Capacity reduction factor missing in Equation 8.2.7(2) AS 5100.5-2017 ?

[kww2008]

Is there a capacity reduction factor missing from Equation 8.2.7(2) AS 5100.5-2017 ?

I think this equation represents phi R > = F*, where phi is the capacity reduction factor, R represents capacity and F* represents "design action effects" at ULS. This is a common way to show the concept of Limit State design.

If we move all the resistance terms of Equation 8.2.7(2) to the left and keep all the action effect terms on the right, I think the equation should read (after I have inserted the missing capacity reduction factors to terms without them):

DeltaF_cd + [0.5*phi_v*V_us + phi_p*P_v]*cot(theta_v) > = 0.6 N* + V* cot(theta_v)

So if we compare the above equation with Equation 8.2.7(2) AS 5100.5-2017, with that equation rearranged:


DeltaF_cd + [0.5 *V_us + P_v ]*cot(theta_v) > = 0.6 N* + V* cot(theta_v),

it can be seen that phi_v and phi_p are missing,
phi_v =0.7 is the shear capacity reduction factor from Tsble 2.3.2(c) AS 5100.5 (I have included a subscript v for clarity)
phi_p is included just for the discussion and is 1.0 (if we use the same reasoning as per the previous AS 5100.5 where the resisting component from the
prestressed force is factored by 1.0).

If my observation above is correct, then a possible fix is to introduce a symbol phi into Equation 8.2.7(2) to read:

DeltaF_cd = 0.5 N* + (V* -0.5 phi Vus - Pv) cot(theta_v) where phi is that for shear,

and a few other equations (namely 8.2.7(1), 8.2.8(1) and 8.2.8(2)) also require "0.5 V_us" to read "0.5 phi V_us" where phi is that for shear and is equal to 0.7.

Please discuss.

 

[rapt]

No.

Phi (= .7) comes into it below that equation in calculating the area of reinforcement to provide that tension force.

 

[kww2008]

Thank you Rapt for getting the ball rolling for the discussion to this.

Equation 8.2.7(2) of AS 5100.5-2017 is:
DeltaF_td = 0.5*N* + (V* - 0.5*V_us -P_v) * cot (theta_v)

Equation 8.2.7(3) of the same standard is:
A_s*f_sy + A_p*f_py > = DeltaF_td / phi where phi =0.5

Combining the above two equations gives additional reinforcement required as:
A_s*f_st + A_p*f_py >= 0.5 N*/phi + (V*/phi - P_v/phi - 0.5 *V_us/phi) *cot (theta_v)

A similar equation given in AASHTO-LFRD for total amount of steel is:
A_ps*f_ps + A_s*f_y >= M_u/(d_v*phi_f) + 0.5 *N_u/phi_c + (|V_u/phi_v - V_p| - 0.5*V_s)*cot(theta)

Comparing the relevant part of the AASHTO equation with the AS 5100.5 equation, one can see that AASHTO does not have a phi below 0.5*V_s whereas AS 5100.5 has under 0.5*V_us. The Australian equation, in subtracting (0.5 * V_us/phi) instead of subtracting (0.5*Vus) gives less additional steel and therefore less conservative.

 

[rapt]

The Canadian Code methodology on which both codes are based also applies Phi to Vs in that formula (in A23.3, Vs is the design capacity and includes a phi factor whereas in AS3600 Vus needs phi to be applied to get the design capacity. So yes it probably should be applied there!

 

[kww2008]

Thank you Rapt for reviewing and providing the feedback. The current Canadian Code uses material strength reduction factors which make it very difficult to use for comparison with AS 5100.5-2017. Since both AASHTO and AS 5100.5-2017 use capacity reduction factors, it is easier to compare them. Also the use of symbol Delta_Ftd in the Australian Standard is confusing as it is not consistent with general usage within the standard as terms (e.g.,V_us, V_p) are mainly unfactored (unless shown with an asterisk -for example M*, V*) as DeltaF_td includes a strength reduction factor.

