MOMENT CREE PAR LE GRADIENT THERMIQUE

[FabriceMendelski3705]

Bonjour,

je souhaite savoir qu'ellle est la formule que utilise robot ( module coque) pour déterminer le moment du à la charge des gradients thermique.

moi j'utilise la formule du fascicule 74 pour déterminer ce moment voir piece jointe.

et j'obtiens pas la meme chose que robot .

je travaille en eurocode 2

Pouvez vous m'aider ?

probleme urgent

Merci d'avance

[Pawel.Pulak · Accepted solution]

Hello,

1. In case of bar elements Robot is using the formula coherent with the formula from fascicule 74
2. In case of panels/shells the formula is also very similar - of course considering in this case 2D state of stresses (in this case plate stiffness D is used instead of EI: ). It may result in disturbances especially close to supports.

I have attached the model and the screen capture made for it.

Model contains 2 pairs of bars and panels (each one is 1.5m wide, 0.5m thick and 4m long) loaded with TZ=+10 temperature difference in case 2:

• bar 1 and panel  1 supported as centilevers - so free to move thus no significant moments observed 
• bar 4 and panel 3 fully fixed on both ends - so no displacements but significant moments observed

Substituting to the formula from fascicule 74:

MY = alpha * Δt * E * I / h = 0.000012 * 10 * 25000 * 15625e6 /500 = 93750000 Nmm = 93.75 kNm

As it can be seen it is precisely coherent with the moment in bar element.

In case of panels moment is expressed in Robot in kNm/m but  assuming constant value of 65 kNm/m in the center of the panel and converting it to the width of 1.5 m we receive 1.5*65 = 97.5 kNm - so close to the value from the formula.

---------------------------------------------
If this post answers your question please click "Accept as Solution". It will help everyone to find answer more quickly!

Best regards,

[Anonymous]

dear pp2008

how did you get:

how did you got this result  15625e6  for I  ?

Thank you

[Pawel.Pulak]

Hello,

it is the moment of inertia about horizontal axis for the rectangle 1.5 m wide and 0.5 m thick:

B*H^3/12 = 1500*500^3/12 = 1.5625e10 mm4 = 15625e6 mm4

Regards,

[Anonymous]

Good morning sir,

could you please tell me why Fx=0, Fy=0 and Fz=0 (see doc attached)  for this example?

could you please give me more explanation about temperature load:

   - Variable 2P - When selected, the temperature load on the defined surface elements of the structure (panel) is linearly variable. You must specify the temperature value in two points (A and B).

   - Variable 3P - When selected, the temperature load on the defined surface elements of the structure (panel) changes according to the defined plane. You must define the temperature value in the points (A, B, and C) which define the temperature load change plane.

you can use previous autodesk robot file to define 2 cases with:  Variable  2P and variable 3P.

Thank you.

[Pawel.Pulak]

Hello,

below explanations to your questions.

---

ROSLA wrote:

could you please tell me why Fx=0, Fy=0 and Fz=0 (see doc attached)  for this example?

---

These forces are zero for bar 4 (bar fixed on both ends) in load case 2 because of the load applied to it.

It is thermal load of TX=0 TY=0 TZ=0

TX=0 means that there is no temperature increase in the neutral axis of the bar - so there is no tendence to increase or decrease the length of the bar - so the normal force FX in it is zero.

TZ=10 means that the temperature at the top of the section is 10 degrees higher than at the bottom with the linear distribution along the height and with no temperature change at the neutral axis (see TX=0) - it means that the bar "has the tendence" to deform in arc. But because of fixed supports on both ends it cannot deform and in results in the constant non-zero bending moment MY without any shear forces TZ - pure flexural state in XZ plane.

TY=0 so there are no thermal effects in XY plane - that is why MZ=0 TY=0

---

ROSLA wrote:

could you please give me more explanation about temperature load:

   - Variable 2P - When selected, the temperature load on the defined surface elements of the structure (panel) is linearly variable. You must specify the temperature value in two points (A and B).

   - Variable 3P - When selected, the temperature load on the defined surface elements of the structure (panel) changes according to the defined plane. You must define the temperature value in the points (A, B, and C) which define the temperature load change plane.

you can use previous autodesk robot file to define 2 cases with:  Variable  2P and variable 3P. 

---

I have attached the modification of my previous model, where I have defined Variable 2P load in load case 3 and Variable 3P load in load case 4 and applied both to panel 1. The view of this panel (with grid and coordinates) and these load records are shown on the screen capture below.

As you can see in load case 3 I have defined the temperature gradient TZ with TZ=10 in point (1; 1; 0) and TZ=20 in point (2; 1; 0)

Such temperature gradient field corresponds to the linear change of TZ along X direction with TZ=0 on the left edge of the panel (x=0) and TZ=40 on the right edge of it (x=4)

In load case 4 I have defined the temperature gradient TZ with TZ=10 in point (1; 1; 0), TZ=20 in point (2; 1; 0) and TZ=60 in point (2; 2; 0)

Such temperature gradient field corresponds to the linear change of TZ in such way that TZ=-40 in the left bottom corner of the panel, TZ=20 in the left top corner, TZ=0 in the right bottom corner and TZ=60 in the right top corner.

These values can be graphically interpreted as corresponding to the inclined plane defined by TZ values in 3 points.

Best regards,

[Anonymous]

Hello,

thank you for you answer.

But i'm not agree with you....

 Temperature gradient will generate normal effort =~ 1125 kN !!!!

 we have an bi-embedded beam here so we must have  normal effort and moment, you will find  formula in attached doc.

please tell me what.

Thank you

------------------------------------

These forces are zero for bar 4 (bar fixed on both ends) in load case 2 because of the load applied to it.

It is thermal load of TX=0 TY=0 TZ=0

TX=0 means that there is no temperature increase in the neutral axis of the bar - so there is no tendence to increase or decrease the length of the bar - so the normal force FX in it is zero.

TZ=10 means that the temperature at the top of the section is 10 degrees higher than at the bottom with the linear distribution along the height and with no temperature change at the neutral axis (see TX=0) - it means that the bar "has the tendence" to deform in arc. But because of fixed supports on both ends it cannot deform and in results in the constant non-zero bending moment MY without any shear forces TZ - pure flexural state in XZ plane.

TY=0 so there are no thermal effects in XY plane - that is why MZ=0 TY=0

[Pawel.Pulak]

Hello,

in my previous response I have written:

                 "TX=0 means that there is no temperature increase in the neutral axis of the bar - so there is no tendence to increase or decrease the length of the bar - so the normal force FX in it is zero."

In load case 2 only TZ=10 was used - it means that the difference of temperature between the top and bottom of the section was 10 degrees and, because the section was symmetrical (neutral axis in the middle of the height), it meant that the temperature at the top was +5 and and at the bottom was -5 in relation to the starting temperature during construction.

In the calculations contained in your screen captures it can be noticed that you assumed:

(Ts + Ti)/2 = 5

and 

Ts - Ti = 10

where:

Ts - temperature at the top of the section

Ti - temperature at the top of the section

It means that you assumed Ti=0 and Ts=10

But in such case the temperature in the neutral axis is 5, so to obtain in Robot the situation coherent with it you should define:

TX=5 TZ=10

Best regards,