Pipe support & restraint introduction. This PPT explaine the variation of pipe suuport and it's function.

1. Pipe support and
restraint
By : Kurniawan

2. Pipe support
Function
• Prevent stresses in excess of permitted stresses.
• Prevent joint leakage.
• Prevent excess forces on equipment to which pipe is
connected.
• Prevent excess stresses in the elements of the support system.
• Avoid resonance from vibrations.
• Avoid interference from the thermal movement with other
pipes or structures.

3. Pipe support
• Anchor • support

4. Pipe support
• Hanger • restraint

5. Basic concept
Jika :
F = 10 N
L = 20 m
A = 17
B = 3
Berapakah Fa dan Fb ?
Jawab :
F = Fa + Fb => 10 ton = Fa + Fb. Sehingga
Fb = 10 N – Fa ….(1)
∑Mb = 0, maka
∑Mb = F.b – Fa.20 => 0 = 10 N.3 – Fa.20
atau 30 N = 20Fa;
sehingga Fa = 1.5 N dan Fb = 8.5 N
• berat pipa
• Insulation
• Fluida
• Tekanan
• Peralatan (valve, flange,
dll)

6. Basic concept
Jika :
F = 45KN
L = 3m
a = 2m
b = 1m
EI = 4.16.10^5 Nm2
Berapakah Defleksi
maks (∆max)?
Jawab :

7. Application on pipe system
• Given: A 6 in Sch 40
carbon steel pipe, 10
ft long, anchored at
both ends
• Find: The amount of
force generated by
the expansion of the
pipe from 70°F to
250°F

8. Application on pipe system
• F = Force [lb]
• A = Metal cross-sectional area [in2]
5.581 in2 for 6 in Sch 40 pipe
• E = Young’s Modulus 30 106 lb/in2
• ε =Strain [in/in]
• σ = Stress [lb/in2]
• δ = Displacement due to thermal
growth [in]
• L = Length of pipe [in]
• A = Coefficient of Linear Expansion =
0.000 0065 in/in/°F for carbon steel
• The thermal growth is given by
• L(T) Equation 9.1
• (0.0000065 in/in/°F) (120 in) (250°F
70°F) 0.14 in
Solution ..

9. Span for pipe
• For uniform load • For non uniform load

10. Application on pipe system
• L= [y.E.I / 22.5.W]¼
L = pipe support spacing, feet,
y = permissible mid-span deflection,
inches
E = modulus of elasticity at design
temperature, lb/in (TABLE C-6(
I = moment of inertia of pipe.
W = weight of supported pipe, including
pipe, contents, insulation, lb/ft
• What is the span of a seamless
ASTM A106 Grade B, 6.625 inch
OD, 0.28 inch wall thick, water
filled pipe with 3 inch of insulation
with a design temperature of 400
F? The specifications limit the mid-
span deflection to 0.25 inch.
Solution:
Pipe = 19.0 lbs per ft
Water = 12.5 lbs per ft
Insulation = 7.6 lbs per ft ( 85 %
Magnesia Calcium
Silicate)
then ,W = 39.1 lb per ft
I = ( π /64)(Do
4 – Di
4)
Do = 6.625, Di = 6.065
I = 28.14 in4
E = 27.7 x 106 psi, Table C-6, C ≤
0.3 at 400°F.
finally, L = [ 0.25x27.7x106 x28.14/(17.
1x39. 1)]1/4
L = 23 feet span
The pipe support spacing would be 23
feet with a mid span deflection of 1/4
inch.

11. ANALYSISOF PIPING SYSTEMSRESTING
ON SUPPORTS

12. Support Selection & Design
• Basic information has to collected before proceeding with
calculations and detailing of pipe supports, as follows;
• A complete set of piping drawings
• A complete set of steel and structural drawings/ data.
• A complete set of drawings showing locations of ventilating ducts, electrical
cable trays, equipment locations (pumps, tanks, etc)
• A complete set of piping specifications and data.
• Insulation specification.
• Movement of all critical equipment connections such as boiler headers, steam
drums, turbine connections, etc..
• The results of stress, flexibility, and movement calculations performed for
critical systems.

13. Spring/hanger
Variable Spring Hanger
• Maksimum variation pada variable
spring hanger adalah 25%.
Constant Spring Hanger

14. Application on pipe system
load is calculated as 2500 lbs (1134 kg) at the
position where
• we need a hanger, and it has a movement of 1.2
in. (30.5 mm) at that location.
• We want the difference between the loads to be
less than 600 lbs (272 kg).
• We have a spring hanger that has an acceptable
range within our movement range. It has a spring
rate of 450 lbs/in. (10 kg/mm). Will
• the hanger meet the load difference
requirement?
a) 1.2 × 450 = 540 lbs−−meets the 600-lb
criteria
b) 30.5 × 10 = 350 kg−−does not meet the
600-lb criteria
Solution
a) (hot load/cold load).100 < 25 %; cold
load = 2500 lbs. then (540/2500 ).100=
22 %. OK
b) Because case “b“ doesn’t meet criteria
so the spring rate must be…?
We want the variablity 24% (less than 25%) so
(272/x) = 24%; x = 1134
x
x

15. Thank You