VERTICAL SHEAR AT H/2

Calculations are done for the section at H/2 from the left support, and at H/2 from the right support (111.65 ft. from left support), the latter one controls. Therefore calculations are shown for the section at H/2 from the right support:

_d_{self wt}_topping_DL-comp

(- 43.26 k) + (- 9.75 k) + (- 47.21 k) + (- 17.1 k) = - 117.3 k

_d_{self wt}_topping_DL-comp

(149.6 k - f t) + (33.70 k - f t) + (163.25 k - f t) + (- 277.661 k) = 67.9 k - f t

M₁=-1342.6k-ft

_u $1.3 (D L + 1.67 L L) = 1.3 (117.3 k + (1.67) (79.9 k)) = 326.0 k$ (Table 3.22.1A)

M_u=1.3[67.9 k-ft + (1.67)(-1342.6 k-ft)]=-2825.3k-ft

V_mu = Shear occurring simultaneously with M

= 1.3[117.3 k + (1.67)(63.0 k)]=288.7k

M_max = M_u-M_d

=-2825.3 k-ft - (67.9 k-ft)=-2894.2k-ft

V_i =V_mu-V_d

=289.2 k - 117.3k=-171.9k

f_pe= Compressive stress in concrete due to effective prestress forces only at extreme fiber of section where tensile stress is caused by externally applied loads. (Art. 9.1.2)

As stated above, the interior end of the beam is where shear will be critical. At that end, the beam is under net negative flexure. Thus, fpe must be evaluated at the top of the deck. Prestress has no effect on the deck. Thus,

f_pe = 0

f_d = Stress due to unfactored dead load, at extreme fiber of section where tensile stress is caused by externally applied loads.

As above, fd must be evaluated at the top of the deck since section is in region of negative moment.

f_{d} = \frac{(- 277.66 k - f t) (12000 \frac{l b - i n}{k - f t})}{55882.7 {i n}^{3}} = - 59.6 p s i

M_cr = Moment causing flexural cracking of section due to externally applied loads (beyond dead load).

_cr

- 6 \sqrt{{f `}_{c t}} = - 6 \sqrt{3500} = - 354.96 p s i

At H/2, total dead load stress in top of slab = - 59.8 psi

Thus, net stress to $- 6 \sqrt{{f `}_{c t}}$ is

- 354.96 p s i - (- 59.8 p s i) = - 295.76 p s i

_cr

\frac{(55882.7 {i n}^{3}) (- 295.16 p s i)}{(12000 \frac{l b - i n}{k - i n)})} = - 1375.4 k - f t

d = Distance from extreme compressive fiber to centroid of negative moment reinforcing for precast girder bridges made continuous. (Art. 9.1.2)

Negative moment reinforcement is assumed to be located at the mid-height of the slab. Thus,

80.50 i n - \frac{8.50 i n}{2} = 76.25 i n

_ci-com_c

0.6 \sqrt{{f `}_{c}} (b ` d) + V_{d} + \frac{V_{i} M_{c r}}{M_{\max}} = 0.6 (\sqrt{6000}) (6 i n) (76.25 i n) + 1173000 l b + \frac{(171900 l b) (1374400)}{2884100 l b - f t} = 220235 l b = 220.2 k

Eq.9-27

_ci-min_c

1.7 \sqrt{{f `}_{c}} (b ` d) = 1.7 (\sqrt{6000}) (6 i n) (76.25 i n) = 60244 l b = 60.2 k

Art 9.20.2

V_ci=maximum of V_ci-com and V_ci-min=220.2k

f_pc = Since the centroid of composite section is not located within the flange of the precast section, fpc is compressive stress at centroid of composite section, rather than at junction of web and flange. (Art. 9.1.2)

Presstress

Eccentricity of strands at H/2:

e = 16.31 i n + \frac{(31.57 i n - 16.31 i n) (0.50 f t + 3.35 f t)}{(0.5 f t + 45.9 f t)} = 17.58 i n

F_{p r e s t r e s s} = (924.65 k) (\frac{1}{767.0 {i n}^{2}} - \frac{17.58 i n}{30177 {i n}^{3}}) = 0.6669 k s i = 666.9 p s i

Non-composite loads

f_{p c} = \frac{(149.60 k - f t + 33.70 k - f t + 163.25 k - f t) (12000 \frac{l b - f t}{k - f t})}{30177 {i n}^{3}} = 137.81 p s i

f_{p c} = 666.9 p s i + 137.6 p s i = 804.5 p s i

V_p = Vertical component of effective prestress force at section. (Art. 9.1.2)

FIGURE 9. Component Forces of Total Prestress Force

θ = \tan^{-1} (\frac{31.57 i n - 16.31 i n}{6.00 i n + 550.8 i n}) = 1.5699̊

V = (924.65 k) \sin (1.5699̊) = 25.1 k

V_{c w} = (3.5 \sqrt{{f `}_{c}} + 0.3 f_{p c}) b ` d + V_{p}

Eq.9-29

d need not be taken less than 0.8h

0.8(80.50 in) = 64.40 in

Therefore, d = 76.25 in

V_{c w} = (3.5 \sqrt{6000} k s i + 0.3 (804.5 k s i)) (600 i n) (76.25 i n) + 25100 l b = 259.5 k

V_c = Minimum of V_ci and V_cw=220.2k

V_{u} = ϕ (V_{c} + V_{c w})

V_{s} = \frac{V_{u}}{ϕ} - V_{c}

ϕ = 0.9

V_s-req'd = Force to be taken by shear reinforcement

V_{s} = \frac{326.0 k}{0.90} - 220.2 k = 142.0 k

V_{s - \max} = 8 \sqrt{f ` c} (b ` d) = 8 \sqrt{6000} k s i (6.00 i n) (12.00 i n) = 283502 l b = 283.5 k

V_{s} = \frac{A_{v} f_{s y} d}{s}

(Eq.9-30)

Set s = 12 in to have units of in²/ft. for A_v

A_{v - c o m} = \frac{V_{s} s}{f_{s y} d} = \frac{142.0 k \times 12 i n}{60 k s i \times 76.25 i n} = 0.37 {i n}^{2}

A_{v - \min} = \frac{50 b ` s}{f_{s y}} = \frac{50 k s i \times 6.00 i n \times 12 i n}{60 k s i} = 0.06 {i n}^{2}

A_v=Maximum of A_v-com and A_v-min= 0.37 in²(per foot)

V_{s - c r i t} = 4 (\sqrt{{f `}_{c}}) b ` d = 4 \sqrt{6000} p s i \times 6.00 i n \times 76.25 i n = 141751 l b = 141.8 k

Art 9.20.3.2

Since V_s > V_s-crit, spacing of stirrups will have to be reduced by one-half. (Art. 9.20.3.2)

Max spc = 0.75h or 24 inches (Art. 9.20.3.2)

0.75h = (0.75)(76.25 in) = 57.19 in

Therefore,

M a x s p c = \frac{24 i n}{2} = 12 i n