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Structural Concrete - Hassoun

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948 Chapter 23 Review Problems on <strong>Concrete</strong> Building Components<br />

Zone 2 (zone for V u >φV c ) and V s < V c1<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)<br />

s 4 = A v f y d (2 × 0.11)(60)(26)<br />

= = 104.7in.<br />

V s (3.28)<br />

Using similar triangles from Figures 23.3<br />

48<br />

180 = 19.30<br />

x 1<br />

x 1 = 72.40 in.<br />

48<br />

180 = 38.60<br />

x 2 + 72.40<br />

x 2 = 72.40 in.<br />

First stirrup at face of support<br />

Second stirrup at s/2 = 13/2 = 6.5 in.<br />

8 stirrups at 13 in.→ 110.5 in.<br />

x 3 = 180 − 72.40 − 72.40 = 35.20 in.<br />

V u = 48 K<br />

ϕV c = 36.60 K<br />

Zone II<br />

Zone I<br />

ϕV c<br />

= 19.30 K<br />

2<br />

35.20″<br />

72.40″<br />

180″<br />

72.40″<br />

Figure 23.3<br />

beam.<br />

Shear reinforcement distribution diagram from the face to midspan of<br />

c. Check development length<br />

Calculate the development length for tension bar l d .<br />

Check if conditions for spacing and cover are met to select an equation: (ACI Code,<br />

Section 25.4.2.2)<br />

d b = 1in.<br />

clear cover = 2.5in.>d b<br />

clear spacing = 14 − 6 − 1 = 1in. ≥ d<br />

4<br />

b

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