jacking force 2
DESCRIPTION
Calculation of jacking forces for pipe jacking techniqueTRANSCRIPT
Jackin force
the tunnel path will be devided through 5 partsPart (1,3,5) case (A)Part (2) case (B)Part (4) case (C)
V0
V0(A) 58.4 Pev(A) 6.23 Pv= 2V0(B) 24 Pev(B) 2.1V0( C) 86 Pev( C) 9.6
Peh(A) 2.94Peh(B) 1.288Peh( C) 4.288
V(A) 12.28206 L+V0(A)V(B) 5.924415 L+V0(B)V( C) 17.46977 L+V0( C)
Actual jacking force path
dist force0 V 0
22.3 V1 273.8947.1 V2 420.82
106.6 V3 1151.60115.6 V4 1308.83135.6 V5 1554.47
Critical jacking force path
dist force0 V 0
20 V1 245.6429 V2 402.87
88.5 V3 1133.65113.3 V4 1280.58135.5 V5 1554.47
after 50m where the resisting force is about 750 ton
V=V0+(m*(Pev+Pv+Peh/2)*D*p)L
p*Di2*P1/4
From the jacking force critical path the intermediate jacking station to be placed
0 20 40 60 80 100 120 140 160
0
200
400
600
800
1000
1200
1400
1600
1800
0
273.89420.82
1151.60
1308.83
1554.47
Actual Jacking force diagram
ditance(m)
forc
e(to
n)
2.4
7.5
10.9
7.57.56.0
22.3 24.8 59.59
20
-1.01
23 4 5
0 20 40 60 80 100 120 140 160
0
200
400
600
800
1000
1200
1400
1600
1800
0
245.64
402.87
1133.651280.58
1554.47
CriticalJackin force diagram
distance (m)
forc
e (t
on
)Engineer and Consultants
10 Samir Morsi Street-Block 6 – Nasr City- CairoTel/Fax: 2736393
0 20 40 60 80 100 120 140 160
0
200
400
600
800
1000
1200
1400
1600
1800
Jacking force
ditance(m)
forc
e(to
n)
0 20 40 60 80 100 120 140 160
0
200
400
600
800
1000
1200
1400
1600
1800
Jacking force
ditance(m)
forc
e(to
n)
2.4
7.5
10.9
7.57.56.0
22.3 24.8 59.59
20
-1.01
23 4 5