COMPUTED LOSSES

Release

Many engineers assume a value of prestress loss at release (e.g., 10%). However, the two components of loss that are significant at release, namely the loss due to elastic shortening of the member and relaxation of the prestressed strand from the time of tensioning to the time of detensioning (cutting), can be computed with a reasonable degree of accuracy. Precast/Prestressed Girder performs this computation with reference to [10].

Steel relaxation (RET)

For stress-relieved steel (see [10]) for low-relaxation steel:

Where:

\frac{f_{s t}}{f_{p y}} - 0.5 \geq 0.05

f_{s t} = S t r e s s i n s t e e l a t t i m e t_{1} = (0.75) (270 k s i) = 202.5 k s i

f_{p y} = (0.90) ({f `}_{s}) = (0.90) (270 k s i) = 243.0 k s i

\frac{f_{s t}}{f_{p y}} - 0.5 = \frac{202.5}{243.0} - 0.55 = 0.2833 > 0.05

R E T = 202.5 (\frac{\log 18}{45}) 0.2833 = 1.600 k s i = 1600 p s i

Elastic Shortening

See Theory section

u n E S = \frac{(\frac{P}{A_{c}}) + (\frac{P_{e}^{2}}{I}) - (\frac{M_{s w}^{e}}{I})}{(\frac{E_{c i}}{E_{s}}) + (\frac{A_{s}}{A_{c}}) - (\frac{A_{s} e^{2}}{I})}

where

P	=	$(f_{s t} - R E T) A_{s} = (202.5 k s i - 1.600 k s i) (5.814 {i n}^{2}) = 1168.8 k i p s$
M_SW	=	$\frac{w l^{2}}{8} = \frac{(0.7990 k l f) {(116.00 f t)}^{2} (12 i n / f t)}{8} = 16126 k i p s$

E S = \frac{(\frac{1168.80 k}{767.0 {i n}^{2}}) (\frac{(1168.80 k) {(31.574 i n)}^{2}}{545894 {i n}^{4}}) - (\frac{(16126 k i p * i n) (31.574 i n)}{545894 {i n}^{4}})}{(\frac{4286.8 k s i}{28000 k s i}) + (\frac{5.814 {i n}^{2}}{767.0 {i n}^{2}}) + (\frac{(5.814 {i n}^{2}) {(31.574 i n)}^{2}}{545894 {i n}^{4}})} = \frac{1.5239 + 2.1345 - 0.9327}{0.1531 + 0.0076 + 0.0106} = 15.897 k s i = 15897 p s i

Total release loss = 1600 psi + 15897 psi = 17497 psi

As a percentage of initial prestress:

= \frac{17497}{202500} = 8.65 %

Final

Long-term concrete shrinkage (SH) Eq.9-4

S H = 17000 - 150 \times R H = 17000 - 150 \times 75 = 5750 p s i

Long-term concrete creep (CRc) Eq.9-9

C R c = 12 (f_{c i r}) - 7 (f_{c d s})

f_{c i r} = \frac{P_{i}}{A_{c}} + \frac{P_{i} e^{2}}{I} - \frac{M_{s t} e}{I}

where

P_i = Prestress force immediately after release= (1178105 lb)(1- 0.0865)= 1076199 lb

f_{c i r} = \frac{1076199 l b}{767.0 {i n}^{2}} + \frac{(1076199 l b) {(31.574 i n)}^{2}}{545894 {i n}^{4}} - \frac{(1612000 l b i n) (31.574 i n)}{545894 {i n}^{4}} = 1076199 l b

F_{c d s} = \frac{(M_{S D L - c o m p}) (e_{c})}{I_{c}} + \frac{(M_{S D L - \Pr e c} + M_{t o p g}) (e_{p})}{I}

where:

M_SDL-Comp= Moment due to superimposed dead load on composite

M_{S D L - C o m p} = 266.7 k i p f t = 266700 l b f t = 3200400 l b f t

e_c= Distance from composite c.g. to c.g. strands

e_{c} = 54.64 i n - 5.03 i n = 49.66 i n

M_{t o p g} = \frac{(1670.8 p l f - 799.0 p l f) {(1150 f t)}^{2} (12 i n / f t)}{8} = 17294333 l b - i n

f_{c d s} = \frac{(3200400 l b i n) 49.66 i n}{1101619 {i n}^{4}} + \frac{(3570750 l b i n + 17294333 l b i n) (31.574) i n}{545894 {i n}^{4}} = 1351 p s i

{C R}_{c} = (12) (2434 p s i) - (7) (1351 p s i) = 19751 p s i

Long-term steel creep (CRs):

{C R}_{s} = 5000 - 0.1 \times 15894 - 0.05 \times (5750 + 19772) = 2135 p s i

Total Final loss

5750 p s i + 19751 p s i + 2135 p s i + 15897 p s i = 43533 p s i

As a percentage of initial prestress

100 (\frac{43533}{202500}) = 21.5 %