## Beams

3.1 Introduction

This Section covers the design of beams for shear and torsion, and supplements the examples given in Section 2. The requirements for adequate safety against lateral buckling are also examined.

3.2 Design methods for shear

### 3.2.1 Introduction

EC2(1) differs from BS 8110(2) because the truss assumption used in shear design is explicit. Leading on from this, two alternative methods are given in the Code.

(1) Standard

(2) Variable Strut Inclination (VSI).

The standard method assumes a concrete strut angle of 45° (cotG = 1) and that the direct shear in the concrete, V ., is to be taken into account. This cd contrasts with the VSI method which permits the designer to choose strut angles between the limits set in the NAD(1), as shown in Figure 3.1, but ignores the direct shear in the concrete.

cot e EC2 limits

Figure 3.1 Limits of cote (VSI method)

cot e EC2 limits

Figure 3.1 Limits of cote (VSI method)

Because the direct shear in the concrete is not taken into account in the VSI method, no savings in shear reinforcement can be achieved until the applied shear exceeds three times the concrete shear (\/ > 3\/cd).

A further disadvantage of this method is that with increasing values of cotG, i.e., reductions in the concrete strut angle, the forces in the tension reinforcement increase significantly and may well outweigh any notional savings in shear reinforcement. These forces are, it should be noted, explicitly checked in EC2 but not in BS 8110. Given special circumstances the VSI method may be required but for most practical situations, the standard method will provide the most economic design.

3.2.2 Example 1 - uniformly distributed loading

The beam shown in Figures 3.2 and 3.3 is to be designed for shear.

= 30 N/mm2 (concrete strength class C30/37)

/ywk = 250 N/mm2 (characteristic yield strength of links)

The beam will be checked for shear reinforcement at three locations using both 4.3.2.4.3

the standard and VSI methods for comparison. These are 4.3.2.4.4

(2) Where l/Sd = l/Rdl, i.e., the point beyond which only minimum shear 4.3.2.2(2) reinforcement is required

(3) An intermediate point between 1 and 2.

The shear force diagram is shown in Figure 3.4.

1155 kN

1155 kN

Figure 3.4 Shear force diagram - example 1

The design shear resistance of the section, V is given by

Rd1 l Rd 1 'r cpJ w rc. = 0.34 Nimm2 for r = 30 N/mm2

Rd ck

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