beam design – CMDC

What is the difference between the length, clear span, bearing length, and design length of a masonry beam in MASS?

Brad Crumb — Mon, 11 Sep 2017 15:03:46 +0000

When creating a new beam in MASS, it can initially seem confusing to be shown so many different terms regarding lengths and spans. This post breaks down the difference between a beam’s length, clear span, bearing length, and design length, explaining the background and purpose for each.

Length

A beam’s length represents the total length of the entire modeled assemblage including any overhanging length on the outside edges of the supports. This value is typically entered in first and must accommodate the length of the opening above which the beam is spanning as well as the bearing plates on either side. Any additional masonry outside of the primary span is not used when distributing loads and determining the factored moment or shear that must be resisted by the beam. Once specified for a new beam design, a clear span,explained below, is then assumed based on the length needed for bearing on either end.

Clear Span

The clear span refers to the length of the opening above which the beam is spanning. While a beam’s length is typically entered first into MASS, it is also possible to start a MASS design by specifying a clear span and let the software automatically fill in the total beam length based on the length required for bearing on either side, rounded to the nearest modular cell length.

Bearing Length

The bearing length is defined as the length along the beam under which a high concentration of stresses due to concentrated loads is transferred to the supporting structure below. It can be spread over a steel plate or an area of masonry under compression. The default bearing length of 300mm was chosen for MASS because it is the longest allowable bearing length (CSA S304-14: 7.14.1.2) that can use a triangular load distribution and not require additional detailing (ie. using a rocker plate) to ensure a rectangular load distribution. For triangular reaction distributions, the centre of reaction is at one third of the bearing length away from the edge of the clear span and for rectangular distributions, the distance to centre of reaction is at half of the bearing length.

Design Length

Design length is the distance between the centre of reactions between beam supports. It is less than the beam length and greater than the clear span, used to determine the factored moment and shear. For example, checking MASS results by hand and looking to replicate the maximum factored moment at mid-span for a simply supported beam, the design length is used in M_f = wL²/8.

Quick demonstration – From masonry elevation to design using MASS

Starting with an elevation containing an opening with masonry extending above and on either side, a portion must be designated as part of the modeled beam. This example where two courses are assumed for the beam’s height, the full beam length extends one full masonry unit to either side which allows room for the bearing area in addition to the clear span.

For the same elevation, it is also possible to design a single course beam or go all the way up to four courses which can all result in acceptable solutions. Smaller beams have reduced moment capacity mainly from a smaller moment arm between coupling tension and compression forces while taller beams can have intermediate steel requirements (S304-14: 11.2.6.3) and may also have to satisfy additional provisions for deep beams (S304-14: 11.2.7) and deep shear spans (S304-14: 11.3.6). Choosing how a beam is modeled is left to the discretion of the designer.

For all masonry beam designs, a load path for vertical loads must be assumed for transfer to the supporting structure below. The default bearing area with a bearing length of 300mm is shown below resulting in a triangular distribution of the reaction force. Had the bearing length been longer than 300mm, the reaction would have been spread over a rectangular distribution (S304-14: 7.14.1.2).

In order to determine the design length, the centre of each reaction force must be determined. For a triangular reaction distribution, this location is one third of the bearing length away from the edge of the clear span for each support. (Rectangular distributions have a centre of reaction point half way through the bearing area)

The locations of the reaction forces expressed as point loads is then used to determine the design span. They are also where MASS draws the support points underneath the bearing plates in beam drawings.

Note that the difference in unit arrangement between the figures above and MASS has no impact on the design as fully grouted masonry. MASS always starts an assemblage with a full unit however starting with a half unit as was done in the illustrations above is functionally the same design.

Click to expand all referenced CSA S304-14 clauses

Taken from the 2014 CSA S304 masonry design standard, Clause 7.14 dictates the stress distribution that is to be used for transferring loads from a beam support to the wall below.

Clause 11.2.6.3 specifies the placement of intermediate reinforcement which is handled automatically by MASS. There is no option to disable intermediate reinforcement as that would result in designs that are not in compliance with the CSA standards.

In evaluating whether the “deep beam” classification is warranted for a design in MASS, the clear span is used in determining the span-to-depth ratio.

The edges of the clear span are also used in checking clause 11.3.6.

