
GJ v"' Ms Mp
831060000 2.06441E-08 488474.5955 1855275.404
831060000 2.27704E-08 538787.4789 1804962.521
831060000 2.95829E-08 699984.0954 1643765.905
831060000 4.25061E-08 1005769.192 1337980.808
831060000 6.41618E-08 1518179.043 825570.957
831060000 9.90523E-08 2343750 0
831060000 6.41618E-08 1518179.043 825570.957
831060000 4.25061E-08 1005769.192 1337980.808
831060000 2.95829E-08 699984.0954 1643765.905
831060000 2.27704E-08 538787.4789 1804962.521
831060000 2.06441E-08 488474.5955 1855275.404
Gambar Ms dan Mp
( –)
( + )
Tugas analisa struktur lanjutan 5
0 22515075 300
375 Mp
Mp
Ms Mp = 1855275.404 Ms = 2343750
Nuriaman 060404108
b.1. Gambar tegangan akibat Torsi
Diagram tegangan τW
τ w = -E . . ν ″′ = 139 kg/cm2
τw
b.2. Gambar tegangan akibat Momen Lentur
Tegangan lentur (σW)
σ = =
σ =
u = v .
=
Mf = - E If .
= - E If . .
= - E If . . ν ″
Mf = - . ν ″
Tugas analisa struktur lanjutan 6
b2 . h16
M1
h Mf
WMf
If /X
Mf . XIf
h2
d2 udx2
- Mf
E If
d2 udx2
h2
d2 νdx2
h2
E . Cwh
Nuriaman 060404108
σmax pada x =
Mf .σmax =
If
- E If . . ν ″. =
If
σmax = - . ν ″
dimana ν ″ = -
z = 375 cm ( di tengah Bentang)
ν ″ =
ν ″ = 1.7214 x 10-6
Gambar diagram Tegangan Lentur (σw)
σmax = - . ν ″ = 1.545,373 kg/cm2
Tugas analisa struktur lanjutan 7
b2
b2
h2
b2
E b h4
Mt
G . Jλ Sinh λz
Cosh λL
2
E b h4
Nuriaman 060404108
b.3. Gambar tegangan akibat Lintang
Akibat M2 :
τ = dimana M2 = G J. ν ′
maka : τ = = G. ν ′.t
dimana ν′ = 1 -
z = 0 cm
ν′ =
ν′ = 2,291
Diagram τ :
Tugas analisa struktur lanjutan 8
M2 . tJ
( G J. ν ′ ) tJ
MT
GCosh λx
Cosh λL /2
τ = G. ν ′.t = 55659375 Kg/cm2
Nuriaman 060404108
Tugas analisa struktur lanjutan 9