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SAP2000 – 34 Concrete Shell Reinforcement Design: Watch & Learn

SAP2000 – 34 Concrete Shell Reinforcement Design: Watch & Learn


this tutorial will illustrate the
approach used by SAP2000 to calculate concrete shell reinforcing intensities
these shells might be used to model slabs domes tanks or numerous other
structures and we will illustrate how SAP2000 resolves shell stresses from an
analysis into forces and the forces into reinforcing greatly simplified the
process looks like this 1) we start with the shell section that is subjected
to a membrane stress and a moment stress 2) the moment is resolved into a moment
couple and then combined with the membrane to 3) give two pure membrane
forces these membrane forces are then used to determine the design forces and
required reinforcing the procedure discussed in this tutorial
assumes that there are two layers of reinforcing that is there is a top layer
and a bottom layer of reinforcing in two principal directions SAP2000 offers concrete shell
reinforcement design following the guidelines as prescribed in Eurocode 2
1992 it assumes the section to be cracked and the design is intended to
satisfy strength requirements only no code minimums or serviceability checks
are included additional information on the design approach used in SAP2000 is
contained in the technical note provided with the program SAP2000’s shell design determines the
amount of reinforcing required on a per unit width basis needed to carry
membrane and moment forces our example we’ll use a 32 foot by 32 foot slab 12
inches thick simply supported around the perimeter although this is a two-way
slab and SAP2000 calculates reinforcing in both directions we will simplify our
tutorial by only concerning ourselves with results in one primary direction
and in this case it will be the local one axis which is the red axis shells typically have eight force
resultants three membrane components f11 f22 and f12 two flexural moment
components m11 and m22 and one twisting moment m12 along with two
transverse shear components the thin shell is idealized as a sandwich model
comprised of two outer layers that contain the reinforcing and an uncracked
center core the design assumes that the outer layers carry the moments and
membrane forces and that the uncracked center core carries the transverse shear
although transverse shear is neither checked nor designed for in this shell
design procedure the design in the program resolves membrane stress and
flexural stress resultants into pure membrane forces which are then combined
into reinforcement forces in this image the equations shown deal with the
stresses in just the one direction the primary direction we are concerned with
although in a typical slab there would be similar components in the two
direction as well we are only dealing with the one direction to make our
discussion simpler in addition to the thickness of the slab the key piece of
information to be entered by the user is a cover for the reinforcing needed at
the top and bottom locations in each direction for instance ct1 is the cover
over the top reinforcing and is the distance from the outside of the shell
to the center of the reinforcing cb1 is similarly the cover for the bottom
reinforcing we can set the cover for our concrete shell or area object by going
to the define section properties area sections command and selecting our shell
section clicking the modify section button note
that the thickness is set to 12 inches and then click the modify show shell
design parameters button we select a rebar material which for this model is
A615 grade 60 and for the rebar layout options we will select the 2 layers
option we will assume that Direction one has the outer bars and will assign a
cover of one and a half inches to these bars we will assign a cover of two
inches to the inside or direction 2 bars now that we have entered the slab
thickness and reinforcing covers the program has the geometric data required
for design the program will now determine the thickness of each of the
outer layers in the sandwich model as the lesser of the following 1) twice
the cover measured to the center of the outer reinforcement which is three
inches or 2) twice the distance from the center of the slab to the center of
the outer reinforcement which is nine inches three inches is less than nine
inches and thus the thickness of the outer layers will be three inches the membrane stress resultants and moment
stress resultants that impact the reinforcing in the one direction are m11
f11 m12 and f12 to convert the moment stress resultants into a membrane
force couple the moments need to be divided by the moment or lever arm in
this case the lever arm is the distance between the top and bottom reinforcing
in the one direction shown as d1 thus for the m11 moment the resulting force
couple we’ll be m11/d1 however converting the moment stress
resultants into pure membrane forces is only half of the equation next we need
to convert the direct membrane stress resultants into pure membrane forces
apportioned to each of the outer layers we will distribute the pure membrane forces
such that the moment equilibrium is maintained that is dt1 times ft1
equals db1 times fb1 where dt1 is the distance from the center of the shell to
the top steel and db1 is the distance from the center of the shell to the
bottom steel we know that dt1 plus db1 equals d1 and f11 equals ft1 plus
fb1 substituting these values into the first equation we see that for the top
layer the resulting pure membrane force is f11 times db1 divided by d1 thus
the total membrane force in the top layer due to m11 and f11 resultant
stresses is N11 equals minus m11 plus f11 times db1 divided by d1 as
shown a similar approach is taken when converting the m12 and f12 stress
resultants into membrane force N12 the next step is to convert the pure
membrane forces into design forces there are a number of different conditions
that need to be checked but it may be as straightforward as N design equals f11
plus the absolute value of f12 once the design forces have been computed for top
and bottom and both directions calculating the rebar intensities that
is the required rebar area per unit width is straightforward in this case
the top reinforcing in the one direction is equal to the design force divided by
phi times Fy where phi is always equal to 0.9 now let’s finish the model
and go ahead with the design we will select the entire slab and in addition
to the self-weight dead load we will add a live load of 50 pounds per square foot
or .00035 Kip’s per inch squared to our slab we will
also define a new load combination called slab design and for our study
will define a single combination of 1.2 times dead plus 1.6 times live although
any number of appropriate design combinations could be specified with the loads assigned we will switch
to a plan view and we can now run the analysis next we can move on to the design
however the design is already done it is automatically performed when running the
analysis no additional steps are required to get to the design results
all we need to do is go to the display show forces shell command choose the
slab design combo and select the concrete design option here we can
select the face for the output or maximums and minimums and we will select
the bottom face as it will be in tension and require the most reinforcing next is a list of components items such
as the pure membrane forces the design forces and the area of required
reinforcing are all available first let’s view N11 and we get a value of
approximately 0.59 at the outer corner for the outside bottom layer now we will
switch to N12 and get a value of approximately 0.72 at the same location
combining these two gives us a value of approximately 1.31 this is what we
would expect the design force in the one direction to be and if we select NDes1 we find that it indeed does match the
value next we will display the area of steel
required on a per unit width basis for the bottom outside layer in the one
direction this is a ASt1 this gives a value of approximately 0.024
this should be equal to the design force divided by phi and Fy if we
do the calculation it does indeed match going to the display once more the
component Fc represents the principal compressive force per unit
width and again there are a number of different conditions that need to be
checked but it may be as straightforward as Fc top equals -2 times F12 if we
switch to the top face since it should be in compression and select the N12
component we see that it has a value of approximately 0.72 if we select the Fc component we
see that Fc is indeed approximately twice the N12 value of 0.72 lastly Sc
is simply the principal compressive stress and is obtained by dividing Fc
by the thickness of the layer which should give us a value for the top layer
of approximately 0.48 and again the value matches in conclusion it is a simple process to
obtain required Eurocode reinforcement intensities for concrete shell objects
using SAP2000 note that these reinforcement values are based purely on
strength requirements and no serviceability or code minimums are
checked this concludes this tutorial on concrete shell reinforcement design in
SAP2000

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