Civil Engineering app for ANDROID

Civil Engineering app for Android

Most Views

  • Bearing walls that resist out-of-plane lateral loads, and shear walls, must be designed to transfer lateral loads to the floors above and below. Examples of such connections are shown in Figs. 6.30 through 6.34. These connections would have to be strengthened for regions subject to strong earthquakes or strong winds. Section 1604.8.2 of the 2009 IBC has additional requirements for anchorage of diaphragms to masonry walls. Section 12.11 of ASCE 7-05 has additional requirements for anchorage of structural walls for structures assigned to seismic design categories C and higher.

    Wall-to-Foundation Connections

    As shown in Fig. 6.31, CMU walls (or the inner CMU wythe of a drainage wall) must be connected to the concrete foundation. Bond breaker should be used only between the outer veneer wythe and the foundation.

    Wall-to-Floor Details

    Examples of a wall-to-floor detail are shown in Figs. 6.31 and 6.32. In the latter detail (floor or roof planks oriented parallel to walls), the planks are actually cambered. They are shown on the outside of the walls so that this camber does not interfere with the coursing of the units. Some designers object to this detail because it could lead to spalling of the cover. If it is modified so that the planks rest on the face shells of the walls, then the thickness of the topping must vary to adjust for the camber, and form boards must be used against both sides of the wall underneath the planks, so that the concrete or grout that is cast into the bond beam does not run out underneath the cambered beam.

    Wall-to-Roof Details

    An example of a wall-to-roof detail is shown in Fig. 6.33.

    Typical Details of Wall-to-Wall Connections

    Typical details of wall-to-wall connections are shown in Fig. 6.34.

    Tags: , , , ,

  • Reinforced masonry shear walls, like unreinforced ones, are relatively easy to design by either strength or allowable-stress approaches. Although shear capacities per unit area is small, the available area is large.

    With either strength or allowable-stress approaches, it is rarely necessary to use shear reinforcement. In this sense, the best shear design strategy for shear walls is like that for shear design of beams—use enough cross-sectional area to eliminate the need for shear reinforcement. Seismic requirements may still dictate some shear reinforcement, however.

    Tags:

  • Minimum Flexural Reinforcement by 2008 MSJC Code

    The 2008 MSJC Code has no global requirements for minimum flexural reinforcement for shear walls.

    Maximum Flexural Reinforcement by 2008 MSJC Code

    The 2008 MSJC Code has a maximum reinforcement requirement (Sec. 3.3.3.5) that is intended to ensure ductile behavior over a range of axial loads. As compressive axial load increases, the maximum permissible reinforcement percentage decreases. For compressive axial loads above a critical value, the maximum permissible reinforcement percentage drops to zero, and design is impossible unless the cross-sectional area of the element is increased. For walls subjected to in-plane forces, for columns, and for beams, the provisions of the 2008 MSJC Code set the maximum permissible reinforcement based on a critical strain condition in which the masonry is at its maximum useful strain, and the extreme tension reinforcement is set at a multiple of the yield strain, where the multiple depends on the expected curvature ductility demand on the wall. For “special” reinforced masonry shear walls, the multiple is 4; for “intermediate” walls, it is 3. For walls not required to undergo inelastic deformations, no upper limit is imposed.

    The critical strain condition for walls loaded in-plane, and for columns and beams, is shown in Fig. 6.29, along with the corresponding stress state. The multiple is termed “α.” The parameters for the equivalent rectangular stress block are the same as those used for conventional flexural design. The height of the stress block is 0.80 fmâ€Č, and the depth is 0.80c. The stress in yielded tensile reinforcement is assumed to be fy. Compression reinforcement is included in the calculation, based on the assumption that protecting the compression toe will permit the masonry there to provide lateral support to the compression reinforcement. This assumption, while perhaps reasonable, is not consistent with that used for calculation of moment-axial force interaction diagrams.

