Masonry Subject to Compression and Flexure – Stability – ASD

References

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2005 version however the method does not change up to the 2013 version. Found here

Ref 2 Masrony Structures Behavior and Design by Robert G. Drysdale, Hamid, and Baker 2nd Edition. (the Third edition found here)

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

Overview / Discussion

This pertains to stability of reinforced masonry subject to compression and flexure. Now when I first started to post about this subject I was a bit confused as I started digging into the code. Here’s what I mean. In my experience it seems that most engineers Most of us are used to some sort of moment magnifier when axial and moment forces are present such as in ACI 318 and AISC’s Steel Construction Manual. However, when we are designing masonry with ASD provisions we do not seem to find this when we read through the reinforced masonry section of ACI 530 (section 2.3.2 in the ’05 code). We don’t seem to find any adjustment or magnifier for second order / slenderness effects. This had confused me for sometime. I was setting up calculations using force equilibrium and compatibility equations similar to reinforced concrete (Only instead of an approximate rectangular stress block a triangular shape is used assuming a linear elastic stress distribution). Then in the analysis of the section you can directly solve for or iterate to find the location of the neutral axis. Well in doing this you are checking that the masonry is not crushing and also checking that there is adequate tension strength in the steel reinforcement, both of these are what I would say are ‘material’ checks. Meaning that you are checking the capacity of the local material not the overall member which may have less of a capacity due to buckling (stability check). You may say that there is a check for buckling, and that would be true. The required axial force vs the allowed axial force (eqn 2-17 and 2-18 in Ref 1). This however does not account for any moment which may be present. This to me did not seem right as there was no interaction in these equations for axial, moment and buckling. So I dug a little deeper and here is what I found.

Some Quick Background To Clarify My Point

I just want to provide some comparison and clarity for what I am referring to when I’m talking about second order / slenderness / stability effects. These effects are the results of the axial force and the deflection of the member which create ‘secondary’ moments. We can account for these effects in a number of approximate ways. If we go back to Timoshenko’s Theory of Elastic Stability we see that we can account (approximately) this additional bending moment by multiplying the first order moment, M by where is the required axial force as found from the results of a first order analysis and is the critical buckling load. We find this amplification in a number of design manuals – AISC 9th edition for a member subject to combined forces. However this formula was removed from the ‘design side’ of the AISC equations and is now found in the ‘analysis side’ in the form of B1 (see chapter c of the AISC 13th edition). This is also found in ACI 318 in the moment magnification procedure. We even find it in the strength design section of the ACI 530 (ref 1). However the procedure if for slender walls and differs from the ACI and AISC approach. In ACI 530 the deflection due to the applied loads is found, then the moment due to axial is found which causes additional deflection. The process is repeated with the new moments until successive trial results in less than 2% error (convergence).
Knowing this now and then reading through Ref 1 ASD design for reinforced masonry we start to think ‘hey something seems to be missing’. Well it is, sorta. Lets take a look.

Design – ASD

UPDATE – this was my first attempt to reason that there was some sort of provision in the code that was considering stability, but ultimately I was wrong. I am leaving this in, for reference.

Whether you are designing reinforced masonry or unreinforced masonry you basically are going back to the unreinforced masonry equations anyway so lets look at reinforced masonry.

————————————————————————————————————————-

1. Reinforced Masonry
   1. Members must satisfy a buckling check given by eqn 2-17 or 2-18 in Ref 1, depending on the h/r ratio.
      1. This is basically a pure axial buckling check, no secondary moments
   2. “The compressive stress in the masonry due to flexure or due to flexure in combination with axial load shall not exceed f’m/3 ”
      1. This is what I was referring to as more of a ‘local’ material failure check as stability does not come into play with these equations.
   3. “The axial load component fa does NOT exceed the allowable stress Fa given in section 2.2.3.1” of Ref 1 which is the unreinforced masonry allowable compressive stress section
      1. Notice that you are only checking the axial stress component not the combined stress.
      2. This is ultimately where stability is check but it is not obvious at first. So we look further.
2. Unreinforced Masonry (section 2.2.3 in Ref 1)
   1. fa/Fa + fb/Fb < 1
      1. Unity check
   2. P<1/4 Pe
      1. Where Pe is simply a buckling equation limit to safegaurd against a premature stability failure caused by eccentricity of an applied axial load. Therefore in equation 2-15 (Ref 1) e is the actual eccentricity of the applied load (min value typically = 0.1t) not M/P where M is caused by other than eccentric load.
      2. Does this e consider deflection?

