Pipe Thrust at Bends: Pressure, Not Flow, Sizes the Anchor

Ask what pushes on a pipe bend and most people reach for the flow: water comes in one way, leaves another, and something has to turn it. True, and almost irrelevant. On a DN300 water main at 10 bar carrying 2 m/s, turning the flow through 90° takes 0.4 kN. Holding the pressure in the bend takes 100 kN. The anchor, the thrust block or the restrained joints are sized by the pressure, and by the test pressure at that, which the line sees with no flow at all.

Here is the same DN300 line at 10 bar(g), 141 L/s, on each fitting it might contain, straight from the pipe thrust calculator:

Why a bend is pushed

Take a straight run of pipe under pressure. The pressure acts on every square millimetre of the inside wall, but for every patch of wall pushed one way there is a patch opposite pushed the other way, and the two cancel. The flow carries momentum, but the same momentum enters each section as leaves it. Nothing is out of balance, which is why a straight buried main needs no anchors along its length.

Now bend it. The pressure acting on the outside of the bend has no matching inside wall to cancel against, and the momentum that came in along one axis leaves along another. Both leave a net force on the fitting, pointing out of the bend. The cleanest way to get its size is the control-volume momentum balance: draw a box around the bend, apply the pressure to the fluid at the two ports (which is where the fluid inside the box touches the fluid outside it), account for the momentum entering and leaving, and the remainder is the force the bend must exert on the fluid, equal and opposite to the thrust on the bend:

• $F$ — net force of the fluid on the bend (N), along the outward bisector of the deflection
• $P$ — gauge pressure (Pa)
• $A$ — flow area on the inside diameter, $\pi D^2/4$ (m²)
• $\rho$ — fluid density (kg/m³)
• $Q$ — volume flow (m³/s)
• $V$ — mean velocity, $Q/A$ (m/s)
• $\theta$ — deflection angle, the change in direction (90° for a quarter bend)

Each port contributes a force $P A + \rho Q V$ along its own axis; the two are the same size and $\theta$ apart, so the vector sum is $2 \sin(\theta/2)$ times one of them, on the line that bisects the angle between the legs. In the plane of the bend, with the inlet flow along $x$ and $y$ across it pointing to the outside of the bend (away from the side the flow turns to):

• $F_x$ — component along the inlet axis, in the direction the inlet flow was travelling (N)
• $F_y$ — component across it, towards the outside of the bend (N)

For the 90° bend the two components are equal and the resultant sits at 45° between them, 45° from the inlet flow direction (90° − θ/2 in general): squarely on the outside of the bend, which is where the thrust block goes.

The numbers: pressure versus momentum

The ratio of the two terms is the whole argument of this article:

• $\rho V^2$ — twice the dynamic pressure of the flow (Pa)
• $P$ — gauge static pressure (Pa)

Water at 2 m/s has $\rho V^2 = 4\,000$ Pa. Against 10 bar that is 0.4 %; against 3 bar 1.3 %; it only reaches 10 % at 0.4 bar, which is a drain, not a main. Push the velocity to 5 m/s and the ratio is 25 kPa over the pressure: 2.5 % at 10 bar, 25 % on a pump discharge at 1 bar. So the momentum term matters on low-pressure, high-velocity lines — pump discharges, cooling-water headers, slurries, gases at high velocity — and on nothing that would be called a water main. On the worked line:

The last row is the one that sizes the anchor. The hydrostatic test is applied with the line full and still, so there is no momentum term, but at 1.5 × the operating pressure the pressure term alone is 49 % larger than everything the line sees in service. Unless a surge analysis gives a higher transient pressure, the test governs.

Gauge, not absolute. The atmosphere presses on the outside of the fitting with the same 1.013 bar it adds to the inside, and the two cancel. Using absolute pressure in the bend formula over-states a DN300 90° bend by $101\,325 \times 0.0707 \times 1.414 = 10$ kN, a tenth of the real thrust.

The other fittings

Reducer

A concentric reducer is pushed towards its small end: the pressure acts on the large area at the inlet and on the smaller area at the outlet, and the fluid leaves faster than it arrived. Taking forces positive towards the small end,

• $P_1, P_2$ — gauge pressures at inlet and outlet (Pa); with no loss, $P_2 = P_1 - \tfrac{1}{2}\rho(V_2^2 - V_1^2)$
• $A_1, A_2$ — inlet and outlet flow areas (m²)
• $V_1, V_2$ — inlet and outlet velocities (m/s)

With no flow it collapses to $P (A_1 - A_2)$, the pressure on the annulus the reducer closes off — and that is the test-pressure value. On the DN300 → DN200 reducer at 10 bar the static force is 39.3 kN; at 141 L/s the flow accelerates from 2.0 to 4.5 m/s, the outlet pressure drops by 8 kPa, and the net force is 39.2 kN. The momentum term is negative here (the fluid leaving faster pulls the fitting back), and the lower outlet pressure pushes the other way; they nearly cancel, and the static value is the safe one.

