masalah kontak sederhana suatu balok

masalah kontak sederhana (tanpa gesekan) dapat dianalisa dgn program bantu SAP2000, berikut ditinjau balok kantilever (kanan) yang menerima beban terpusat P. Akibat beban tersebut terjadi defleksi sebesar ~3.5cm, apabila jarak serat tepi bawah balok kanan dgn serat tepi atas balok kiri ada gap lebih kecil dari nilai diatas, maka balok kiri tersebut akan menerima gaya reaksi akibat kontak dari balok kanan.
.
tanpa peninjauan pengaruh kontak

(lendutan -Z)

.

(tegangan lentur S11)

.

tinjauan gap sebesar 0.0 cm

(lendutan -Z)


(tegangan lentur S11)

.
tinjauan gap sebesar 1.0 cm


(lendutan -Z)


(tegangan lentur S11)

.
tinjauan gap sebesar 2.0 cm


(lendutan -Z)


(tegangan lentur S11)

.

Diatas terlihat hasil sesuai dgn yg diprediksikan berdasarkan pendekatan perhitungan dibawah mungkin ini karena bidang kontak yg rata, agak berbeda saat meninjau kontak pane stress model lug/pin bidang kontak lengkung . Terlihat kontur yg tidak menerus pada lingkaran dalam (pin) dgn bidang kontak, perlu peninjaun ulang ini.

.

‘ Calculation pads
‘ ——————-

‘ Material props. (kgf,cm)
Es = 2000000
nu = 0.30
G = Es/(2*(1+nu)) = 769230.77
‘ Section props. (cm)
bs = 5
hs = 10
Is = 1/12*bs*hs^3 = 416.67
Ss = 1/6*bs*hs^2 = 83.33

‘ Cantilever beams (1)

‘ concentrate loads (kgf)
P = 1000
‘ beam length (cm)
l = 3*100
a = 1.5*100
b = a = 150.00
‘ Bending moment (kgf.cm)
M = P*b = 150000.00
‘ Shear force (kgf)
R = P = 1000.00
V = R = 1000.00
‘ flexural tress (kgf/cm^2)
Sigma = M/Ss = 1800.00
‘ deflection @ free end (cm)
Delta = ((P*b^2)/(6*Es*Is))*(3*l-b) = 3.37


‘ Cantilever beams (2)

‘ concentrate loads (kgf)
P = 1000
‘ beam length (cm)
l = 3*100
a = 1.5*100
b = a = 150.00
‘ Bending moment (kgf.cm)
M = 3/16*P*l = 56250.00
‘ Shear force (kgf)
R1 = 5/16*P = 312.50
R2 = 11/16*P = 687.50
V1 = R1 = 312.50
V2 = R2 = 687.50
‘ flexural tress (kgf/cm^2)
Sigma = M/Ss = 675.00
‘ deflection @ point load (cm)
Delta = (7*P*l^3)/(768*Es*Is) = 0.30

‘ overhang beams

‘ concentrate loads (kgf)
P = R1 = 312.50
‘ beam length (cm)
l = 3*100 = 300.00
a = 1.5*100 = 150.00
x1 = a = 150.00
x = l/2 = 150.00
‘ Bending moment (kgf.cm)
M1 = P*a = 46875.00
M2 = P*a*x/l = 23437.50
‘ Shear force (kgf)
R1 = P*a/l = 156.25
R2 = P/l*(l+a) = 468.75
‘ flexural tress (kgf/cm^2)
Sigma = M1/Ss = 562.50
‘ deflection @ free end (cm)
Delta = ((P*a^2)/(3*Es*Is))*(l+a) = 1.27

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