Design the required shear reinforcement for the simple beam shown in Figure 15.21a. Use fc′ = 3 ksi [20.7 MPa] and fs = 20 ksi [138 MPa] and single U-shaped stirrups.

Question

Design the required shear reinforcement for the simple beam shown in Figure 15.21a. Use [latex]f_c^′[/latex] = 3 ksi [20.7 MPa] and [latex]f_s[/latex] = 20 ksi [138 MPa] and single U-shaped stirrups.

Accepted Answer

The maximum value for the shear is 40 kips [178 kN], and the maximum value for shear stress is computed as

[latex] u = \frac{V}{bd}=\frac{40,000}{12×24}[/latex]= 139 psi [957 KPa]

Now construct the shear stress diagram for one-half of the beam, as shown in Figure 15.21c. For the shear design, the critical shear stress is at 24 in. (the effective depth of the beam) from the support. Using proportionate triangles, this value is

[latex](\frac{72}{96})(139)=[/latex] 104 psi [718 kPa]

The capacity of the concrete without reinforcing is

[latex] u_c = 1.1\sqrt{f_c^′}=1.1\sqrt{3000}[/latex] = 60 psi [414 KPa]

At the point of critical stress, therefore, there is an excess shear stress of 104 – 60 = 44 psi [718 – 414 = 304 kPa] that must be carried by reinforcement. Next, complete the construction of the diagram in Figure 15.21c to define the shaded portion, which indicates the extent of the required reinforcement. Observe that the excess shear condition extends to 54.4 in.[1.382 m] from the support.

In order to satisfy the requirements of the ACI Code, shear reinforcement must be used wherever the computed unit stress exceeds one-half of [latex] u_c[/latex]. As shown in Figure 15.21c, this is a distance of 75.3 in. from the support. The code further stipulates that the minimum cross-sectional area of this reinforcing be

[latex]A_ u= 50 (\frac{bs}{f_y})[/latex]

Assuming an [latex]f_y[/latex] value of 40 ksi [276 MPa] and the maximum allowable spacing of one-half the effective depth, the required area is

[latex]A_ u = 50(\frac{12×12}{40,000})[/latex]= 0.18 in²

which is less than the area of 2 × 0.11 = 0.22 in² provided by the two legs of the No. 3 stirrup.

For the maximum ν′ value of 44 ksi, the maximum spacing permitted is determined as

[latex]s=\frac{A_νf_s}{ν^′b}=\frac{0.22×20,000}{44×12}[/latex]= 8.3 in.

Since this is less than the maximum allowable of one-half the depth or 12 in., it is best to calculate at least one more spacing at a short distance beyond the critical point. For example, at 36 in. from the support, the stress is

[latex] u = (\frac{60}{96}) (139) = 87 [/latex] psi

and the value of ν′ at this point is 87 – 60 = 27 psi. The spacing required at this point is thus

[latex]s=\frac{A_νf_s}{ν^′b}=\frac{0.22×20,000}{27×10}[/latex]= 13.6 in.

which indicates that the required spacing drops to the maximum allowed at less than 12 in. from the critical point. A possible choice for the stirrup spacings is shown in Figure 15.21d, with a total of eight stirrups that extend over a range of 74 in. from the support. There are thus a total of 16 stirrups in the beam, 8 at each end. Note that the first stirrup is placed at 4 in. from the support, which is one-half the computed required spacing; this is a common practice with designers.

Question 15.6: Design the required shear reinforcement for the simple beam ...

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