All nodes (or better surface) of one side are in contact with table weight of the plate that are responsible for friction. How to put in into inp?

better use a linear isostatic equivalent (Creating Iso-static Restraints), that means you block 1-2-3 dofs in a vertex and 1-2 and 3 in 2 other, be careful to select them. The reactions in all those nodes should be very small, so they don’t change the solution you are seeking. Trying to make the model non singular using contact and friction is very tricky.

looking at your model, the long edges are the free edges and the short edges are both welded?

The weld is done along long edge on one side.

And that’s exactly what I give a thought! To get things more complicated. And thank you very much for the iso-static hint.

Were your hand calculations done assuming cantilever support of the plate ?

Calculation did not take any support into account.

So how did you do it ? Can you share your hand calcs ?

Give me some time. Give me some time.

maybe you should start with a 2D model (plane strain or plane stress) and use symmetry. That will restrain the model without further need of constrains (except the needed for your analysis).

To understand what you suggested here I need more experience in CalculiX which I do not have. I stick to your suggestion of iso-static restraints as I get the point and approximately know what to do.

Plane stress/strain modeling is simple. You just have to draw a surface in the XY plane (instead of a volume), mesh it with triangular/quad elements of proper type (CPSx/CPEx - check the documentation) and specify the thickness under `*SOLID SECTION`

. But of course you can use the 3D model for now, just make sure that boundary conditions are correct - otherwise, you won’t have agreement with calculated deflection.

Column 1 | Column 2 | Column 3 | Column 4 | E |
---|---|---|---|---|

Welding surfacing of the roller on the plate edge | ||||

Length of plate | ℓ | 1060.5 | mm | |

Width of plate | h | 48 | mm | |

Thickness of plate | δ | 5 | mm | |

Specific heat capacity | с | 0.16 | cal/g ºС | |

Specific gravity | γ | 7.8 | g/сm^3 | |

Modulus of elasticity | Е | 2039400 | kgf/сm^2 | |

Volumetric heat capacity | сγ | 1.25 | cal/сm^3 °C | |

Length of plate | ℓ | 106.05 | сm | |

Width of plate | h | 4.8 | сm | |

Thickness of plate | δ | 0.5 | сm | |

Metal yield strength | σy | 1275 | kgf/сm^2 | |

Welding current | I | 150 | А | |

Voltage | U | 69.65 | V | |

ηu | 0.7 | |||

Velocity of heat source | v | 40 | m/h | |

1.1 | сm/sec | |||

I. Active stress zone bn. | ||||

Specific heat flow | q | 0.24 * ηu * I * U | 1755.18 | cal/sec |

The area of heating to a plastic state b1 | b1 | (0.484 * q) / (v * δ * c * γ * 550°) | 2.2 | cm |

Specific heating energy | q0 | q/(v * δ) | 3191 | cal/сm^2 |

We set the value | m | For welding heating modes, the value of m ranges from 0.6 to 0.9, and the average values will be approximately 0.7 – 0.8. | 0.77 | |

bn | h / (1 + ((σт * h) / (9.68 * q0 * m))) | 3.82 | cm | |

We check the accepted value of the value | m | 1 – 0.008 * (bn^2 / δ) | 0.77 | |

Cross section of the active voltage zone | Fc | bn * δ | 1.9 | cm^2 |

Active internal effort | Р | σy*Fc | 2422.5 | kgf |

Plate cross section | F | h * δ | 2.4 | cm^2 |

Reactive axial compression stress | σ2 | P / (F – Fc) | 4845 | kgf/сm^2 |

Bending moment | М | (P * h) / 2 | 5814 | kgf сm |

The moment of inertia of the rectangle | J | (δ * h^3) / 12 | 4.608 | cm^4 |

Residual deflection of the plate | f | (M * ℓ^2) / (8 * E * J) | 0.87 | сm |

W | (δ * h^2) / 6 | 1.92 | cm^3 | |

Bending stress | σ | M / W | 3028 | kgf/сm^2 |

In the hope your curiosity is satisfied.

so…how do you want to apply your model to the FEM analysis? the whole journey must be something like this but you can shortcut at several points…do you want to apply a predefined stress distribution in the welded edge or something else?

PS: 127 MPa seems a low yield strength for almost any steel.

I’m glad you posted this in such a well-organized form. I wanted to see your hand calcs to give you better advice regarding CalculiX analysis.

That’s the formula for the deflection of a beam subjected to pure bending, right ?

Then you will have to apply proper BCs. Here’s an example from Abaqus:

Just without this offset of the supports because that was 3-point bending. Kinematic coupling constraints were used to apply opposite rotations (of course, they could be also replaced with moments).

I am very sick these days. You have all of my calc stuff so if you have time and small amount of wish to apply iso-static restraints on my plate to see the result that would be something.

actually predefined stress distribution is applied on the ends of welding surfacing of the roller on the plate edge. As I do in fbd file.

I tried my best

beam.fbd

```
valu Etyp he20r
# Length of the beam in meters, converted from mm
valu L / 1065 1000
valu divL 80
# Width of the beam in meters, converted from mm
valu B / 5 1000
valu divB 4
# first partial height in meters, converted from mm
valu H1 / 38.2 1000
valu divH1 8
# Second partial height in meters, converted from mm
valu H2 / 9.8 1000
valu divH2 4
# Pressure in Pa (N/m2), coverted from MPa (N/mm2)
valu P * 127.5 1e6
seto beam
pnt p0 0 0 0
pnt p1 0 L 0
line l0 p0 p1 divL
seta se0 l l0
swep se0 se1 tra B 0 0 divB
seta ses0 A001
swep ses0 ses1 tra 0 0 H1 divH1
seta rear A005
seta front A006
comp rear e
rep rear
comp rear f
rep rear
comp front e
rep front
comp front f
rep front
seta extend A002
swep extend extended tra 0 0 H2 divH2
setc beam
elty beam Etyp
mesh beam
send beam abq
seta a n 2858
seta b n 2415
seta c n 2859
send a abq spc 123
send b abq spc 3
send c abq spc 13
send rear abq surf
send front abq surf
send rear abq pres P
send front abq pres P
plot f rear g
plus f front r
```

beam.inp

```
*HEADING
Model: CalculiX Beam Input File for welding experiment
*INCLUDE,INPUT=beam.msh
*BOUNDARY
*INCLUDE,INPUT=a_123.bou
*INCLUDE,INPUT=b_3.bou
*INCLUDE,INPUT=c_13.bou
*MATERIAL,NAME=EL
*ELASTIC
200e9,0.3
*SOLID SECTION,ELSET=Ebeam,MATERIAL=EL
*STEP
*STATIC
*DLOAD
*INCLUDE,INPUT=front.dlo
** *NODE PRINT
** U
*NODE FILE ** It adds to Datasets -> 1 DISP
U
*EL FILE ** It adds to Datasets -> 2 STRESS
S
*END STEP
*STEP
*STATIC
*DLOAD
*INCLUDE,INPUT=rear.dlo
** *NODE PRINT
** U
*NODE FILE ** It adds to Datasets -> 1 DISP
U
*EL FILE ** It adds to Datasets -> 2 STRESS
S
*END STEP
```

and I am still perplexed as what the beam deflection is.

2 mm…if too much, that depends on how realistic your loads are.

My paper calculation

and experiment gave same residual deflection of the plate of 8 mm and calculix not, May I miss something? Strange, such a discrepancy between paper, experiment and calculix results.