Thermal Stresses and Forces

[ZeroStress]

Hello all,

I am trying to calculate the thermal stresses and forces developed into the concrete deck right at the centre line of columns, as attached in the picture.

- The difference in temperature is taken as 30 deg Celsius.
- Coefficient of thermal expansion of concrete is 12e-6
- Width of the slab is 30 metres.
- Height of slab is 200mm.
- There are two circular columns on each side of the deck. 4 total in plan.
- Youngs modulus for concrete is 30 kN/mm2
- The columns are hinged into the deck.

Now if I am to find the stresses due to temperature change, I would calculate as:

Stress = Young’s modulus x Coeff. Thermal Exp. x Temperature Difference
Stress = 30 kN/mm2 x 12e-6 x 30
Stress = 10.8 N/mm2

The area of slab cross section = Width x Height = 6 sq.m

Therefore the force on the face would be: Cross sectional area x stress
Which is F = 6 sq.m x 10.8 N/mm2 = 64,800 kN

Because there are two columns in the row, this force will be halved for each column.

F = 32,400 kN

Now the problem is, this force is huuugeee… it shouldn’t be that much as it doesn’t make sense. Surely there is something wrong I am doing. Can anyone help me?

Thanks

 

[slickdeals]

If the column can flex and let the slab move, then you won't develop the force. The force developed in the column is a restraining force. If it flexes, the restraint goes away and the slab can "breathe".

 

[JoshPlumSE]

Thermal forces are self limiting forces. Meaning that if the slab is free to expand/contract, then the force that develops is ZERO. So, it's only when the slab or member is rigidly restrained that the force develops. The solution then is to allow the slab or member to expand.

Also, if you ran an FEM model of the structure, you would want to make sure that the supports are modeled with realistic rigidity..... When you do that, you'll see that the magnitude of the forc is much, much less.

 

[bkal]

In doing this, you are assuming that your columns are infinitely stiff. But as soon as they start to deform, the force would drop. Could you treat it as a displacement driven problem; your slab expands by ~11mm which means that in an ideal case each column needs to move by ~5mm at the top - can it break your columns?

 

[ZeroStress]

My mistake, I didnt clarify earlier.

Lets put it this way, I am trying to find the defection of the columns. This deflection would be outwards, if the concrete slab in the middle is expanding (as in picture).

So, to find the deflection of the column I would need the force generated from the thermal stresses in the slab. The columns are 13m high with same stiffness as the concrete slab.

So if I assume that 32,400 kN is the force at the tip of the column, it would deflect it by a stupid amount.

Deflection = F L3 / 3 E I

E = 30 kN/mm2
I = 2.64 e11 mm4 (columns are 1.524m in dia)
L = 13,000 mm

Deflection = 32,400 x 13,000^3 / 3 x 30 x 2.64 e11

Deflection = 2995 mm <-- This is stupid value, cant be that high?

 

[TXStructural]

The slab will expand a certain amount with temperature increase, and unless restrained sufficiently by reinforcement and other structure (which will not restrain it completely.) The expansion is not an unlimited force vector like gravitational weight, it is a displacement.

 

[ZeroStress]

Thanks.

I think I am sort of getting the hang of the problem here. Or perhaps I am coming to terms with the fact that thermal stresses can be disastrous if not taken care of properly.

I am pretty sure the stresses I calculated in my first post are true. But the deflection in the columns is not real. The columns have to allow the thermal movement to stop those huge stresses being developed.

Thanks everyone!

 

[bkal]

ZeroStress, the key point is that this is a displacement controlled problem. The dispalcement in the column should be slightly less than the displacement of the slab (assuming it was free to deform). You can find the force in the column as:

your columns will expand by:

dl=30*12E-6*30=10.mm

assume that top of each columns moves half of that distance (two columns moving in opposite directions)

d=5.4mm


F=3EId/(L^3)=58kN.

 

[rb1957]

i think the key is accommodating the dispalcements that arise ...

how much will the slab expand ? if the columns are standing on the slab then they'll see this dispalcement. what happens at the other end of the columns ? does it constrain this displacement ?

the columns will expand, as noted in the previous post. can they ? or are they constrained by the rest of the structure ?

Quando Omni Flunkus Moritati

 

[johnbridge231]

bkal is right, you should consider a stiffness provided from the columns, like a spring in order to get "real" displacement values and thus column forces.

Analysis and Design of arbitrary cross sections
Reinforcement design to all major codes
Moment Curvature analysis

http://www.engissol.com/cross-section-analysis-design.html