How to extract forces from the pad footing model made using brick modelling?


I have done a brick modelling of the pad footing & unable to extract the forces. Gone through your link showing how to extract the forces using local direction force sum in brick model ( but unable to replicate the same in my model.

I have attached my model for your review. The image below shows the load application at pier location

The results needs to be extracted at the face of the pier, see below snap showing the plane for results extraction.


Question: Ignore my previous query. I have fixed it.
Please see the revised model attached. Here I have extracted the forces at face of the pier using "Local direction force sum"

From the above snap - transverse bending at the face of the pier is 3657.9 kN-m & shear is 6519.4 kN, these forces are over the entire transverse width. Please reconfirm if I am reading the results correctly?
Yes, the interpretation of the forces at the face of the pier is correct as Local direction force sum( LDFS) provides internal forces along the selected plane or line in the local coordinates system. 

To verify the output from the solid brick model I have prepared 20m long rectangular solid beam (1m deep x 0.5m wide) & applied an UDL of 10kN/m (20kN/m2) on the solid beam.
On either ends I have given a pin support & verified the moments @ centre & shear @ ends.

Please see the below snaps-
1. 20m solid beam (1m deep x 0.5m wide) with 20kPA pressure, which is around 10kN/m 

2. LDFS at centre-

I am getting moment as 250kN-m. Whereas I should get 10x20x20/8 = 500 kN-m. So incorrect.

3. If you see the above image we can see that bending is worked at CG of the plane. Now I have revised the Z coordinate to -1 using user input option.

Here we can see 500 kN-m bending moment - which is the expected value.

So based on the extreme tension face do we need to change the Z coordinates every time? For example- if the beam is sag z coordinate will be at the bottom of the plane and if the beam is in hog z coordinate will be at the top of the plane! Am I right? 

4. Coming to shear @ ends

Here shear from LDFS is 95 kN, where as it should be 10x20/2= 100kN. Why there is a small difference? And also why I am getting My moments here? It should be 0 ideally.
I have checked the LDFS for sections at sag & hog. Irrespective of the location of tension fibre we need to consider z coordinate at the bottom the plane. This gives us exact results. 
Please confirm if this is the correct technique in extracting the forces.

As both ends of the beams are pinned (restrained from movement in x-direction), the internal axial force is generated. 
So, either change that to pinned-roller system and same results will be observed across the section. So, auto calculate option can be used as it calculate the forces at the centroid.

When both ends are fixed, Fx is obtained which is causing the different value of My at top and both end.

but when system is changed to pinned-roller, Fx=0, thus no variation in My value. 

Now, this is happening as the pinned-pinned is generating Fx force at the nodes of that plane. As the section is pinned at bottom fiber, the moment obtained at bottom fiber is 500 kNm, now due to internal axial force (500 kN)  generated, the moment at top node will be  
be reduced to 0 (500 - 500x1) and at the centre will be (500-500x0.5 = 250).  
Similarly, when we provide supports at top fiber of solid, top fiber will obtain the moment 500 kNm, and now due to internal force generated, the moment at bottom node will be reduced to 0.

2) Regarding the extraction of the forces, when selecting the nodes it has to be selected in the counter-clockwise direction with 5 coordinates to be entered.  (Start and end point of the polygon) 

Kindly find the attached video for your reference. 

Creation date: 5/27/2021 11:46 AM      Updated: 7/5/2023 10:47 AM
20m rectangular beam for load extraction.mcb
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Footing brick model_R1.mcb
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Footing brick model_R4.mcb
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IRS code.pdf
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