Design of Masonry Shear Walls – ASD

There are a few methods that we can use to design masonry shear walls, see here for another ‘simplified’ method. Here we will present a method based on force equilibrium and an adjusted bar spacing. Also included is an example and a spreadsheet. This method is a bit more involved than the simplified method but we can better analyse the shear wall and provide an economic design. We will distribute the tension due to bending stress to steel reinforcement based on a linear strain distribution. Steel in compression will be neglected (it must be tied to be considered effective in compression) per MSJC (for lateral tie requirements refer to Ref 1 Section 2.1.6.5) . Note that this procedure assumes that there is net tension on the section. If there is no net tension than only jamb steel is needed and axial needs to be checked. If you iterate the section and Kd is greater than the length of wall then the entire section is in compression and there is no net tension.

References:

Ref 1. MSJC 2005 and 2008 Building Code Requirements and Specification for Masonry Structures and Related Commentaries (AKA TMS 402 or ACI 530). Found here

Ref 2. Masonry Structures Design and Behavior, Second Edition by Drysdale, Hamid and Baker (the Third edition found here)

Ref 3. Reinforced Masonry Engineering Handbook 6th Edition by Max Porter. Found here

Notes/Examples:

Reinforced Masonry Shear Wall – Force Equilibrium Adjusted Spacing Notes

Masonry Shear Wall Design ASD – Force Equilibrium Example

Shear Wall ASD – Force Equilibrium Adjusted Spacing Spreadsheet

Procedure:

Masonry Shear Wall Design ASD – Force Equilibrium Example Sketch

1.  Calculate Section Properties.
   • Calculate out-of-plane or weak axis section properties based on the bar spacing and grout. Then divide these properties by the bar spacing so that they you have Area/ft of wall, I (moment of inertia)/ft of wall and then r (radius of gyration).
2. Adjust Bar Spacing
   • Because we normally specify the on center spacing of the bars, find the number of bars based spacing and wall length N’ = L/S + 1 and round down. This will be 1 bar less than placed in the field.  Find the adjusted spacing S’ = L’ / (N’-1). We will assume bars will be placed in the first cell and then placed at the on center spacing. Use this spacing/layout in the analysis.
3. Wall Analysis – Combined Axial and Moment Forces
   • First find the eccentricity of the required load er=Mr/Pr. Then to maintain equilibrium the ratio of the section must have the same eccentricity ed=Mn/Pn. We assume a triangular compressive stress distribution and only the reinforcement resists tension.
   1. Start by assuming the location of the neutral axis (NA) from the end of the wall this distance is Kd.
   2. Assuming that there is net tension in the section, find the compressive force in the masonry, Cm = 0.5*A/ft * Kd (based on triangular stress distribution).
   3. Find the tension force in each rebar. Based on a linear strain distribution the the stress can be found as fs=n(di-Kd)*fm/Kd. Where n=modular ratio (Es/Em), di=distance from extreme compression face to the rebar under consideration. This stress should be less than the allowable stress (minimum (24ksi, Fy/2.4)). However it should not carry any compression stress unless ties are provided.
   4. Find the compressive strength of the section. Pn=Cm-Ttotal.
   5. Find the moment strength of the section. Mn=Mcm+Mtot
      • Where Mcm is the masonry compression x moment arm (L/2-Kd/3) and Mtot = Ti (tension force of rebar) x moment arm (di-L/2).
   6. Find ed = Mn/Pn. Iterate Kd if this does not match er.
4. Shear Design
   • See here for shear design
5. Axial check for slenderness.
   • See here for axial design. The area used in this check should be the total net area of the wall (not kd) but area should be subtracted out for partial grout.

• Perform Checks

1. 1. Check that the section axial load strength is greater than the required, Pn=Pc > Pr
   2. Check that the section moment strength is greater than the required, Mn=Mc > Mr
   3. Check axial – slenderness. Check Pcs > Pr the area used

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Masonry Shear Design

For right now I am going to cover basic shear design in accordance with MSJC 2005 / 2008 (AKA TMS 402/ACI 530). I will keep updating this for strength design and include some further discussion in the near future.

Ref 1. MSJC 2008 (2005 is very similiar however these provisions change for 2011) Building Code Requirements and Specification for Masonry Structures and Related Commentaries (AKA TMS 402 or ACI 530). Found here

Design and Analysis of Masonry Subject to Shear Forces

This will be a more indepth look at shear forces and how they are handled in masonry design and analysis.

