Checking flexural shear in a two-way slab

[ipswitch]

I'm looking at example 1.30 on pg 1-45 of SERM 6th Edition. Seems to me the critical section for flexural shear analysis is where the two critcal shear planes intersect.

The cirical section is located from the center of the panel using l1 = 28' and then you find applied shear and shear capacity with l2 = 24'

Is that what they're saying? Please confirm.

 

[ipswitch]

The attached PDF seems to explain it a little more clearly.

 

[MA_PE]

sorry I don't see any attachment.

 

[ipswitch]

http://nptel.iitm.ac.in/courses/IIT-MADRAS/PreStressed_Concrete_Structures/pdf/9_Special_Topics/Section9.4.pdf

The file is too big to attach, so here's the link.

 

[ipswitch]

It says to do an analysis for each span direction separately. CERM 6 is doing something else I need to digest.

 

[dakota_79]

Unless I'm reading it wrong, SERM is doing it correctly (and in a way that jibes with your attachment), although they seem to have made a mistake in flipping their L1 and L2 terms in the equations vs what they show in Figure 1.25 (but proceed to use the correct numbers in the calcs).

Just walking through the SERM example to see if I'm missing something...along the 28-ft long span, the "beam" for checking shear is 24' wide (which means they're taking the entire width of the frame tributary to that column, correctly). Similar to an actual beam, the critical section is d from the face of the column, or "0.5*span - 0.5*column width - d" from the center of the span, and the shear force is the total trib load on the portion from there to midspan. This matches with your attachment and with how you'd do a true beam, from what I'm seeing. Then they seem to calculate capacity correctly, using the 24' width.

The other direction only has a 24' long span (smaller shear load than the 28' span), and has a 28' width (higher shear capacity than the 24' wide span), so that direction does not control by inspection.

 

[MA_PE]

right. If you are checking "flexural" shear, it gets checked like a one-way slab for each direction. I don't see reference to any L1 or L2 terms just Ln versus L2 in figure 9-4.1 (I don't see a Figure 1.25 either).

Flexural shear is taken as the shear load from 1/2 a span versus the section a distance "d" from the support. Once you know the slab can handle the shear from each direction, you look at the total load as punching shear surrounding the column.

Hope that helps.

 

[ipswitch]

It should be x= l2/2 - (d+c1/2) = 28'/2 + (7.5"+10") = 12.54'

Vu = qu l1 x = 0.2 kip/ft2 *24'*12.54' = 60 kips

phi * Vc = 2*phi*l1 *lamda*sqrt of f'c = 2 * 0.75 * 24' * 63.2 = 205 kips

 

[dakota_79]

Your effective beam is 24' wide, per discussion above (not 28').

So Vu = 0.2*24*12.54 = 60k, as SERM has it.

 

[ipswitch]

Your effective beam is 24' wide, per discussion above (not 28'). So Vu = 0.2*24*12.54 = 60k, as SERM has it.

Right, it's not intuitively obvious. Your turning the kips/ft2 loading into a strip load (kip/ft) along l2 (28') by multiplying it by the width (24') then you multiply it by 12.54' to get your shear at the critical plane.