Forces Pipe Bends

Overview

Engineering reference data for Forces Pipe Bends in piping systems.

Key Formulas

Continuity

$A_1 v_1 = A_2 v_2$

Mass conservation in pipe flow.

Pressure Drop

$\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}$

Darcy-Weisbach equation.

Pipe Area

$A = \frac{\pi D^2}{4}$

Cross-sectional area of a pipe.

Variables

Symbol	Description	Unit
$D$	Pipe diameter	m
$v$	Flow velocity	m/s
$\Delta P$	Pressure drop	Pa
$f$	Friction factor	—

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 = \dot{m} v (1 - \cos\beta) \tag{1}

R_x = \rho A v^2 (1 - \cos\beta) \tag{1b}

R_x = \rho \pi \left(\frac{d}{2}\right)^2 v^2 (1 - \cos\beta) \tag{1c}

The resulting force in y-direction due to mass flow and flow velocity:

R_y = \dot{m} v \sin\beta \tag{2}

R_y = \rho A v^2 \sin\beta \tag{2b}

R_y = \rho \pi \left(\frac{d}{2}\right)^2 v^2 \sin\beta \tag{2c}

The combined resulting force on the bend:

R = \sqrt{R_x^2 + R_y^2} \tag{3}

Resulting Force Due to Static Pressure

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

R_{px} = p A (1 - \cos\beta) \tag{4}

R_{px} = p \pi \left(\frac{d}{2}\right)^2 (1 - \cos\beta) \tag{4b}

R_{py} = p \pi \left(\frac{d}{2}\right)^2 \sin\beta \tag{5}

The combined resulting force due to static pressure:

R_p = \sqrt{R_{px}^2 + R_{py}^2} \tag{6}

Worked Examples

Example 1: Force from Mass Flow

A 45° bend with internal diameter 102 mm carrying water ($\rho = 1000\ \text{kg/m}^3$) at 20 m/s:

• X-direction: $R_x = 1000 \times \pi \times (0.051)^2 \times 20^2 \times (1 - \cos 45°) = 957\ \text{N}$
• Y-direction: $R_y = 1000 \times \pi \times (0.051)^2 \times 20^2 \times \sin 45° = 2311\ \text{N}$
• Resultant: $R = \sqrt{957^2 + 2311^2} = 2501\ \text{N}$

Example 2: Force from Pressure

A 45° bend with internal diameter 102 mm at 100 kPa gauge pressure:

• X-direction: $R_{px} = 100 \times 10^3 \times \pi \times (0.051)^2 \times (1 - \cos 45°) = 239\ \text{N}$
• Y-direction: $R_{py} = 100 \times 10^3 \times \pi \times (0.051)^2 \times \sin 45° = 578\ \text{N}$
• Resultant: $R_p = \sqrt{239^2 + 578^2} = 625\ \text{N}$

Engineering Notes

• When there is no flow and no pressure, there is no force on the bend.
• When $\beta = 90°$, the resulting forces in x- and y-directions are equal in magnitude.
• The total force on a bend support must account for both momentum and pressure contributions combined.

References

• Original Source