 

[rapt]

I am stating where the original theory came from. You would need to check back to the reasoning for this in the development of A23.3 as it is the original theory. Who is to say that AASHTO got the conversion right!

Where does AS5100 DeltaF_td include a strength reduction factor (other than the one applied to Vus)? I do not know what you mean by this.

The AS5100 says specifically that a strength reduction factor of phi = .7 should be used to calculate the Area of steel required to resist DeltaF_td. This could be a good reason why the phi was left out, as Canadian code uses .85 or .9 for the steel capacity factor. And they use it twice, once in calculating the force in its effect on Vs in reducing the reduction due to this and a second time in calculating the area of steel to resist the extra longitudinal tension force. AS5100 using .7 for this is much more conservative than the Canadian code.

You mentioned a factor of .5 in an earlier post. This is wrong. phi = .7 for this calculation in AS5100.

 

[kww2008]

Rapt,

If we use a symbol T to represent the unfactored tensile force from the additional reinforcement:

T = A_s*f_st + A_p*f_py, and

AS 5100.5-2017 Equation 8.2.7(3) reproduced below:
A_s f_sy + A_p f_py > = DeltaF_td / phi,

Combining the above two gives:
DeltaF_td = phi * T, and
this shows that the parameter DeltaF_td has already included (implicitly) a capacity reduction factor of 0.7.

Refer to Equation 8.2.7(2). the terms with action Effects listed below are all factored up:
0.5N*, V* cot(theta_v)

The terms with capacities are:
DeltaF_td, 0.5V_us cot(theta_v), P_v cot(theta_v).

Of the three, only DeltaF_td is factored down, the other two terms are not factored down. For the term with P_v, this is okay as this resisting force is unfactored at ULS, but not factored down V_us is not okay as this does not satisfy the limit state criterion of phi R > = F*.

I could not find in my earlier posting where I have used a phi of 0.5.

 

[kww2008]

Rapt,

Please note that all the constants equal to value 0.5 are not capacity reduction factors. They are constants in the equation.
In Equation 8.2.7(2), only one term has a capacity reduction factor and it is the term DeltaF_td, as pointed out in my posting above.

 

[rapt]

In your post on 7 Aug 17 04:41

Equation 8.2.7(3) of the same standard is:
A_s*f_sy + A_p*f_py > = DeltaF_td / phi where phi =0.5

Re
"Combining the above two gives:
DeltaF_td = phi * T, and
this shows that the parameter DeltaF_td has already included (implicitly) a capacity reduction factor of 0.7."

Sorry, I do not understand your comment about "implicit inclusion" of a capacity factor. DeltaF_td is the force to be supplied. A Capacity reduction factor is then applied in calculating the area of reinforcement required to provide this force.

 

[kww2008]

Rapt,

DeltaF_td is not the unfactored resisting force in the reinforcement to be resisted. It is, as the way it is used in the standard:
unfactored resisting force * phi where phi = 0.7 (apology in my earlier post of the typo error of this phi=0.5)

If we represent the unfactored resisting force required as R_t:
R_t=As*f_sy + A_p*f_py. --- (1)
Equation 8.2.7(3) AS 5100.5-2017 reproduced below:
As*f_sy + A_p*f_py = DeltaF_td / phi --- (2)
where phi = 0.7 not 0.5 (apology for my previous post typo error)

Subtitute (2) into (1)
R_t = DeltaF_td / phi
DeltaF_td = phi * R_t

Therefore DeltaF_td as shown in the standard is not an unfactored force to be supplied. It is the unfactored force to be supplied * phi.

 

[rapt]

Have it your way. It is not how I think of it.

As long as you put in sufficient reinforcement to provide a tension force of M/dv + DeltaF_td where the force supplied by the reinforcement = (As*f_sy + A_p*f_py) * phi then we will all be happy.

I will have a word with the code committee to clarify the M/df bit!