    Locate the neutral axis using the critical strain condition:

    Compute the tensile and compressive forces acting on the section, assuming uniformly distributed flexural reinforcement, with a percentage of reinforcement ρ =A bd s/ .On each side of the neutral axis, the distance over which the reinforcement, in the elastic range, is ÎČc, where ÎČ is given by proportion as ÎČ = Δ Δ y mu / .
    The compressive force in the masonry is given by

     

    Tags: , , , , ,

  • Consider the masonry shear wall shown in Fig. 6.26.

    Design the wall. Unfactored in-plane lateral loads at each floor level are due to earthquake, and are shown in Fig. 6.27, along with the corresponding shear and moment diagrams.

    Assume an 8-in. nominal clay masonry wall, grouted solid, with Type S PCL mortar. The total plan length of the wall is 24 ft (288 in.), and its thickness is 7.5 in. Assume an effective depth d of 285 in.

    Unfactored axial loads on the wall are given in the table below.

    Shear design is satisfactory so far, even without shear reinforcement.
    Code Sec. 1.17.3.2.6.1 will be checked later.
    Now check flexural capacity using a spreadsheet-generated momentaxial force interaction diagram. Try #5 bars @ 4 ft. Neglecting slenderness effects, the diagram is shown in Fig. 6.28.

    At a factored axial load of 0.9 D, or 0.9 × 360 kips = 324 kips, the design flexural capacity of this wall is about 4000 ft-kips, and the design is satisfactory for flexure.

    We have designed the wall for the calculated design shear, which is normally sufficient. Now suppose that the wall is a special reinforced masonry shear wall (required in areas of high seismic risk), so that the capacity design requirements of Code Sec. 1.17.3.2.6.1.1 apply. First try to meet the capacity design provisions of that section. At an axial load of 324 kips, the nominal flexural capacity of this wall is the design flexural capacity of 4000 ft-kips, divided by the strength reduction factor of 0.9, or 4444 ft-kips. The ratio of this nominal flexural capacity to the factored design moment is 4444 divided by 3000, or 1.48. Including the additional factor of 1.25, that gives a ratio of 1.85.

     

    Tags: , ,

  • Introduction to Strength Design of Reinforced Shear Walls

    In this section, we shall study the behavior and design of reinforced masonry shear walls. The discussion follows the same approach used previously for unreinforced masonry shear walls.

    Design Steps for Strength Design of Reinforced

    Shear Walls Reinforced masonry shear walls must be designed for the effects of:

    1. Gravity loads from self-weight, plus gravity loads from overlying roof or floor levels.

    2. Moments and shears from in-plane shear loads.

    Actions are shown in Fig. 6.22.

    Flexural capacity of reinforced shear walls using strength procedures is calculated using moment-axial force interaction diagrams as discussed in the section on masonry walls loaded out-of-plane. In contrast to the elements addressed in that section, a shear wall is subjected to flexure in its own plane rather than out-of-plane. It therefore usually has multiple layers of flexural reinforcement. Computation of momentaxial force interaction diagrams for shear walls is much easier using a spreadsheet.

    From the 2008 MSJC Code, Sec. 3.3.4.1.2, nominal shear strength is the summation of shear strength from masonry and shear strength from shear reinforcement:

    The nominal resistance from reinforcement is taken as the area associated with each set of shear reinforcement, multiplied by the number of sets of shear reinforcement crossing the hypothetical failure surface. Because the hypothetical failure surface is assumed to be inclined at 45°, its projection along the length of the member is approximately equal to dv, and number of sets of shear reinforcement crossing the hypothetical failure surface can be approximated by (dv/s):

    The actual failure surface may be inclined at a larger angle with respect to the axis of the wall, however. Also, all reinforcement crossing the failure surface may not yield. For both these reasons, the assumed resistance is decreased by an efficiency factor of 0.5. From the 2008 MSJC Code, Sec. 3.3.4.1.2.2,

     

     

     

    Tags: ,

« Previous Entries   

Recent Posts

Recent Comments