Learn: Prestress friction loss along tendon
A tendon loses force along its length as friction develops against its duct. This exponential model combines intended curvature and unintended wobble to estimate the remaining force at distance x from the stressing end.
What the formula is saying
The dimensionless attenuation is μθ + kx. The remaining fraction is exp[−(μθ + kx)], so positive friction terms reduce P0 rather than subtracting a fixed force per metre.
Read the symbols in plain language
- P0
- Jacking force
Jacking force. Scale the initial force by the remaining fraction without adding unrelated loss mechanisms.
kNKilonewtons measure force; 1 kN = 1000 N.
- μ
- Curvature friction coefficient
Curvature friction coefficient. The friction coefficient multiplies the accumulated angular change along the tendon.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- θ
- Cumulative angle
Cumulative angle. The friction coefficient multiplies the accumulated angular change along the tendon.
radRadians measure angle; a full turn is 2π rad. Do not enter degrees in this base unit.
- k
- Wobble coefficient
Wobble coefficient. The wobble coefficient times distance represents the additional distributed friction effect.
1/mUse 1/m as the base unit shown here. P0 and P(x) are in kN. θ is the accumulated angular change in radians, not degrees; μ is dimensionless, k is 1/m and x is m so kx is dimensionless.
- x
- Tendon length
Tendon length. The wobble coefficient times distance represents the additional distributed friction effect.
mMetres measure length; 1 m = 1000 mm.
- P(x)
- Result to find
Prestress friction loss along tendon. Scale the initial force by the remaining fraction without adding unrelated loss mechanisms.
kN
Sort out the units first
P0 and P(x) are in kN. θ is the accumulated angular change in radians, not degrees; μ is dimensionless, k is 1/m and x is m so kx is dimensionless.
Assumptions before calculating
The supplied curvature is accumulated along the tendon from the stressing end, and μ and k are appropriate constant friction parameters. Use nonnegative P0, μ, θ, k and x for this one-end loss model.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find P(x) and explain the result in the stated output unit.
- P0 · Jacking force
- 1500 kN
- μ · Curvature friction coefficient
- 0.2
- θ · Cumulative angle
- 0.1 rad
- k · Wobble coefficient
- 0.001 1/m
- x · Tendon length
- 30 m
Find the curvature contribution
The friction coefficient multiplies the accumulated angular change along the tendon.
(0.2) × (0.1) = 0.02Find the wobble contribution
The wobble coefficient times distance represents the additional distributed friction effect.
(0.001) × (30) = 0.03Find the remaining force fraction
The negative exponential converts total attenuation into a fraction between zero and one.
exp(-((0.02) + (0.03))) = 0.9512294245Find prestress remaining at x
Scale the initial force by the remaining fraction without adding unrelated loss mechanisms.
(1500) × (0.9512294245) ≈ 1426.844137 kN
Does this worked answer make sense?
With zero curvature and zero wobble distance, the force equals P0. Increasing either loss contribution lowers the remaining force, which should not exceed P0 for nonnegative inputs.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- P0 · Jacking force
- 1200 kN
- μ · Curvature friction coefficient
- 0.18
- θ · Cumulative angle
- 0.2 rad
- k · Wobble coefficient
- 0.0015 1/m
- x · Tendon length
- 40 m
Find the curvature contribution
The friction coefficient multiplies the accumulated angular change along the tendon.
(0.18) × (0.2) = 0.036Find the wobble contribution
The wobble coefficient times distance represents the additional distributed friction effect.
(0.0015) × (40) = 0.06Find the remaining force fraction
The negative exponential converts total attenuation into a fraction between zero and one.
exp(-((0.036) + (0.06))) ≈ 0.9084640161Find prestress remaining at x
Scale the initial force by the remaining fraction without adding unrelated loss mechanisms.
(1200) × (0.9084640161) ≈ 1090.156819 kN
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- P0 · Jacking force
- 1800 kN
- μ · Curvature friction coefficient
- 0.2
- θ · Cumulative angle
- 0.15 rad
- k · Wobble coefficient
- 0.001 1/m
- x · Tendon length
- 50 m
Find: Learn: Prestress friction loss along tendon
A hint, not the answer
The dimensionless attenuation is μθ + kx. The remaining fraction is exp[−(μθ + kx)], so positive friction terms reduce P0 rather than subtracting a fixed force per metre.
P0 and P(x) are in kN. θ is the accumulated angular change in radians, not degrees; μ is dimensionless, k is 1/m and x is m so kx is dimensionless.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the curvature contribution
The friction coefficient multiplies the accumulated angular change along the tendon.
(0.2) × (0.15) = 0.03Find the wobble contribution
The wobble coefficient times distance represents the additional distributed friction effect.
(0.001) × (50) = 0.05Find the remaining force fraction
The negative exponential converts total attenuation into a fraction between zero and one.
exp(-((0.03) + (0.05))) ≈ 0.9231163464Find prestress remaining at x
Scale the initial force by the remaining fraction without adding unrelated loss mechanisms.
(1800) × (0.9231163464) ≈ 1661.609423 kN
Avoid the common trap
Do not enter an angle in degrees while selecting radians. Keep the negative sign in the exponent, and do not replace exp with multiplication or use signed curvature cancellation instead of accumulated angular change.
When this method applies — and when it does not
Anchorage seating, elastic shortening, creep, shrinkage, steel relaxation, two-end stressing and varying duct properties are not included. This is a friction-loss component, not the final effective prestress at all ages.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Prestress friction loss along tendon. First-generation EN 1992 teaching: material properties and the relevant bending, shear, serviceability, detailing or prestress relationship. Read the applicability conditions as well as the expression.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.