SHEAR DESIGN

Shear design will be carried out at the critical section near the left pier. Transverse shear reinforcement shall be provided when:

V_{u} > 0.5 ϕ (V_{c} + V_{p})

LRFD Eq. 5.8.2.4-1

where:

V_u = total factored shear force, kN.

V_c = shear strength provided by concrete, kN.

V_p = component of the effective prestressing force in the direction of the applied shear, kN.

Φ=resistance factor. [LRFD Art. 5.5.4.2.1]

Critical section near the supports is the greater of 0.5 d_v cot and d_v from the face of the support. [_{LRFD Art. 5.8.3.2}]

where:

d_v = Effective shear depth = distance between resultants of tensile and compressive forces, but not less than the greater of 0.9 de and 0.72 h, mm. [LRFD Art. 5.8.2.9]

d_e = the corresponding effective depth from the extreme compression fiber to the centroid of the tensile force in the tensile reinforcement = 1705 mm.

Note: d_e is calculated considering the non-prestressed reinforcement in the slab as the main reinforcement and neglecting the prestress reinforcement. This is because this section lies in the negative moment zone.

a = depth of the equivalent compression block = 108 mm

h = total height of the section = 1800 mm

θ = angle of inclination of diagonal compressive stresses, assume = 36.9º

It is obvious that the shear design at any section depends mainly on the angle of diagonal compressive stresses at this section. The shear design is an iterative process that begins with assuming a value for β. In the following example, only the final cycle of calculations are shown:

d_v = d_e - a/2= 1705 - 0.5 (108)= 1651 mm

> 0.9 d_e= 0.9 (1705)= 1534.5 mm

> 0.72 h= 0.72 (1800)= 1296.0 mm

Therefore, d_v= 1651 mm

Critical section near the supports is the greater of 0.5 d_v cot θ = 0.5 (1651) cot 36.9 ° = 1099.5 mm and d_v = 1655.4 mm. Therefore d_v controls.

Since the width of the bearing is not yet determined, it is conservatively assumed to be equal to zero for the computation of the critical section of shear. Therefore, the critical section in shear is at a distance of 300 + 1651 = 1951 mm from the centerline of

the pier. Check that b_v satisfies Eq. 5.8.3.3-2:

\frac{V_{u}}{ϕ} \leq V_{n} = 0.25 {f `}_{c} b_{v} d_{v} + V_{p}

LRFD Eq. 5.8.3.3-2

b_{v} \geq \frac{\frac{V_{u}}{ϕ} - V_{p}}{0.25 {f `}_{c} d_{v}}

Where:

V_u = applied factored shear force at the specified section, taken as positive quantity, kN.

Using load combination Strength I:

V_u = 1671.1 kN

Φ= resistance factor = 0.9 [LRFD Art. 5.5.4.2.1]

V_p = Component of the effective force in the direction of the applied shear

= (Force for 4 draped strands)(sin Ψ) where force for 4 strands

= 394.8 {m m}^{2} \times 0.75 \times 1861.58 M P a (1 - 0.1857) = 448853.43 N

Ψ = \tan^{-1} (\frac{1600 - 75 - 75}{9810}) = 8.4

V_{p} = (\frac{448853.43}{1000}) \sin (8.4) = 65.2 k N

f`_c = 45 MPa

d_v = 1.651 m

b_v = effective web width of the beam = 150 mm.

150 m m \geq \frac{(\frac{1671.1}{0.9} - 65.2) \times 1000}{0.25 \times 45 \times 1651} = 96.46 m m

0.5 cot θ = 1 - this convenient simplification is to take the flange force due to shear as V_u - V_p

Compute the factored shear force and bending moment at the critical section for shear according to Strength I load combinations. Note that in determining ε_xat a particular section, M_u is taken as positive quantity of the moment corresponding to maximum V_u, but not less than V_ud_v.[LRFD Art. 5.8.3.4.2]

V_u = 1671.1 kN

M_u = - 936.1 kN-m

V_ud_v = 1671.1 (1.651) = 2758.99 kN-m

That means that M_u<V_ud_vtherefore the value which will be used for computing ε_x is V_ud_v

Compute the Strain in the Reinforcement, ε_x

ε_{x} = \frac{\frac{M_{u}}{d_{v}} + 0.5 N_{u} + 0.5 (V_{u} - V_{p}) \cot (θ) - A_{p s} f_{p o}}{2 (E_{s} A_{s} + E_{p} A_{p s})}

LRFD Eq. 5.8.3.4.2-1

where:

N_u = applied factored normal force at the specified section = 0 kN.

f_po=a parameter taken as modulus of elasticity of prestressing tendons multiplied by the locked in difference in strain between the prestressing tendons and the surrounding concrete (ksi). For the usual levels of prestressing, a value of 0.7 f_pu will be appropriate for both pretensioned and post tensioned members, MPa.

Within the transfer length, f_po shall be increased linearly from zero at the location where the bond between the strands and concrete commences to its full value at the end of the transfer length.

Compute the C.G. of Prestressing Strands at the Critical Location for Shear:

The c.g. of the 4 draped strands= $75 + \frac{1450}{9810} (9810 - 1805) = 1258$ measured from the bottom fiber of the beam.