——————————————————————————————————————-

I ended up contacting MSJC and received an excellent response from a Mr. Art Schultz from the University of Minnesota. I would like to add that I am very appreciative of Mr. Schultz and MSJC’s assistance. Also the “masonry community” in general seems to be very helpful.

Here the response:MSJC Stability Treatment ASD Axial and Flexure

To summarize: The code does not address stability / second order effects for reinforced masonry design using ASD.

I would say use the LRFD approach, it’s not so bad, see here.

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Masonry Columns, Piers and Pilasters

How to analyze and design reinforced masonry columns, piers and pilasters and clarifying effective spacing.

For analysis of masonry members utilizing flanges and compression reinforcement see Masonry – Compression Reinforcement and Effective Flanges

References

Ref 1: MSJC – 2005, TMS 402 (AKA ACI 530) – 05 (2008 version is very similar). Found here

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

Ref 3: Reinforced Masonry Engineering Handbook Clay and Concrete Masonry 6th Edition, Max L Porter (James E. Amrhein original author). Found here

Definitions

Based on MSJC 05 (all should be similar for 08)

Column

(MSJC-05 Section 1.6) – An isolated vertical member whose horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness and whose height is greater than 4 times its thickness.

Pier

(MSJC-05 Section 1.6) – An isolated vertical member whose horizontal dimension measured at right angles to its thickness is not less than 3 times its thickness nor greater than 6 times its thickness and whose height is less than 5 times its length. In reference to MSJC these are typically part of wall frames (see Ref 3 page 250) and is why there dimensions are defined as such. This however may be confusing as many times a pier is referred to a column which is built integrally with a wall (see pg 403 of Ref 2). Having said that I believe if the column is built integrally with the wall it would be considered a “flush pilaster” (see Ref 3).

Pilaster

–  While Ref 1 does not give an explicit definition of a pilaster, it does infer that a pilaster is built integrally with the wall. Ref 2 states that a pilaster is a column that is built integrally with a wall and interacts with the wall to resist an out-of-plane lateral load, it is called a pilaster. Pilasters maybe ‘flush’ (in the plane of the wall) or project out from the wall in one or both directions.

Pedestal

– Upright compression member with a ratio of unsupported height to average least lateral dimension not exceeding 3. This term is typically found in concrete design and

refers to short foundation elements. Many may refer to pedestals as piers or foundation piers.

Foundation Piers

– (Ref 1) An isolated vertical foundation member whose horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness and whose height is equal to or less than 4 times its thickness.

Wall

– (Ref 1) A vertical element with a horizontal length to thickness ratio greater than 3, used to enclose space.

Loadbearing Wall

– (Ref 1) a Wall carrying vertical loads greater than 200 plf in addition to its own weight.

Discussion

As you can see these elements may be referred to one another as they all are used to resist axial and flexure. I would need to check the 2011 version of the MSJC but it should also be noted that Piers are found in the Strength Design of Masonry only and Pilaster are only found in the ASD method. However lets look at how the analysis and design differs for each. I think some main points to consider are – Is the element isolated or part of wall construction? If the element is isolated it would generally be considered a column or pier. If it is part of a wall system then it would generally be considered a pilaster or flush pilaster. The main difference then between these is that for column/pier elements there is – added vulnerability due to isolation, typically compression reinforcement and potential for bi-axial bending. Another key to remember is that if the reinforcement is going to be considered to be effective in compression, the enforcement must be tied (confined) for any of the above mentioned elements.

Now lets look at how each classification effects the design.

Design

Dimensional Constraints / Effective Widths

This can be hard to keep track of as well. How much of the wall should be taken as effective to resist axial and flexural loads? What are the limits on effective flange widths? It is interesting to note that Ref 3 pg 177 limits the effective flange for pilasters to 3x the wall thickness each side of the pilaster however allows for 6x wall thickness for effective flange width on flanged masonry shear walls.