As always, feel free to contact us if you have any questions at all. CMDC is the authorized service provider for the MASS software which is a joint effort of between CCMPA and CMDC.

Masonry Beam Design: How to design beams with what might seem like less than the minimum reinforcement ratio

Brad Crumb — Mon, 10 Apr 2017 19:22:47 +0000

When a design seems like it might not meet minimum steel, all hope is not lost

While there is no way around satisfying the minimum steel requirements in the CSA S304, there is a clause that can be taken advantage of that is not used by MASS. By following the procedure outlined in this article, beams can be designed within the restraints of the CSA standards while also containing less than the minimum reinforcement area as prescribed by clause 11.2.3.1. This often overlooked clause can be very helpful when it comes to masonry beam design, especially those that would seem at face value to be the easiest to design due to nominal loads and short spans.

July 2020 Update: This clause has been incorporated into the release of MASS Version 4.0.

Disclaimer: This post is exclusively intended to provide insight into the approach taken by the MASS design software in interpreting a CSA S304-14 code compliant design. It is up to the professional discretion of the designer to input an appropriate layout, boundary and loading conditions, interpret the results, and determine how they should be incorporated into their designs. As per the end user license agreement (and also recommended within PEO’s guidelines for using engineering software), a tool cannot be considered competent and reliance on a tool does not relieve the user of responsibility.

To jump straight to a summary further down the post, click here.

The smallest beams with only nominal loading can also be the trickiest to design. One of the reasons is the need to satisfy minimum reinforcement ratio requirements in S304-14: 11.2.3:

Currently in the MASS software, all beam designs not meeting ρ_min in clause 11.2.3.1 will fail moment and deflection design, shown below in the simplified moment results tab:

While these designs are failed before attempting an incrementally larger design with more reinforcement, there is an option at the engineer’s disposal outside of MASS: invoke the mighty power of clause 11.2.3.2 which often goes overlooked.

There was some initial due diligence behind including this clause in the MASS software. It was not included as programming costs would be very high. The change would involve functionally designing two beams; the first being the actual beam used for the design, and the second beam being the theoretical beam containing one third less reinforcement which would be used to resist the factored moment. While at first glance, it might seem acceptable to instead ensure that the beam has additional moment resistance by factor of 4/3. However, while it is very close, there is not quite a directly proportional relationship between a beam’s moment resistance and its longitudinal reinforcement area so this approach would not adequately reflect the phrasing of clause 11.2.3.2.

How to use this clause

There are two different approaches that can be taken to satisfy clause 11.2.3 and design a beam that contains less than the required reinforcement ratio, or ρ_min, of clause 11.2.3.1:

Method 1

Compare the area of reinforcement to the area required by analysis increased by one third

Method 2

Compare the factored load to the load resisted by the beam if the reinforcement area were reduced by a third

Unfortunately, there is no way around completing a separate analysis to determine “the area of reinforcement required by analysis” quoted directly from the standard. Since the minimum reinforcement is a function of loading, one cannot be solved without assuming the other. Method 1 is likely more intuitive as it returns an alternative minimum reinforcement area to clause 11.2.3.1. However, Method 2 can be more straight forward to calculate as there is no need solve a quadratic or cubic function.

Method 1: Comparing areas of reinforcement

The approach taken in method 1 is to answer the question: “How much reinforcement is required in this beam design?”. This can be expanded to “What area of steel results in factored moment being equal to moment resistance?”. For beams with compression steel or intermediate steel which may not necessarily be yielding, it is likely easiest to solve for this value using a spreadsheet and GoalSeek or Solver since a simplified expression would likely require finding the roots of a cubic function and would also rely upon an assumed strain profile if the steel is yielding. Minimum reinforcement failure messages do not tend to present themselves for beams with several layers of steel so this article will focus on the simple beam designs where these issues arise.

The area of steel required, or A_{s_req}, can be solved for using force and moment equilibrium for any beam configuration where the beam’s moment resistance is equal to the maximum factored moment (M_r = M_f). The expression below can be used calculate this minimum area of reinforcement for beams with only tension reinforcement:

Note that this expression is only applicable to beams that exclusively contain primary tension steel to resist flexure (ie. no intermediate or compression steel). This formula also assumes that the tension steel is yielding which is a requirement in clause 11.2.2. Also, while the area of reinforcement required is typically a function of a beam’s moment resistance, it is possible that other factors such as cracking and deflection may govern the design and as such, should never be assumed to be satisfied and always checked manually (in addition to using this formula).

