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Assume 0.5″ diameter strand has cross-sectional area of 0.153 sq.in. and weight of 0.525 lbs/ft.

Assume 0.6″ diameter strand has cross-sectional area of 0.217 sq.in. and weight of 0.740 lbs/ft.

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Minimum Ultimate Tensile Strength (MUTS) = (Grade of Steel) x (Cross-Sectional Area)

0.5″ inch diameter = (270 ksi) x (0.153 sq.in.) = 41.3 kips

0.6″ inch diameter = (270 ksi) x (0.217 sq.in.) = 58.6 kips

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Minimum Yield Strength = 90% of MUTS = MUTS x 0.90 (per ASTM-A416)

0.5″ inch diameter = (41.3 kips) x (0.90) = 37.2 kips

0.6″ inch diameter = (58.6 kips) x (0.90) = 52.7 kips

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Jacking Force = 80% of MUTS = MUTS x 0.80 (per ACI Code)

0.5″ inch diameter = (41.3 kips) x (0.80) = 33.0 kips

0.6″ inch diameter = (58.6 kips) x (0.80) = 46.9 kips

“Jacking Force” is the force that tendons are stressed to.

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Allowable Initial Force = (Jacking Force) minus (Short-Term Losses) = 70% of MUTS = MUTS x 0.70 (per ACI-318)

Short-Term Losses include:

1. Angular Profile of Tendon
2. Horizontal sweeps in Tendon
3. Wedge-Seating (typically 0.25 inch)
4. Wobble due to installation (CLICK HERE to view the video on how to calculate Angular and Wobble Coefficients in unbonded post-tensioning tendons.)

0.5″ inch diameter = (41.3 kips) x (0.70) = 28.9 kips

0.6″ inch diameter = (58.6 kips) x (0.70) = 41.0 kips

“Initial Force” is the force at the anchorage after the wedges are seated and stressing jack is removed.  The calculated values above are approximate since the actual short-term losses may differ from the theoretical values.

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Final Force = (Initial Force) minus (Long-Term Losses)

Long-Term Losses include:

1. Creep of concrete (permanent deflection due application of constant load)
2. Elastic Shortening of concrete
3. Relaxation of steel prestressing strand
4. Shrinkage of concrete during curing

0.5″ inch diameter = approx 26.9 kips

0.6″ inch diameter = approx. 38.1 kips

“Final Force” is the force at the anchorage after the long-term losses are accounted for.  The calculated values above are approximate since the actual long-term losses may differ from the theoretical values.

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Average Tendon Elongation (approx.) = (P x L) / (A x E)

P = Prestress jacking force (70% of MUTS)

L = Length of steel (inches)

A = Cross-Sectional Area of steel (sq.in.) on mill certificates

E = Modulus of Elasticity of steel (ksi) on mill certificates

For example, using a 100-foot tendon (L = 100 x 12 inches) with Modulus of Elasticity of 28,500 ksi.

0.5″ inch diameter = (28.9 kips x 1,200 inches) / (0.153 sq.in. x 28,500 ksi) = 7.95 inches

0.6″ inch diameter = (41.0 kips x 1,200 inches) / (0.217 sq.in. x 28,500 ksi) = 7.95 inches

***Notice that the 0.5″ and 0.6″ have the same Avg. Elongation***

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Post-Tensioning Institute recommends an allowable elongation range of plus/minus 7% of the Average Tendon Elongation for unbonded post-tensioning tendons.

Min. Allowable Elongation = 93% of Avg. Elongation = 0.93 x (Avg.El.)

Max. Allowable Elongation = 107% of Avg. Elongation = 1.07 x (Avg.El.)

If we use the same 100-foot tendon with average elongation of 7.95 inches, then Min.El. = 0.93 x 7.95 inches = 7.40 inches and Max.El. = 1.07 x 7.95 inches = 8.51 inches.

- Rattan Khosa, President, AMSYSCO