Effective Width

1. Walls – (ref 1 Sect. 2.3.3.3)
   1. Out-of-plane Forces – Running Bond Effective width shall be the lesser of:
      1. Center – Center spacing of bars
      2. 6 x wall thickness
      3. 72 inches
      4. H/6 in the case of a flanged shear wall. Where H is the total wall height (between lateral supports transferring load to the load to the shear wall).
   2. Axial Concentrated Load – Effective width
      1. 4 x wall thickness + bearing width or length of wall
   3. Design for Axial and Flexure
      1. The reinforcement is rarely tied and typically only resists tension forces.
      2. For ASD if the load is concentrated then the 4x the wall thickness limit would apply otherwise the limits presented in point 1 would apply and the reinforcement would need to be tied if it going to be relied upon to carry compression forces.
   4. In-Plane Forces – Flanged Shear Walls
      1. ASD – The flange is limited to 6x wall thickness each side of the web (Ref 3 pg 199)
      2. Wall intersections shall meet Ref 1 – Section 1.9.4.1
2. Flush Pilasters  
   1. Effective Wdith
      1. Tpcially taken as 4 x wall thickness (Ref 3 page 178)
3. Pilasters
   1. Effective width
      1. Ref 1 and 2 – 6 x Wall thickness each side of web
      2. Ref 3 – 3 x Wall thickness each side of web
   2. Consideration for flanges to be effective (see Ref 1 – 1.9.4.2).
      1. Flanges  (wall intersections) must be capable to transfer the required shear stress
      2. Atleast 50% of the masonry units at the interface shall interlock; or Steel connectors grouted into the wall (1/4″x1.5″ x28″ inculding 2″ long 90 deg bend at each end to form a U or Z shape; or Intersecting bond beams shall be provided in intersecting walls at a maximum spacing of 48″ oc (As = 0.1 in^2 and shall be developed on each side of intersection)
4. Piers
   1. Limits on Width
      1. The width should be greater than 3 x thickness but less than 6 times its thickness
         1. I believe if it were less than 3 x thickness it would then be a column and if it were greater than 6 it would then be a wall.
   2. Design limitations – Strength Design
      1. Max factored axial load (per Ref 1 3.3.4.3.1)
      2. Max effective height should not exceed 25 x nominal thickness unless the pier is designed according to provisions of Ref 1 Section 3.3.5 (walls, must consider P-Delta effects)
      3. Nominal thickness should not exceed 16″
      4. Provide (1) bar in the end cells and minimum area of longitudinal reinforcement should be 0.0007bd
      5. Uniformly distribute reinforcement
5. Column
   1. Limit on Width (Ref 1 – 2.1.6)
      1. Horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness. Minimum side dimension should be 8″
   2. Limits on Height
      1. Height is greater than 4 times its thickness
      2. Ratio of Effective Height / least nominal dimension < 25 (Ref 1 – 2.1.6.2)
   3. ASD Requirements (Ref 1 Section 2.1.6
      1. Min Side Dimension – 8″
      2. Height / least nominal dimension <= 25
      3. Min Moment should be 0.1 x each side dimension
      4. Min longitudinal reinforcement and shall at least 4 bars
      5. Max long reinf
      6. Compression reinforcement shall met lateral tie requirements (as always)
   4. SD (strength design) Requirements (Ref 1 Section 3.3.4.4)
      1. Same as ASD except that the provisions limit the distance between lateral support should be limited to 30 x the nominal width where as ASD uses ‘effective height’ (meaning considering end restraints) should be < 24 x thickness. Also max reinforcement shall be per Ref 1 Section 3.3.5 but not to exceed  
6. Lateral Ties
   1. Requirements (Ref 1 Section 2.1.6.5) If reinforcement is to be considered effective in resisting compressive forces lateral ties must be provided.
      1. 1/4″ diameter (min)
      2. Vertical spacing < smaller of –
         1. 16″
         2. 48 lateral tie bar diameters
         3. least cross-sectional dimension
      3. Every corner and alternate longitudinal bar shall have lateral support provided by the corner of a lateral tie (angle should be less than 135 deg). Except for a circle arrangement which is permitted. No bar bar shall be farther than 6″ (each side) from a laterally supported bar. (i.e. the middle bar should be less than 6″ from 2 bars that are laterally supported by a lateral tie ‘corner’)
      4. Place lateral ties in mortar joint or grout
      5. Lap length shall be 48 tie diameters (min)
      6. Locate the first and last lateral tie at 1/2 spacing.  Also above and below single beams framing into the vertical member.
      7. Where beams or brackets frame into a column from four directions ties may be terminated 3″ (max) from the lowest reinforcement in the shallowest of the beam members framing into the column.

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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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Partially Grouted Reinforced Masonry Shear Wall Design

We will cover a simplified design method for designing masonry shear walls. I don’t have a direct reference at this time however the Reinforced Masonry Engineering Handbook 6th edition and MSJC TMS402 -05/08/11 could be referenced. The TMS 402-08 is referenced for allowable stress values. This method is generally conservative when used for low rise structures with generally small lateral loads and relatively longer shear walls. When designing shear walls for taller buildings or shear walls with high aspect ratios (height/length) and/or large overturning moments a different method of analysis/design is recommended.

