RAM CONCEPT: Shear reo in slabs according to AS3600-2009

At this point the program is designing the shear reinforcement and getting the bar area from the design strip properties. As to the heart of the question, check this wiki for the common reasons you can see unexpected one way shear reinforcement from Concept:

https://communities.bentley.com/Products/Structural/Structural_Analysis___Design/w/Structural_Analysis_and_Design__Wiki/ram-concept-shear-reinforcement-faq.aspx 

If it doesn't answer your question, reply back, and attach the model if you can. If it's too large you can use "Secure File Upload". Otherwise the full audit would work too.

Hi Seth,

Thanks for your reply. Unfortunately it doesn't answer my question, which remains the same: why software provides shear ties where V*=64kN<phiVuc=272kN ?

The section I refer was formed correctly, but even if it wouldn't the question still remains as according to AS3600 shear ties in slabs to be provided only if V*>phiVuc (8.2.5 II).

I've attached the model using secure file upload, name "GL_North_A" . Please, take a look at strips 35-1, 40-1, etc.

Strip 35C-1, Section 1 seems like a representative case and controls for one way shear design in that strip. The V* value there is more like 323 kN. Other sections in the strip may also require shear bars, others may not, but keep in mind transverse shear bars always extend at least one section beyond the point at which they are needed.Which section were you auditing in the posted data before?

Seth,

Thanks for your reply.

A: If you look at strip 35-1:

Column strip: yes, max shear is 323 kN but the next section is located 0.67 m away and shear is 174 kN which is clearly less than phiVuc, shear in next sections is definitely less than 174 kN but shear ties are provided in every section! (see picture attached)

Middle strip: max shear is 47 kN but shear ties are provided in every section (see picture attached)

 

B: Also, in the Tech note it says (https://communities.bentley.com/Products/Structural/Structural_Analysis___Design/w/Structural_Analysis_and_Design__Wiki/ram-concept-shear-reinforcement-faq):

"One of the biggest factors in thick slabs (especially in mats) is the fact that the program designs for shear at each cross section, while the design codes generally allow the designer to ignore shear some distance from the face of the support (typically "d"). In cases where shear reinforcement is only required in the last section at the face of the support it can be ignored." - in your comment you've mentioned that "transverse shear bars always extend at least one section beyond the point at which they are needed" - does it mean that if shear ties required in section 'n' they will be extended to section 'n+1'? - if yes, then designer has to recheck every strip in order to avoid over design as it wouldn't be clear on the screen that shear ties are required only at the face of the column.

You might find the section design plot for "Shear Density" to be informative. In the case os your right middle strip, the shear demand is indeed quite small, but a plot of Strength Design - Shear density indicates ties are required at several stations (2,3,4,5, and 8)

I glanced at the audit for section 2 in the right middle strip and I see the following comment, "Vuc is set to zero due to cracking in compression zone caused by reversal loads (8.2.7.4)" which seems to be the root cause for the ties.

I don't have AS 3600-2009 setting in front of me, but from what I gather, load reversal or high torsion cause cause cracking in the zone that is normally the compression zone and when that happens all of the shear capacity has to come from the steel. 

The same image shows how the ties are extended slightly.

You might find the section design plot for "Shear Density" to be informative. In the case os your right middle strip, the shear demand is indeed quite small, but a plot of Strength Design - Shear density indicates ties are required at several stations (2,3,4,5, and 8)

I glanced at the audit for section 2 in the right middle strip and I see the following comment, "Vuc is set to zero due to cracking in compression zone caused by reversal loads (8.2.7.4)" which seems to be the root cause for the ties.

I don't have AS 3600-2009 setting in front of me, but from what I gather, load reversal or high torsion cause cause cracking in the zone that is normally the compression zone and when that happens all of the shear capacity has to come from the steel. 

The same image shows how the ties are extended slightly.

Thanks Seth,

The thing is that there's no load reversal (DL=1 and LL=20 kPa) in the model and torsion is transferred into bending by wood-armer method and section audit clearly calculates it by "envelopes after torsion conversion" table where T=0

From section Audit it is very unclear why Vuc is actually set to 0 (see below the result from section 2 in the middle strip):

Designing for Envelope 4

Starting 8.2 Shear Design

Considering Effect of Torsion on Shear:

No torsion to consider

No torsion reinforcement required

General Shear Design Parameters

bv = 1834 mm

do = 202 mm

bvd = 370500 mm²

Ac = 458500 mm²

Ast = 1659 mm²

Apt = 0 mm²

Ac = 458500 mm²

fc = 40 N/mm²

fsyf = 500 N/mm²

V* = 4.622 kN (absolute value)

