## AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

## AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

(OP)

The thread744-499791: AS 3600 AMDT 2 Equation 8.2.7(2) additional longitudinal tension forces caused by shear is closed. To continue discussing, I created this new thread.

In a former posting by Doug (IDS(Civil/Environmental)in the thread, he stated:

"Case 3 is just using the AS5100.5 code formula for the longitudinal force due to shear, and finding the adjusted bending capacity, allowing for the force. Increasing the shear reinforcement reduces the longitudinal force, so increases the bending capacity."

as a response to my query:

"Could you please describe the steps in your calculation for Case 3 which resulted in an increase in bending capacity from an increase in the area of shear reinforcement ?"

My understanding of the analysis described in Doug's blog (link provided in thread744-499791: AS 3600 AMDT 2 Equation 8.2.7(2) additional longitudinal tension forces caused by shear) is as below.

The total longitudinal steel A

V

V*

A

A new moment effect M*

Owing to M*

Doug,

Before I discuss further, I would like to check whether my description of the analysis above is accurate.

In a former posting by Doug (IDS(Civil/Environmental)in the thread, he stated:

"Case 3 is just using the AS5100.5 code formula for the longitudinal force due to shear, and finding the adjusted bending capacity, allowing for the force. Increasing the shear reinforcement reduces the longitudinal force, so increases the bending capacity."

as a response to my query:

"Could you please describe the steps in your calculation for Case 3 which resulted in an increase in bending capacity from an increase in the area of shear reinforcement ?"

My understanding of the analysis described in Doug's blog (link provided in thread744-499791: AS 3600 AMDT 2 Equation 8.2.7(2) additional longitudinal tension forces caused by shear) is as below.

The total longitudinal steel A

_{st1}is first determined for shear reinforcement A_{sv1}/s for a section with design shear V*_{1}, ultimate shear strength V_{u1}such that the axial force in A_{st1}is close to f_{sy}under combined action effects M*_{1}and V*_{1}. The shear reinforcement is the increased to A__{sv2}/s with the following values kept the same:V

_{u2}= V_{u1},V*

_{2}= V*_{1}A

_{st2}= A_{st1}A new moment effect M*

_{2}is then determined through trial and error (or iteration) so that A_{st2}is close to f_{sy}.Owing to M*

_{2}determined being greater than M*_{1}, it was concluded that an increase in the shear reinforcement caused an increase in "reduced moment capacity"Doug,

Before I discuss further, I would like to check whether my description of the analysis above is accurate.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

- When the ultimate shear capacity is exactly equal to the design shear force the AS 3600 equation gives exactly the same result as AS 5100.5.

- If the shear reinforcement has spare capacity the angle of the shear failure plane would have to be increased for the section to fail in shear.

- If the failure plane angle is increased, the longitudinal force due to shear is reduced, so the steel area available to resist bending is increased.

- You don't need iteration to find the moment capacity allowing for shear effects if you assume the applied shear is exactly equal to the shear capacity. You know the steel area available to resist bending effects, so you can find the moment capacity in the usual way.

- If you want to find the maximum shear and moment capacity for a given ratio of moment/shear, then iteration is required for that.

Another link with more details:

https://newtonexcelbach.com/2022/12/15/more-on-com...

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

For design to both AS 3600 and AS 5100.5, the design action effects M* and V* are known, and they have safety factors prescribed by these standards. If we provide shear reinforcement larger than the amount required for shear adequacy, this gives a value of φV

_{us}such that φV_{u}is greater than V*. The requirement for design is φV_{u}>= V* [Eqn 8.2.1.6(1),AS3600]. There is no requirement for the shear reinforcement provided to yield at V*.Angle Θ

_{v}[8.2.4.2(5), AS3600) is determined using M*, V* and A_{st}. It is not dependent on the amount of shear reinforcement A_{sv}/s provided. Increasing the amount of shear reinforcement increases φV_{u}but has no effect on Θ_{v}.According to the MCFT, the shear capacity can be determined for any combination of applied shear and moment. For design to AS 5100.5 and AS 3600, and for design action effects M* and V*, the reduced ultimate shear capacity φV

_{u}is calculated directly. For load rating of a load rating vehicle, at any stage of movement of the vehicle along the bridge, M* and V* can be calculated. The capacity φV_{u}also can then be calculated without iteration.There is no single "actual shear capacity" for a section, even for a test structure with an increasing test load. For a test structure with an increasing applied load P such that both M

_{n}and V_{n}increases proportionally with the load, at any stage (say i) of loading with action effects M_{ni}and V_{ni}, the strength V_{nui}can be determined. To check experimentally the accuracy of the equation used for MCFT, one has to iterate to predict the load for shear failure P_{ui}with shear V_{uni}where V_{uni}=V_{ni}. This shear capacity V_{uni}=V_{ni}is not the actual shear capacity of the section:It is the shear capacity of the section only for action effects V_{ni}(=V_{uni}) and the coexisting M_{ni}.Iteration (to obtain V*=phiVu) is not necessary for design (to AS 3600 and AS 5100.5) and for the determination of load rating factors (to AS 5100.7) because for both activities the reduced ultimate shear capacity phiVu is calculated directly from M* and V*. The phiVu calculated directly without iteration is the capacity of the section with V*. This capacity is the "actual shear capacity" if one wanted to call it that. The capacity obtained using iteration does not give any useful information as the design requirement is V* =< phiVu, not V*=phiVu.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The main points I have made on this topic are:

1) The AS 5100 equation for longitudinal force due to shear is highly unconservative because it doesn't place any limit on Vus.

2) The AS 3600 version may be very conservative if used with the minimum value of Theta because it doesn't allow for any effect of shear steel area being greater than required.

3) Adjusting the theta value with the AS 3600 formula removes this conservatism, and is consistent with results from the AASHTO and Canadian codes.

4) If you are checking a known section with known reinforcement and known V* and M* values, no iteration is required.

5) For checking sections with many different possibly critical load cases it is straightforward to generate a shear/moment interaction diagram.

Do you disagree with any of that?

Have I missed any problems with designing for shear/moment?

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Thank you for your comments.

Point 1In your blog, you recommended that the value of ØV

_{us}to be limited to V*. This restriction of "ØV_{us}not to be taken greater than V*" was accidentally left out in AS 3600 with amendment 1 and in AS 5100.5 since a similar restriction is present in AASHTO and the Canadian design standard. In a recent paper presented in the Concrete2023 conference held in Perth (which you attended), reference listed below, we also recommended that this missing requirement be included in AS 5100.5 and to the relevant equation in AS 3600 with amendment 1 (if reinstated). So we both agree that the equation of AS 3600 Amendment 2 can be conservative.Although I could not find the reason for the restriction similar to "ΦV

_{us}not to be taken greater than V*" in AASHTO and the Canadian standard, it is prudent to include it in AS 5100 to limit the amount of reduction of longitudinal steel.(Vannisorn V, Wong KW and Mendis PA, Increase in longitudinal steel reinforcement to account for shear in RC beam design to AS 3600, Concrete Institute of Australia Conference Concrete2023 held in Perth 10-13 September 2023)

"The AS 5100 equation longitudinal force due to shear is highly unconservative" is not strictly accurate. In design to AS 5100.5, one has to meet all requirements. The three key requirements for a section with shear and bending are

(a) flexure: φM

_{u}> = M*(b) section shear: φV

_{u}> = V*, and(c) block rotation : stress from the force in the longitudinal steel <= f

_{sy}.If A

_{sv}/s is increased and the amount of longitudinal steel is reduced to satisfy (c), one must go back to check whether the total steel provided still satisfies (a) and (b). If we provide total A_{st}which satisfies all three requirements currently in AS 5100.5 and AS3600 with amendment 1, we expect that the section under the design action effects has adequate safety to these standards for bending, section shear and block rotation.If the design is carried out in sequence (i.e. not iteratively) by reducing the longitudinal steel in (c), increases A

_{sv}/s and fails to go back to check that both (a) and (b) are met, then the design is unconservative.Point 2We both agree that the equation in AS 3600 amendment 2 may result in excessive longitudinal reinforcement.

