CHECK OF STRENGTH LIMIT STATE

Positive Moment Section

Total factored bending moment for Strength I is:

M_u = 1.25 (DC) + 1.5 (DW) + 1.75 (LL + IM)

(LRFD Tables 3.4.1-1&2)

At midspan of center span:

M_u = 1.25(1619.6 + 2153.7 + 111.3 + 87.4) + 1.5(148.2) + 1.75(2485.2) = 9536.4 kN-m

Average stress in prestressing steel when

f_{p e} \geq 0.5 f_{p u}

f_{p s} = f_{p u} (1 - k \frac{C}{d_{p}})

(LRFD Eq. 5.7.3.1.1-1)

where

f_ps	=	average stress in prestressing steel, MPa
k	=	$2 (1.04 - \frac{f_{p y}}{f_{f u}})$ (LRFD Eq. 5.7.3.1.1-2) Note: k = 0.28 for low relaxation strands
d_p	=	distance from extreme compression fiber to the centroid of the prestressing tendons. $= h - y_{b s} = 190 + 10 + 1600 - 75 = 1725 m m$
c	=	distance between the neutral axis and the compression face, mm

To compute c, assume rectangular section behavior, and check if c is equal to or less than t_s. (LRFD C5.7.3.2.2).

c = \frac{A_{p s} f_{p u} + A_{s} f_{y} - {A `}_{s} f_{y}}{0.85 {f `}_{c} β_{1} b + k A_{p s} \frac{f_{p u}}{d_{p}}}

where

A_ps	=	area of prestressing steel = 36 × 98.7 = 3553.2 mm²
f_pu	=	specified tensile strength of prestressing steel = 1861.580 MPa
A_s	=	area of mild steel tension reinforcement = 0.0 mm²
f_y	=	yield strength of tension reinforcement, MPa
A`_s	=	area of compression reinforcement = 0.0 mm²
f`_y	=	yield strength of compression reinforcement, MPa
f`_c	=	compressive strength of deck concrete = 28.0 MPa
β₁	=	stress factor of compression block (LRFD Art. 5.7.2.2) $= 0.85 f o r {f `}_{c} \leq 28 M P a$ $= 0.85 - \frac{0.05 {f `}_{c} - 28}{7} \geq 0.65 f o r {f `}_{c} \geq 28 M P a$
b	=	effective width of compression flange = 2892.5 mm

c = \frac{3553.2 \times 1861.580 + 0.0 - 0.0}{(0.85 \times 28 \times 0.85 \times 2892.5 + 0.28 \times 3553.2 \times \frac{1861.580}{1725.0})} = 111.0 m m

We check if βx c < t_s as per 2006 Interims Art5.7.3.2.3

= 0.85 x 111.0 = mm OK

Therefore, rectangular section behavior assumption is valid. Therefore, average stress in prestressing steel:

f_{p s} = 1861.580 (1 - 0.28 \frac{111.0}{1725}) = 1828.0 M P a

a = depth of the equivalent stress block

= β_{1} c = 0.85 \times 111.0 = 94.4 m m

Nominal flexural resistance (LRFD Art. 5.7.3.2.3):

M_{n} = A_{p s} f_{p s} (d_{p} - \frac{a}{2})

The above equation is a simplified form of Eq. 5.7.3.2.2-1 because no compression reinforcement or mild tension reinforcement are considered and the section behaves as a rectangular section.

m = \frac{f_{y}}{0.85 {f `}_{c}} = \frac{413.7}{0.85 \times 45} = 10.816

M_{n} = \frac{3553.2}{10^{6}} (1828.0) (1725 - \frac{94.4}{2}) = 10899.1 k N - m

Factored flexural resistance:

M_{r} = Φ M_{n}

LRFD Eq. 5.7.3.2.1-1

where:

As per2006 Interims, the c/dt value will be used in computing the resistance factor

dt - the distance from the extreme compression fiber to the extreme tension steel element.

dt = h - ysteel = 190 + 10 + 1600 - 50 = 1750 mm

The report c/dt is = 110/1750 = 0.063 < 0.375

Therefore, the section is Tension Controlled and the resistance factor Φ= 1

M_r = 10899.1 kN-m > M_u = 9536.4 kN-m OK

Negative Moment Section

Design of the Section

Total ultimate bending moment for Strength I is:

Mu = 1.25(DC) + 1.5(DW) + 1.75(LL + IM) (LRFD Tables 3.4.1-1&2)

At the pier section:

Mu = -5986.1 kN-m

Note that at negative moment section, the compression face is the bottom flange of the beam, which is 975 mm wide and 135 mm thick. This section is being designed as non-prestressed reinforced concrete section, thus Φ = 0.9 for flexure. The bottom flange of the beam is in compression so concrete compressive strength of precast at final is used, f`_c = 45 MPa

Assume the deck reinforcement is at mid height of the deck, so the effective depth, d, is:

d = 1600 + 10 + 0.5 (190) = 1705 mm

R_{u} = \frac{M_{u}}{Φ b d^{2}} = \frac{5986.1 \times 10^{6}}{0.9 \times 975 \times 1705^{2}} = 2.347 M P a

ρ = \frac{1}{m} (1 - \sqrt{(1 - \frac{2 R_{u} m}{f_{y}})}) = \frac{1}{10.816} (1 - \sqrt{1 - \frac{2 \times 2.347 \times 10.816}{413.7}}) = 0.005858

As = (ρbd) = (0.005858) (975) (1705) = 9738.19 mm²

This is the amount of mild reinforcement required in the slab to resist the negative moment. Compute the depth of the compression block:

a = \frac{A_{s} f_{y}}{0.85 b {f `}_{y}} = \frac{9409.04 \times 413.7}{0.85 \times 975 \times 45} = 108.03 m m

where

a	=	108.03 mm
β₁	=	$0.85 - \frac{0.05 ({f `}_{c} - 28)}{7} = 0.85 - \frac{0.05 (45 - 28)}{7} = 0.729$
c	=	$\frac{a}{β_{1}} = \frac{108.03}{0.729} = 148.19 m m$

As per 2006 Interims, Art 6.7.3.2.3 we compare a = b1 x c = 108.03mm < hf = 135mm, therefore the rectangular section behavior assumption is valid.

ϕ M_{n} = ϕ (A_{s} f_{y} (d - \frac{a}{2})) = 0.9 \times (9.738 \times 413.7 (1705 - \frac{108.09}{2}) 975) = 5986.1 k N - m