Info Shear Friction (Sliding Shear). The shear design performed in the previous section is intended to prevent diagonal tension failures rather than direct shear transfer failures. Direct shear transfer failure, also referred to as sliding shear failure, can occur by the sliding of two vertical segments of a wall at weak sections such as at construction joints. The shear resistance is verified by using the equation


Avf = area of shear friction reinforcement, in.2, that crosses the potential sliding plane m = coefficient of friction = 1.0 for a normal weight concrete surface roughened to 1/4 -inch amplitude.

Additionally, ACI 318-02 permits permanent net compression across the shear plane as additive to the resistance provided by shear friction reinforcement. For the example shear wall we will conservatively ignore the beneficial effect of compression.

As will be seen presently, the vertical reinforcement Avf required to satisfy the governing axial load and moment combination is equal to

Avf = 32 # 11 plus 36 # 7 = 32 x 1.56 + 36 x 0.60 = 71.5 in.2

The sliding shear resistance Vn = 71.5 x 60 x 1 = 4290 kips fVn = 0.65 x 4290 = 2788 kips > 1350 kips

Therefore the wall is OK for sliding shear.

Section 11.7.5 limits the shear friction strength to 0.2 f'Ac or 800 Ac inch-lb, where Ac is the area of concrete resisting shear transfer. For the example wall

800 x 16 x 366

(a) Y

366x 16 in.

Figure 4.43. (a) Shear wall interaction diagram; (b) cross section of wall. Longitudinal Reinforcement. Design of vertical reinforcement to resist a given set of axial loads and bending moments is typically a trial and error procedure. Given a wall section and an assumed reinforcement layout, the section is checked for the governing axial load and bending moment combinations. Although hand calculations and spreadsheet approaches are possible, the most desirable and expedient method is to use a computer program such as PCACOL developed by Portland Cement Association.

Figure 4.43 shows an interaction diagram for the wall with 16 #11 placed near the wall boundaries and #7 @ 9, each face, in between the boundaries for a total Avf = 71.5 in.2 The figure is a printed screen output of the PCACOL run. The points 1 and 2 that lie within the interaction curve represent the governing loads. Point 1 is for Pu = 2604 kips and Mu = 71,000 kip-ft, and 2 is for Pu = 1056 kips and Mu = 71,000 kip-ft. Since both 1 and 2 lie within the interaction curve, the example wall is OK for the ultimate axial load and moment combinations. Web Reinforcement. Section requires a uniform distribution of both horizontal shear reinforcement pn and vertical reinforcement pv. Further, to control width of inclined cracks due to shear, a minimum reinforcement ratio equal to 0.0025 and a maximum spacing of 18 inches is specified for both pn and pu. However, a reduction in the reinforcement ratio is permitted if the design shear force Vu is less than Ac^f'.

Minimum ratios of pu if Vu < Ac^jf' (see Section 14.3) are

• 0.0020 for #5 and smaller bars, with fy > 60,000 psi.

• 0.0020 for welded fabric not larger than W31 or D31.

The minimum ratios of pv (vertical reinforcement) for the same condition are

• 0.0012 for #5 and smaller bars with fy > 60,000 psi.

• 0.0012 for welded wire fabric not larger than W31 or D31.

In seismic design, the vertical reinforcement at the bottom few stories of a shear wall is typically controlled by bending requirements. The upper levels are likely to be controlled by the ACI 318-02 minimum reinforcement ratio of 0.0025. For the example wall, f = 366 x W5000 = 4i4 kips < = i400 kips CVS 1000 u

The minimum horizontal reinforcement

2 x 0.31 x 12 , Use #5 @ 15 giving a steel area = -—-= 0.496 in.2

Section requires at least two curtains of reinforcement if the factored shear force Vu exceeds 2 A^^f.

For the example wall,

22 X 366 X 16V5000 CV^ c 1000

= 828 kips

Since Vu = 1350 kips is greater than 828 kips, we use two layers of #5 @ 15. The reason for two layers of reinforcement is to place web reinforcement close to the wall surface to inhibit fragmentation of concrete in the event of severe cracking of concrete during an earthquake. Boundary Elements. Stress Index Procedure. This method is quite straightforward (Section A stress index of 0.2 f is used as a benchmark for the maximum extreme fiber compressive stress corresponding to factored forces that include gravity and earthquake effects. If the calculated compressive stress is less than the index value, special boundary elements are not required. If not, detailing of boundary elements in accordance with Section is required. The compressive stresses are calculated for the factored axial forces and bending moments using a linear elastic model and gross section properties. For the example wall,

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