Engineering Stiffness of a Concrete Core in a Multi-Storey Building

AI Thread Summary
The discussion focuses on calculating the stiffness of a concrete core in a multi-storey building to accurately assess load shares and deflections of structural elements. The user has successfully approximated stiffness for moment-resisting frames but is struggling to incorporate the concrete core into their calculations. They seek a reliable method to determine the core's stiffness, currently using E*I/H, but suspect this may be incorrect due to poor results when combined with other structural elements. The conversation highlights the need to consider the interaction between the core and rigid frames, as they work together to resist horizontal forces. Accurate stiffness calculations are crucial for understanding the building's overall structural behavior under wind loads.
Tygra
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Homework Statement
Loading Sharing/Deflection of Structural Elements in a Multi-Storey Building
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Hi there, everyone,

I have been working on a problem in which I would like to calculate the load shares and deflections of structural elements in a multi-storey building. These elements include moment-resisting rigid frames, diagonally braced frames, concrete cores, and shear walls.

Thus far, I have achieved very good approximations for irregularly shaped buildings that contain just moment-resisting rigid frames. However, now I am trying to do the same but with the addition of concrete cores. I need to know the stiffness of the concrete core so that I can include it in my calculations for stiffness of the entire structure?? This is going to be a long post, so if you can answer my question you can stop here. If you are interested in the procedure read on.

Because my building is asymmetric it will experience a torsion and a translation

Here is the building I am working on:


L-Shaped Building Image.png


I start with calculating the shear stiffness of each sub frame that makes up the entire structure in the X and Y directions. Look at the following plan of my building:

Plan of L-Shaped Building.png

The X direction is from left to right, the Y direction is from top to bottom, with the wind pressure acting on the structure from right to left.

I hope you can see the frames in the X direction (parallel to the wind load) are made up of 5 frames. Going from bottom to top in the plan we have 3 frames that have 4 columns and 3 beams; then there are 2 more frames that have 2 columns and 1 beam.

To compute the shear stiffness of a frame we can use the formula:

Kframe1 = Kb*(Kc/(Kb + Kc))

where:

Kc = sum (12*E*Ic)/h^2

Kb = sum (12*E*Ib)/(h*l)

E = modulus of elasticity of the steel

Ic = second moment area of the columns

Ib = second moment of areas of the beams

h = storey height

l = bay width

Thus, for the first frame in the X direction, we would get:

Kc = 4*12*E*Ic/h^2

Kb = 3*12*E*Ib/(h*l)

This procedure is reapeated for all the frames in the Y direction.

So we have the shear stiffness of 5 frames in the X direction and 4 frames in the Y direction.

Lets call these frames:

K1x

K2x

K3x

K4x

K5x

K1y

K2y

K3y

K4y

Now, a coordinate system must be defined. Taking the bottom left corner as the origin, we find the centroids of all the frames. This is easy as its just the centre of each frame. Hence in the Y direction:

K1x = 0m

K2x = 4m

K3x = 8m

K4x = 12m

K5x = 16m

And in the X direction:

K1y = 0m

K2y = 6m

K3y = 12m

K4y = 18m

From this I can calculate the centre of stiffness

ybar = (K1x*0 + K2x*4 + K3x*8 + K4x*12 + K5x*16)/(K1x + K2x + K3x + K4x)

xbar = (K1y*0 + K2y*6 + K3y*12 + K4y*18)/(K1y + K2y + K3y + K4y)

For my building, these coordinates are:

ybar = 6m

xbar = 10.67m

Now the torsional stiffness is required. For this a new cooordinate system must be defined, where the origin is the centre of stiffness. Thus:

t1x = ybar - 0

t2x = ybar - 4m

t3x = 8m - ybar

t4x = 12m - ybar

t5x = 16m - ybar

t1y = xbar - 0

t2y = xbar - 6

t3y = 12m - xbar

t4y = 18m - xbar

The torsional stiffness is obtained by:
q1x = K1x*t1x^2

q2x = K2x*t2x^2

q3x = K3x*t3x^2

q4x = K4x*t4x^2

q5x = K5x*t5x^2

q1y = K1y*t1y^2

q2y = K2y*t2y^2

q3y = K3y*t3y^2

q4y = K4y*t4y^2

qw = (q1x + q2x + q3x + q4x + q5x + q1t + q2y + q3y + q4y)

The total force coming from the wind is:

Fwind = 10 kN/m^2*H*B

Where:

H = height of the building

B = width of the building face that the wind is acting upon


The moment induced is simply:

Mwind = Fwind*(8 - ybar)

Now the load share of frame1 can be computed by:

F1 = Fwind*(K1x/(K1x + K2x + K3x + K4x) - Mwind*(K1x*t1x/qw)

For the other frames use their stiffnesses and plug them into the above equation.

Sorry for this post being so long, but I thought it was a good idea to show you what I am doing. Again, I need to find out how stiff a concrete core is. I have the modulus of elasticity and second moment of area for it, but the equations I have tried like: (8*E*I)/H^4 are not working!

Many thanks in advance,

Tygra
 
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Depending on the direction from which the wind blows, the horizontal cross-sections of that concrete core will be experiencing bending and shearing loads alone (zero summation of horizontal moments), or a combination of those plus horizontal torsion.

Those wind loads will be transferred to our core via the metal structure surrounding the core.
The deformation of those metal elements will always be much greater than the deformation of our concrete core for all types of loads (which is its reason to exist via design).

You could calculate that relatively small deformation, exactly as you would do it for a comparable metal structural element or vertical tubular beam, only that using the values associate to concrete.
 
Hi there Lnewqban,

Yes I can calculate the small deflection of a core by itself. Under a uniformly distributed horizontal load you can use:

delta = w*H^4/(8*E*I)

But the rigid frames and the core are working as one system. Hence, the deflection "delta" will be less because the core and frames together provide more resistance to the horizontal force.

This is the formula to compute the load share on each structural element

Load Share Forumla.jpg

Kyy is the stiffness of all the structural elements in the building in the Y direction. This could be rigid frames, braced frames, shear walls and cores. I have the equations to compute the stiffness for the rigid frames in my building, and when I use rigid frames on their own I am getting very good results. Its when I try to add the core that the bad results come. So, the stiffness of the core that I am currently using must be incorrect!

I am currently computing the stiffness of the core as E*I/H

Where:
E = modulus of elasticity of the concrete
I = second moment of area of the core cross-section
H = core height

Similarly Kxx is the stiffness of all the structural elements in the building in the X direction.

y and x are the the coordinates from the centre of stiffness to the centroid of the structural element - frames, cores, shear walls etc.
 

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