Introduction in the Design of Thrust block and Anchor blockFull description
Introduction in the Design of Thrust block and Anchor block
Thrust Block Design AWWA M45Descripción completa
How to position thrust blocks in a pipeline
thrust block excel ,AWWA M45,ACI
civil engineering and hydraulic works
Thrust-Block-Calculation
Thrust Block Design
Design of thrust block foundation for pipe lines
thrust block excel ,AWWA M45,ACIFull description
Thrust Block Design AWWA M45
civil engineering and hydraulic worksFull description
tata cara deign thrust blockDeskripsi lengkap
ign
Thrust Block Design sheet
Design and analysis of Thrust Block for pipelines Geo tecnical stability check for Anchor BlocksFull description
Design and analysis of Thrust Block for pipelines Geo tecnical stability check for Anchor Blocks
Thrust blocks for horizontal elbows designed for 90, 60, 45, 30 22.5, and 11.25 degrees angle of bend.
nbme 12
Full description
DESIGN OF THRUST BLOCKS #REF!
A) 150 DIA 90 DEGREE BEND Dia of pipe
D
=
Internal Pressure
P
=
Degree of deviation
θ
=
Area of cross section of Pipe
A
=
Thrust on Pipe
150 mm 4 kg/cm 90 176.71 sqcm
= 2*P*A*Sinθ/2
To
=
999 kg
tcb =
0.15 m
Concrete cover at top
tct =
0.45 m
Concrete cover to sides of pipe
tcs =
Concrete cover at bottom
Үw
Density of concrete
2
0.45 m
=
2500 kg/m
=
L
3
SOLUTION : Length of block
m
Width of block
B
=
1.05 m
Height of block
H
=
0.75 m
a) Weight of pipe
=
b) Weight of water
=
18 kg
c) Weight of gross concrete
=
1968.75 kg
d) Deduct portion of pipe Total weight per meter
W
100 kg
=
-44 kg
=
2131 kg
As the resultant is to lie within the kern zone of the base of block, X should be either equal to or greater than one third of base width The resultant of the force 'R' would cut the base at distance of X from A such that
X
=
L*W*b - To*h W
b
=
0.525 m
h
=
0.225 m
As the resultant is to lie within the kern zone of the base of the block, X should be either to or greater than 1/3 of the base width Therefore
X
= =
Therefore , length of block
L
=
B/3 0.350 m W*B/3 + To*h W*b
W*B/3 To*h W*b
=
746 kg
=
225 kg
=
1119 kg
=
0.87 m
L
X
B
X
H
0.90
X
1.05
X
0.75
Therefore provide block of size
B=
L
m
1.05
tct
D 0.75
To tcs h
tcb
A b
X
B) 150 DIA 45 DEGREE BEND Dia of pipe
D
=
Internal Pressure
P
=
Degree of deviation
θ
=
Area of cross section of Pipe
A
=
Thrust on Pipe
To
150 mm 4 kg/cm 45 176.71 sqcm
= 2*P*A*Sinθ/2 =
541 kg
tcb =
0.15 m
Concrete cover at top
tct =
0.45 m
Concrete cover to sides of pipe
tcs =
Concrete cover at bottom
Үw
Density of concrete
2
0.45 m
=
2500 kg/m
=
L
3
SOLUTION : Length of block
m
Width of block
B
=
1.05 m
Height of block
H
=
0.75 m
a) Weight of pipe
=
b) Weight of water
=
18 kg
c) Weight of gross concrete
=
1968.75 kg
d) Deduct portion of pipe Total weight per meter
W
100 kg
=
-44 kg
=
2131 kg
As the resultant is to lie within the kern zone of the base of block, X should be either equal to or greater than one third of base width The resultant of the force 'R' would cut the base at distance of X from A such that
X
=
L*W*b - To*h W
b
=
0.525 m
h
=
0.225 m
As the resultant is to lie within the kern zone of the base of the block, X should be either to or greater than 1/3 of the base width Therefore
X
= =
Therefore , length of block
L
=
B/3 0.350 m W*B/3 + To*h W*b
W*B/3 To*h
746 kg
=
122 kg
W*b
=
1119 kg
L
=
0.78 m
L
X
B
X
H
0.90
X
1.05
X
0.75
Therefore provide block of size
B=
=
m
1.05
tct
D 0.75
To
tcs h tcb
A b
X
B) 150 DIA 22 1/2 DEGREE BEND Dia of pipe
D
=
Internal Pressure
P
=
Degree of deviation
θ
=
Area of cross section of Pipe
A
=
Thrust on Pipe
To
150 mm 2 4 kg/cm
45 176.71 sqcm
= 2*P*A*Sinθ/2 =
541 kg
tcb =
0.15 m
Concrete cover at top
tct =
0.45 m
Concrete cover to sides of pipe
tcs =
0.45 m
Concrete cover at bottom
Үw
Density of concrete
=
2500 kg/m
=
L
3
SOLUTION : Length of block
m
Width of block
B
=
1.05 m
Height of block
H
=
0.75 m
a) Weight of pipe
=
b) Weight of water
=
18 kg
c) Weight of gross concrete
=
1968.75 kg
d) Deduct portion of pipe Total weight per meter
W
100 kg
=
-44 kg
=
2131 kg
As the resultant is to lie within the kern zone of the base of block, X should be either equal to or greater than one third of base width The resultant of the force 'R' would cut the base at distance of X from A such that
X
=
L*W*b - To*h W
b
=
0.525 m
h
=
0.225 m
As the resultant is to lie within the kern zone of the base of the block, X should be either to or greater than 1/3 of the base width Therefore
X
= =
Therefore , length of block
L
=
B/3 0.350 m W*B/3 + To*h W*b
W*B/3 To*h
Therefore provide block of size
B=
=
746 kg
=
122 kg
W*b
=
1119 kg
L
=
0.78 m
L
X
B
X
H
0.90
X
1.05
X
0.75
1.05
tct
D 0.75
To tcs
m
h
tcb
A b
X
1415
1550
1015
1150
160 1310
1000 200
250 1400
200
160 (OD) 150 SAND BEDDING
760
SAND BEDDING
CROSS SECTION OF TRENCH 160 (OD) H.D.P.E. PIPE LINE
CROSS SECTION OF TRENCH 250 (OD) H.D.P.E. PIPE LINE
1415
1550
1015
1150
160 1310
850
1000 200
250 1400
200
160 (OD) 150 SAND BEDDING
760 CROSS SECTION OF TRENCH 160 (OD) H.D.P.E. PIPE LINE
SAND BEDDING
850 CROSS SECTION OF TRENCH 250 (OD) H.D.P.E. PIPE LINE