In the conference paper, it was stated that the AS 3600 Amendment 2 equation unnecessarily "removes the flexibility in setting the proportion of the amount of shear reinforcement to additional reinforcement for cost optimisation and to ease congestion of longitudinal reinforcement". Also "it may result in the provision of an excessive amount of additional reinforcement in comparison with the amount determined using the previous equation".

Not sure what you meant by "minimum value of theta" as Θ

_{v}is determined in AS 3600 from M* and V* and the total A_{st}provided.Point 3I do not follow what you meant by "adjusting the theta value". The only limit in AS 3600 to Θ

_{v}is from ε_{x}<= 3E10-3 [Eqn 8.2.4.2(5), AS 3600] which is a requirement for section shear. This limits Θ_{v}to 50 degrees.Point 4Agree

Point 5Disagree.

The interaction diagram using the section-shear equation for MCFT is an MCFT interaction diagram. It is not necessary for design or load rating as one determines the reduced ultimate shear strength ∅V

_{u}directly from M* and V* (without the use of iteration).The M* on the MCFT interaction line is not a "moment capacity". It is the coexisting action effect with V* for shear adequacy, i.e V*=∅V

_{u}. The term "reduced moment capacity" should be limited to ∅M_{u}.If the calculated ∅V

_{u}<> V*, this value cannot be obtained from an interaction diagram. This calculated value is the "actual" shear strength of the section with M* and V*. An interaction diagram is not required to check shear adequacy of a section with action effect M* and V*. if V* <= ∅V_{u}, the section is adequate (inside or on the interaction line) for section shear. If V* > ∅V_{u}, not adequate (outside the interaction line).## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The top line complies with all the current requirements of AS 5100.5, but if the restriction on Vs was included it would come down to the light blue line (Canadian code). How is that not unconservative?

The reason is that the force in the shear steel cannot be more than the applied shear force. But it seems that you agree that AS 5100.5 should be changed anyway.

For AS 3600 I agree that the equation can be over-conservative if applied without adjusting theta, but with adjustment of theta it is less conservative than the Canadian and AASHTO codes, whilst still complying with the requirements of MCFT.

The procedure for adjusting Theta is:

1) Find the angle (and associated strain) such that phi.Vu = V*.

2) If that is greater than 50 degrees then set Theta to 50 degrees and strain to 0.003.

3) Calculate the longitudinal force due to shear with the new Theta.

4) Find the moment capacity allowing for this force.

Note that AS 3600 explicitly allows the longitudinal strain to be increased above the calculated minimum value (Cl. 8.2.4.2.2), up to a maximum of 0.003. The only reason I can think of for doing that is to reduce longitudinal steel forces by taking advantage of available vertical reinforcement capacity, by increasing Theta.

I don't follow your objection to using an interaction diagram. For any point on the line the moment is equal to the section moment capacity with the associated applied shear force. Any point inside the line will have a greater shear and bending capacity than the applied actions.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

If you look at the theoretical development of the MCFT by Collins etc,

in determining the longitudinal tension force, they came up with a formula that included a term

2Vc + Vs.

They decided to then use the equation V* = Vc + Vs, reworked to Vc = V* - Vs

to remove the Vc from that longitudinal tension force calculation with the assumption that the shear reinforcement is at yield.

In this Vs is not the capacity of the vertical stirrups, it is the force that can be carried by the stirrups under the loading. The stirrups cannot create vertical force.

If more stirrups are provided than are required for the vertical force at the section, the stress in those stirrups will be less.

Your error is in assuming that Vus = Asv fsv

It should be fs.

In AS3600 formula for Vus, that does not matter in the shear capacity calculation. But when Vus from the code formula based on fsv is used in the longitudinal force calculation, it does matter, and Vus in that case should be based on fs, not fsv

Vus can never be greater than the applied shear force the stirrups are carrying.

In AS3600, it was decided, instead of removing Vuc from the initial formula above, and then requiring designers to calculate the actual stress in the vertical reinforcement, we would remove Vus. The result is exactly the same, if fs is used correctly in the Vus formulation

Part of the reason was exactly the mistake you are making, assuming the stirrups are at yield. There were instances seen in design checks where designers had reduced the longitudinal tension force to zero by providing large quantities of vertical reinforcement.

I have explained this several times. Other codes have not picked up the difference between their formulae and the original MFCT development and formulation.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Thank you for your comments.

The equation for the block rotation (a term I have been using for this potential failure mode) is from the AASHTO bridge standard as described in one of my earlier posts. It was developed using a free body diagram, and used a few simplifying assumptions. One of these is that the contribution of the resistance from the concrete to the rotating body is small owing to its small lever arm about the point where the block rotates and can be ignored.

Much of the work of Collins et al is for the developments of the CFT and MCFT, and it was for research purposes to come up with equations to check the accuracy of equations from these theories through laboratory testings. These equations are able to predict forces and strains over the entire range of loading to enable the researchers to verify the accuracy of their equations.

The free body diagram used for block rotation appears to be similar to those used for "strut and tie" design to ensure sufficient amount of reinforcement is provided to prevent failure. For shear, the design requirement is adequate capacity at V*. There is no ductility requirement for shear so both the longitudinal steel and the shear reinforcement do not have to yield at V*.

It was pointed out in the conference paper by Vannisorn, Wong and Mendis (reference given in my earlier post) that "the longitudinal steel determined using Equation 8.2.7.1 in AS 3600 with amendment 1 is based on the available resistance from the yielding of the provided shear reinforcement and the design shear V*. It is also not expected to fail with the failure mode depicted in the free body diagram before V equal to V*". This equation uses V

_{us}to determine the amount of longitudinal reinforcement that has to be provided to prevent failure through block mechanism at V*. The term φV_{us}is the potential resistance available from the shear reinforcement provided. It is not the force in the stirrups at V*. By providing the steel determined using this equation, the design is adequate for block rotation.You stated "There were instances seen in design checks where designers had reduced the longitudinal tension force to zero by providing large quantities of vertical reinforcement.". I think you meant the additional tension force required for the effect of shear. This is because the restriction of "φV

_{u}s is not to be taken to be larger than V*" found in AASHTO and CSA standards (to prevent excessive reduction in the amount longitudinal reinforcement) was missing. It is recommended in the conference paper that this restriction be included in AS 5100.5.It should be pointed out that after reducing the total amount of longitudinal reinforcement to satisfy block rotation, designers should should go back to check that the section-shear requirement is still met if design checks are carried out sequentially.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The ASSHTO Diagram is not complete. We started with that. There are a few actions missing to satisfy statics!

You cannot base the calculation on the assumption of steel yield if the stresses are never sufficient to cause yield. Compatibility and Statics have to be satisfied.

I meant what I said, longitudinal tension force for shear was calculated as zero at a location where the longitudinal tension force for flexure was also 0, so the total longitudinal tension force was 0, so no development was required!

