PRESTRESS LOSSES [LRFD ART. 5.9.5]

Total prestress losses $Δ f_{p T} = Δ f_{p E S} + Δ f_{p S R} + Δ f_{p C R} + Δ f_{p R 2}$

where:

Δf_pES = loss due to elastic shortening, MPa

Δf_pCR = loss due to creep, MPa

Δf_pSR = loss due to shrinkage, MPa

Δf_pR2 = loss due to relaxation of steel after transfer, MPa

Elastic Shortening [LRFD Art. 5.9.5.2.3a]

Δ f_{p E S} = \frac{E_{p}}{E_{c i}} f_{c g p}

where:

E_p= modulus of elasticity of prestressing reinforcement = 197000 MPa

E_ci = modulus of elasticity of beam at release = 29966.3 MPa

f_cgp = sum of concrete stresses at the center of gravity of prestressing tendons due to prestressing force at transfer and the self-weight of the member at sections of maximum moment, MPa

where:

Force per strand immediately after transfer = (area of strand) (prestress after transfer) = (98.7)(0.70)(1861.58 MPa) = 128628 N

Note: LRFD Art. 5.9.5.2.3 states that …fcgp may be calculated on the basis of a prestressing steel stress assumed to be 0.65 fpu for stress-relieved strands and 0.70 fpu for low-relaxation strands.

P_i = Total prestressing force at release = (36 strands)(128628) = 4630642 N

f_cgp will be computed based on Mg using the overall beam length at release.

f_{c g p} = \frac{P_{i}}{A} + \frac{P_{i} e_{c}^{2}}{I} - \frac{M_{g} e_{c}}{I} = 13.42 M P a

Therefore, the loss due to elastic shortening is:

Δ f_{p E S} = \frac{197000}{29966.3} (13.42) = 88.24 M P a

Shrinkage [LRFD Art. 5.9.5.4.2]

f_{p S R} = 117 - 1.03 H

LRFD Eq. 5.9.5.4.2-1

where:

H = relative humidity (assume 70%)

Relative humidity varies significantly from one area of the country to another, see LRFD Fig. 5.4.2.3.3-1.

where

where:

H = relative humidity (assume 70%)

Relative humidity varies significantly from one area of the country to another, see LRFD Fig. 5.4.2.3.3-1.

Δ f_{p S R} = 117 - 1.03 \times 70 = 44.900 M P a

Creep of Concrete [LRFD Art. 5.9.5.4.3]

Δ f_{p C R} = 12 f_{c g p} - 7 Δ f_{c d p}

where:

Δf_cdp = Change in concrete stress at center of gravity of prestressing due to permanent loads except the loads acting at time of applying prestressing force, calculated at the same section as f_cgp, MPa

Δ f_{c d p} = \frac{M_{s} e_{c}}{I} + \frac{(M_{b} + M_{w s}) (y_{b c} - y_{b s})}{I_{c}} = \frac{(2267.8 \times 10^{6}) (647)}{1.908 \times 10^{11}} + \frac{(87.4 + 148.2) (10^{6}) (1171.5 - 75)}{4.23 \times 10^{11}} = 8.301 M P a

Now for the total final losses, fcgp will be conservatively computed based on Mg using the design span length

Δ f_{c g p} = \frac{4630642}{523107} + \frac{4630642 \times 647^{2}}{1.908 \times 10^{11}} - \frac{1618 \times 10^{6} \times 647}{1.908 \times 10^{11}} = 13.524 M P a

Therefore, the loss due to creep is:

Δf_pCR = 12 (13.524) – 7 (8.301) = 104.19 MPa

Relaxation of Prestressing Strands [LRFD Art. 5.9.5.4.4]

Relaxation at Transfer [LRFD Art. 5.9.5.4.4b]

For low relaxation strands, loss due to relaxation at transfer is:

Δ f_{p R I} = \frac{\log (24 t)}{40.0} (\frac{f_{p j}}{f_{p j}} - 0.55) f_{p j} = \frac{\log (24.0 \times 0.75)}{40.0} (\frac{0.75 f_{p u}}{0.90 f p u} - 0.55) \times 0.75 \times 1861.6 = 12.41 M P a

LRFD Eq. 5.9.5.4.4b-2

Relaxation after Transfer [LRFD Art. 5.9..4.4c]

For low-relaxation strands, loss due to strand relaxation after transfer is

Δ f_{p R 2} = 30 % (138 - 0.4 {Δ f}_{p E S} - 0.2 (Δ f_{p S R + Δ f_{p C R}}))

where: 30% as per LRFD Art. 5.9.5.4.4c

Therefore, 0.30 [138 –0.4 × 88.24 – 0.2 (44.900 + 104.19)] = 21.87 MPa

Total Losses At Transfer

Δf_pi = Δf_pES + Δf_PRI = 88.24 + 12.41 = 100.65 MPa

where:

Stress in tendons after transfer, f_pt = f_pi – Δfpi = (1396.328 – 100.65) = 1295.6 MPa

Force per strand = (f_pt) (strand area) = 1295.6 × 98.7 = 127875 N

Therefore, total prestressing force after transfer is

P_i = (127875) (36) = 4603500 N

I n i t i a l L o s s % = \frac{T o t a l l o s s e s a t t r a n s f e r}{f_{p i}} = \frac{100.65}{1396.238} 100 = 7.21 %

Total Losses at Service Loads

Δ f_{p T} = Δ f_{p E S} + Δ f_{p S R} + Δ f_{p C R} + Δ f_{p R 2} = 259.20 M P a

Stress in tendon after all losses, fpe is

= f_{p i} - Δ f_{p T} = 1396.328 - 259.20 = 1137.413 M P a

LRFD Table 5.9.3-1 states that the prestressing stress limit after all losses should be such that f_pe< 0.80 f_py.

f_pe = 1137.13 MPa < 0.80(1675.485) = 1340.388 MPa. OK

Force per strand = (f_pe) (strand area) = (1137.13) (98.7) = 112234.53 N

Therefore, the total prestressing force after all losses is:

P_pe = (112234.5) (36) = 4040443 N

I n i t i a l L o s s % = \frac{Δ f_{p} T}{f_{p i}} = \frac{259.20}{1396.238} 100 = 18.57 %