Queensland University of Technology · FACULTY OF ENGINEERING

EGB375 · Design of Concrete Structures

Q: Why is the tendon stress at ultimate (σ_pu) less than the breaking stress f_pb?

Because failure of the beam is governed by the concrete reaching its ultimate strain, not by the tendon snapping. At that point the bonded tendon has only developed part of its capacity, so AS3600 gives a reduced stress σ_pu = f_pb(1 − k1·k2/γ). The more heavily reinforced the section (larger reinforcement index k2), the lower σ_pu — the concrete crushes before the steel is fully stressed.

Q: How do losses feed into the number of strands?

The Magnel serviceability design gives the force needed at transfer, P_i (already after the ~10% immediate loss). You undo that loss to get the jacking force P_j = P_i/(1 − immediate loss), then divide P_j by the maximum allowed force per strand (for example 0.85 f_pb·A_p for stress-relieved post-tensioning) and round UP. The ~20% time-dependent loss then reduces P_i to the long-term effective force P_e but does not change the strand count.

Q: Can AI help me with PSC losses and ultimate flexural capacity in EGB375?

Yes — Sia is an AI tutor that explains the method step by step: how losses cascade from jack to effective force, why σ_pu = f_pb(1 − k1·k2/γ), and how to assemble d_n and M_u for a fully or partially prestressed section, with the right AS3600 factors and units. It is a study aid to build your understanding and check your reasoning; it does not sit your exam, do your assessed work, or promise any particular grade, and you should confirm every clause and factor against AS3600-2018 and your unit materials.

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Chapter 5 of 12 · EGB375

PSC — Losses, Tendon Sizing & Ultimate Flexural Capacity

This chapter of EGB375 Design of Concrete Structures at Queensland University of Technology closes the prestressed-concrete (PSC) design loop: from a required prestress force it sizes the tendon and then checks ultimate flexural strength to AS3600-2018. You will account for losses (about 10% immediate and 20% time-dependent, roughly 30% total), turn the jacking force into a real number of strands within the code stress cap, and compute the ultimate tendon stress σ_pu = f_pb(1 − k₁k₂/γ), the neutral-axis depth d_n, and the moment capacity M_u for fully and partially prestressed sections. Every result ends on the adequacy check φM_u ≥ M* with φ = 0.85.

In this chapter

What this chapter covers

01Why prestress is lost: immediate (~10%) vs time-dependent (~20%) losses, ~30% total
02Maximum jacking force per AS3600-2018 §17.3.4.6 — the 0.80 / 0.85 / 0.75 f_pb·A_p caps
03Jacking force P_j = P_i/(1 − immediate loss) and converting it to a strand count
04Total tendon area A_pt from the number of strands (always round UP)
05Ultimate tendon stress σ_pu = f_pb(1 − k1·k2/γ) and the reinforcement index k2
06AS3600 rectangular stress-block factors α₂ and γ as functions of f′c
07Neutral-axis depth d_n from equilibrium and the ductility check k_uo = d_n/d_o ≤ 0.36
08Ultimate moment M_u for fully vs partially prestressed sections
09The bending adequacy check φM_u ≥ M* with φ = 0.85

Worked example · free

Ultimate moment of a partially prestressed beam

Q [12 marks]. A rectangular post-tensioned beam has b = 350 mm, D = 800 mm and f′c = 40 MPa. The tendon area is A_pt = 1183 mm² (12 × 12.7 mm strands, f_pb = 1870 MPa) at d_p = 700 mm, plus non-prestressed A_st = 3N16 = 600 mm² at d_s = 750 mm (so d_o = 750 mm); there is no compression steel. The beam is simply supported over L = 12 m carrying G = 35 kN/m and Q = 12 kN/m. Take k1 = 0.4. Compute σ_pu, d_n and M_u, and check whether φM_u ≥ M*.

