5. Flexural, Shear, and Torsion Design

5.1. Flexural Design

For this beam, the moment at the fixed end governs the design as shown previously.

Use #8 bars with 1.50 in. concrete cover per ACI 318-14 (Table 20.6.1.3.1) and #4 stirrups. The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:

To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the beam section (jd). In this example, tension-controlled section will be assumed so the reduction factor ϕ is equal to 0.9, and jd will be taken equal to 0.9253 × d. The assumptions will be verified once the area of steel is finalized.

The required reinforcement at initial trial is calculated as follows:

Where:

Therefore, the assumption that section is tension-controlled is valid.

The minimum reinforcement shall not be less than

And not less than

Torsion requirements for longitudinal steel have to be determined and combined with reinforcement area required for flexure.

5.2. Torsion Design

Check if torsional effects can be neglected:

Where:

Check if the factored design torsion can be reduced:

It is permitted to reduce T_u to ϕT_cr; due to redistribution of internal forces after torsional cracking; if the exterior continuous beam meet the following requirements:

ACI 318-14 (22.7.3.2)

1. The beam is statically indeterminate (continuous beam).

2. T_u ≥ ϕT_th.

The Torsion is needed for equilibrium; therefore, design for:

T_u = 28.00 kip-ft (Figure 2)

The torsional properties of the beam are the following:

5.2.1. Adequacy of Cross-Sectional Dimensions for Torsion

For solid sections, the limit on shear and torsion is given by:

Using d = 21.50 in., the factored shear force at the critical section located at a distance d from the face of the support is:

V_u = 57.14 kip-ft (Figure 2)

Also, the nominal shear strength provided by the concrete is:

Using a 1.50 in. clear cover to # 4 closed stirrups.

5.2.2. Transverse Reinforcement Required for Torsion

Where:

Therefore,

5.2.3. Transverse Reinforcement Required for Shear

V_u = 57.14 kip-ft (Figure 2)

Shear strength provided by concrete

5.2.4. Total Required Transverse Reinforcement for Combined Shear and Torsion

Minimum transverse reinforcement for shear and torsion is calculated as follows:

Calculate the required spacing:

Maximum spacing of transverse torsion reinforcement:

Maximum spacing of transverse shear reinforcement:

Check whether the required spacing based on the shear demand meets the spacing limits for shear reinforcement per ACI 318-14 (9.7.6.2.2).

Therefore, maximum stirrup spacing shall be the smallest of d/2 and 24.00 in.

For #3 stirrups, A_v+t (two legs) = 0.22 in.², and the required s = 0.22 / 0.0704 = 3.13 in. < 10.75 in.

For #4 stirrups, A_v+t (two legs) = 0.40 in.², and the required s = 0.40 / 0.0704 = 5.69 in. < 10.75 in.

Use No. 4 closed stirrups at 5.00 in. on centers, placing the first stirrup at 2.50 in. from the face of the support.

5.2.5. Additional Required Longitudinal Reinforcement for Torsion

Where:

Therefore,

The minimum total area of longitudinal torsional reinforcement:

Since A_l > A_l,min, use A_l = 1.265 in.²

The longitudinal reinforcement is to be distributed around the perimeter of the stirrups, with a maximum spacing of 12.00 in. There shall be at least one longitudinal bar in each corner of the stirrups.

ACI 318-14 (9.7.5.1)

Longitudinal bars shall have a diameter at least 0.042 times the stirrup spacing, but not less than 3/8 in.

ACI 318-14 (9.7.5.2)

Provide 4 - #4 bars in the lower portion of the beam (two at mid-depth and two at bottom) and add 1.265 in.² (Section 5.2.5) - (4 × 0.20) = 0.465 in.² to the flexural steel.

The total area of steel required in top face of the beam is 2.550 in.² (Section 5.1) + 0.465 in.² = 3.015 in.². Thus, we have:

6 - #7 bars: A_s = 3.600 in.², will not fit in one layer.

4 - #8 bars = A_s = 3.160 in.², will fit in one layer.

Provide 4 - #8 bars at the top of the beam. (see Figure 4).

ACI 318-14 (9.5.4.5) allows a designer to subtract M_u / (0.9 × d × f_y) from the area of longitudinal steel in the flexural compression zone. We shall not do this, for three reasons:

1. M_u varies along the length of the beam.

2. We need two bars of minimum diameter 0.21 in. in the corners of the stirrups.

3. We need bars to support the corners of the stirrups.

Figure 4 – Reinforced Concrete Cantilever Beam Cross Section