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Piping Elbows - Thrust Block Forces

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In a piping structure without adequately support fluid flow and internal pressure may create intolerable forces and tensions. 

The resultant force - or required support force - on a thrust block - or an anchor - for a bend depends on

  • the fluid mass flow, or flow velocity
  • the change of flow direction
  • the internal pressure

With no flow and no pressure there is no force.

Online Pipe Bend Resulting Force Calculator

The calculator below can used to calculate the resulting force in a piping bend:

Metric Units


Imperial Units


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Resulting force due to Mass flow and Flow Velocity

The resulting force in x-direction due to mass flow and flow velocity can be expressed as:

Rx = m v (1 - cosβ)                          (1)

    = ρ A v2 (1 - cosβ)                            (1b)

    = ρ π (d / 2)2 v2 (1 - cosβ)                           (1c)

where

Rx = resulting force in x-direction (N)

m = mass flow (kg/s)

v = flow velocity (m/s)

β = turning bend angle (degrees)

ρ = fluid density (kg/m3)

d = internal pipe or bend diameter (m)

π = 3.14...

The resulting force in y-direction due to mass flow and flow velocity can be expressed as:

Ry = m v sinβ                            (2)

    = ρ A v2 sinβ                            (2b)

    = ρ π (d / 2)2 v2 sinβ                           (2c)   

Ry = resulting force in y direction (N)

The resulting force on the bend due to force in x- and y-direction can be expressed as:

R = (Rx2 + Ry2)1/2                            (3)

where

R = resulting force on the bend (N)

.

Example - Resulting force on a bend due to mass flow and flow velocity

The resulting force on a 45o bend with

  • internal diameter 102 mm = 0.102 m
  • water with density 1000 kg/m3
  • flow velocity 20 m/s

can be calculated by as 

Resulting force in x-direction:

Rx = (1000 kg/m3) π ((0.102 m) / 2)2 (20 m/s)2 (1 - cos(45))

    = 957 N

Resulting force in y-direction:

Ry = (1000 kg/m3) π ((0.102 m) / 2)2 (20 m/s)2 sin(45)

    = 2311 N

Resulting force on the bend

R = (957 N)2 + (2311 N)2)1/2 

    = 2501 N

Note - if β is 90o the resulting forces in x- and y-directions are the same.

Resulting force due to Static Pressure

The pressure "acting" on the end surfaces of the bend creates resulting forces in x- and y-directions.

The resulting force in x-direction can be expressed as

Rpx = p A (1- cos β)                             (4)   

    = p π (d / 2)2 ( 1- cos β)                            (4b)

where

Rpx = resulting force due to pressure in x-direction (N)

p = gauge pressure inside pipe (Pa, N/m2)

The resulting force in y-direction can be expressed as

Rpy = p π (d / 2)2 sinβ                               (5)

where

Rpy = resulting force due to pressure in y-direction (N)

The resulting force on the bend due to force in x- and y-direction can be expressed as:

Rp = (Rpx2 + Rpy2)1/2                           (6)

where

Rp = resulting force on the bend due to static pressure (N)

Example - Resulting force on a bend due to pressure

The resulting force on a 45o bend with

  • internal diameter 102 mm = 0.102 m
  • pressure 100 kPa

can be calculated by as 

Resulting force in x-direction:

Rx = (100 103 Pa) π ((0.102 m) / 2)2 (1 - cos(45))

    = 239 N

Resulting force in y-direction:

Ry = (100 103 Pa) π ((0.102 m) / 2)2 sin(45)

    = 578 N

Resulting force on the bend

R = ((239 N)2 + (577 N)2)1/2 

    = 625 N

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