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

Masonry Shear Wall – Simplified Design Method General Force Diagram

Steps for the simplified design method:

1. Determine forces on wall: Axial (Pr), Moment (Mr) and Shear (Vr)
2. Determine overturning tensile reinforcement needed:
   •  Use a force couple to represent the moment: Tension = Compression = Mr/d where d is the moment arm of the force couple.
   • Consider dead load to resist uplift – I have seen anywhere from 6*t (6 x wall thickness) to half the wall length used. The 6*t comes from the allowable effective wall width when considering out of plane loads (Ref 1 Section. If there is a return wall (masonry wall perpendicular to shear wall) at the end of the wall, the wall may be used as a flange. The effective width of the flange for tension should limited to 0.75*wall height or the length of the flange.

   • 

     Masonry Shear Wall – Flange Diagram

3. Overturning compression:
   • There is some engineering judgment required. Usually the compression force is considered to act uniformly over a certain area. Similar to concrete design (ACI 318) and the rectangular stress block. I have seen different effective areas used however commonly 6*t or 0.8*half the wall length is used.
4. Pure axial – For axial design see here (post coming soon)
   • Find the reduction factor for axial load and check Pr vs Pa (allowed axial load) (MSJC Ref 1 Section 2.3.3)
   • Reinforcement is not considered effective in compression unless ties are provided.
5. Shear reinforcement: For shear design see here
   • Determine if shear reinforcement is required. If shear reinforcement is required the reinforcement is design for the full shear force.
   • Note that even though the shear reinforcement is designed for the shear force this only increase the allowable shear stress and is still limited i.e. shear reinforcement only provides an increased allowable stress and cannot keep adding reinforcement for further shear strength. Similar to concrete ACI318 design.
   • Determine vertical shear reinforcement required = 1/3 of the horizontal shear reinforcement.
6. Deflection
   • We will save this for another post. Usually this is not a major concern for short buildings with relatively small loads and long shear walls.

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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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Reinforced Masonry Walls – Partial Grout Out of Plane ASD

Most text decide to conveniently use a fully grouted wall. Well this is great but fully grouting walls can lead to unnecessary expenses and also add to seismic weight.  Therefore we will look at how to design and analyse a partially grouted wall. We’ll first cover ASD solutions for out-of-plane forces. See here for Strength Design.

References

The notes and spreadsheet below are based on MSJC 2008 (2005 is similar)

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2005 version however the method does not change up through the 2013 version. Found here

Notes and Tedds Calc:

Reinforced Masonry Out of Plane ASD – Equilibrium Method Notes

Reinforced Masonry Wall Out-of-Plane Axial Bending and Shear – ASD Tedds

Design/Analysis:

1.) Determine a required eccentricity, ed=M/P based on loading.
2.) Assume actual masonry stress is equal to allowable, fm=Fm=1/3f’m
3.) Find stress in steel based on compatiblity which must be less than allowable. fs<Fs
4.) Based on equilibrium sum forces P+Cf+Cw-T=0 solve for Pns.
5.) Based on equilibrium sum moments and solve for Mns.
6.) find actual eccentricty ea=Mns/Pns.
7.)ea=ed
8.) Check Pns>P, Mns>M
9.) check fa<Fa

Questions/Comments:

1.) The shear area is subject to engineering judgment. Becasuse the section may have net tension it may be more conservative to only use the area in compression. However the are is subject to compression may have a higher shear capacity.

2.) Notice that the axial stress (fa) is based on the axial load and the grouted area + mortared area of the full section (not just the area in compression). This value (fa) is then checked against the allowable axial stress (Fa) that has been reduced for slenderness. We have made sure that the masonry compression stress Fm is not exceeded by setting the maximum compressive stress (fm) equal to the allowable (Fm) and then starting our analysis by iterating the calculation and finding the neutral axis.
Examples:

Will Update Soon!

Some further discussion:
http://eng-tips.com/viewthread.cfm?qid=313424

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Masonry Walls – Partial Grout Out of Plane STR

It seems that most text decide to conveniently use a fully grouted wall. Well this is great but fully grouting walls can lead to unnecessary expenses and also add to seismic weight.  Therefore we will look at how to design and analyze a partially grouted wall. We did this once before for ASD so now we will look at Strength design for out-of-plane forces.

References

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2008 version however the method does not change up through the 2013 version. Found here

Ref 2 Masrony Structures Behavior and Design by Robert G. Drysdale, Hamid, and Baker 2nd Edition. (the Third edition found here)

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

Hand calc and TEDDS Calcs added.

Reinforced Masonry Out of Plane STR – Strain Compatibility Notes

Reinforced Masonry Wall Out-of-Plane Axial Bending and Shear – STR Tedds

Strength Design of masonry walls subject to axial and out of plane bending forces.

