Deflection & Cracking of Prestressed Members Notes

Last reviewed 30 Sept 2026 · 6 min read

Deflection is a serviceability limit

A prestressed beam is stiff and its prestress produces an upward deflection (camber). The net deflection is the difference between the downward deflection from loads and the upward camber, and it changes with time because of creep, shrinkage and loss of prestress. Excessive deflection damages partitions, causes ponding, upsets the riding surface of a bridge and looks bad; too much camber does the same in reverse.

Short-term deflection

Uncracked, elastic analysis is used at service load: $E_{c}$ and the gross moment of inertia $I$ .

FormulaStandard deflections (simply supported, span

L

)

Uniform load $w$ (downward): $δ_{w} = \frac{5 w L ^{4}}{384 E _{c} I}$
Central point load $W$ : $δ = \frac{W L ^{3}}{48 E _{c} I}$
Parabolic tendon, central eccentricity $e$ (upward camber): $δ_{P} = \frac{5 P e L ^{2}}{48 E _{c} I}$
Straight tendon, constant eccentricity $e$ : $δ_{P} = \frac{P e L ^{2}}{8 E _{c} I}$
Harped (bent) tendon, dip $e$ at mid-span: $δ_{P} = \frac{P e L ^{2}}{12 E _{c} I}$

The parabolic-tendon formula comes from the equivalent upward load $w_{b} = 8 P e / L^{2}$ : substituting into $5 w_{b} L^{4} / (384 E_{c} I)$ gives $5 P e L^{2} / (48 E_{c} I)$ . Where the tendon has different eccentricities at the ends and the centre, the deflection is found by the moment-area method applied to the moment diagram $P e (x)$ .

Net short-term deflection = camber from prestress − deflection from self-weight and other loads.

Worked ExampleExample — camber and deflection

A simply supported beam, $L$ = 12 m, $E_{c}$ = 31 500 N/mm², $I = b h^{3} /12 = 300 \times 60 0^{3} /12 = 5.4 \times 1 0^{9}$ mm⁴. Parabolic tendon $P$ = 800 kN, $e$ = 150 mm; self-weight $w$ = 4.5 kN/m; live load $q$ = 6 kN/m.

Camber: $δ_{P} = \frac{5 \times 800 000 \times 150 \times 12 00 0 ^{2}}{48 \times 31 500 \times 5.4 \times 1 0 ^{9}} = \frac{8.64 \times 1 0 ^{16}}{8.165 \times 1 0 ^{15}} = 10.6 mm$ upward.

Self-weight (take $w$ = 4.5 N/mm so that all units are N and mm): $δ_{g} = \frac{5 \times 4.5 \times 12 00 0 ^{4}}{384 \times 31 500 \times 5.4 \times 1 0 ^{9}} = \frac{4.666 \times 1 0 ^{17}}{6.532 \times 1 0 ^{16}} = 7.1 mm$ downward.

Net at transfer/permanent: 10.6 − 7.1 = +3.5 mm (still upward). The live load adds 6/4.5 × 7.1 = 9.5 mm downward → net at full load ≈ 6.0 mm downward, well within the span/250 = 48 mm limit.

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Deflection is a serviceability limit

Short-term deflection

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