CHECK CONCRETE STRESSES AT TRANSFER (RELEASE)

Total prestressing force after transfer:

P_i= 727.89 kips

Stress Limits for Concrete [LRFD Art. 5.9.4]

Compression = 0.6 f`_ci = 0.6(4.5) = 2.7 ksi

Tension:

Without bonded auxiliary reinforcement
$= - 0.0948 \sqrt{{f `}_{c i}} = - 0.0948 \sqrt{4.5} = - 0.201 < - 0.200 k s i$

Therefore, -0.200 ksi controls.
With bonded auxiliary reinforcement which is sufficient to resist 120% of the tension force in the cracked concrete
$- 0.22 \sqrt{{f `}_{c i}} = - 0.22 \sqrt{4.5} = - 0.467 k s i$

Check of Stresses at Transfer Length Section

This section is located at a distance equal to the transfer length from the end of the beam. Stresses at this location need only be checked at release, because losses with time will reduce the concrete stresses making them less critical.

Transfer length = 60 times the strand diameter [LRFD Art. 5.8.2.3]

=60(0.5)=30in=2.5ft

The transfer length is located at 2.5 ft from the end of the beam or at 2.0 ft from centerline of the bearing. Due to the camber of the beam at release, the self-weight of the beam acts on the overall release beam length, 101 ft. Using simple statics, bending moment at the transfer length due to beam is:

M_{g} = 0.5 w (L - x) = 0.5 (0.878) (2.5) (101 - 2.5) = 108.1 k i p s - f t

Concrete stress at the top fiber of the beam, f_t:
$f_{t} = \frac{P_{i}}{A} - \frac{P_{i} e_{c}}{S_{t}} + \frac{M_{g}}{S_{t}} = \frac{804.465}{843} - \frac{804.465 \times 18.48}{9571} + \frac{108.1 \times 12}{9571} = 0.954 - 1.553 + 0.135 = - 0.464$

Tension stress limit for concrete with bonded reinforcement = -0.467 ksi OK
Concrete stress at the bottom fiber of the beam, f_b:
$f_{b} = \frac{P_{i}}{A} - \frac{P_{i} e_{c}}{S_{b}} + \frac{M_{g}}{S_{b}} = \frac{804.465}{843} + \frac{804.465 \times 18.48}{9773} - \frac{108.1 \times 12}{9773} = 0.954 + 1.521 - 0.133 = - 2.343$

Compression stress limit for concrete = 2.700 ksi OK
Calculate As-top, required steel at top of precast section:
- Depth of neutral axis,c₁* is :
  $h \times \frac{f_{t}}{f_{b}} = 42 \times \frac{- 0.464}{- 0.464 - 2.343} = 6.94 >$ _ft
  
  Therefore, the neutral axis is within the web.
- Stress of bottom of top flange:
  $f_{f t} = \frac{c_{1}^{*} - h_{f t}}{c_{1}^{*}} \times (- f_{t}) = \frac{6.94 - 5.5}{5.5} \times (- 0.412) = - 0.097 k s i$
- Steel required at top of precast sections:
  $A_{s - t o p} = \frac{\frac{f_{t} + f_{f t}}{2} \times b_{f t} \times h_{f t} + \frac{f_{f t}}{2} (c_{1}^{*} - h_{f t}) \times b_{v}}{f_{s}}$
  
  As per 2005 Interims, Art. 5.9.4.1.2: fs = min (30ksi, 0.5fy)
  
  Therefore, fs = min (30ksi, 0.5 x 60) = 30ksi
  
  $A_{s - t o p} = \frac{\frac{0.464 + 0.097}{2} \times 48 \times 5.5 + \frac{0.097}{2} (6.94 - 5.5) \times 10}{30} = 2.492 {i n}^{2}$

Check of Stresses at Midspan

Bending moments due to beam self-weight at midspan, based on overall length, is 1119.6 ft-kips

f_{t} = \frac{P_{i}}{A} - \frac{P_{i} e_{c}}{S_{t}} + \frac{M_{g}}{S_{t}} = \frac{804.465}{843} - \frac{804.465 \times 18.48}{9571} + \frac{1119.6 \times 12}{9571} = 0.954 - 1.553 + 1.403 = 0.804

Tension stress limit for concrete with bonded reinforcement = - 0.467 ksi OK

f_{b} = \frac{P_{i}}{A} - \frac{P_{i} e_{c}}{S_{b}} + \frac{M_{g}}{S_{b}} = \frac{804.465}{843} + \frac{804.465 \times 18.48}{9773} - \frac{1119.6 \times 12}{9773} = 0.954 + 1.521 - 1.375 = 1.100 k s i

Compression stress limit for concrete = + 2.700 ksi OK