Shear Design

Shear design will be carried out at the critical section near the supports.

Transverse shear reinforcement shall be provided when

V_{u} > 0.5 ϕ (V_{c} + V_{p})

where:

V_u = total factored shear force, kips

V_c = shear strength provided by concrete, kips

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

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

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

where:

d_v= Effective shear depth, in

distance between resultants of tensile and compressive forces, but not less than the greater of 0.9 d_e and 0.72 h [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 = d_p, since only prestressing steel is used= 42 - 2.30= 39.70 in.

a = equivalent compressive block depth = 2.1 in.

h = total height of the section = 42.00 in.

θ = angle of inclination of diagonal compressive stresses. assume = 20.6°

For a rectangular section, the centroid of the compression block is at a/2, so d_v=d_e -a/2

For a flanged section, the centroid needs to be determined.

Shear design is an iterative process that begins with assuming a value for θ.

For this example, only the final cycle of calculations is shown.

d_{v} = d_{e} - a / 2 = 39.70 - 0.5 \times 2.1 = 38.65 i n

> 0.9 d_{e} = 0.9 \times 39.70 = 35.73 i n

> 0.72 h = 0.72 \times 42 = 30.24 i n

Therefore, d_v = 38.65 in.

Critical section near the supports is the greater of 0.5d_v cot θ = [0.5 (38.65)] cot 20.6° = 51.41 in. and d_v = 38.80 in.

Since the width of the bearing is not yet determined, it was conservatively assumed to be equal to zero for determining of the critical section for shear. Therefore the critical section for shear is at a distance of $\frac{51.41}{12 f t}$ = 4.29 ft. from centerline of support.

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}

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, kips.

Using load combination Strength I: V_u = 152.3kips

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

V^p = component of the effective prestressing force in the direction of the applied shear = 0 kips (because no draped strands are used).

f`_c = 5.5 ksi

d_v = 38.65 in

b_v = effective web width = 2 × 5 = 10 in.

10 i n \geq \frac{\frac{152.3}{0.9} - 0}{0.25 \times 5.5 \times 38.65} = 3.184 i n

Calculate the Strain In the Reinforcement On the Flexural Tension Side, 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}}{E_{s} A_{s} + E_{p} A_{p s}}

Where:

M_u = applied factored bending moment at the specified section which occurs simultaneously with V_u, taken as positive quantity, but not to be less than V_ud_v, kips-ft.

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

V_p = component of the effective prestressing force in the direction of the applied shear = 0 kips (because no draped strands are used).

Using load combination Strength I:

V_u = 152.3 kips

d_v = 38.65 in=3.22ft

M_u = 498.0 kips-ft

V_ud_v = 152.3 (3.22) = 490.406 kips-ft

That means that M_u>V_ud_v therefore the value which will be used for computing ε_x is M_u

N_u = applied factored axial force at the specified section = 0 kips.

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.

Within the transfer length, fpo 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.

f_{p o} = 0.7 (f_{p u}) = 0.7 \times 270 = 189 k s i

A_ps = area of prestressing steel on the flexural tension side of the member= 0.0 in²

As per 2005 Interims, Art. 5.8.3.4.2:

0.5cotθ =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,
$ε_{x} = \frac{\frac{M_{u}}{d_{v}} + 0.5 N_{u} + 0.5 (V_{u} - V_{p}) \cot θ - A_{p s} f_{p o}}{2 \times {(E}_{s} A_{s} + E_{p} A_{p s})} = \frac{\frac{498 \times 12}{38.65} + 0 + (152.3 - 0) - (4.131) \times 189}{2 (0 + 28500 (4.131))} = - 2 \times 10^{- 3}$

Because the value of ε_x from the 2003 Interim, Eq. 5.8.3.4.2-1 is negative, the strain should be calculated with the following equation (with the same convenient simplification: 0.5cotθ =1):

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 \times {(E}_{s} A_{s} + E_{p} A_{p s} + E_{c} A_{c})}

where:

A_c = area of concrete on the flexural tension side, the flexural tension side should be taken as one-half of the composite section depth

A_c = (0.5 × 843) = 421.5 in²

Therefore,

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 \times {(E}_{s} A_{s} + E_{p} A_{p s})} = \frac{\frac{498 \times 12}{38.65} + 0 + (152.3 - 0) - (4.131) \times 189}{2 (0 + 28500 (4.131) + 4496.06 \times 421.5)} = - 1.2 \times 10^{- 4}

Note: The sign of ε_x should be maintained.

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

where:

V_u = factored shear stress on the concrete, ksi.

V_p = component of the effective prestressing force in the direction of the applied shear = 0 (because no draped strands are used).

Φ = resistance factor = 0.9

b_v = effective web width = 2 × 5 = 10 in.

