INTERFACE SHEAR AND SHEAR FRICTION Shear-Friction

There are situations where shear has to be transferred across a defined plane of weakness, nearly parallel to the shear force and along which slip could occur (Fig.

6.11). Examples are planes of or potential cracks, interface between dissimilar materials, interfaces elements such as webs and flanges, and interface between concrete placed at different In such cases, possible failure involves sliding the of weakness rather than diagonal tension.

it would be appropriate to consider resistance developed along such planes in the form of resistance to the tendency to slip. The concept is a method to do this.

When two bodies are in contact with a normal reaction, R, across the surface of contact, the frictional resistance, F, acting tangential to this surface and resisting relative slip is to be where is the of friction (Fig.

Figure shows an cracked concrete specimen loaded in shear. In such a specimen, a clamping between the two faces of the crack can be induced by providing reinforcement (shear-friction reinforcement, perpendicular to the crack surface. Any slip between the two faces of rough irregular crack causes the

'The is tied to the longitudinal bar using 'binding wire'

faces to ride upon each other, which opens up crack (Fig. This in turn induces tensile forces in the which ultimately yields (Fig. If the of reinforcement is A,, and yield at ultimate, the clamping force between the two is R = and the frictional rcsistance is

patentia crac

crack

...

Corbel (b) Precast beam seat Column face plate

Fig. 6.11 Typical cases friction is applicable (adapted from Ref. 6.10) In reality, the actual resistance to shear, is of this frictional force

the resistance to shearing off of the on the irregular surface of the crack, the force developed in the transverse reinforcement, and when there are no cracks developed yet, the cohesion between the two parts as well. The nominal or without safety factors) shear resistance, due to the friction the crack is given 6.30. Other less simple methods of calculation have been proposed 6.11, 6.12) which result in of shear transfer rcsistance in substantial with comprehensive test results.

For shear-friction reinforcement placed to the shear plane,

(6.30) where, nominal shear resistance due to the assumed friction part

alone contributed by reinforcement stress

area of shear-friction reinforcement, placed normal to the plane of possible slip

coefficient of friction.

Shear-friction reinforcement may also be placed at an angle q t o the shear plane, such that the shear force produces tension in the reinforcement, as shown in Fig. As the shear-friction reinforcement yields, the tensile force in the reinforcement is which has a component to the shear plane of and a normal to the plane equal to sin The latter produces the clamping force. total force resisting shear is then obtained

cos sin and the nominal shear rcsistance is given by:

sin (6.31)

(a) Fictional farce

(b) Direct shear crack ^{Slip and} separation Fig. 6.12 Shear-friction

(d) Internal stresses

If the area of concrete section at the interface resisting shear transfer is A,,, the nominal resistance per area can be expressed (from Eq. 6.31) as:

where A,, and nominal shear resistance due to the reinforcement.

If there is a load, normal to the interface, this will either increase or decrease the effective normal pressure across the interface, and correspondingly affect the shear resistance associated with shear-friction, depending on whether it is compressive or tensile (Fig. Taking if compressive, the effective normal pressure R across the interface will then be:

R = sin

+

Reinforcement inclined at an angle is ineffective in resisting interface shear, because, as relative slip between the two parts occurs and the reinforcement deforms, the effect is to separate the two parts farther rather than to introduce any clamping forces (Fig. Hence reinforcement with 90" placed such that shear force produces tension in the bar) only is effective as shear-friction reinforcement. [Indeed, this type of inclined bars will be more effective than perpendicular to the interface as tensile strains are initiated in the and more effectively than in the latter].

254 REINFORCED CONCRETE DESIGN general case of inclined shear-friction reinforcement and normal force can be obtained as:

The Code IS 456 : 2000 does not give any guidance related to shear friction concepts.

T h e Canadian standard CSA A23.3 recommends the following formula for determining the factored interface shear resistance, v,, based on the shear-friction concept:

DESIGN FOR SHEAR 255

where, = (6.34)

A,, area of concrete section resisting shear A, gross section transferring N

area of reinforcement

inclination of reinforcement with shear plane ratio of shear-friction reinforcement

coefficient of friction

, material resistance factors for concrete and steel reinforcement and A factor to account for low density concrete

The to the material strengths as

multipliers, used in the Canadian Code format correspond to the of the factors (see Sections 3.5.4 and 3.6.2) used in the IS Code and have values of 0.60 0.85 rcspectivcly. These compare with corresponding values of 111.5 0.67 and 111.15 = 0.87 used in IS 456 for concrete and steel. The design Eq. 6.33 is obtained from the nominal strength Eq. 6.32, by introducing the safety factors und for concrete and steel and, in addition, a density factor to allow for low density concrete (which has lower shear strength) when used. Recommended values for A are 1.00 for density concrete, 0.85 for structural density concrete and 0.75 for structural low density concrete. The CSA recommends the following values for and

6.2 of c and to be used with Eq. 6.33

Any direct tension, across plane must be provided for by additional reinforcement having an area equal to Such tensile forces be caused by restraint of deformations to temperature change, creep and etc.

Although there is a effect of a permanently net compressive force across the shear plane that the amount of shear-friction reinforcement required, it is prudent to this effect. When there is a bending moment acting

concrete, clean and intentionally roughened

on the shear plane, the flexural tensile and compressive forces balance each other, and the ultimate compressive force across plane induces the frictional resistance) is equal to Hence, the flexural reinforcement A,, can be included in area computing V,. When there is no bending moment acting on the sliear plane, shear-friction reinforcement is best distributed uniformly along the shear plane in order to minirnise crack widths. When a bending moment also exists, most of the shear-friction reinforcement is placed closer to the tension face to provide required effective depth. Since it is assumed that shear-friction reinforcement yields at the ultimate strength, it must be anchored on sides of the shear plane so as to develop the specified yield strength in tension

Equation 6.32 can be adapted to the IS code format by the partial safety factors. Thus introducing the factors given in Section 3.6.2, the interface shear resistance may be taken as:

where, =

Values for c and given in Table 6.2 may be

.

6.10 SHEAR CONNECTORS IN FLEXURAL MEMBERS