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Thank you for your comments. Several queries.

(1) Are all your plotted points optimally adequate for section shear (i.e having V*=φV

_{u}) ?(2) In your procedure for adjusting theta, you mentioned about getting theta greater than 50 degrees for φV

_{u}= V*. How is it possible to get theta > 50 degrees when the standards do not allow ε_{x}to be larger than 3E-3 in the calculation of φV_{u}?(3) Does any of you points satisfy optimal adequacy for both block rotation (f

_{s}= f_{sy}) and section shear V*=φV_{u}?## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I have just gone back through the development of the method, where they determine all forces from statics of the actions on the member. As the Collins work did not show the full derivation, and there were some questions that could not be answered without it, especially regarding the effects at end supports compared to sections along the member, we developed it again from 1st principles to confirm their results and answer our questions and achieved the same result as the original Collins et al work.

From this they determined,

Longitudinal tension force at the tension face

Asy fsy = (Vc + .5 Vs) cot theta

In this original form developed from the static equilibrium of the actions at the section, if you were to use the logic that Vs = Asv fsv, then adding extra vertical reinforcement would actually increase the longitudinal reinforcement requirement.

They then re-arranged the formula, using V* = Vc + Vs, so Vc = V* - Vs to substitute for Vc in the formula.

This resulted in Asy fsy = (V* - .5Vs) cot Theta.

They used the further logic that V* = Vu / phi to come up with their final formula

Ast fsy = (Vu / phi - .5 Vs)cot theta

The only way both of the substitution assumptions they made are correct is if V* = Vc + Vs.

Your interpretation of the reliance on Vs to reduce the longitudinal reinforcement is exactly the opposite of the actual calculation.

The AASHTO diagram does not even have a Vc force component from memory and neither does it include any moments at the section.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Thank you for your comments.

The AASHTO equation is described in a published paper titled "Longitudinal Steel Stresses in Beams Due to Shear and Torsion in AASHTO-LRFD Specifications" by Rahal N, published in ACI Structural Journal, 2005. This paper can be found on ResearchGate. Results from test beams show that the equation can predict accurately the longitudinal steel strains (with the term V_s in the equation). There is no mention of the equation requiring an assumption of V_c = V_u/phi - V_s to derive it. The use of V_s directly in the equation used to predict strains for the tests shows that there is no need to have the term V_s restricted to 'V_u/phi -V_c' for the equation to be accurate.

The AASHTO has the force V_c component shown next to the failure surface but not labelled since not used in developing the equation. It diagram does not include any moments at the section because this free body is for the determination of the force from shear. The force from moment is determined separately using the term M_u /(phi dv).

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

No, the whole point of the exercise is to investigate the effect of increasing the area of shear steel so that V* < φVu. When V*=φVu the AS 3600 and AS 5100 equations give the same result.

No, εx can vary between the minimum value given by the equations and 0.003, as stated in the code, so the maximum value of Theta is 50 degrees. That is why the moment capacity line using AS3600 + adjustment of Theta flattens out at 14 mm in the graph plotted in my previous post, but the AS 5100 lines continue to increase.

Aa above, we are looking at the cases where AS 5100 gives unconservative results for bending when the shear steel has been increased so that V* < φVu, but for a shear steel diameter of 10 mm V*=φVu and all the codes give the same result.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The formula in MCFT for longitudinal tension force is derived from the same force diagram and compatibility equations that were used to derive the formulas for the calculations of the strains for the calculation of Vuc and uses a combined effect of applied moment and shear, which is the basis of MCFT.

It is not the same force diagram as is shown in AASHTO for longitudinal tension force into an end support as it includes both co-existing applied moments and applied shears and Vc and is applicable at any point along the member, not just at the end support. And from memory the best way to resolve the forces was moments about the support point, in which case , as you investigate points away from the support, the effect of Vc is not insignificant as the AASHTO discussion suggests.

And produces equals longitudinal tension force due to shear in the top and bottom as is defined in MCFT and AS3600 and even Eurocode 2.

AASHTO appears to not have a longitudinal tension force from shear at the compression face.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Thank you for your prompt response.

Since you did not assume φV

_{u}= V*, I wonder what assumption(s) you used to determine M* for a new A_{sv}and how optimally you have designed for section shear and block rotation.If I can see the values for the listed variables below for the first two points of the AS 5100.5 line, that will help me to understand better your study.

M*

V*

φV

_{u}(since not equal to V* for the second point)f

_{st}, the force in the longitudinal steel from combined action M* and V*Also if possible,

A

_{st}width B

depth D

effective depth d

A

_{sv1}used## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I should clarify what I meant there. I was looking at cases where φVumax > V*. In that case the strain and associated Theta were calculated such that φVureduced = V*. If the adjusted Theta was <= 50 then that value was used, otherwise 50 degrees was used.

Either way, a new longitudinal force due to shear was calculated using the AS 3600 formula, and this was used to find the reduced moment capacity. If φMuadjusted >= M* then the section is OK, and if it is < M* it isn't.

Note that for the interaction diagram an iterative calculation is required since the points on the curve are found for specified values of M*/V*, so after φMuadjusted is calculated the calculation needs to be repeated for a new V*.

I'll have a look and either post the values or the complete spreadsheet.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Considering more than 1 mode together is confusing when using an interaction diagram. I therefore limit my discussion below to section shear effect.

Say, for the solution for point 2, you use an interaction diagram for A_sv2/s. And you iterate along the line from the origin of the diagram maintaining the relation M*/V* for the point (V1*,M1*), and this line intersects with the interaction diagram. Say, the values you get from the intersection point are M*_intersect,V*_intersect and phiVu_intersect. M*_intersect is not a "reduced moment capacity" of (V1*,M1*). It is the coexisting design moment with design shear of V*_intersect, and for these effects, the ultimate shear capacity is phiVu_intersect.

Before you spend more time on this, I better check with you whether the above steps were used to get point 2 since you mentioned M*/V*. If you did, the coexisting moment (your "moment capacity") you get for point 2 is for a different design shear, not for V1*.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I mentioned M*/V* with respect to creating an interaction diagram, where iteration is required.

For checking a section for a specified V* and M* the procedure is:

1) Find the mid-height strain for V* and M* and then find the associated Phi.Vu.

2) If Phi.Vu < V* then the section doesn't work.

3) If Phi.Vu > V* then find the mid-height strain such that Phi.Vu = V*, and the associated adjusted Theta.

4) Using the lesser of the adjusted Theta or 50 degrees, find the section Phi.Mu, including allowance for longitudinal load due to shear.

5) If Phi.Mu >= M* the section is OK. If it isn't the section doesn't work.

6) Finished.

I don't know what you mean by "iterate along the line from the origin". If I create an interaction diagram I then plot all the M*, V* points, and those inside the line are OK, those outside are not.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

In your step 3, you have to assume a relation between M* and V* to be able to get V* = phiVu. I think you used the ratio of M*/V*. At the end of this step, the design shear is same as the ultimate shear strength V*_intersect = phiVu_intersect. The design moment is

M*_intersect = M*/V* x V*_intersect. This adjustment is what I mean by "iterate along the line from the origin". At the end of the adjustment, M* and V* are on the interaction line.

My understanding of MCFT and the use of AS 5100 for design of your point 1 and 2 AS5100case is :

When you increase the shear reinforcement from Asv1 to Asv2, for a set of design actions (M*,V*),.