+1Stress-block factors (f′c = 40 MPa): α₂ = 0.85 − 0.0015(40) = 0.79; γ = 0.97 − 0.0025(40) = 0.87 (both ≥ 0.67 ✓).
+2Reinforcement index k2 = [A_pt·f_pb + (A_st − A_sc)·f_sy] / (b·d_p·f′c) = [1183×1870 + 600×500] / (350×700×40) = 2 512 584 / 9 800 000 = 0.256.
+2Ultimate tendon stress σ_pu = f_pb(1 − k1·k2/γ) = 1870×(1 − 0.4×0.256/0.87) = 1870×(1 − 0.1179) = 1650 MPa (< f_pb = 1870 ✓).
+2Neutral axis from equilibrium d_n = (σ_pu·A_pt + A_st·f_sy) / (α₂·f′c·γ·b) = (1650×1183 + 600×500) / (0.79×40×0.87×350) = 2 251 800 / 9622 = 234.0 mm.
+1Ductility check k_uo = d_n/d_o = 234.0/750 = 0.312 ≤ 0.36 → under-reinforced / ductile ✓. Lever-arm term γ·d_n/2 = 0.87×234.0/2 = 101.8 mm.
+2Ultimate moment (partially prestressed) M_u = σ_pu·A_pt(d_p − γd_n/2) + f_sy·A_st(d_s − γd_n/2) = 1 951 800×(700 − 101.8) + 300 000×(750 − 101.8) = 1167.6 + 194.5 = 1362 kN·m (keep N and mm, then ÷10⁶ for kN·m).
+2Adequacy. w* = 1.2(35) + 1.5(12) = 60 kN/m; M* = w*L²/8 = 60×12²/8 = 1080 kN·m. φM_u = 0.85×1362 = 1158 kN·m ≥ 1080 → the section is ADEQUATE in bending ✓.

σ_pu = 1650 MPa, d_n = 234 mm (k_uo = 0.31 ≤ 0.36, ductile), M_u = 1362 kN·m and φM_u = 1158 kN·m ≥ M* = 1080 kN·m, so the beam is adequate in bending.

Sia tip — Work in N and mm inside M_u and divide by 10⁶ for kN·m — mixing kN with mm inflates the answer 1000×. Always finish with both the ductility check (k_uo ≤ 0.36) and the φM_u ≥ M* statement; an M_u with no adequacy conclusion is an unfinished answer.

Glossary

Key terms

Immediate vs time-dependent losses: Immediate losses (~10%) occur at transfer — elastic shortening of the concrete, anchorage slip and duct friction. Time-dependent losses (~20%) develop over months to years from concrete shrinkage and creep plus steel relaxation. Together they total about 30% from jack to long-term effective force.
Jacking force (P_j): The prestress force applied at the jack, before any loss. It is found by undoing the immediate loss on the required transfer force: P_j = P_i/(1 − immediate loss). AS3600-2018 §17.3.4.6 caps it at 0.80 f_pb·A_p (pre-tensioning), 0.85 f_pb·A_p (stress-relieved post-tensioning) or 0.75 f_pb·A_p (other post-tensioning).
Number of strands / A_pt: Strands = P_j divided by the maximum force per strand (e.g. 0.85 f_pb·A_p for stress-relieved post-tensioning), rounded UP so no strand is overstressed past the cap. The total tendon area is A_pt = (number of strands) × A_p, where A_p is the area of one strand (98.6 mm² for a 12.7 mm strand).
Ultimate tendon stress σ_pu: The stress developed in a bonded tendon at the ultimate limit state — below the breaking stress f_pb. AS3600 gives σ_pu = f_pb(1 − k1·k2/γ), with k1 = 0.4 (0.28 only if f_py/f_pb ≥ 0.9) and reinforcement index k2 = [A_pt·f_pb + (A_st − A_sc)·f_sy]/(b_ef·d_p·f′c). More reinforcement raises k2 and lowers σ_pu.
Rectangular stress-block factors α₂ and γ: AS3600-2018 idealises the concrete compression as a uniform stress α₂f′c over a depth γ·d_n. Both vary with concrete strength: α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67 (e.g. at f′c = 40 MPa, α₂ = 0.79 and γ = 0.87). They are not fixed at 0.85.
Neutral-axis depth d_n and k_uo: Horizontal equilibrium (concrete compression = tendon + bar tension) gives d_n = (σ_pu·A_pt + A_st·f_sy)/(α₂·f′c·γ·b). The ductility measure k_uo = d_n/d_o (d_o = depth to the outermost tension layer) must satisfy k_uo ≤ 0.36 for a ductile, under-reinforced section.
Fully vs partially prestressed M_u: For a fully prestressed section M_u = σ_pu·A_pt(d_p − γd_n/2). A partially prestressed section also carries ordinary tension steel, adding a term: M_u = σ_pu·A_pt(d_p − γd_n/2) + f_sy·A_st(d_s − γd_n/2). Each lever arm is the steel depth minus γd_n/2 (the depth to the compression resultant).
Bending adequacy check (φM_u ≥ M*): The section is adequate in bending when the design capacity φM_u is at least the factored design moment M*, using φ = 0.85 for bending. M* comes from the ultimate load combination w* = 1.2G + 1.5Q.

