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    Ce 479 Reinforced Masonry Fall 2004

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    Reinaldo Andrei Salazar
    The document discusses the design of reinforced masonry shear walls. It describes how shear walls must be designed to resist both gravity and lateral loads. The design process involves checking the wall's capacity for flexure and shear using strength design procedures outlined in building codes. As an example, the document analyzes the design of a clay masonry shear wall with #5 bars placed vertically at 4 feet and horizontally at 2 feet to resist the wall's shear, flexure, and code requirements.

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    Ce 479 Reinforced Masonry Fall 2004

    Uploaded by

    Reinaldo Andrei Salazar
    34 views9 pages
    The document discusses the design of reinforced masonry shear walls. It describes how shear walls must be designed to resist both gravity and lateral loads. The design process involves checking the wall's capacity for flexure and shear using strength design procedures outlined in building codes. As an example, the document analyzes the design of a clay masonry shear wall with #5 bars placed vertically at 4 feet and horizontally at 2 feet to resist the wall's shear, flexure, and code requirements.

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    Lecture5_masonry

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    Lecture5

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    The document discusses the design of reinforced masonry shear walls. It describes how shear walls must be designed to resist both gravity and lateral loads. The design process involves checking the wall's capacity for flexure and shear using strength design procedures outlined in building codes. As an example, the document analyzes the design of a clay masonry shear wall with #5 bars placed vertically at 4 feet and horizontally at 2 feet to resist the wall's shear, flexure, and code requirements.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    34 views9 pages

    Ce 479 Reinforced Masonry Fall 2004

    Uploaded by

    Reinaldo Andrei Salazar
    The document discusses the design of reinforced masonry shear walls. It describes how shear walls must be designed to resist both gravity and lateral loads. The design process involves checking the wall's capacity for flexure and shear using strength design procedures outlined in building codes. As an example, the document analyzes the design of a clay masonry shear wall with #5 bars placed vertically at 4 feet and horizontally at 2 feet to resist the wall's shear, flexure, and code requirements.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 9

    Ce 479

    Reinforced Masonry
    Fall 2004

    DESIGN OF REINFORCED MASONRY SHEAR WALLS

    Introduction

    Next, let us consider the behavior and design of reinforced masonry shear walls.

    Design Steps

    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; and
    2) moments and shears from in-plane shear loads
    Actions are shown below. Either strength or allowable-stress design can be used.

    DESIGN OF REINFORCED SHEAR WALLS USING STRENGTH PROCEDURES

    Flexural capacity of reinforced shear walls using strength procedures is calculated using moment-
    axial force interaction diagrams as discussed in the lecture on masonry beam-columns. In contrast
    to the elements addressed in that lecture, 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 moment-axial force interaction diagrams for shear walls is much easier using a
    spreadsheet.

    From the 2002 MSJC Code, Section 3.2.4.1.2, nominal shear strength is the summation of shear
    strength from the masonry and shear strength from the shear reinforcement.

    From the 2002 MSJC Code, Section 3.2.4.1.2.1,

    Instructor: Julio A. Ramirez 1


    Ce 479
    Reinforced Masonry
    Fall 2004

    As (M/Vdv) increases, Vm decreases. Because (M/Vdv) need not be taken greater than 1.0 (2002
    MSJC Code, Section 3.2.4.1.2.1), the most conservative (lowest) value of Vm is obtained with
    (M/Vdv) equal to 1.0.

    Just as in reinforced concrete design, this model assumes that shear is resisted by reinforcement
    crossing a hypothetical failure surface oriented at 45 degrees:

    Instructor: Julio A. Ramirez 2


    Ce 479
    Reinforced Masonry
    Fall 2004

    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 degrees,
    its projection along the length of the member is approximately equal to d, and number of sets of
    shear reinforcement crossing the potential failure crack can be approximated as (d/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 2002 MSJC
    Code, Section 3.2.4.1.2.2 (Page C-34)

    Finally, because shear resistance really comes from a truss mechanism in which horizontal
    reinforcement is in tension, and diagonal struts in the masonry are in compression, crushing of the
    diagonal compressive struts is controlled by limiting the total shear resistance Vn , regardless of
    the amount of shear reinforcement:

    Instructor: Julio A. Ramirez 3


    Ce 479
    Reinforced Masonry
    Fall 2004

    For (M/Vd) ≤ 0.25

    and for (M/Vd) > 1.00,

    Interpolation is permitted between these limits.

    If these upper limits on Vn are not satisfied, the cross-sectional area of the section must be
    increased.

    Instructor: Julio A. Ramirez 4


    Ce 479
    Reinforced Masonry
    Fall 2004

    Example: Strength Design of Reinforced Clay Masonry Shear Wall

    Consider the masonry shear wall shown below:

    Design the wall. Unfactored in-plane lateral loads at each floor level are due to earthquake, and
    are shown below, 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.

    Instructor: Julio A. Ramirez 5


    Ce 479
    Reinforced Masonry
    Fall 2004

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

    Use ASCE 7-02 -Basic Strength Load Combination 7- : 0.9D + 1.0E

    Check shear for assumed wall thickness. By Section 3.2.4.1.2 of the 2002 MSJC Code,

    V =V +V
    n m s

    M = 3,000 x 12 x 1,000 in.-lb = 36.0 x 106 in.-lb

    V = 120,000 lb and dv = 285 in

    Instructor: Julio A. Ramirez 6


    Ce 479
    Reinforced Masonry
    Fall 2004

    The 2002 MSJC Code requires that this check be carried using unfactored loads. The MSJC is
    considering changing this to factored loads for the 2005 Code.

    Shear design is satisfactory so far, even without shear reinforcement. Code Section 3.1.3 will be
    checked later. Now check flexural capacity using a spreadsheet-generated moment-axial force
    interaction diagram. Try #5 bars @ 4 ft.

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

    Instructor: Julio A. Ramirez 7


    Ce 479
    Reinforced Masonry
    Fall 2004

    Now check Code Section 3.1.3. of MSJC 2002. 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 4000 ft-
    kips, divided by the strength reduction factor of 0.9, or 4,444 ft-kips. The ratio of this nominal
    flexural capacity to the factored design moment is 4,444 divided by 3,000, or 1.48. Including the
    additional factor of 1.25, that gives a ratio of 1.85.

    The wall is satisfactory without shear reinforcement. Prescriptive seismic reinforcement (for
    example, for SDC D) will probably require #5 bars horizontally @ 24 in.

    Prescriptive seismic reinforcement is sufficient for shear. Use #5 bars at 2 ft.


    Summary:

    Use #5 @ 4 ft vertically, #5 @ 2 ft horizontally.

    Instructor: Julio A. Ramirez 8


    Ce 479
    Reinforced Masonry
    Fall 2004

    MINIMUM AND MAXIMUM REINFORCEMENT RATIOS FOR FLEXURAL DESIGN BY THE


    STRENGTH APPROACH

    The strength design provisions of the 2002 MSJC Code include requirements for minimum and
    maximum flexural reinforcement.

    Minimum Flexural Reinforcement by 2002 MSJC Code

    The 2002 MSJC Code has no global requirements for minimum flexural reinforcement. For
    design of shear walls for in-plane loads, however, Section 3.2.6.2 requires that the vertical
    reinforcement be not less than one-half the horizontal reinforcement.

    Maximum Flexural Reinforcement by 2002 MSJC Code

    In an innovative departure from previous codes for masonry and concrete, the 2002 MSJC
    Code has a maximum reinforcement requirement (Section 3.2.3.5) 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 2002
    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 at 5
    times its yield strain. For walls subjected to out-of-plane forces, the critical strain gradient has a
    strain in the extreme tension reinforcement of 1.3 times its yield strain. Less restrictive
    requirements are imposed on members not required to undergo inelastic deformations.

    The critical strain condition for walls loaded in-plane, and for columns and beams, is shown
    below, along with the corresponding stress state. 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 0.80fm ′ , and the depth is 0.80 c . The reinforcement is conservatively assumed to be at
    1.25 fy, to allow for possible steel overstrength and strain hardening. This assumption is
    conservative because its effect is to reduce the maximum amount of reinforcement that can be
    used.

    Instructor: Julio A. Ramirez 9

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