M* = -10.44 kN-m

Setting Vo to zero for RC cross section

Setting Pv and Apt to zero for RC cross section

Calculating Web-Shear Cracking Vuc (8.2.7.2(b))

Precompression Stress = 0.009894 N/mm² (at centroid)

Failure Tensile Stress = 2.277 N/mm²

Shear Stress Limit = 2.282 N/mm² (at centroid)

I = 2388000000 mm⁴

Q = 14330000 mm³

Vt = 697.5 kN [(limit stress) * bv * (I/Q)]

Vuc (web cracking) = 697.5 kN (Vt + Pv)

Calculating Flexure-Shear Cracking Vuc (8.2.7.2(a))

Beta1 = 1.538

Beta2 = 1

Beta3 = 1

Shear Stress Term = 0.5637 N/mm² ([..]¹´³)

Vuc (flexure-shear) = 321.2 kN

Vuc is set to zero due to cracking in compression zone caused by reversal loads (8.2.7.4)

Vuc = 0 kN

Vu,max = 2964 kN

Original assumption of no ligs required was incorrect. Repeating shear design considering minimum ligs required...

Considering Effect of Torsion on Shear:

No torsion to consider

No torsion reinforcement required

General Shear Design Parameters

bv = 1834 mm

do = 202 mm

bvd = 370500 mm²

Ac = 458500 mm²

Ast = 1659 mm²

Apt = 0 mm²

Ac = 458500 mm²

fc = 40 N/mm²

fsyf = 500 N/mm²

V* = 4.622 kN (absolute value)

M* = -10.44 kN-m

Setting Vo to zero for RC cross section

Setting Pv and Apt to zero for RC cross section

Calculating Web-Shear Cracking Vuc (8.2.7.2(b))

Precompression Stress = 0.009894 N/mm² (at centroid)

Failure Tensile Stress = 2.277 N/mm²

Shear Stress Limit = 2.282 N/mm² (at centroid)

I = 2388000000 mm⁴

Q = 14330000 mm³

Vt = 697.5 kN [(limit stress) * bv * (I/Q)]

Vuc (web cracking) = 697.5 kN (Vt + Pv)

Calculating Flexure-Shear Cracking Vuc (8.2.7.2(a))

Beta1 = 1.538

Beta2 = 1

Beta3 = 1

Shear Stress Term = 0.5637 N/mm² ([..]¹´³)

Vuc (flexure-shear) = 321.2 kN

Vuc is set to zero due to cracking in compression zone caused by reversal loads (8.2.7.4)       -      WHY??

Vuc = 0 kN

Vu,max = 2964 kN

Less than minimum ligatures required, using large shear lig spacing (0.75D or 500 mm)

Maximum lig spacing = 187.5 mm (considers torsion lig spacing also)

Calculating Compression Strut Angle

Using 0.6062 instead of 0.6 for compatibility of 8.2.8, 8.2.9 and 8.2.10

Asv,min = 261 mm² (at max spacing)

Vu,min = 243.5 kN

Vu,max = 2964 kN

V*/phi = 6.602 kN

Theta-v = 30 degrees

Adjusted Vu,max = 2964 kN (after considering 8.3.3)

Design Vu = 6.602 kN (considering Vu,max)

Design Vus = 6.602 kN

Design Asv = 7.076 mm² (at max lig spacing)

Asv/s = 0.03774 mm (shear lig density)

Ast/s = 0 mm (torsion lig density)

As/s = 0.03774 mm (shear + torsion lig density)

As,min/s = 1.392 mm (minimum lig density)

As/s = 1.392 mm (lig density considering minimum)

Lig Spacing = 187.5 mm (lig spacing, if ligs required)

One of my colleagues reviewed the second section in the right middle strip from your model and noted that Envelopes 4 and 12 have small negative moments, and there are large enough positive moments in other envelopes to cause cracking on the bottom face, so 8.2.7.4 applies for those 2 envelopes.

Hi Seth,

thanks, I've seen this moment. I tried to find where this moment could come from and couldn't find any load combination which has negative moment in this section (I went through max and min M)

could you advise how this moment appeared in the envelope?

The final design envelopes are after torsion and zero-axial conversions. You won’t be able to plot the exact stresses we are using in this check because they aren't exposed externally. The strength load case just includes the secondary forces, and for stress checks you want the balanced forces so we do a special calculation for this internally.

Thanks Seth. What do you mean "we do a special calculation for this internally"? Are you talking about wood-armer conversion of twisting moments? What an ordinary structural engineer shall do in order to find out where the moment comes from?