The units of each input are shown below as well as the derivation which can be expanded.

Units - click to expand

A_{s_min}, A_{s_req} are both areas in mm²

Φ_s, Φ_m, χ are unitless

f_s, f’_m are strengths in MPa (or N/mm²)

d, b are both lengths in mm

M_r is the moment resistance (set equal to factored moment) in kN*m

Note: M_r is multiplied by 10⁶ to convert the value from kN*m to N*mm

Expression derivation - click to expand

Starting with force equilibrium of a simple beam in bending, an expression for c can be rearranged based on all other inputs.

This expression can be substituted into a moment equilibrium equation which can be rearranged into ax² + bx + c = 0 format.

Now, recall (from grade 10 math) how to solve for the roots of a quadratic. In this case, it is the positive root which gives the simplified expression for A_{s_req} that we can use (The other root is not a valid solution since it is orders of magnitudes larger, violating strain profile assumptions which are used to generate the expression).

Quick note regarding the other root: The positive root in the original quadratic solution end up with the expression above with a negative sign in front of the square root term once substitutions have been made and then simplified. Had the expression been arranged on the left side of the equal sign rather than the right (as was done above) then we would be using the negative root, with a positive sign in front of the square root term with all other terms having flipped to negative.

The simplified expression no longer contains the plus/minus symbol to omit the other root that is not relevant in checking minimum steel. This is because it results in a required area of steel which is order of magnitudes larger than the value we are interested and the internal cross section forces as well as the strain profile used to derive this expression are no longer valid when the area of steel is so large. For example, in the scenario in the collapsible section below, a 2 course beam requires either an area of steel at the balanced condition of 8.64mm² or 105,633mm². If we were to entertain the second solution as possibly being valid, this would result in a tension force coupling with almost 36,000,000kN from the masonry. This would require a beam with a compression depth of more than 70m so hopefully by now it is clear how far outside the scope of our original beam we have wandered. With that new ridiculous new neutral axis location, the steel is no longer in the tension side of the beam which is one of the areas where things fall apart. From a pure mathematical perspective, it does satisfy the equation but the conditions behind that expression are no longer valid at such large values of A_s,req.

Method 1 Example (with MASS input)

Consider 1.2m long opening within a concrete block elevation that is being spanned by a masonry beam which rests upon 200mm supports on either side. Note that the bearing length has been shortened from the default 300mm further explained here.

Assuming that the condition for arching are present (see Section 5.6. Load Distribution on lintel Beams in our textbook for more), the loading can be modeled as two triangularly distributed loads plus the self-weight of the beam itself. In this case, the maximum magnitude of the masonry supported by arching (assuming hollow masonry in the three courses above the grouted beam) is 1.672 kN/m, or 800mm (length of beam divided by 2) multiplied by 2.19kN/m² (the weight per square metre of wall which can be found on page 751, here, for all unit sizes and types)

If using a 20cm, 15MPa unit, this beam will initially fail moment design, citing the following failure message:

Keep in mind that when a design in MASS fails, the displayed error corresponds to the most recently attempted design. In this case, the error message is based on a beam with several No. 25 longitudinal bars placed in tension. Earlier sections that would have been attempted could have failed for a number of reasons and in this case where the loading is nominal, sections not containing the minimum steel ratio in clause 11.2.3.1 result in the error message displayed at the top of this post. It can be triggered in MASS by deselecting all reinforcement options other than one No. 10 bar placed at the bottom of the beam.

The actual reinforcement ratio of this “failing design” can be found by clicking on the Detailed Moment Results tab and scrolling down to the bottom:

While it fails according to MASS, designs such as these are prime candidates for invoking the mighty power of clause 11.2.3.2.

This beam with a single No. 10 bar (100mm²) has more than eleven times the area of reinforcement required by analysis (8.64mm²) and easily exceeds requirement of having an additional one-third. Compared to the limit from clause 11.2.3.1, which would require 117.6mm² (0.002ρ = 0.002*190mm*309.35mm), this is the S304’s way of taking light loading into account for these types of designs.