There are several design methods / configurations to consider.

1. Unreinforced Masonry – ASD (Allowable Stress Design)
   • All members (flexural and shear walls)
2. Reinforced Masonry – ASD
   • Flexural members
   • Shear walls
3. Unreinforced Masonry – STR (Strength Design)
   • All members (flexural and shear walls)
4. Reinforced Masonry – STR
   • Flexural members
   • Shear walls

ASD Methods:

MSJC-08 provides a flowchart for shear design in the commentary (Figure CC-2.3-2) which is very helpful. Also allowable stress may be multiplied by 4/3 for short term loads – wind and earthquake per ASCE7-05 C2.4.1. However make sure that you are only multiplying the end result and not f’m found in the equations and the end result.

Here are the basics – Is the section subjected to a net flexural tension stress (i.e. P/A-M/S, is the stress negative)? If no then calculate the shear stress using fv = VQ/In*b where V=shear force, Q=first moment of area, In= net section moment of inertia and b=width or thickness of section. Then base the allowable stress on an unreinforced masonry section (not to be confused with a reinforced section w/out reinforcement). If there is net tension and the wall is reinforced (with longitudinal reinforcement for bending stress)  then find the shear stress using fv=V/(b*d). Base the allowable stress a reinforced wall section (flexural member or shear wall as applicable).

Unreinforced Masonry ASD (MSJC Section 2.25)

 In-plane Shear Forces (shear walls):

Allowable shear stress without any reinforcement (Ref 1 Section 2.2.5.2):

• Fv = Minimum of
  • 1.5x
  • 120 psi
  • 37 psi + 0.45 (For running bond masonry not grouted solid)
  • 37 psi + 0.45 (For stack bond masonry with open end units and grouted solid)
  • 60 psi + 0.45 (For running bond masonry grouted solid)
  • 15 psi

Out-of-plane Shear Forces (Ref 1 Commentary 2.2.5):

In MSJC 2005 and 2008 the commentary suggested that section 2.2.5.2 be used due to the absence of suitable research data.  This is shown above for In-Plane shear forces. However, in 2011 the MSJC, the code states (section 8.2.6.3) that the minimum normalized web area shall be 27 in^2/ft which provides sufficient web area so that hte shear stresses between the web and face shell of a unit will not be critical for out of plane loading.

Allowable shear stress without any reinforcement (Ref 1 Section 2.3.5.2.2):

• Fv = See above (same as in plane)

Reinforced Masonry ASD (MSJC Section 2.3.5):

Flexural members:

Allowable shear stress without shear reinforcement (Ref 1 Section 2.3.5.2.2):

• Fv= minimum of:
  • sqrt(f’m)
  • 50 psi

Allowable shear stress with shear reinforcement (Ref 1 Section 2.3.5.2.3):

• Fv= minimum of:
  • 3*sqrt(f’m)
  • 150 psi

Shear Wall:

Allowable shear stress is based on the ratio of M/(V*d) where M=moment V=shear force and d=depth from extreme compression fiber to tension reinforcement

Allowable shear stress without shear reinforcement (Ref 1 Section 2.3.5.2.2):

• Fv for M/(V*d) < 1 = minimum of:
  • 1/3*(4-(M/(V*d))*sqrt(f’m)
  • 80-45*(M/Vd)
• Fv for M/(V*d) >= 1 = minimum of:
  • sqrt(f’m)
  • 35 psi

Allowable shear stress with shear reinforcement (Ref 1 Section 2.3.5.2.3):

• Fv for M/(V*d) < 1 = minimum of:
  • 1/2*(4-(M/(V*d))*sqrt(f’m)
  • 120-45*(M/Vd)
• Fv for M/(V*d) >= 1 = minimum of:
  • 1.5*sqrt(f’m)
  • 75 psi

Shear Reinforcement Required (MSJC 2.3.5.3):

Shear reinforcement is designed for the entire shear force.

• Area of reinforcement parallel to shear force
  • Avpar=V*S/(Fs*d)
  • V=Shear force, S=vertical spacing of horizontal reinforcement  Fs = Allowable steel stress (with 1/3 increase as applicable) and d=distance from extreme compression to tension reinforcement
  • Max spacing is the lesser of d/2 or 48″
• Area of  reinforcement perpendicular to shear force:
  • Avv = 1/3 * Avh
  • Max spacing is 8′ o.c.

Strength Design Methods:

Will update soon.

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