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Replies

Hi , In all cases, the bursting reinforcement in the X direction, for example, should be divided by 2 to distribute it between two faces. The resulting number should then be divided by the wall height, and that number will be the reo needed in square millimetres per metre of height. For example, if the default wall height is 3 metres and the total bursting reo is 3000, you need 1500 sq. mm per face. Divide this by 3, and you will get 150 sq. mm per metre of height. However, if you provide N12-200, you will be providing 565 sq. mm, which is higher than the result. Note that the spacing should not exceed the maximum spacing specified in the code, and fys is set by default in the program as 250 MPa. You can reduce this depending on the crack control specifications mentioned in the code (moderate, high, low degree of crack control, etc.).

Replied by: RKALC Admin | Date: 2/1/2025 11:00:04 PM

Hi ,
In all cases, the bursting reinforcement in the X direction, for example, should be divided by 2 to distribute it between two faces. The resulting number should then be divided by the wall height, and that number will be the reo needed in square millimetres per metre of height.

For example, if the default wall height is 3 metres and the total bursting reo is 3000, you need 1500 sq. mm per face. Divide this by 3, and you will get 150 sq. mm per metre of height. However, if you provide N12-200, you will be providing 565 sq. mm, which is higher than the result.

Note that the spacing should not exceed the maximum spacing specified in the code, and fys is set by default in the program as 250 MPa. You can reduce this depending on the crack control specifications mentioned in the code (moderate, high, low degree of crack control, etc.).


Follow up by the user: Does the application allow for 1 support or minimum 2 supports are required - this is for a case that pile is eccentric (moved in perpendicular direction of capping beam due to construction > tolerance of 75mm)?, and there is additional shear due to torsion. What if there are tension reversal forces on pile - I could not do for uplift loads?

Replied by: RKALC Admin | Date: 2/1/2025 11:01:06 PM

Follow up by the user:

Does the application allow for 1 support or minimum 2 supports are required - this is for a case that pile is eccentric (moved in perpendicular direction of capping beam due to construction > tolerance of 75mm)?, and there is additional shear due to torsion.

What if there are tension reversal forces on pile - I could not do for uplift loads?


Further clarification by RKALC: Hi, The application requires 2 supports. You may look at the corbel calculator, that one could help on your case. More generally if you have (equilibrium) torsion on capping beam due a pile moving out of the capping beam line, I assume load comes from columns above capping beam, you can simply draw the torsion diagram of the beam (just like shear diagram) and reinforce it with closed ties + longitudinal bars. See RC beam calculator for working beam reo. The decision that the torsion can't be taken by any element but the capping beam needs some judgement. I am not exactly sure i am following in relation to "uplifting", how does that happen if the pile is moving perpendicular to capping beam? I can only see an eccentricity moment Pile itself must have been designed to take the moment resulting from the 75mm out of position.

Replied by: RKALC Admin | Date: 2/1/2025 11:02:15 PM

Further clarification by RKALC:

Hi,

The application requires 2 supports. You may look at the corbel calculator, that one could help on your case.
More generally if you have (equilibrium) torsion on capping beam due a pile moving out of the capping beam line, I assume load comes from columns above capping beam, you can simply draw the torsion diagram of the beam (just like shear diagram) and reinforce it with closed ties + longitudinal bars. See RC beam calculator for working beam reo.

The decision that the torsion can't be taken by any element but the capping beam needs some judgement.

I am not exactly sure i am following in relation to "uplifting", how does that happen if the pile is moving perpendicular to capping beam? I can only see an eccentricity moment

Pile itself must have been designed to take the moment resulting from the 75mm out of position.


further question by the user: Hi, I carried out an example from text book - the STM application gives different output. Please verify.
Replied by: RKALC Admin | Date: 2/1/2025 11:37:54 PM
further question by the user:

Hi,

I carried out an example from text book - the STM application gives different output. Please verify.