For 2

Bending capacity phiMu no change from 1 since no change in Ast.

Shear capacity phiVu increases as a result of the increase of shear reinforcement from N10 to N12

phiVuc and theta_v - no change since they depend on M*, V* and Ast only, not Asv.

phiVus increases owing to the increase in Asv from Asv1 to Asv2

MCFT is a theory for shear capacity, and this capacity is determined using the applied action effects (not factored up [or down] action effects), so no iteration (adjustment of strain to give V*=phiVu) is required. The theory cannot determine bending capacity.

I reproduced below a block of text from one of my earlier posting (with correction) to illustrate this:

"There is no single "actual shear capacity" for a section, even for a test structure with an increasing test load. For a test structure with a monotonically increasing applied load Pi such that both Mni and Vni increase with the load, at any stage (say i) of loading, the shear strength Vnui can be determined. To check experimentally the accuracy of the equation used for MCFT, one has to iterate (numerically) to predict the load for Pui when the shear Vni where Vuni=Vni. Note that this iteration uses the characteristic of the loading, i.e. Mni/Vni is assume constant. This strength Vni=Vuni is not the actual shear strength of the section:It is the shear capacity of the section only for the level of loading where the action effects are Vni (where Vni=Vuni) and the coexisting Mni."

The subscript "n" indicates nominal or unfactored.

For design (and for load rating of a rating vehicle),and for a section with M* and V*, the only ultimate shear capacity of relevance is the phiVu determined without iteration. The action effects are factored up. So if V* < phiVu, section-shear is adequate.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

No, that is not the procedure I am following. The graph of Moment Capacity against bar diameter shows the design moment capacity (Phi.Mu) for a specified and unchanging V*. That is why all the lines other than AS 5100.5 flatten out when phi.Vu with Theta = 50 degrees > V*.

For generating an interaction diagram you need to iterate to find each point on the line but having done that you just plot each load point from the analysis using M* and V*, which seems to be entirely consistent with your final statement:

... subject to M* < phiMU of course.

So given that we agree on that, I have no idea what your point is.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Imagine a situation where the design actions on a particular beam are very small, say V* ~ 0 kN and M* ~ 0 kNm, just for argument's sake.

Now take the exact same beam and apply some substantial combination of V* >> 0 and M* >> 0, perhaps near the brink of shear failure.

According to the MCFT formulation in the Australian codes, which calculate epsilon_x, k_v and theta_v based on the ULS design actions (V*, M*, N*, etc.), these two beams would be predicted to have two completely different shear strengths, which is impossible. I understand the Australian codes don't want to get into the iteration process because it becomes very unwieldy, but in my opinion it needs to be done.

For design purposes, at best, you can use the ULS design actions to satisfy V* < phi*Vu, but merely satisfying that inequality will only give a lower bound on the shear capacity, and may give the impression that there is more reserve capacity than there actually is. For a load rating exercise of an old bridge, on the other hand, using the ULS design actions to calculate the shear strength of a beam may give an overly-conservative estimate of its shear strength, because you may be assessing the beam for a combination of loads that it can't actually reach. In this case, you would have to scale the design actions down.

In my opinion, the shear strength of a concrete beam is simply the shear force that it can resist right up to the brink of shear failure. It has nothing to do with the magnitude of the design actions, only their ratios (e.g. M*/V*, N*/V*, etc.). I believe the iteration approach is given more consideration in AASHTO for example.

As a side question, I am interested to know everyone's approach when it comes to iteration. Do we try to match V* and phi*Vu directly, or are we trying to match V* and Vu, then apply the phi factor separately after the iterations are complete? So far I have been leaning towards applying phi after the iterations, but there has been some debate each way in my office. I understand that applying phi after the iterations tends to give (very) slightly more conservative results, though probably nothing meaningful.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

So the concrete shear capacity, Vc is dependent on both the co-existing moment and shear actions on a member and will be different for every load condition.

The problem with older code shear design methods was that they were basically independent of flexural effects. So a member with the same design shear, but very different moments had the same shear strength if they had the same flexural reinforcement, even though the principal strains were very different. So a section at the free end of a member where M is basically zero had the same capacity of the critical section at the face of an internal support with a large M if the flexural steel quantities were the same.

MCFT recognises that the very different principal strains in these 2 cases gives very different principal stress and therefore very different concrete shear capacities.

In older design methods for PT, most codes required a principal shear design check as well as a flexural shear design check. But not for RC members. MCFT combines them into one shear check, but they are dependent on M and V.

SMCFT introduced a simplified expression for determining the principal strains due to this combination to make application of the theory codifiable and more simple to use (if you every looked at even the initial MCFT it was basically impossible to apply in a design situation, CFT was impossible) which is in the codes (Epsilonx in AS3600), but I think all codes then allow the designer to determine the strains by more rigorous methods. The strain calculated in SMCFT is basically at the point when the flexural steel reaches Yield, not at ultimate.

We have been saying for years that a more logical approach considering the combined flexural and shear actions was needed. MCFT provides that and is by far the most logical method of determining shear capacity in the design phase which is why codes are adopting it.

This may present problems for people trying to do the reverse and are attempting to determine a load rating for an existing structure. But that does not mean there is a problem with AS3600 or other design codes that have adopted the solution for the design of concrete structures.

Maybe codes used in areas where load ratings are required to be calculated on a regular basis should introduce a method for doing this. Or engineers doing load rating exercises need to go back to the fundamental principles of CFT and determine a methodology based on that for themselves.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

My main points in this thread were:

1) The current procedure in AS 5100.5 can be un-conservative.

2) The procedure in AS 3600 complies with the MCFT it is based on, and is not over-conservative if applied with adjustment of Theta where necessary.

kww2008 seems to disagree with one or both points, but so far it isn't clear to me why.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I have just realised that in the Draft Amendment 2 for AS 5100.5, there is a new statement in Clause 8.2.4.1: "For the calculation of the capacity of a predefined member and the assessment of members of existing structures, iteration shall be used to determine the value of the longitudinal strain (epsilon_x)."

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Cross section and longitudinal steel:

Shear steel and applied loads:

The only inputs that were changed were:

The shear steel diameter.

For AS 3600 with adjustment of Theta the "Compression Strut Angle" was adjusted to reduce the shear capacity to 280 kN, with a maximum of 50 degrees.

For AS 3600 the moment capacity results were multiplied by 0.8/0.85 so that they had the same Phi value for bending as AS 5100.5

The spreadsheet used for the calculations can be downloaded from:

https://interactiveds.com.au/software/RC%20design%20functions9.zip

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Dia, mm Phi.Vu, kN

10.0 274.6

12.0 350.8

14.0 440.8

16.000 544.6

20.0 793.9

Yes, ThetaV is 37.56 degrees. That is why it can be adjusted when using the AS 3600 provisions.

I have just used the lowest possible Phi values across all the codes, to keep everything the same, other than the equation for longitudinal force.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Disagree that there is an "actual section capacity" for a section subjected to a given M*/V*. According to MCFT, there is no "actual section capacity" for this section. There are multiple section ultimate shear strengths, each corresponding to a different level of V*. There is a phiVu for V*=phiVu. This ultimate shear strength has optimal shear adequacy (ie. V* is not greater or less than phiVu) for the corresponding level of design shear V*.

Iterations are only required to determine shear=strength for investigative studies relating to MCFT where a test to failure is carried out by applying a monotonic load, and not necessary for design or load rating. Iterations is often used to seek the level of shear loading Vun = Vn for tests. Vun the the shear level we expect to see signs of shear failure. We use mean material strengths obtained from testing and to exclude all safety coefficients. The subscript "n" shows the use of unfactored values. At intermediate levels of shear(Vn < Vun), we can check accuracy of the equation using strain measures since no visible shear distress is expected.

To show the inappropriateness of using a phiVu determined for another level of loading for the calculation of load rating factor for shear, let us look into the meaning of V*= phiVu. phiVu is smaller than Vun as it has safety factors (from using φ, characteristic values for steel and concrete). V* is larger than Vn (from using load factors). Thus, phiVu has moved to the right on the magnitude line and Vn to the left. Since these values have safety coefficients included, V*= phiVu has meaning only for shear adequacy to the standards. phiVu (=V*) is the reduced ultimate shear strength at V* (for this monotonic load) for optimal (not more or less than required for the level of safety of the standards) shear adequacy. It is not an "actual shear capacity).

Since the MCFT equations in AS5100.5 and AS3600 are expected to be accurate, and safety coefficients have been included as required by standards, the phiVu calculated using M* and V* (without iterations) for the most critical (expected to give lowest factor for the bridge) vehicle position is suitable for calculating load factor for shear. Also the use of phiVu determined without iteration is consistent with rating the loading of the nominated vehicle on a bridge as intended in AS 5100.7. When we iterate, the phiVu determined is for a scaled vehicle.

By using the phiVu determined without iterations, the load rating factor determined gives an accurate measure of the degree of shear adequacy for the bridge vehicle system. Using the strength from iteration, on the other hand, will not give an accurate measure since the strength is not consistent with the design actions M* and V*, and with the load.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

It seems you also agree that the equation for the longitudinal force in AS 5100.5 is unconservative, because you agree that Phi.Vs should be limited to V*, as in the AASHTO and Canadian codes.

The only outstanding question is the validity of the AS 3600 equation. When Phi.Vu = V* it exactly agrees with the AS 5100.5, and with adjustment of the Theta angle it is not over-conservative when Phi.Vu > V*, in fact slightly less conservative than the American versions, so I still see no problem with it.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

When we changed over from using an ultimate shear capacity which depends on action effects, the relationship between strength and shear is nonlinear. While the load rating factor can be taken as a load scaling factor (to determine the ultimate shear from design shear) in the past with earlier design standards because of usage of a non-varying strength with shear, this is no longer the case when we use MCFT to determine shear strength. Not using iterations gives a rating factor for the rating vehicle. Using iterations (on the live load component only) gives a scaling factor (such that the scaled vehicle has a load rating factor of one). Note that the scaling factor is not a rating factor.

The load rating factor should continue be determined without the use of iterations. This is the current practice and should not be changed in AS 5100.7 as the load rating factor gives an accurate measure of shear adequacy. If a scaling factor (applied to the rating vehicle to give a scaled vehicle with a load rating factor of unity) is required, that should be requested by individual road authority owning the bridge since they are the users of the rating information to manage bridges.

_______________________________________________________

Doug,

Thank you for your comments.

I agree with including a restriction "The term phiVus should not be taken to be greater than V*" . Designers can still use phiVus > V* for section shear design.

This should be changed to using phiVus to be consistent with the equations in other national standards and to add the restriction above.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I strongly disagree. AS 3600 is already consistent with MCFT, and gives results that are consistent with the N. American codes based on MCFT.

AS 5100.5 is currently under final stages of revision following public comment. It needs to be revised and should be revised to use the same equation as AS 3600, so the national concrete codes are consistent on this issue.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

In their model,( assuming it is complete even though we disagree), The only vertical forces are Vu, Vs and Vc (which for some reason you think you can ignore).

The whole derivation if done properly requires

Vu = Vs + Vc .

You cannot then ignore part of this in deciding on the Vs limit. You cannot create vertical force. You cannot ignore one force. If you put in more vertical steel, the stress in the steel reduces. The force stays basically the same to maintain vertical equilibrium. The vertical reinforcement cannot attract more force than exists.

Just because 2 other codes have decided to ignore statics does not mean we should. Otherwise other designers around the world could be saying " but AS5100 says Vs is unlimited" so it must be.

Do we then accept that as correct?

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Agree to disagree with "It needs to be revised and should be revised to use the same equation as AS 3600". Thank you for your comments.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Suggest you get hold of a recent copy of AASHTO bridge Specification to view the diagram and the assumptions made, and to see the equation is derived. In AASHTO 2010 5th Edition, this diagram has a label "Forces Assumed in Resistance Model Caused by Moment and Shear". An assumption is made that Vc (which is acting along the assumed diagonal crack) has a negligible moment (owing to having a small lever arm) about point 0, a point near the top of the diagonal crack.

The failure mode of interest is rotation of the block, so Vu = Vs + Vs (the requirement for section shear) has no relevancy to this "determination of the requirement for tension force in the longitudinal reinforcement". Section shear is a mode which has to be designed for separately.

The calculation of longitudinal force in the longitudinal reinforcement is based on the assumption that the full resistance from the vertical steel Vs (yielded) is mobilised and the acting force is Vu. If the vertical steel is not at yield ( if excess vertical steel is provided), that does not invalidate the use of the model because failure in this mode requires both the horizontal and vertical steels to have reached yield. That is why one has the flexibility to adjust the amount of vertical steel provided and the amount of horizontal steel required.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

_{p}should read P_{v}.## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

If you iterate using the live load portion V*(LL) only and match the varying V*(DL+LL) with phiVu to give phiVu(iteration), the factor determined using V*(DL&LL) (design shear), phiVu(iteration) and V*(DL) is a scaling factor. This factor applied to the full vehicle gives a scaled vehicle with a load rating factor of one (no iterations). If not convinced that the factor obtained is a scaling factor for a scaled vehicle with optimal shear adequacy, you can carry out simple analyses to verify this.

Using a phiVu (no iterations) consistent with the load of the rating vehicle, the factor calculated using the rating equation satisfies the definition of rating factor in AS5100.7. This factor is correct as it gives an accurate measure of the degree of adequacy for section shear. It is useful even for phiVu < V* as it shows the degree of under-strength. A scaling factor does not give this information. Additionally, determining scaling factors for rating vehicles with fixed axle loads, eg. SM1600 and T44 does not make sense as the scaled vehicles do not represent real vehicles. Also, scaling variable axle loads of a platform of a road train changes fixed axle loads of its prime mover.

To ensure compliance with MCFT, the strength must be consistent with the applied load. Since you are trying to get optimal adequacy (V*=phiVu) for a scaled vehicle, the Vu used must be for the same loading (i,e, V*,M*). Matching V* with phiVu gives a scaled vehicle with optimal shear adequacy (i.e, V* equals phi.Vu(V*,M*)).

It is not possible to get V*=phiVu if you iterate to satisfy V*=Vu(V*,M*), and then apply phi.

Scenario V*= Vu, say 10 =10, phiVu is 7 with phi=0.7. Therefore V* will always be less than phiVu.

In a design situation, to satisfy V*=phiVu for (M*,V*), the Vu(V*,M*) calculated using V* and M* has to be larger than V*. Not easy to visualise as they are factored values and we are dealing with nonlinear behaviour. Maybe, Vu=V* has no specific meaning in design and load rating since they are factored values.

