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Structural Concrete - Hassoun

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Review Problems on <strong>Concrete</strong> Building Components 947<br />

Beam details are shown in Figure 23.2.<br />

26″<br />

3″<br />

Figure 23.2<br />

5 no. 8<br />

14″<br />

Detail of the beam.<br />

b. Design for shear reinforcement<br />

Calculate the factored shear from external loading:<br />

w u = 1.2w DL + 1.6w LL = 1.2(2)+1.6(0.5) =3.2K∕ft<br />

V u = w u L = (3.2)(30) = 48K<br />

2 2<br />

Calculate V u at distance d from face of support: (ACI Code, Section 9.4.3.2)<br />

( ) ( )<br />

d 26<br />

V ud = V u − w u = 48 − 3.2 = 41.07K<br />

12<br />

12<br />

Calculate φV c , φV c<br />

, V 2 c1, v c2 : (ACI Code, Section 22.5.5.1)<br />

(<br />

φV c = φ 2λ √ )<br />

√<br />

f c<br />

′ b w d = 0.75(2)(1) 5000 (14)(26) =38.60K<br />

φV c<br />

2 = 19.30K<br />

V c1 = 4 √ √<br />

f c ′ b w d = 4 5000(14)(26) =102.9K<br />

V c2 = 8 √ √<br />

f c ′ b w d = 8 5000(14)(26) =205.9K<br />

Calculate V s : (ACI Code, Section 9.5.1.1)<br />

V s = V ud − φV c<br />

φ<br />

=<br />

41.07 − 38.60<br />

0.75<br />

= 3.28K<br />

Calculate maximum bar spacing for no. 3 stirrups: (ACI Code, Section 9.7.6.2.2)<br />

Zone 1 (zone for V u between φV c , and φV c<br />

) 2<br />

s 1 = 24 in.<br />

s 2 = d 2 = 26 2<br />

= 13 in. → Controls<br />

s 3 = A v f y (2 × 0.11)(60000)<br />

= = 18.9in.<br />

50b w (50)(14)

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