To further illustrate this relationship, As,min plotted as a function of Mf is shown below:

Since the area of a No. 10 bar is less than the area required by 11.2.3.1, it is necessary to use clause 11.2.3.2 to satisfy minimum steel. The relationship between A_{s_min} and M_f is very close to linear (R2 = 0.997) so for each 1kN*m increase in factored moment, roughly 11.3mm² of steel is required (Note that this relationship is specific to a particular beam configuration and cannot be applied to others).

Method 2: Comparing Loads

Rather than ask the question, “What is the minimum area of reinforcement needed?” for a beam design, the other way to approach satisfying clause 11.2.3.2 is to assume that the area of steel present is equal to the minimum allowable and determine the largest possible factored moment that the assumption is valid for. Since this clause takes loading into account to evaluate minimum steel, rather than use a load to check the area of steel, method 2 involves using an area of steel to check the load. As mentioned earlier, this method may appear less intuitive as it is not as simple as comparing the reinforcement present to another minimum value. However, the steps used to determine the maximum allowable are the same as those used to calculate the moment resistance of any other beam design which is why method 2 may be preferable.

The table below shows a summary of the maximum allowable factored moments resisted by beams which contain less reinforcement than allowable by clause 11.2.3.1 for four possible beam geometries:

Maximum applied factored moment for a beams having less than the required reinforcement ratio in accordance with S304-14: 11.2.3.1 to satisfy S304-14: 11.2.3.2

As stated under the simplified method 1 expression earlier, although the area of steel required is typically governed by moment resistance, it is possible that other factors such as cracking and deflection may govern the design and as such, should never be assumed to be satisfied and always checked manually.

Table assumptions and background information

This table is only meant to be a guide to assist the engineer in designing a beam that does not satisfy minimum steel requirements in MASS which only checks against clause 11.2.3.1. An example outlining the exact process can be found further below by clicking on the expandable heading: “Method 2 Example“.

Masonry unit properties such as length, height thickness, and face shell thickness were set to the default values used by MASS according to the default masonry unit database. These values can be found on p751 to p753 of our textbook. Reinforcement placement was also assumed to be the same as the default bar placement used in MASS for each unit and reinforcing bar size which is based on a 75mm vertical clearance from the bottom face of the beam to the closest face of the bar. A yield strain of 0.002 was used for all bars with an elastic modulus of 200,000 MPa.

Obviously there are other properties such as masonry unit size and strength which will also affect these values and can be found by expanding the detailed table below. The table was simplified relatively late in this investigation exercise after observing how little variance there was in maximum factored moment as a function of unit size and strength. Factored moment was selected as the limiting variable for this table so that it would be independent of a beam’s span or load magnitude (a large span would be limited to a considerably smaller uniformly distributed load compared to that of a smaller span).

Expanded table taking unit size and strength into account

Maximum applied factored moment for a beams having less than the required reinforcement ratio in accordance with S304-14: 11.2.3.1 to satisfy S304-14: 11.2.3.2

Within each beam arrangement (ie. “2 Course Beam” with a “No. 15 Bar”), there is a smaller table where the column headings refer to masonry unit sizes (10cm, 15cm, 20cm, etc.) and the row headings refer to masonry unit strength (15MPa, 20MPa, etc.). As mentioned earlier, the variance within each beam arrangement was found to be relatively low.

All “N/A” values refer to configurations which satisfy clause 11.2.3.1 and will currently pass using MASS, independent of loading. Designs that did not result in strain profiles with yielding reinforcement were also checked against and removed as they are not allowable in accordance with S304-14: 11.2.2. There were not found to be any configurations that did not yield that also did not contain the reinforcement ratio required by clause 11.2.3.1. This was expected as yielding errors are a result of a beam containing too much reinforcement, increasing the coupled compression zone and lowering the location of the neutral axis while reducing the strain of reinforcement in tension.

Note that for designs using 2 No. 10 bars in tension (same bar area as 1 No. 15 bar), the placement is affected as the distance from the compression face of the beam is slightly further away from the vertical centroid of the bars. This change increases the moment arm separating the coupled internal forces and slightly increases the maximum allowable moment for these designs. The lower (and more conservative) values are shown in the table for simplicity however the comparison can be expanded below.