The Basics:

The design methodology is similar to reinforced concrete design. We assume the following:

1. Strain compatibility (linear) with a maximum allowable compressive strain.
2. Assume a compressive stress distribution (Compression Block) of the section.
3. The masonry has zero tensile strength and the reinforcement carries all tensile stress. The steel reinforcment has zero compressive strength unless it is tied.

References:

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2005 version however the method does not change up through the 2013 version. Found here

Ref 2 Masrony Structures Behavior and Design by Robert G. Drysdale, Hamid, and Baker 2nd Edition. (the Third edition found here)

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

 Design/Analysis:

Masonry Out of Plane ASD Diagram

0.) Given or assumed: Section Geometry, spacing /size of reinforcement and material strengths.

1.)    Assume depth to neutral axis (iterate this value), c.

• Because the wall is subject to a moment and axial force we can replace the moment and axial force with a single axial force that is eccentric from the centroid of the section. The eccentric distance is equal to the  Moment (M)/ Axial (P) force. We know that the internal forces must equal the external forces for the section to be in equilibrium. therefore because the external M and P are known we find er=M/P. We then set out to find the position of the neutral axis that will yield a section capacity strength with the same Moment / Axial eccentricity (ratio) as the applied forces.

2.)    Assume masonry strain is equal to ultimate, em=0.0025 (concrete masonry) em=0.0035 (clay masonry).

• The depth of the compression block is 0.8*c (depth of the neutral axis).

3.)    Find stress in steel based on compatibility which must be less than allowable. fs<Fs

• Draw the strain diagram and find the strain in the reinforcement from similar triangles.

Depth of compression block;           a=0.8*c
Effective flange thickness (strength , deflection);      t’_f=min(t_f,a)    t’_fD=min(t_f,c)
Effective height of web (strength, deflection);            y_cw=max(0in,a-t’_f)    y_cwD=max(0in,c-t’_fD)
Strain in steel based on compatibility;                         e_s=e_m/c*(d-c)
Stress in steel;                                             f’_s=E_s*e_s
Usable steel stress;                                         f_s=min(F_y,f’_s)
Compressive force on flange;                                       C_f=0.8*f’_m*t’_f*b
Compressive force on web;      C_w=0.8*f’_m *y_cw *b_w

5.)    Based on equilibrium sum moments and solve for Mns.

Nominal compressive force based on assumed c location and Maximum masonry strain;       P_nsT =C_f+C_w-T;

Nominal moment strength;                                                                                             M_nsT=M_cf+M_cw-M_s1;

Eccentricity of section;                                                                                             e_s=M_nsT/P_nsT;

6.)    Find cracked and effective moment of inertia Icr and Ie

7.)    Assume a deflection (iterate this value)

8.)    Find actual deflection.

• Deflection should be found for ultimate forces and service level forces. The service level deflection should be less than 0.007H (however 0.01 has been found acceptable by SEAOSC/ACI).

9.)    Find moment caused by deflection (iterate till convergent <5%)

10.) Find actual eccentricity of section ea=Mns/Pns.

11.) Find new eccentricity of loads e=(Mu+Mudeflection)/P

12.) Check if eccentricity of section = eccentricity of design loads (e=ea) if not iterate location of N.A. and deflection.

13.) Check Pns>P, Mns>M

14.) Check fa<Fa < where Fa<0.2*f’m or 0.05*f’m (H/d>30)

15.) Note that a slenderness reduction factor is not applied

Questions/Comments:

1.       The axial stress (Pu/Ag ) must be less than 0.2f’m. If the wall’s slenderness ratio exceeds 30 than the axial stress must be less than 0.05f’m.

2.       All walls are designed by an iterative process by checking the moment caused by the deflection due to out-of-plane loads and eccentricity of the DL/LL no matter what the slenderness ratio.

3.       The slenderness reduction factor does not need to be applied (MSJC eqn (3 – 16) and ( 3 – 17)  Discussion – It doesn’t seem like it should be applied as we are accounting for P-d and P-D effects. Also here are my reasons (I only have an older code in front of me ’02) Walls are addressed in section 3.2.5 where they address the limit to axial loads as 0.2f’m and 0.05f’m then stability is addressed with the P-D iterative method and they do not refer to eqn 3-16 and 3-17. However in section 3.2.6 they address walls for in-plane load where they do refer to eqns 3-16 and 3-17. Also in the reinforced masonry engineering handbook’s design example of a wall subject to out of plane forces they do not use the slenderness ratio. They use the iterative P-D Method.

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