LIMITS OF REINFORCEMENT

Check Minimum Reinforcement [LRFD Art. 5.7.3.3.2]

At any section, the amount of prestressed and non-prestressed tensile reinforcement should be adequate to develop factored flexural resistance, M_r, equal to the lesser of:

1.2 times the cracking strength determined on the basis of elastic stress distribution and the modulus of rupture, and
1.33 times the factored moment required by the applicable strength load combination.

Check at Midspan:

In 2005 Interims, Art. 5.7.3.3.2. was introduced the procedure to compute the cracking moment

M_{c r} = γ_{3} ((γ_{1} f_{r} + {γ_{2} f}_{p b}) S_{b c} - M_{d / n c} (S_{b c} / S_{b} - 1))

Where:

= 0.24 \sqrt{{f `}_{c}} = 0.24 \sqrt{5.5} = 0.5628 k s i

f_pb = compressive strength in concrete due to effective prestress force only (after allowance for all prestress losses) at extreme fiber of section where tensile stress is caused by externally applied loads

\frac{P_{p e}}{A} + \frac{P_{p e} e_{c}}{S_{b}} = \frac{736.465}{843} + \frac{736.456 \times 18.48}{9773} = 2.267

M_d/nc = non-composite dead load moment at the section, kips-ft

S_bc = composite section modulus for the extreme fiber of section where the tensile stress is caused by externally applied loads = 9773 in³

S_b = non composite section modulus for the extreme fiber of section where the tensile stress is caused by externally applied loads = 9773 in³

γ₁ = flexural cracking variability factor =1.6 for all other concrete strcutures

γ₂ = prestress variability factor = 1.1 for bonded tendons

γ₃ = ratio of specified munumum yield strength to ultimate tensile strength of the reinforcement= 1.00 for prestressed concrete structures

Since there is no cast-in-place topping, S_bc = S_b, therefore, the second part of the equation is zero.

M_{c r} = \frac{1.0 \times (1.6 \times 0.5628 + 1.1 \times 2.267) 9773}{12} = 2764.2 f t - k i p s

1.2 M_cr = 1.2 X 2764.2=3317.04 ft-kips

At midspan, factored moment required by Strength I load combination is

M_{u} = 3164.6 k i p s - f t

Thus, $1.33 M_{u} = 1.33 \times 3164.9 = 4209.3 k i p s - f t$

Since M_cr < 1.33 M_u, the M_cr requirement controls.

M_r = 3232.3 kips-ft > M_cr OK

Note: Contrary to AASHTO Standard Specifications, LRFD requires that this criterion be met at all sections