Piping Elbows - Thrust Block Forces

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

ρ - density of fluid (kg/m³)

d - int. diam. pipe or bend (m)

v - velocity of fluid (m/s)

β - turning angle of bend (^o)

p - gauge pressure (kPa)

Imperial Units

 SG - specific gravity of fluid 

d - int. diam. pipe or bend (inches)

v - velocity of fluid (ft/s)

β - turning angle of bend (^o)

p - gauge pressure (psi)

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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:

R_x = m v (1 - cos(β))                          (1)

    = ρ A v² (1 - cos(β))                            (1b)

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

where

R_x = 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:

R_y = m v sin(β)                            (2)

    = ρ A v² sin(β)                            (2b)

    = ρ π (d / 2)² v² sin(β)                           (2c)   

R_y = 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 = (R_x² + R_y²)^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 45^o 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:

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

    = 957 N

Resulting force in y-direction:

R_y = (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 90^othe 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

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

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

where

R_px = 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

R_py = p π (d / 2)² sin(β)                               (5)

where

R_py = 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:

R_p = (R_px² + R_py²)^1/2                           (6)

where

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

Example - Resulting force on a bend due to pressure

The resulting force on a 45^o bend with

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

can be calculated by as 

Resulting force in x-direction:

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

    = 239 N

Resulting force in y-direction:

R_y = (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