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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