V_{u} = \frac{152.3}{0.9 \times 10 \times 38.65} = 0.4378

\frac{V_{u}}{f_{c}^{t}} = \frac{0.4378}{5.5} = 0.081

From LRFD Figure 5.8.3.4.2-1, using linear interpolation, are obtained the values for α and β using (V_u/f`_c) and ε_x

Therefore, θ=20.5° and β=4.64

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} = 0.0316 β \sqrt{{f `}_{c}} b_{v} d_{v} = 0.0316 \times 4.64 \times \sqrt{5.5} \times 10 \times 38.65 = 133.0 k i p s

Check if $V_{u} > 0.5 ϕ (V_{c} + V_{p}) = 0.5 \times 0.9 \times 133.0 = 59.85 k i p s$

Therefore, transverse shear reinforcement should be provided.

\frac{V_{u}}{ϕ} \leq V_{n} = V_{c} + V_{s} + V_{p}

where:

V_s = shear force carried by transverse reinforcement, kips

V_{s} = (\frac{V_{u}}{ϕ}) - V_{c} - V_{p} = \frac{152.3}{0.9} - 133.0 = 39.20 k i p s

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

where:

A_v = area of shear reinforcement within a distance s, in²

s = spacing of stirrups, in

f_y = yield strength of shear reinforcement, ksi.

α = angle of inclination of transverse reinforcement to longitudinal axis = 90°

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

\frac{A_{v}}{s} = \frac{F_{s}}{f_{y} d_{v} \cot θ} = \frac{39.2}{60 \times 38.65 \times \cot 20.5ͦ} = 0.064 {i n}^{2} / i n = 0.016 {i n}^{2} / f t

Check Maximum Spacing Of Transverse Reinforcement [LRFD Art.5.8.2.7]

Check if $V_{u} < 0.125 {f `}_{c}$ then $s \leq 0.8 d_{v} \leq 24 i n$

= 0.125(5.5) = 0.6875 ksi

Since v_u = 0.4372 ksi < 0.6875 ksi, then

s \leq 24 i n

\leq 0.8 d_{v} = 32.72 i n

Check Minimum Reinforcement Required [LRFD Art. 5.8.2.5]

The area of transverse reinforcement shall not be less than

A_{v} / s = 0.0316 \sqrt{{f `}_{c}} \frac{b_{v}}{f_{y}} = 0.0136 \sqrt{5.5} \frac{10}{60} = 0.012 {i n}^{2} / i n

If s = 18 in, then A_v = 0.012 × 18 = 0.216 in²

Use 2 legged #3 stirrup @ 18 in., A_v provided = 0.22 in²

Check Longitudinal Reinforcement [LRFD Art. 5.8.3.5, Interim 2005]

Longitudinal reinforcement should be proportioned so that at each section the following equation is satisfied:

T = \frac{M_{u}}{d_{v} ϕ} + 0.5 \frac{N_{u}}{ϕ} + (| \frac{V_{u}}{ϕ} - V_{p} | - 0.5 V_{s}) \cot θ

Where:

M_u = factored moment at the section corresponding to the factored shear force, kips-ft.

N_u = applied factored axial force = 0, kips.

V_u = factored shear force at section, kips.

V_s = shear resistance provided by shear reinforcement shall not be taken as greater than Vu/ Φ, kips

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

d_v = effective shear depth, in.

Φ = resistance factor as appropriate for moment, shear and axial resistance.

θ = angle of inclination of diagonal compressive stresses

Check Longitudinal Reinforcement [LRFD Art. 5.8.3.5]

For this example, the values at the critical location for shear will be used. Using load combination Strength I, the factored shear force and bending moment are:

V_u = 154.9 kips

M_u = 500.7 kips-ft

T = \frac{M_{u}}{d_{v} ϕ} + 0.5 \frac{N_{u}}{ϕ} + (| \frac{V_{u}}{ϕ} - V_{p} | - 0.5 V_{s}) \cot θ

= \frac{498 \times 12}{38.65 \times 1.0} + 0 + (| \frac{152.3}{0.9} - 0 | - 0.5 \times 39.2) \cot 20.5 = 552.82 k i p s

Therefore, reinforcement must be developed beyond the inside face of the bearing to resist 552.82 kips longitudinal force.

Shear Design

Calculate the Strain In the Reinforcement On the Flexural Tension Side, εX:

As per 2005 Interims, Art. 5.8.3.4.2:

Shear Force Carried by the Concrete

Check Maximum Spacing Of Transverse Reinforcement [LRFD Art.5.8.2.7]

Check Minimum Reinforcement Required [LRFD Art. 5.8.2.5]

Check Longitudinal Reinforcement [LRFD Art. 5.8.3.5, Interim 2005]

Check Longitudinal Reinforcement [LRFD Art. 5.8.3.5]

Calculate the Strain In the Reinforcement On the Flexural Tension Side, ε_X: