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Structural Concrete - Hassoun

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5.11 Examples Using SI Units 221<br />

5. Distance from the face of the support at which 1 φV 2 c = 65 kN is<br />

x ′ 291 − 65<br />

= (3) =2.33 m (from triangles)<br />

291<br />

6. Design of stirrups:<br />

a. Choose stirrups 10 mm in diameter with two branches (A s = 78.5 mm 2 ).<br />

A v = 2 × 78.5 = 157 mm 2<br />

Spacing S 1 = A y f yt d 157 × 280 × 550<br />

= = 168.5mm< 600 mm<br />

V s 143.5 × 10 3<br />

b. Check for maximum spacing of d/4:<br />

S max =<br />

{ (<br />

1 d for V 2 s ≤ 0.33 √ )<br />

f c<br />

′ bd<br />

1<br />

d for If V 4 s > (0.33√ f c ′ )bd<br />

bd(0.33 √ √<br />

f c ′ )=0.33 28 × 350 × 550 = 336.1kN<br />

Actual V s = 143.5 kN < 336.1 kN. Therefore, S max is limited to d/2 = 275 mm.<br />

7. The shear reinforcement, stirrups 10 mm in diameter and spaced at 160 mm, will be needed<br />

only for a distance d = 0.55 m from the face of the support. Beyond that, the shear stress V s<br />

decreases to zero at a distance x = 1.66 m when φV c = 130 kN. It is not practical to provide<br />

stirrups at many different spacings. One simplification is to find out the distance from the face of<br />

the support where maximum spacing can be used and then only two different spacings may be<br />

adopted:<br />

Maximum spacing = 1 d = 275 mm<br />

2<br />

V s (for s max = 275 mm) = A v f yt d<br />

S<br />

=<br />

φV s = 87.9 × 0.75 = 65.94 kN<br />

157 × 0.280 × 550<br />

275<br />

= 87.9kN<br />

The distance from the face of the support where S max = 275 mm can be used (from the triangles):<br />

291 −(130 + 65.94)<br />

x 1 = (3) =0.98 m<br />

291<br />

Then, for 0.98 m from the face of the support, use stirrups 10 mm in diameter at 160 mm, and<br />

for the rest of the beam, minimum stirrups (with maximum spacing) can be used.<br />

8. Distribution of stirrups:<br />

One stirrup at S 2 = 160 = 80 mm<br />

2<br />

Six stirrups at 160 mm = 960 mm<br />

Six stirrups at 270 mm = 1620 mm<br />

Total = 1040 mm = 1.04 m > 0.98 m<br />

Total = 2660 mm = 2.66 m < 3m<br />

The last stirrup is 3 − 2.66 = 0.34 m = 340 mm from the centerline of the beam, which is adequate.<br />

A similar stirrup distribution applies to the other half of the beam, giving a total number of<br />

stirrups of 28.<br />

The other examples in this chapter can be worked out in a similar way using SI equations.

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