The c.g. for all the strands $\frac{4 (1258) + 32 (75)}{36} = 206.44$ mm measured from the bottom of the beam. Thus, eccentricity of strand group at critical section of shear is 722 - 206.5 = 515.5 mm.

f_{p o} = 0.7 f_{p u} = 0.7 \times 1968.58 = 1303.1 M P a

LRFD Art. C5.8.3.4.2

A_ps = area of prestressing steel on the flexural tension side of the member. The flexural tension side of the member should be taken the half-depth containing the flexural tension zone as illustrated in LRFD Fig. 5.8.3.4.2-3.=4(98.7)=394.8mm²

A_s = area of non prestressing steel on the flexural tension side of the member=6531mm²

V_p = 65.2 kN

As per 2003 Interims, Art. 5.8.3.4.2:

0.5 cot θ = 1 - this convenient simplification is to take the flange force due to shear as V_u - V_p
the initial value of ε_x should not taken greater than 0.001.

Therefore,

ε_{x} = \frac{\frac{M_{u}}{d_{v}} + 0.5 N_{u} + (V_{u} - V_{p}) \cot (θ) - A_{p s} f_{p o}}{2 (E_{s} A_{s} + E_{p} A_{p s})} = \frac{2758.99 (10^{6}) + (1671.1 - 65.2) 10^{3} - 394.8 (1303.1)}{2 (207000 (6531) + 197000 (394.8))} = 9.79 \times 10^{- 4} \leq 0.001

V_{u} = \frac{V_{u} - ϕ V_{p}}{ϕ b_{v} d_{v}}

where:

V_u= factored shear stress on the concrete, MPa.

b_v = Effective web width of the beam= 150 mm.

V_{u} = \frac{71 \times 10^{3} - 0.9 (652 (56.8) \times 10^{3})}{(0.9) (150) (1651)} = 7.234 M P a

(V_{u} / {f `}_{c}) = \frac{7.234}{45} = 0.161

From LRFD Fig. 5.8.3.4.2-1, using linear interpolation, are obtained the values for θ and β using (v_u/f`_c ) and ε_x.

Therefore θ = 36.9° and β= 2.04

where:

β=factor indicating ability of diagonally cracked concrete to transmit tension (a value indicating concrete contribution).[LRFD Art. 5.8.3.4]

Shear Force Carried by the Concrete:

V_{c} = 0.083 β \sqrt{f `} b_{v} d_{v} = \frac{0.083 \times 2.04 \times \sqrt{45} \times 150 \times 1651}{10^{3}} = 281.4 k N

LRFD Eq. 5.8.3.3-3

Check if $V_{u} > 0.5 ϕ (V_{c} + V_{p})$

V_{u} = 1671.1 k N > 0.5 ϕ (V_{c} + V_{p}) = 0.5 \times 0.9 \times (281.4 + 65.2) = 155.97 k N

Therefore, transverse shear reinforcement is required.

V_{n} = (V_{u} / ϕ) \leq V_{c} + V_{s} + V_{p}

LRFD Eq. 5.8.3.3-1

Where:

V_s = shear force carried by transverse reinforcement

= (V_{u} / ϕ) - V_{c} - V_{p} = \frac{1671.1}{0.9} - 281.4 - 65.2 = 1510.1 k N

= \frac{A_{v} f_{y} d_{v} (\cot (θ) + \cot (α)) \sin (α)}{S}

LRFD Eq. 5.8.3.3-4

where:

A_v = area of shear reinforcement within a distance s, mm².

s = spacing of stirrups, mm.

f_y = yield strength of shear reinforcement, MPa.

α= angle of inclination of transverse reinforcement to longitudinal axis = 90º.

Therefore, required area of shear reinforcement within a spacing s is:

A_{v} / s = \frac{V_{s}}{f_{y} d_{v} \cot (θ)} = \frac{1510.1 \times 10^{3}}{413.7 \times 1651 \times \cot (36.9 \deg)} = 1.6602 {m m}^{2} / m

Check Minimum Reinforcement Required:

The area of minimum transverse reinforcement should satisfy the following equation:

A_{v} \geq 0.083 \sqrt{{f `}_{c}} \frac{b_{v} s}{f_{y}}

LRFD Eq. 5.8.2.5-1

M i n i m u m (A_{v} / s) = 0.083 \sqrt{45} \frac{150}{413.7} = 0.20189 {m m}^{2} / m m

Thus, minimum shear reinforcement does not control.

Check Maximum Spacing of Transverse Reinforcement [LRFD Art. 5.8.2.7]:

Check if

V_{u} < 0.125 {f `}_{c}

LRFD Eq. 5.8.2.7.-1

V_{u} \geq 0.125 {f `}_{c}

LRFD Eq. 5.8.2.7.-2

where:

0.125 f`_c = 0.125 (45) = 5.625 MPa

v_u = 7.234 MPa

Since v_u > 0.125 f`_c,

then $S_{\max} = 0.4 d_{v} = 0.4 \times 1651 = 660.4 m m > 300 m m$

Therefore, maximum spacing of the stirrups = 300 mm.

SHEAR DESIGN

Compute the Strain in the Reinforcement, εx

Compute the C.G. of Prestressing Strands at the Critical Location for Shear:

As per 2003 Interims, Art. 5.8.3.4.2:

Shear Force Carried by the Concrete:

Check Minimum Reinforcement Required:

Check Maximum Spacing of Transverse Reinforcement [LRFD Art. 5.8.2.7]:

Compute the Strain in the Reinforcement, ε_x