FAQ

PSC — Losses, Tendon Sizing & Ultimate Flexural Capacity FAQ

Why is the tendon stress at ultimate (σ_pu) less than the breaking stress f_pb?

Because failure of the beam is governed by the concrete reaching its ultimate strain, not by the tendon snapping. At that point the bonded tendon has only developed part of its capacity, so AS3600 gives a reduced stress σ_pu = f_pb(1 − k1·k2/γ). The more heavily reinforced the section (larger reinforcement index k2), the lower σ_pu — the concrete crushes before the steel is fully stressed.

How do losses feed into the number of strands?

The Magnel serviceability design gives the force needed at transfer, P_i (already after the ~10% immediate loss). You undo that loss to get the jacking force P_j = P_i/(1 − immediate loss), then divide P_j by the maximum allowed force per strand (for example 0.85 f_pb·A_p for stress-relieved post-tensioning) and round UP. The ~20% time-dependent loss then reduces P_i to the long-term effective force P_e but does not change the strand count.

Can AI help me with PSC losses and ultimate flexural capacity in EGB375?

Yes — Sia is an AI tutor that explains the method step by step: how losses cascade from jack to effective force, why σ_pu = f_pb(1 − k1·k2/γ), and how to assemble d_n and M_u for a fully or partially prestressed section, with the right AS3600 factors and units. It is a study aid to build your understanding and check your reasoning; it does not sit your exam, do your assessed work, or promise any particular grade, and you should confirm every clause and factor against AS3600-2018 and your unit materials.

Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.

Study strategy

Exam move

Drill PSC ultimate design as one fixed routine so it becomes automatic under exam pressure. (1) Losses → force: undo the ~10% immediate loss to get the jacking force P_j = P_i/0.9, and note the ~20% time-dependent loss gives the long-term P_e (~30% total). (2) Size the tendon: strands = ⌈P_j / (factor × f_pb·A_p)⌉, rounding UP with the right cap (0.80 pre-tensioned, 0.85 stress-relieved PT, 0.75 other PT), then A_pt = strands × A_p. (3) Factors: recompute α₂ and γ from f′c — never leave them at 0.85. (4) Stress: k2 then σ_pu = f_pb(1 − k1·k2/γ). (5) Neutral axis: d_n from equilibrium, then check k_uo = d_n/d_o ≤ 0.36. (6) Moment: M_u, adding the f_sy·A_st term only if the section is partially prestressed. (7) Finish on φM_u ≥ M* with φ = 0.85. Keep every force in N and length in mm inside M_u and divide by 10⁶ for kN·m. Put the loss percentages, the three jacking caps, the α₂/γ formulas and the σ_pu / d_n / M_u equations on your permitted note sheets so exam time goes to clean substitution, not recall.

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