Summary

To determine a scaling factor when applied to the rating vehicle gives adequate shear for the scaled vehicle, use the reduced strength obtained by iterating the live load to meet the requirement phi.Vu(M*,V*)=V*.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

My post above is consistent with how we do design now. For a set of (V_d*,M_d*) calculate V_ud, so this Vu_d is consistent with M_d*/_dV*. check whether V_d*=phiV_ud if we want optimal adequacy. So we have consistency since M_d*,V_d* and phiV_ud is for the same level of loading.

I am able to understand now how you can determine phiV_u by matching V* with V_u when not considering the iteration as scaling the vehicle to get an optimal scaled vehicle for shear. Iterate V* maintaining ratio M_d*/V_d* till V_u = V*. Then determine phiV_u. Doing this way is not using the iterations to scale the vehicle to get optimal adequacy. This way of calculating phiV_u is assuming a fictitious loading path of M_d*/V_d* to determine phiV_u. If you do it this way, a new design with optimal adequacy (using the current design method) will not give a factor of one.

The load rating factor we get using AS 5100.7 gives an accurate measure of shear adequacy, consistent with MCFT. (V_d*,M_d*) is just a point on the MCFT interaction diagram. It cannot be used to scale the load to check if a scaled vehicle has adequate ductility owing the nonlinearity in MCFT. Therefore, in addition having this rating factor, having a scaling factor is useful for vehicle-bridge systems with critical shear from a set of variable axle loads. A new bridge with phiVu_d = V_d* gives a rating factor of unity as expected.

If the iteration is not a numerical method to scale the vehicle for adequacy, then matching V*=phiVu may not be suitable. Unless we know the reason for using the numerical procedure (iterations) to determine strength, difficult to decide which approach is suitable.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

If you iterate using the live load portion V*(LL) only and match the varying V*(DL+LL) with Vu to give Vu(iteration), the factor determined using V*(DL&LL) (design shear), Vu(iteration) [Use this Vu to replace phiVu in the load rating equation] and V*(DL) is a scaling factor. This factor when applied to the full vehicle gives a scaled vehicle, and this scaled vehicle (on the bridge at the same position as the full vehicle) gives a load rating factor less than one (no iterations). Iterating this way does not give any useful information since the scaled vehicle does not give optimal shear adequacy.

If not convinced that the factor obtained is a scaling factor when applied to the full vehicle gives a scaled vehicle with less than optimal shear adequacy,you can carry out simple analyses to verify this.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Just been looking at the latest FIB Model Code 2020 in its shear section.

They appear to be making the same mistake as us in AS3600, requiring horizontal and vertical force equilibrium as well as rotational equilibrium.

They have exactly the same equation for longitudinal tension force from shear as AS3600-2018 Amendment 2 for their Level III result, which is based on MCFT, in equations 30.1-30 and 30.1-31!

deltaT = .5(Ved + Vrd.c) cot (theta)

when you combine the 2 equations!

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The equation for additional force with the term φV

_{us}in longitudinal steel caused by shear in Clause 8.2.7 of AS 3600-2018 with Amdt 1 and AS 5100.5-2017 with Amdt 1 is not based on MCFT. This can be seen in previous published papers mentioning this equation.In the paper "Background to the general method of shear design in the 1994 CSA-A23.3 standard" by Khaldoun N. Rahal and Michael P. Collins, Canadian Journal of Civil Engineering Volume 26, Number 6, December 1999, they stated that "Collins et al. (1996) and Collins and Mitchell (1991) calculated the longitudinal force at a cracked section due to the shear force as cot θ (V

_{f}– 0.5V_{s}– V_{p})."The first cited reference is a published paper:

Collins, M.P., Mitchell, D., Adebar, P.E., and Vecchio, F.J. 1996. "A general shear design method", ACI Structural Journal, 93(1):pp 36–45.

The second cited reference is the book:

Collins, M.P., and Mitchell, D. 1991. Prestressed concrete structures. Prentice-Hall, Inc., Englewood Cliffs, N.J.

I do not have access to a copy of the book and will try to get hold of a copy to see the calculation. I think the calculation is similar to that described in the first cited reference.

The first cited reference shows how this equation is derived. It is not based on MCFT which is the theory for the determination of shear strength for a section subjected to a set of load effects. The MCFT-based equation used for design is applicable to section behaviour only, not block behaviour. As shown in the first cited reference, the equation for force is derived using a free body diagram of a block of concrete with a diagonal crack (similar to the one used in AASHTO). While this equation is quite often mentioned by researchers in technical papers on design for section shear using MCFT, it is not based on MCFT.

I cannot comment on the equation in the model code since I do not have access to a copy. In your posting, you mentioned that the equation is based on MCFT. This suggests that the force in the model code (for section behaviour) is not the same as that used in the AASHTO standard, AS 3600 Amdt1 and AS 5100.7 (for block behaviour).

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I have updated the graphs I posted previously to compare AS3600 with AS5100.5, with and without the limit. In the graphs below the loadcases are:

AS3600-1: Longitudinal force due to shear to AS 3600 with Theta values set to minimum.

AS3600-2: As above with Theta adjusted so that shear capacity = applied shear, up to a maximum of 50 degrees.

AS5100-1: Longitudinal force due to shear to AS 5100.5, with Theta set to minimum value and no limit on Vus.

AS 5100-2: As above but phi.Vus set with a maximum value of V*.

AS 5100-3: As ASS5100-1, but with Theta adjusted.

AS5100-4: As AS5100-2 with Theta adjusted.

With AS3600 reduction factors applied to both codes:

With applicable code reduction factors applied to both codes:

More details at:

https://newtonexcelbach.com/2023/10/09/longitudina...

In summary, if the limit on φVus is applied with the current code reduction factors then, at least for this one case, AS 5100.5 results are always equal to or less than AS3600, but if it is not applied they can be unconservative.

How this will change in the next amendment of AS5100.5 I don't know, but in the draft for public comment there was no change to the equation proposed.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Doug,

If you refer back to your earlier posting dated 21 Sep 23 (see above quote) and go through my previous postings on https://www.eng-tips.com/viewthread.cfm?qid=499791, you will notice that I did not ignore the missing restriction in the AS 5100.5 equation. The paper presented at the Concrete2023 conference held recently in Perth (see the reference Wong and Vimonsatit(2022) cited earlier in this thread also recommended that this missing restriction be included in AS 5100.5. It is possible that the AS 5100.5 Committee is already aware of this shortcoming which requires addressing before it is raised by us. If not, it is likely that they are aware of it now owing to the conference paper and our discussions in this thread.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

And it uses the equation

Vus = V* - Vuc

in it derivation. That is the restriction that should be applied.

I am sure Doug can confirm this.

The AS5100.5 committee has been told the errors in their shear rules many times over the last about 7 years since the first release with MCFT.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

The paper by Vannisorn V, Wong KW and Mendis PA, Increase in longitudinal steel reinforcement to account for shear in RC beam design to AS 3600, Concrete Institute of Australia Conference Concrete2023 held in Perth 10-13 September 2023, stated that:

(1) the equation in Amendment 2 to limit phiVus = V* -phiVuc is not necessary as the amount of longitudinal steel determined in the old version is adequate to resist V*.

(2) Clause 8.2.7 AS 3600 could be revised and is not necessary to be included in AS 5100.5. The missing restriction "phiVus shall not be taken as greater than V*" found in other major standards should be included in AS 5100.5.

