Structural Concrete - Hassoun

220 Chapter 5 Shear and Diagonal Tension
Example 5.5
A 6-m clear-span simply supported beam carries a uniform dead load of 47.5 kN/m and a live load
of 25 kN/m (Fig. 5.21). The dimensions of the beam section are b = 350 mm, d = 550 mm. The beam
is reinforced with four bars 25 mm diameter in one row. It is required to design the necessary shear
reinforcement. Given: f c ′ = 28 MPa and f y = 280 MPa.
C L
3 m
291 kN
237.65 kN
S max
130 kN
65 kN
0
x 1 = 0.98 m
x = 1.66 m
x' = 2.37 m
3 m
0.67 m
C L
Solution
Figure 5.21 Example 5.5.
1. Factored load is
2. Factored shear force at the face of the support is
1.2D + 1.6L = 1.2 × 47.5 + 1.6 × 25 = 97 kN∕m
V u = 97 × 6 = 291 kN
2
3. Maximum design shear at a distance d from the face of the support is
V u (at distance d) =291 − 0.55 × 97 = 237.65 kN
4. The nominal shear strength provided by the concrete is
V c =(0.17λ √ √
f c ′ )bd =(0.17 28)×350 × 550 = 173.2kN
V u = φV c + φV s
φV c = 0.75 × 173.2 = 130 kN
1
2 φV c = 65 kN
φV s = 237.65 − 130 = 107.65 kN
V s = 107.65
0.75 = 143.5kN