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0.0030 for mild steels in members where the characteristic compressive

strength of the concrete (5~ fractile) is between 5800 and 7250 psi. In

members where the concrete has a characteristic compressive strength

between 3600 and 5000 psi the minimum percentage of web reinforcement

must be equal to 0.0011 for web reinforcement made out of high strength

steel or 0.0024 for mild steels. The maximum stirrup spacing is 0.5*d

if Vu ~ (2/3) Vn or 0.3*d if Vu ~ (2/3) Vn• The transverse spacing of

legs in each stirrup group under no circumstance should be greater than

"d" or 800 mm <32 in.) whichever is the smaller.

In the case of members subjected to torsion, the minimum

percentage of web reinforcement is the same as in the case of shear.

The minimum area of longitudinal reinforcement must be 0.0015 btd for

high strength steels and 0.0025 btd for mild steel where bt is the width

of the member in the tension zone. However, the total tension area

should not exceed 0.04*Ag, where Ag is the cross-sectional area of the

member. The stirrup spacing shall not exceed the value of u/8. The

longitudinal bars can be uniformly distributed around the interior

perimeter formed by the stirrups, but spacing shall not exceed 350 mm

(14 in.), and at least one bar must be placed at each corner.

2.2.2 Swiss Code. The design procedure in the case of

reinforced and prestressed concrete members in the Swiss Code,

Structural Design Code SIA 162 (10), is based on the truss model with

variable angle of inclination of the diagonal compression struts.

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In the Swiss Code (10) the actions not considered directly in

the truss model are introduced in a manner similar to the one followed

in the CEB-Refined method:

a. In the geometry of the truss model: The angle of inclination of the diagonal strut is allowed to vary between the limits suggested in Eq. 2.1.

b. In an allowance for an additional diminishing concrete contribution to the shear and torsion carrying capacity of the member. The concrete contribution approaches zero as the nominal shear stress due to shear and/or torsion increases.

In the case of shear, the design procedure followed in the Swiss

Code speci fies that the shear force at calculated ultimate load minus

the vertical component of the prestressing force under service load

conditions when inclined tendons are utilized, must be equal or less

that the sum of the nominal resistance Vs carried by the truss action,

and the resistance Vc attributed to the concrete in the transition

state. The shear carried by the truss is computed using Eq. 2.15.

Vs = [AvfyZcot O']/s (2.15)

where fy is the yield stress of the stirrup reinforcement (2~ permanent

strain), "z" is the distance between the centroids of the top and bottom

longitudinal reinforcement enclosed by the stirrups andQ' is the angle

of inclination of the compression diagonal. Under the Swiss Code 3/5 <

tan 0' < 5/3.

Due to the inclination of the concrete compression field in the

truss model an area of longitudinal steel in addition to that required

for flexure must be provided. The horizontal component of this diagonal

compression field is assumed to act at the web center (z/2). If the

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resultant from the combined action of bending and shear is compressive

in the chord of the truss model where the applied moment induces

compression, then the following additional reinforcement should only be

placed on the truss chord where the applied moment induces tension.

(2.16)

where fy is the yield stress of the longitudinal reinforcement (2~

perm anent strain).

The concrete contribution Vc in the transition range between the

uncracked shear resistance of the member and its shear resistance with

the fully developed truss action is assumed to vary linearly as the

nominal shear stress increases. The Swiss Code proposed values for this

concrete contribution are shown in Fig. 2.6. The nominal shear stress

" "is taken to be Vu

or

Vu = Vu/[bw*O.8*H]

(2.17)

(2.18)

where bw is the minimum web thickness, z is the distance between centers

of the top and bottom longitudinal reinforcement enclosed by the stirrup

reinforcement, and H is the member depth.

The proposed values for the concrete contribution shown in

Fig. 2.6 are based on the limits originally suggested by Thl.irlimann for

the different stages in the behavior of a beam subjected to shear, i~.

1/6 vmax for the uncracked range and 1/2 vmax for the limit between the

transition and the full truss action stages. In determining the values

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Vcu

concrete contri bution

uncracked .. ~ Transition state ~ • Full truss

I Action

Vcu ~ Vc= 1(3Vcu-vu) >0

----+--.1 I I "-', I I ......... I , , ...

Vcu 3Vcu 6Vcu

(~ ~ox) ( ~ Vmax) (V.mox)

Vcu: Shear Stress Centri bution by the Concrete

Fig. 2.6 Concrete contribution in the case of reinforced concrete members

of the shear stress that the concrete can carry in the transition state

of the member, the following values of the uncracked shear stress

carrying capacity of the section are recommended.

If f~ = 1400 psi then vcu = 112 psi

- f~ = 2100 psi then vcu = 140 psi

f~ = 3000 psi then vcu = 168 psi

fb > 3500 psi then vcu = 196 psi

If the preceding values for vcu are put in a k.['fj form, k

would be found to vary between 3.0 and 3.3, which is in the higher range