(3) The change in Clause 8.2.7 of AS 3600 to use "V*-phiVuc" instead of phiVus could be revisited. The changed equation is not suitable for load rating of shear particularly with loading resulting in shear deficiency, therefore need not be included in AS 5100.5.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

My analysis fully conforms with MCFT for section shear in the determination of φV

_{u}since I did not adjust the mid-depth strain. The strain used for my calculation of φV_{u}is equal to the strain calculated using the load effects of M* = 258 kNm and V* = 275 kN. To ensure consistency between the shear strength and the corresponding load effects (a key feature of MCFT),the strain 8.967E-4 is not increased in the analysis. The calculation for force in longitudinal steel uses the equation in the standard which does not have the restriction "φV_{us}not to be taken greater than V*".Below are the findings.

(1) For a section (see Case 2) adequate for section shear and force in the the longitudinal steel for M*=258 kNm and V* = 275 kN, increasing the diameter of the shear reinforcement increases φV

_{u}. The moment capacity phiMU (which is for pure flexure) remains the same and equals 472 kNm.(2) Sections of Case 3, 4 and 5 are conservative for both the design to AS 5100.5 of section shear and the design of force in the longitudinal steel since φV

_{u}/V* and A_{st.provided}/A_{st.required}are both greater than 1.0. Note that when stating whether a design is conservative or not, it is important to state clearly what it relates to.## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

1) It isn't clear what the point of the exercise was. The question being discussed is how the longitudinal force due to shear is affected by increasing the area of shear reinforcement, but in your example the most lightly reinforced section was adequate for both bending and shear, so of course the more highly reinforced sections will also be adequate. That does not show that the code is conservative.

2) You have compared your results with an M* of 258 kNm with my results with an M* of 400 kNm, so of course the results are different. If the AS 3600 equation for longitudinal force is used with that section, with all other factors according to AS 5100.5, then it finds that the shear capacity with 10 mm shear steel is as your calculation, and the longitudinal steel is adequate with a factor of 1.28. Note that your calculation of Ast_required is wrong. See details below.

3) The section capacity (PhiMu) to AS 5100.5 with zero applied shear is 445 kNm. I don't know where 472 kNm comes from. I don't agree that the moment capacity is not affected by increasing the shear reinforcement. The code is quite clear that the longitudinal steel area available for bending must be reduced by the area required to resist the longitudinal force due to shear, using a reduction factor of 0.7 (in AS 5100.5).

4) The result of the above is that the longitudinal steel has two different factors applied: 0.8 for the force due to bending and 0.7 for the force due to shear. In your calculation it appears that you have used 0.7 for the combined force. Note that in AS 3600 the reduction for bending (0.85 or 0.8) is applied to the combined force, and in overseas codes using partial factors there is of course only one factor applied to all forces in steel.

In response to general comments:

1) You have not provided any reasons for not adjusting the strain at mid-height. This is explicitly allowed in both AS 3600 and AS 5100.5.

2) Your frequently quoted quotes from your conference paper state that the current AS 5100.5 procedure for longitudinal force due to shear is adequate, and you appear to have used it without any limit in your calculations. It then immediately goes on to say that overseas codes have a limit on the value of Vus used, and this should be applied in AS 5100.5!

3) As stated previously, with the addition of the limit on the value of Vus the AS 5100.5 equation gives very similar results to the AS 3600 equation, applied with adjustment of the mid-height strain where necessary.

4) I don't agree that the AS 3600 equation is unconservative when applied to load rating analysis.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

My response here is to address the above comment. Since so many comments provided by IDS which require addressing, I will address them (most later) in separate posts to not overload readers with too much information in one post and to allow each to be discussed separately.

Your first sentence in the quote is not accurate. The above is one of the reasons for not adjusting the mid-level strain. The other, not previously pointed out by me, is provided below.

While both codes allow the calculated mid-level strain (my value is 8.967E-4) to be increased up to 0.003 from the value calculated for M

^{*}and V^{*}(my values are M^{*}= 258 kNm and V^{*}=275 kN) for calculating Θ_{v}and k_{v}(which is conservative for both load-rating and and design since the resulting φV_{us}and V_{u}are both smaller), it is not the intention of the standards for this adjusted value to be used to determine a new set of action effects. The allowed adjustment is just a simplification to enable a more conservative shear strength to be used. For design and load rating, it is the shear strength at your design load effects M^{*}and V^{*}that is of importance, not the strength for a set of load effects where one or more of the effects is different from the design values.## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I have not came across any published technical papers providing a reason for the inclusion of the mentioned restriction in standards. This restriction appears to be a "good practice" requirement. If this is correct, this means that a structural system designed to AS 5100.5 and not satisfying the requirement does not necessary fail to have the safety required by this standard. A well designed load testing program is required to shed more light on this. However, it is prudent to have this restriction included in AS 5100.5 in the next amendment or version of the standard, which is what the COncrete2023 paper proposed.

I have not came across a suitable analysis (assuming that this is even possible) which shows that a system is inadequate for force in the longitudonal reinfrcement as a result of not satisfying this restriction. Neither have I came across information of reported previous beam tests to show insufficient strength without the restriction.

The paper with the derivation of the equation for the area of longitudinal steel required for design (see paper titled "A general shear design method", by Collins, M.P., Mitchell, D., Adebar, P.E., and Vecchio, F.J. ACI Structural Journal,pp 36–45.1996) showed this equation without the restriction. A few papers and thesis quoting this equation also do not have this restriction which suggests it is very likely that this restriction was included in standards to promote "good practice".

In his blog https://newtonexcelbach.com/2022/11/11/rc-design-f... stated "If the shear reinforcement area is increased well above the area required for the design shear force the AS 5100.5 equation gives bending capacity results that are highly unconservative"

As shown from the result of my excel attachment, for a set of load effects M* and V*, increasing the area required for the design loading has no effect on the bending moment. As described earlier, the resulting design is more conservative in regard to the current requirements of AS 5100.5 with increase shear reinforcement.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Points 3 and 4 noted. When adjusting the spreadsheet for use with AS 5100.5 from a previous one I used based on AS 3600, a few of the equations were not corrected to those given in AS 5100.5. These errors are now corrected. The problem with using the equations of AS 5100.5 for longitudinal steel force is that additional steel is used instead of total steel with a single reduction factor Φ

_{f}. This is the first study I did which uses the equations for force of AS 5100.5. I prefer the format of AS 3600 which uses total steel and a single factor. The AASHTO specification uses an individual factor for each action effect.My findings are updated below. They are still basically the same as my previous findings except incorrect values have been corrected. Also two excel files are attached, one for M*= 297 kNm and another for M*=400 kNm (in response to IDS comment number 1). Also an independent check using the open software WXMAXIMA has been carried out for Case 2 M*=297 kNm and a printout of it is also attached.

---------------------------

My analysis fully conforms with MCFT for section shear in the determination of φV

_{u}since I did not adjust the mid-depth strain. The strain used for my calculation of φV_{u}is equal to the strain calculated for the load effects in each case. To ensure consistency between the shear strength and the corresponding load effects (a key feature of MCFT),the strain is not increased in the analysis. The calculation for force in longitudinal steel uses the equation in the standard which does not have the restriction "φVus not to be taken greater than V*".Below are the findings for the analysis using M* =297 kNm.

(1) For Case 2 adequate for section shear and force in the the longitudinal steel for M*=297 kNm and V* = 275 kN, increasing the diameter of the shear reinforcement increases φVu . The moment capacity ØM

_{u}(which is for pure flexure) remains the same and equals 444 kNm.(2) Case 3, 4 and 5 are conservative for both the design to AS 5100.5 of section shear and the design of force in the longitudinal steel since φV

_{u}/V^{*}and A_{st.provided}/A_{st.required}are both greater than 1.0. Note that when stating whether a design is conservative or not, it is important to state clearly what it relates to, as shown here.Below are the findings for the analysis using M* =400 kNm to address IDS comment that he used a moment of 400 kNm.

