foundation combine

Can you attach your .sfa file here? Also mention the version of STAAD Foundation Advanced you are using.

work14_foundation.sfa

connect edition 

Consider footing 1 and load case 2. There is an uplift force of 87.59 + 99.62 = 187.21 kN from the 2 columns. Since there is no soil on the footing, this entire uplift has to be counteracted by the selfweight of the footing. So, minimum weight of the concrete in the footing has to be 187.21 kN. But this will result in a net downward load of 0.0 kN. A bit more is required to satisfy the sliding check as described next.

In load case 2, there is a lateral load FZ of 12.46 + 12.62 kN = 25.408 kN. Since the factor of safety required in sliding is at least 1.5, and the coefficient of friction is 0.4, minimum additional downward force required is

(25.408 / 0.4) * 1.5 = 94.05 kN

Thus, the weight of the footing has to be at least 187.21 + 94.05 = 281.26 kNs

The program achieves this by keeping the thickness at 150 mm (the minimum value you specified) and increasing the plan dimensions to 13.107 m X 5.75 m, which gives a total weight of (13.107 * 5.75 * 0.15 * 25 = 282.62 kN) which is slightly greater than 281.26 kNs

Subsequently, during the concrete design phase, the 150 mm thickness is found to be insuficient, so it is increased to 250 mm.

That is how you end up with 13.107 m X 5.75 m X 0.25 m

If you start with a larger thickness, the plan dimensions will be smaller. Also, note that the plan dimension increment of 50 mm is more efficient than a 50 mm thickness increment from the standpoint of overall volume of concrete, so reducing the thickness increment to 10 mm will produce a more efficient design.

Consider footing 1 and load case 2. There is an uplift force of 87.59 + 99.62 = 187.21 kN from the 2 columns. Since there is no soil on the footing, this entire uplift has to be counteracted by the selfweight of the footing. So, minimum weight of the concrete in the footing has to be 187.21 kN. But this will result in a net downward load of 0.0 kN. A bit more is required to satisfy the sliding check as described next.

In load case 2, there is a lateral load FZ of 12.46 + 12.62 kN = 25.408 kN. Since the factor of safety required in sliding is at least 1.5, and the coefficient of friction is 0.4, minimum additional downward force required is

(25.408 / 0.4) * 1.5 = 94.05 kN

Thus, the weight of the footing has to be at least 187.21 + 94.05 = 281.26 kNs

The program achieves this by keeping the thickness at 150 mm (the minimum value you specified) and increasing the plan dimensions to 13.107 m X 5.75 m, which gives a total weight of (13.107 * 5.75 * 0.15 * 25 = 282.62 kN) which is slightly greater than 281.26 kNs

Subsequently, during the concrete design phase, the 150 mm thickness is found to be insuficient, so it is increased to 250 mm.

That is how you end up with 13.107 m X 5.75 m X 0.25 m

If you start with a larger thickness, the plan dimensions will be smaller. Also, note that the plan dimension increment of 50 mm is more efficient than a 50 mm thickness increment from the standpoint of overall volume of concrete, so reducing the thickness increment to 10 mm will produce a more efficient design.

i have a same problem in other file i am upolading it. actually in design it shoe zero error but when i take detailed output it shows error and dimesion is also taken to long althoung i have updated the soil load to counter the effect.phulwari sharif3_foundation.sfa10.xml

Here is the output from the current version - 9.3.0.14 that was released at the end of April 2020.

why i am getting such big dimension.

Your footings are subjected to large uplift forces, with nothing except the selfweight to counteract them. Use the approach explained earlier to find out the reason for your current model. If you are still unable to determine the reason, please raise a service request.

how to know there is uplift pressure?