## Fwn

1-cosa z may normally be taken as 0.9 d

(4) To find the least amount of shear reinforcement, for low and intermediate shear stresses, the upper limits given in (1) above for cot 0 will normally govern the design. For higher shear stresses, the largest value of cot 0 (corresponding to the lowest amount of shear reinforcement) may be found by equating the design shear force Vsd to V^d2. The amount of shear reinforcement is then found by equating the design shear force Vsd to V^d3. The value of cot 0 may alternately be selected to optimise the design, for example by minimising the total amount of reinforcement.

(5) The tensile force in the longitudinal reinforcement can be calculated from:

(6) As an alternative to Equation (4.30), the Td-curve may be constructed by displacing the Msd/z-curve a distance (cot0 - cotsa) z/2, z 0.9d (cot0 - cota)/2, in such a direction that Msd/z is always increased ("shift" rule, see 5.4.2).

(7) At indirect supports the longitudinal reinforcement should always resist the tensile force Td, defined by Equation (4.30).

(8) The second term of Equation (4.30) gives the increase of the tensile force above the value determined by taking account of only the bending moment.

4.3.2.4.5 Elements of variable depth

(1) Taking into account the variation in the internal lever arm, the design shear force is given by:

where

Vod is the design shear force in the section

Vccd the force component in the compression zone, parallel to Vod. Vtd the force component in the tensile zone, parallel to Vod.

Vccd and Vtd are taken as positive in the same direction as V(

(2) A reduction of Vod determined by Equation (4.31) Can only be combined with a reduction according to 4.3.2.4.6 if a detailed verification can be given.

4.3.2.4.6 Members with inclined prestressing tendons

(1) Taking into account the effect of inclined prestressing tendons, the design shear force is given by: VSd = Vod - Vpd (4.32)

where

Vpd denotes the force component of the inclined prestressed tendons, parallel to Vod. Vpd is taken as positive in the same direction as Vod.

(2) Concerning the value Vpd in Equation (4.32), two cases should be distinguished: Case 1 : The stresses in the tendons do not exceed the characteristic strength fp 0.1 k:

The relevant prestressing force is the mean value Pmt allowing for losses [see 2.5.4.2 P(1)] multiplied by the relevant safety coefficient (generally Yp = 0.9).

Case 2 : The steel stress in the tendons exceeds fp 0.1 k:

The prestressing force is calculated with fp 0.1 k/Ys.

(4) In shear analysis, the effective depth d is calculated ignoring the inclined tendons. 4.3.2.5 Shear between web and flanges

P(1) The shear strength of the flange may be calculated considering the flange as a system of compressive struts combined with ties in the form of tensile reinforcement.

P(2) The ultimate limit state may be attained by compression in the struts or by tension in the ties which ensure the connection between flange and web. At least a minimum amount of reinforcement shall be provided, see Chapter 5.

(3) The mean longitudinal sliding shear per unit length to be resisted is defined by:

a v where

AFd is the variation of the longitudinal force acting in a section of flange within the distance av. See Figure 4.14.

av is the distance between points of zero and maximum moments (see Figure 4.14)

(4) In the absence of more rigorous calculations, it should be checked that:

vSd r vRd3 (4.35) with vRd2 = 0.2 fcd hf (4.36) TRd is taken from Table 4.8 in 4.3.2.3. For Asf and sf, see Figure 4.14 above.

M-Mmax

Figure 4.14 — Notation for the connection between flange and web

(5) If, at the section with M = Mmax, the flange is subjected to a tensile force, the concrete term 2.5 r^d hf in Equation (4.37) should be neglected.

(6) In the case of combined shear, between the flange and web, and transverse bending, the greater of the areas of steel required to satisfy either Equation (4.37) or transverse bending should be provided.