(1) Case 2, 3, 4 are all inadequate for force in the longitudinal steel for M

^{*}=400 kNm and V^{*}= 275 kN. Increasing the diameter of the shear reinforcement increases φV_{u}. The moment capacity ØM_{u}(which is for pure flexure) remains the same and equals 444 kNm.(2) Only Case 5 is conservative for both the design to AS 5100.5 for section shear and the design of force in the longitudinal steel since φV

_{u}/V^{*}and A_{st.provided}/A_{st.required}are both greater than 1.0.## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Any engineer who understands statics and equilibrium (you cannot have rotational equilibrium without vertical and horizontal force equilibrium) can derive it for themselves. We we do not chase Technical Publication and as this work on Standards is an unpaid sideline have other priorities like earning a living.

Obviously the FIB-Model Code writers have done the same, as they have the same resulting formula as us but have not published the derivation.

Collins & Mitchell derivation does not state the limitation with the final result, but if you follow through the derivation in their original text on it, they decided at one stage in the derivation to leave a Vs term in the equation, rather than Vc. The basis of this substitution was

V* = Vc + Vs

That is published. So Vs = V* - Vc.

It is not Vus as defined in the code as Vus assumes the stress in the shear steel is the yield stress. The maximum force that is available in the shear steel is Vu - Vc, the maximum force it can carry is Asv * fsv which is Vus in the code. But that is not the actual force in the shear steel, it is the maximum it can carry. In the calculation in the code, it is assumed the designer will want to use the minimum area of shear reinforcement so its stress is set at fsv. If Asv is increased above the minimum requirement from the code formula, that does not mean you can generate that extra force in the steel. It means that the vertical steel is at a lower stress than fsv.

For equilibrium, the force in the vertical steel cannot be greater than the force available at the section and that is V* - Vc.

I have no idea why they decided to leave Vs in and remove Vc in the C&M derivation, but it makes the resulting equation misleading in our opinion and open to abuse as to use it properly the engineer would have to realise that they have to calculate the real stress in the shear reinforcement rather than use the yield stress. If they had done the substitution the other way, they would have finished with our equation which is in AS3600. If the equation is to maintain the vertical steel force term, it needs to define a separate term, Vs = Asv fs where fs = (V* - Vc) / Asv (leaving out phi factors).

It is interesting that Looking at Eurocode based rules on it is misleading as Vc is set to 0 as soon as shear reinforcement is required, so Vus = V* in Eurocode logic, but the shear angle is reduced to 21.8 degrees so cot Theta is much larger. In their case, Vus would be limited to V*

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

Firstly, when I say that a procedure is "conservative" in this discussion I mean that it will return section design capacity results no greater than those found using a more refined procedure, or other widely accepted procedures. I am not sure what you mean when you use that word.

Looking at the results with an M* of 400 kNm you agree that the current AS 5100.5 procedure indicates that the section with 20 mm shear steel is satisfactory for this applied moment, with a shear force of 275 kN, whereas the Canadian and AASHTO codes and AS 3600 indicate that it is not.

The current AS 5100.5 procedure is therefore not conservative.

There are many points in your posts above that I disagree with and which could be discussed further, but none of them change the facts listed above, which (based on your provided results) you agree with.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

I disagree with your statement above as my results do not show what you stated in the quote above. AS 3600 and the Canadian and AASHTO codes all have a different set of equations from that of AS 5100.5. AS 5100.5 has no restriction, AS 3600 with Admt 2 restricts φV

_{us}to V*-φV_{uc}and both AASHTO and Canadian Codes restrict φV_{us}to not greater than V*. Owing to this, determining the degree of adequacy across these standards serves no useful purpose. Why use one of AS 3600, the Canadian or AASHTO codes as the reference for your comparison? If you used AS 5100.5 as the reference for comparison, you would have concluded that all three AS 3600, Canadian and AASHTO codes are overly conservative.I have not come across any studies in the literature of suitable experimental studies which show that without the restriction of AASHTO and Canadian codes, the degree of safety required by AS 5100.5 is compromised. The restriction is likely a "good practice" requirement to limit excessive reduction of the amount longitudinal steel since the original equation developed by Collins and Mitchell in their published book "Prestressed Concrete" did not include the restriction of AASHTO and Canadian codes.

In the Concrete 2023 paper (by Vimonsatit, Wong and Mendis) presented in Perth recently, we alerted that the restriction in the current Canadian and AASHTO codes is missing in the current standard and should be included in the next revision of AS 5100.5. Please refer to the paper for more information.

## RE: AS 3600 AMDT 2 Equation 8.2.7(2) -continuing a previous thread

While C&M derivation (Reference: Prestressed Concrete Structures by M P Collins and D. Mitchell, Response Publication Canada), 1997" made use of the equation for nominal shear resistance of a section V

_{n}= V_{c}+ V_{s}+ V_{p}(Eqn 7-42, C&M's book) in the calculation in Section 7.13 of their book, they stated on page 372 "Because we are providing more stirrups than are required, we will determine the actual value of V_s at each section since an increase in V_s will decrease the required force in the longitudinal steel". C&M also pointed out that the equation for the required tension can be derived independently using a free-body diagram (see page 373, C&M's book) which is similar to the one used in AASHTO LFRD Bridge Design Specification (9th Edition). AASHTO LFRD referred to the caption of Figure C 5.7.3.5-1 as "Forces assumed in Resistance Model Caused by Moment and Shear". It is not an equilibrium model as can be seen from the use of the the inequality sign ">=" in the equation for the required tension force. Similarly, for section shear requirement for adequacy in AS 5100.5, the equation φV_{u}>= V^{*}where V_{u}=V_{us}+ V_{uc}is not derived from considering forces using an equilibrium diagram; it is from a resistance model.The magnitude of the additional tension force in the compression half depth of the C&M's approach is the same as the force in the tension half. As stated in your text above "resulting longitudinal tension force top and bottom by dividing the total longitudinal tension force from shear by 2". There is no need for me to repeat this derivation as it has already been described by C&M. This value for force is used in AS 5100.5 Clause 8.2.8 "Proportioning longitudinal reinforcement on the flexural compression side".

The equation for the additional force due to shear required for the longitudinal reinforcement in the flexural tensile chord is Delta F

_{td}= (V_{Ed}/2)cot(theta), equation 7.3(34) for members with shear reinforcement in FIB - Model Code 2010. This equation appears to be derived using a "Variable Angle Truss Analogy" approach. The different types of truss models are described in this web link (https://www.midasbridge.com/en/blog/bridgeinsight/....The equation of C&M for the additional force in the longitudinal reinforcement from shear is not from the truss analogy approach described above. As stated in C&M's 1997 text "Prestressed Concrete Structures", their equation (Eqn 7.62 in the text) is from the consideration of transmission of forces across a crack (Figure 7-37 of the text) and the assumption of yielding of the longitudinal steel and the shear ligatures.

Eurocode 2 uses a Variable Angle Truss Model where concrete contribution to resistance is zero after cracking. The equation ΔF

_{td}= 0.5 V_{ED}(Cot Θ - cot α) is from the model, not from the resistance model of AS 5100.5.