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Longitudinal Shear In Upstand Beam - AS3600 Clause 8.4.3

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Trenno

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Feb 5, 2014
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Hello everyone,

I've never gone through this calculation before, so I would appreciate if someone could comment on the validity of my methods below via an example.

This upstand beam will be cast in two separate pours. Therefore mew will be 0.9 and Kco will be 0.4 (Table 8.4.3).

100_en7hgn.jpg


101_xjw1vk.jpg


To aid me finding the design shear stress, I used my textbook. From what I can gather, when the shear plane is in the tension zone - beta will always be 1.

102_w4wdo8.jpg
 
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Hey Asixth - things are good! I'll message you on Linkedin today.

Anyway, are you able to pdf your link? When I opened it up the formatting was all of the shop.

 
Trenno said:
From what I can gather, when the shear plane is in the tension zone - beta will always be 1.

In your case it will definitely be 1. If you had tension bars on the other side of the shear plane, however, that would not be the case. Strictly speaking, it seems that your z value would be that corresponding to the location corresponding to V. Your way of doing it would be conservative and expedient however.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
Hi Asixth,

I managed to replicate my numbers.

Originally you had 0 kPa for "Gp" (makes a small difference), but there is 0.6x0.9x25 kN / unit length above the shear plane.

The spacing of the ligs in the transverse direction would be more like 275mm rather than 300mm. (900 - 2x40)/3 = 273mm.

Your internal lever arm is an estimate - I got mine from RAPT.

Either way I'm comfortable that this longitudinal shear isn't a problem. Am I correct in saying this check is generally most critical at supports where V* is maximum?

Thanks for the help / verification of numbers.


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I'm sure that it won't make much difference here but I believe the gp value should really be negative. Your load is applied to the bottom of your beam which will create a tension demand on your stirrups additive to any shear demand. In reality, I suspect that your outer stirrups will be partially utilized as hanger bars and thus only partially available for resisting shear.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
I was under the impression the Gp value was more a magnitude or 'clamping' force to create the shear friction resistance.

 
Trenno said:
I was under the impression the Gp value was more a magnitude or 'clamping' force to create the shear friction resistance.

It is. That's why it would be rather important to recognize that your loads are causing tension perpendicular the shear plane rather than compression.

Capture2_zjh30m.jpg


I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
"When it doubt, draw it out." (my manager, 2015)

Thanks KootK, makes sense now.

I crunched the numbers again, the difference is four fifths of not much. The gp term really doesn't make a huge difference - especially when you have an inverted T section.

Design shear capacity is now 1.21 MPa (97% utilised).

 
It still may not make a difference but I feel that the code consistent way to account for the un-clamping would be leave gp alone but eliminate the factored steel area required to hang the factored load. What is your load anyhow?

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
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