4.3.3 Torsion

Ak Area enclosed within the centre-line of a thin-walled cross-section (including inner hollow areas)

As1 Required additional area of longitudinal reinforcement for torsion

TRd1 Maximum torsional moment resisted by compressive struts

TRd2 Maximum torsional moment resisted by reinforcement

VRd1 Design shear resistance of a section in elements without shear reinforcement

VRd2 Maximum design shear force that can be carried without web crushing t Thickness of wall u Outer circumference of a section having an area A

uk Circumference of area Ak

0 Angle between concrete struts and the longitudinal axis of the beam

V Efficiency factor rSd Tangential shear stress due to torsion

### 4.3.3.1 Pure torsion

P(1) Where the static equilibrium of a structure depends upon the torsional resistance of elements of the structure a full design for torsion covering both ultimate and serviceability limit states, will be necessary.

Where, in indeterminate structures, torsions arise from consideration of compatibility only, and the structure is not dependent on torsional resistance for its stability, then it will commonly be unnecessary to consider torsion at the ultimate limit state.

In cases where torsion is not essential for stability, torsion arising from some arrangements of structural elements may need consideration to limit excessive cracking in the serviceability limit state.

(2) In cases where torsion does not require consideration at the ultimate limit state, a minimum reinforcement in the form of stirrups and longitudinal bars should be provided to avoid excessive cracking. The requirements of 4.4.2, 5.4.2.2 and 5.4.2.3 will normally be sufficient for this purpose.

(3) The torsional resistance of sections is calculated on the basis of a thin-walled closed section. Solid sections are replaced by an equivalent thin-walled section. Sections of complex shape, such as "T" sections, are divided into a series of sub-sections each of which is modelled as an equivalent thin-walled section and the total torsional resistance taken as the sum of the capacities of the individual elements. The torsional resistance moment carried by each individual sub-section should not deviate too far from that predicted on the basis of an uncracked elastic calculation. For non-solid sections, the equivalent wall thickness should not exceed the actual wall thickness. The torsional moment carried by the individual elements according to elastic theory may be found on the basis of the St. Venant torsional stiffness. The St. Venant torsional stiffness of a non-rectangular section may be obtained by dividing the section into a series of rectangles and summing the torsional stiffness of these rectangles. The division of the section should be arranged so as to maximize the calculated stiffness.

P(4) Reinforcement for torsion shall consist of closed stirrups combined with longitudinal bars distributed around the periphery of the section. Longitudinal bars shall always be provided at all corners of the section (See 5.4.2.3).

(5) The design torsional moment should satisfy the following two conditions:

TSd r TRd1 TSd r TRd2 where

TRd1 is the maximum torsional moment that can be resisted by the compressive struts in the concrete.

TRd2 is the maximum torsional moment that can be resisted by the reinforcement.

(6) The resisting torsional moment TRd1 is given by:

t r A/u 8 the actual wall thickness. In the case of a solid section, t denotes the equivalent thickness of the wall. A thickness less than A/u can be used provided Tsd r Tr^, where TRd1 is determined from Equation (4.40). A thickness less k than twice the cover, c, to longitudinal bars is not allowed.

u _ outer circumference

A = the total area of the cross-section within the outer circumference, including inner hollow areas. Ak = the area enclosed within the centre-line of the thin-walled cross-section, including inner hollow areas.

This value applies if there are stirrups only along the outer periphery of the member. If closed stirrups are provided in both sides of each wall of the equivalent hollow section or in each wall of a box section, v can be assumed to be 0.7 - fck/200 @ 0.5.

0 = the angle between the concrete struts and the longitudinal axis of the beam and should be chosen so that:

Other values of 0 may be used provided they can be justified. (7) The resisting torsional moment T^d2 is given by:

and the additional area of longitudinal steel for torsion is given by:

Asl ' fyld = (TRd2 * uk/2Ak) * cot 0 where uk = the circumference of the area Ak s = the spacing of the stirrups

= the design yield strength of stirrups

= the design yield strength of longitudinal reinforcement Asl = the cross-sectional area of the bars used as stirrups = the required additional area of the longitudinal steel for torsion

fywd fyld

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