Tee and dead end

A blind flange, a closed valve or a capped end sees the full pressure on the full area, $F = P A$, along the pipe: 106 kN on the DN300 line at 15 bar test. A tee whose branch carries flow is a dead end with a momentum term added:

• $A_b$ — branch flow area (m²)
• $Q_b, V_b$ — flow and velocity leaving through the branch

The DN200 branch at 10 bar taking 50 L/s sees 31.4 kN of pressure thrust and 0.08 kN of momentum thrust along the branch axis — again, the pressure on the branch area and nothing else worth mentioning.

What reacts the thrust

The equation does not care what holds the bend; the installation does.

• Welded steel and fused polyethylene are self-restrained: the thrust goes into the pipe wall as axial stress and is carried along the line to wherever the pipe is anchored, or into the soil by friction on a buried line. Anchors on an above-ground steel line take the net load after the pipe's own stiffness has done its work, together with the thermal expansion forces that the same anchors see.
• Push-on and mechanical joints (ductile iron, PVC, some GRP) carry no axial load. Every bend, tee, reducer and dead end needs either a thrust block or restrained joints over a calculated length on each side.
• Flanged pipe is restrained at the flanges, so the thrust from a flanged bend is carried into the adjacent pipe and the question moves to whatever that pipe is attached to.

Thrust blocks

A block converts the point load into a bearing pressure on undisturbed soil:

• $A_{bearing}$ — horizontal bearing face of the block against the trench wall (m²)
• $SF$ — safety factor on the thrust, 1.5 typical
• $\sigma_{bearing}$ — allowable horizontal bearing strength of the soil (Pa)

For the DN300 bend at its 15 bar test thrust of 149.9 kN, on cemented sand and gravel taken at 192 kPa with a factor of 1.5, that is 1.17 m², roughly a 1.1 m square face. The arithmetic is the easy half. The block must bear on native soil, not backfill; its bearing face must be centred on the line of thrust, the pipe centreline for a horizontal bend, so the resultant passes through the middle third of the face and does not lever the joint; and the soil value must come from the trench, not from a table. The ladder of indicative values that every guide reproduces (soft clay about 50 kPa, sand 100, sand and gravel 150, cemented gravel 190, hard pan 480) is a first estimate for sizing a sketch, and a geotechnical bearing value replaces it before anything is poured.

Restrained joints

Where a block cannot be trusted — a trench wall that will be re-excavated for the next utility, a tight site, soft or made ground — restrained (locked, harnessed or gripping) joints carry the thrust into the pipe wall and shed it to the soil by friction along a length of pipe on each side of the fitting. That length depends on the thrust, the depth of cover, the soil-to-pipe friction, the trench type and the pipe's weight and outside diameter, through the AWWA M41 / DIPRA procedure. The thrust from the formulas above is the input to it; the length is the output, and it is not a number a one-line calculator should hand out.

Vertical bends

The magnitude is the same; the direction changes what is available to resist it. A bend turning the flow downward pushes the pipe up: the "block" is a buried mass, concrete plus the soil above it, heavy enough to hold the pipe down. Below the water table only the submerged weight counts: buoyancy takes concrete from about 24 to 14 kN/m³ and backfill from roughly 18 to 10 kN/m³, so a gravity block sized on dry weights in wet ground can be short by nearly half. A bend turning the flow upward pushes down onto the trench floor and needs a bearing pad sized against the floor's bearing capacity. Horizontal bends get the side-bearing block in the figure.

A checklist

1. Use the inside diameter and the gauge pressure.
2. Take the test pressure (or the surge pressure, if higher) with no flow; take the operating pressure with flow; design for the larger thrust. It is the test in almost every case.
3. Keep the momentum term in the calculation and expect it to be under 1 % on a water main; it earns attention only below a few bar or above 4–5 m/s.
4. Resolve the force along the outward bisector for a bend, along the branch for a tee, along the axis for a reducer or dead end.
5. Decide what reacts it — pipe wall, anchor, block or restrained joints — from the joint type, not from the force.
6. Size a block against a geotechnical bearing value with a safety factor, and detail it: native soil, face centred on the thrust line, middle third.

The pipe thrust calculator does the first four steps for bends, reducers, tees and dead ends, at operating and test pressure, with a first-pass bearing area for a block. The pressure at the fitting comes from the hydraulics: for a single line the friction loss calculator, for a network SimuPipe, which reports the pressure at every node of the solved model. The other load on the same anchors, thermal expansion, is in the pipe thermal expansion table.