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Structural Concrete - Hassoun

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19.8 Design for Shear 773<br />

Use d p = 32 in.<br />

V p = 306.2 × 1<br />

13.2 = 23.2K<br />

where 1/13.2 = slope of tendon profile = 14.5 in./(16 × 12).<br />

3.5λ √ √<br />

f c ′ = 3.5 × 1 × 5000 = 248 psi<br />

Therefore,<br />

V cw =(0.248 + 0.3 + 0.823)×6 × 32 + 23.2 = 118.2K<br />

c. Because V cw < V ci ,thevalueV cw = 118.2 K represents the nominal shear strength at section<br />

h/2 from the end of the beam. In most cases, V cw controls at h/2 from the support.<br />

4. Web reinforcement:<br />

V u = 82.3K<br />

φV cw = 0.75 × 118.2 = 88.65 K<br />

Because V u

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