CHECK OF CONCRETE STRESSES AT TRANSFER (RELEASE)

Force per strand after initial losses = 127875 N

Total prestressing force after transfer is Pi = 4603500 N

Stress Limits for Concrete [LRFD Art. 5.9.4]

Compression = 0.6f`_ci = 0.6 (35.0) = 21.0 MPa

Tension

Without bonded auxiliary reinforcement
$= - 0.25 \sqrt{{f `}_{c i}} \leq - 1.38 M P a$

$- 0.25 \sqrt{35} = - 1.479 M P a < - 1.39 M P a$

Therefore, -1.38 MPa controls.
With bonded auxiliary reinforcement, which is sufficient to resist 120% of the tension in the cracked concrete.

$= - 0.63 \sqrt{{f `}_{c i}} = - 0.63 \sqrt{35} = - 3.727 M P a$

Check of Stresses at Transfer Length Section

Due to the camber of the beam at release, the beam self-weight acts on the overall beam length. Therefore, values of bending moment displayed in the program under Service Limit State I Moments and Shears, cannot be used since they are based on the span between centerline of bearings. Using simple static, the bending moment due to beam self-weight at a distance of 762 mm from the end of the beam is:

M_{g} = \frac{0.5 \times 12.325 \times 762 \times (32700 - 762)}{10^{6}} = 150.0 k N - m

Notice that 4 strands are draped at 0.3L points.

Compute the center of gravity of the prestressing strands at the transfer length using the draped pattern. The distance between the center of gravity of the 4 draped strands at the end of the beam and the top fiber of the precast beam is 75 mm. The distance between the c.g. of the 4 draped strands at the drape point and the bottom fiber of the beam is 75 mm. The distance between the center of gravity of the 4 draped strands and the bottom fiber of the beam at the transfer length section is:

75 + \frac{(1600 - 75 - 75) (9108)}{9870} = 1413.1 m m

The distance between the center of gravity of the bottom straight 32 strands and the extreme bottom fiber of the beam is 75 mm. Therefore, the distance between the center of gravity of the total number of strands and the bottom fiber of the precast beam at transfer length is:

\frac{4 (1413.1) + 32 (75)}{36} = 223.7 m m

Eccentricity of the strand group at transfer length, e = 722 – 223.7 = 498.3 mm

The distance between the center of gravity of the total number of strands and the bottom fiber of the precast beam at the end of the beam is:

\frac{4 (1600 - 75) + 32 (75)}{36} = 236.1 m m

Compute the Top and Bottom Stresses at Transfer Length Using the Draped Pattern.

Concrete stress at top fiber of the beam:
$f_{t} = \frac{P_{i}}{A} - \frac{P_{i} e}{S_{t}} + \frac{M_{g}}{S_{t}} = \frac{4603500}{523107} - \frac{4603500 \times (722 - 223.73)}{2.173 \times 10^{8}} + \frac{150 \times 10^{6}}{2.173 \times 10^{8}} = 8.80 - 10.56 + 0.690 = - 1.07 M P a$

Tensile stress limit for concrete without bonded reinforcement = -10.38 MPa OK

Therefore, no reinforcement is required.

$= \frac{4603500}{523107} - \frac{4603500 \times 647}{2.643 \times 10^{8}} + \frac{1647.6 \times 10^{6}}{2.643 \times 10^{8}} = 8.80 - 13.707 + 7.581 = 2.674 M P a$

Compressive stress limit for concrete = +21.000 MPa OK
Concrete stress at bottom fiber of the beam:
$f_{b} = \frac{P_{i}}{A} + \frac{p_{i} e}{S_{b}} - \frac{M_{g}}{S_{b}} = \frac{460350}{523107} + \frac{4603500 \times (722 - 223.73)}{2.643 \times 10^{8}} - \frac{150 \times 10^{6}}{2.643 \times 10^{8}} = 8.80 + 8.679 - 0.568 = 16.99 M P a$

Compressive stress limit for concrete = +21.000 MPa OK

Check of Stresses at the Drape Points

The strand eccentricity at the drape points, e_c = 722 – 75 = 647 mm

The bending moment due to beam self-weight at a distance of 9810 mm from the end of the beam is:

M_{g} = \frac{0.5 \times 12.325 \times 9810 \times (32700 - 9810)}{10^{6}} = 1383.8 k N - m

Concrete stress at top fiber of the beam:
$f_{t} = \frac{P_{i}}{A} - \frac{P_{i} e}{S_{t}} + \frac{M_{g}}{S_{t}} = \frac{4603500}{523107} - \frac{4603500 \times 647}{2.173 \times 10^{8}} + \frac{1383.8 \times 10^{6}}{2.173 \times 10^{8}} = 8.80 - 13.707 + 6.368 = 1.461 M P a$

Compressive stress limit for concrete = +21.000 MPa OK
Concrete stress at bottom fiber of the beam:
$f_{b} = \frac{P_{i}}{A} + \frac{p_{i} e}{S_{b}} - \frac{M_{g}}{S_{b}} = \frac{460350}{523107} + \frac{4603500 \times 647}{2.643 \times 10^{8}} - \frac{1383.8 \times 10^{6}}{2.643 \times 10^{8}} = 8.80 + 11.269 - 5.235 = 14.834 M P a$

Compressive stress limit for concrete = +21.000 MPa OK

Check of Stresses at Midspan

The bending moment due to beam self-weight at midspan:

M_{g} = 13.25 \frac{{32.700}^{2}}{8}

Concrete stress at top fiber of the beam:
$f_{t} = \frac{P_{i}}{A} - \frac{P_{i} e}{S_{t}} + \frac{M_{g}}{S_{t}} = \frac{4603500}{523107} - \frac{4603500 \times 647}{2.643 \times 10^{8}} + \frac{1647.4 \times 10^{6}}{2.643 \times 10^{8}} = 8.80 - 13.707 + 7.581 = 2.674 M P a$

Compressive stress limit for concrete = +21.000 MPa OK
Concrete stress at bottom fiber of the beam:
$f_{b} = \frac{P_{i}}{A} + \frac{p_{i} e}{S_{b}} - \frac{M_{g}}{S_{b}} = \frac{4647578}{523107} + \frac{4603500 \times 647}{2.643 \times 10^{8}} - \frac{1647.4 \times 10^{6}}{2.643 \times 10^{8}} = 8.80 + 11.269 - 6.233 = 13.836 M P a$

Compressive stress limit for concrete = +21.000 MPa OK