Maximum factored moment using 2 No. 10's compared to 1 No. 15 bar

Maximum factored moment for a 3 course beam with 1 No. 15 Bar (d = 507mm):

Maximum factored moment for a 3 course beam with 2 No. 10 Bars (d = 509.35mm):

Method 2 Example

Having already established that 7mm² of longitudinal steel is sufficient in satisfying 11.2.3, what is the largest moment that can be applied to a beam containing one No. 10 bar?

Since the requirements of 11.2.3.2 are a function of loading where minimum reinforcement area increases with loading, the highest load can be found by assuming that exactly the minimum cross sectional area is present within the beam. If a No. 10 bar has an area of 100mm² which is also equal to the minimum, the moment resistance will be based on a beam with 3/4 the steel, or 75mm².

Note: 3/4 was used rather than 2/3 due to the interpretation of “at least one third greater” from clause 11.2.3.2 being represented as “1 + 1/3” or “4/3” of the area required. The inverse, or 3/4, can be used to use this clause to check the design with a known area of reinforcement.

Continuing with the beam in the earlier example (20cm, 15mPa unit reinforced using a No. 10 bar, loaded up to 0.9 kN*m), the location of the neutral axis can be determined, followed by the moment resistance:

The moment resistance solved for here represents the maximum allowable load for the provisions of clause 11.2.3.2.to be valid. Therefore, the beam satisfies the CSA S304-14:11.2.3 minimum reinforcement requirements as long as the factored moment does not exceed 7.25kN*m (specific to the example beam constructed with 20cm, 15MPa units reinforced using a single No. 10 bar). Since the example beam was loaded to only 0.9kN*m, minimum steel requirements are satisfied.

The same procedure was followed for all of the other entries in the expanded table where unit size and strength are considered. For example, a 2 course beam constructed using 30cm, 25mPa units reinforced with a No. 15 bar can be loaded up to a bending moment or 24.79kN*m. (Note: The simplified table lists 23.83kN*m which is based the most conservative configuration which in this case is a 25cm, 15MPa unit)

Final Summary

All hope is not lost when a beam design fails due to not satisfying minimum steel using MASS, which only checks against CSA S304-14: 11.2.3.1. It is possible to satisfy minimum reinforcement requirements by instead using clause 11.2.3.2 by approaching the design in one of two ways:

Method 1: checking bar area against one third greater than the area resulting in M_f being equal to M_r
- This can be done quickly for beams with only primary tension reinforcement using the following expression:

Method 2: checking factored moment against moment resistance for a beam with a third less steel than what is actually present within the beam.
- This can be quickly checked by comparing the factored moment to the maximum allowable moment in the table below:

Disclaimer: These values should not be relied upon as part of the design process. They are meant to illustrate the concept that very low areas of reinforcement may be acceptable, depending on the applied loads. The full cross section should be analyzed by the engineer by hand (or other tool) to check these requirements in a way that is applicable to the situation. Regardless of which method is used, it is up to the designer to ensure that in addition to satisfying the minimum reinforcement requirements in S304-14:11.3.2, all other provisions must be considered and independently verified. The guides in both methods are based on the area of reinforcement being governed by applied bending moment at any section within the beam which is not true for all cases.

As always, feel free to contact us if you have any questions at all. CMDC is the authorized service provider for the MASS software which is a joint effort of between CCMPA and CMDC.

Why can’t I add stirrups to my masonry beam?

Brad Crumb — Fri, 26 Aug 2016 15:23:46 +0000

Want to place stirrups but the options are greyed out?

This is one of the most common question we get having to do with masonry beam design so if this is something you are stuck on, know that you are not the first! We’ll first break down the reasons for this issue and then quickly outline what you can do about it.

First of all, this is not a bug in the code. This issue is a product of the way the 2004 S304 standard limits stirrup spacing. It has since been addressed in the 2014 edition and MASS Versions 3.0 and newer will allow two course designs with stirrups.

When does this happen?

This issue is specific to beams designs where:

The beam is 2 courses high or less, and
The beam is failing in shear design for having a shear resistance that is too low

You can see in the screenshot below that only the “none” option is selected for stirrup placement and if you try and click the single or double leg selections, nothing happens because the whole input area is greyed out or “disabled”.

The fix seems obvious, right? Just add some stirrups to boost my shear capacity! This is where we run into problems.