Further clarification by RKALC I have had a look at this previously, no the calculations they are doing for the strut length or lb is incorrect, also the tie force is incorrect if you work the truss depth which is 1552 mm corresponding to an angle of 32.6. it also adopts the 2009 code (fi factor) best example found is the Lonnie Oack book. look at the attachment for the input that works with geometry assumptions they made. you need to minus the covers to get the wall and node dimensions right
Replied by: RKALC Admin | Date: 2/1/2025 11:39:20 PM
Further clarification by RKALC

I have had a look at this previously, no the calculations they are doing for the strut length or lb is incorrect, also the tie force is incorrect if you work the truss depth which is 1552 mm corresponding to an angle of 32.6.

it also adopts the 2009 code (fi factor)
best example found is the Lonnie Oack book.


look at the attachment for the input that works with geometry assumptions they made. you need to minus the covers to get the wall and node dimensions right


further quetstion by the user Why changed the width of applied load to 875mm instead column is 600x600 for this example? In the example you have shown in guide online has 420mm width of strut? why 1000mm width of pile cap is not used? further clarification by RKALC: the width of applied load is not changed, remained as in the example you mentioned. at 875 mm. With regards to the column width of 600, this is a question to the example writer. He might have reduced the width of the node to claim development length past the node. In real life the width of the wall should extend past the column a bit, not flush. In all cases the main question you might ask about this example is that why did the writer reduce the node on both sides of the column not only on the outer edge? More generally, reducing the width of the node by some 10 mm won't make tahd difference, you would also design for some utilisation in reo as well as the stresses. As I mentioned in the previous message, the example you qouted includes several problems, i only tried to tweak the input to get the angle and lengths matching. In RKALC we have allowed for 50 mm reduction of the span on each side for covers and the like. You as a user can make all of these judgements in sizing the nodes or span or width of applied load or effective depth etc..this is the whole point of iterating that would take forever if you are doing this manually. In the website example, the width of the wall was reduced to 420 because the support is a 600 dia pile, converted to a 420x420 sq.mm area. In other words the depth into the page is made smaller than the wall's. This is a bit of conservatism and simplification to work out the strut check whcih is having a bottleneck at the node. This software is specifically developed for someone like you who can verify and challenge the output via research and cross checking. Thank you for all of the queries.

Replied by: RKALC Admin | Date: 2/1/2025 11:41:38 PM

further quetstion by the user

Why changed the width of applied load to 875mm instead column is 600x600 for this example?
In the example you have shown in guide online has 420mm width of strut? why 1000mm width of pile cap is not used?

further clarification by RKALC:
the width of applied load is not changed, remained as in the example you mentioned. at 875 mm.
With regards to the column width of 600, this is a question to the example writer. He might have reduced the width of the node to claim development length past the node. In real life the width of the wall should extend past the column a bit, not flush.
In all cases the main question you might ask about this example is that why did the writer reduce the node on both sides of the column not only on the outer edge?
More generally, reducing the width of the node by some 10 mm won't make tahd difference, you would also design for some utilisation in reo as well as the stresses.
As I mentioned in the previous message, the example you qouted includes several problems, i only tried to tweak the input to get the angle and lengths matching.
In RKALC we have allowed for 50 mm reduction of the span on each side for covers and the like.
You as a user can make all of these judgements in sizing the nodes or span or width of applied load or effective depth etc..this is the whole point of iterating that would take forever if you are doing this manually.

In the website example, the width of the wall was reduced to 420 because the support is a 600 dia pile, converted to a 420x420 sq.mm area. In other words the depth into the page is made smaller than the wall's. This is a bit of conservatism and simplification to work out the strut check whcih is having a bottleneck at the node.

This software is specifically developed for someone like you who can verify and challenge the output via research and cross checking. Thank you for all of the queries.


final comment by the user: Thanks you for in depth clarification. Really appreciated your assistance. This was interesting as NZS and AS also specifies to do STM rather than rational simplied formular method of designing RC elements. STM is now mandatory as well as FEM.

Replied by: RKALC Admin | Date: 2/1/2025 11:42:12 PM

final comment by the user:

Thanks you for in depth clarification.
Really appreciated your assistance. This was interesting as NZS and AS also specifies to do STM rather than rational simplied formular method of designing RC elements. STM is now mandatory as well as FEM.