Why MASS won’t let you just add stirrups to your beam

The error message you are seeing reads: “Design fails: There is insufficient steel stirrup area and/or bar spacing according to CSA S304.1-04: 11.3.4.7.1,2 and/or 11.3.4.8”.

For reference, 11.3.4.7.1 and 11.3.4.7.2 each refer to stirrup placement and minimum area requirements. The issue we are running into comes from 11.3.4.8 which covers “Spacing limits for shear reinforcement”.

Click to see full reference

The issue here and the reason MASS will not let you place stirrups in your beam is that clause 11.3.4.8 is specifying a maximum spacing of d/2 which for 2 course beams is less than 200mm. MASS recognizes this and greys out the option to even place stirrups because the modular nature of masonry restricts you to multiples of 200mm (or the space between the centre of adjacent cells in concrete block construction).

Since you cannot physically place stirrups closer than 200mm within a masonry beam, MASS disable this option.

As you can see, there is no way to place stirrups any closer than one per cell. For those curious, these beams were built for university research supported by CMDC. You can read more about that here.

What you can do to get a successful design

You have a couple of options to get your design to pass. You can always send your MASS project file over to us in an email and we can walk through these options together over the phone as well. To do this, visit our contact page to get in touch.

1. Re-examine your loads

Oftentimes when going through a design, there are assumptions that are made along the way that can have a big impact on the final design. In the case of load distribution, are all of the loads being carried straight down to the beam or can arching be assumed as outlined on page 305 of our textbook? Did you manually add the self-weight of the beam in your dead load? If so, you might be double counting if you have the self-weight option selected in the loads window. Are there other loads applied that might not actually be resisted by the beam? All of these can mean the difference between a beam design passing and failing in shear design based on the loads applied.

2. Model an additional course of masonry within your beam

Many masonry beams are supporting more masonry above them so why not take advantage of this by considering a third course to be part of your beam? The real world difference could mean as little as simply grouting an additional course but the key difference is that the increase in height and in turn, d (depth of tension steel from compression face), moves the d/2 spacing restriction to being greater than 200mm. This might not be possible if the clear span is very small since the additional course might bump you into deep beam territory but for most designed masonry beams, adding the third course to your design does the trick!

3. Think about using a high lintel unit

If you can’t get your d/2 value below 200mm, you can approach the problem from the other direction and remove the 200mm restriction by changing to a high lintel beam design. High lintel beams have a continuous grouted area that can fit any reinforcement configuration. They can even be modeled in MASS for those of you who are feeling more ambitious! Give us a call if you are interested and we’ll be happy to walk you through it.

Click here to see high lintel masonry construction in action!

A high lintel beam contains a soldier course of block directly above the opening. High lintel units look like tall U-shaped blocks and can be ordered from the unit supplier.

The nice thing about a high lintel design is that you don’t have to worry about your stirrups lining up with the webs of the concrete blocks. You gain the versatility of being able to drop in any configuration of reinforcement (including using any stirrup spacing) and then simply drop it in and grout around it.

Here you can see several high lintel beams used in one project.

One thing to consider which may affect your decision to use a high lintel in your design is that the running bond pattern present in the rest of of your wall will be visually interrupted by the soldier course. This is essentially a non-issue if there is a veneer covering it up.

4. Consider using a stronger masonry unit

No, this does not mean you also need to increase your grout strength (See note 4 of Table 4 within the S304 masonry standard for reassurance). Using a unit with a higher f’m value increases the shear strength of the masonry itself, possibly giving you enough resistance as to no longer require stirrups in your beam. This is not always practical and is only effective if the failing masonry shear resistance is close to the required factored shear force.

One more thing…

Seeing as this whole issue is born not from a software bug but the way the CSA S304 standard was written, MASS Versions 3.0 and newer will not have this issue. CMDC currently maintains an active role in shaping and developing masonry standards and as a result, the 2014 edition of the S304 standard includes an addition to the “problem” clause, 11.3.4.8, which now includes a minimum value of 200mm for the maximum spacing of stirrups.

It is for this reason that the issue you are dealing with now will no longer be a problem when you start designing with the 2014 S304 standard.

Still have questions? Feel free to call or send us an email! (Including your MASS file is very helpful)