Lecture 31 Oneway Slab Design

1
Lecture 31 - One-way Slab Design

• November 13, 2002
• CVEN 444

2
Lecture Goals

• One-way slab design
• Short Column Biaxial Design

3
Example 3 One-way Slab Design
Design a slab for an interior span of concrete
joist floor system using 30 in forms. LL 80
psf fc 4 ksi
fy 60 ksi DL
20 psf (excluding self-weight)
4
Example 3 One-way Slab Design
Assume an initial thickness of 3 in. For a single
slab without ribs.
5
Example 3 One-way Slab Design
Find the minimum height of the joist from Table
9.5a in ACI Code The slab is 3 in. thick so
the joist must extend 15 in. 3 in. 12 in.
below the slab .
Joist Design
6
Example 3 One-way Slab Design
Including the ribs in the calculation of the
weight with with a depth of 12 in.
7
Example 3 One-way Slab Design
Compute the moments for the sections the
internal length is 24.5 ft. For the rib with a
negative moment The d 15 in. cover db/2
15 in. 0.75 in. 0.25 in d 14 in.
(Guess a 4 bar with minimum cover)
8
Example 3 One-way Slab Design
Find (c/d) for external column, Mu 36.3 k-ft
(435.6 k-in) d 14 in The area can be computed
using the equilibrium equation
9
Example 3 One-way Slab Design
The area of the steel The
minimum area required
10
Example 3 One-way Slab Design
Use 4 and 2 5 bars (As 0.81 in2 ), to find
the neutral axis

11
Example 3 One-way Slab Design
Determine the moment capacity of the
slab Compute the factored nominal moment

12
Example 3 One-way Slab Design
Verification for shear strength

13
Example 3 One-way Slab Design
Verification for shear strength

14
Example 3 One-way Slab Design
Compute nominal V Since fn lt freq we will need
to enlarge the ribs _at_ the ends

15
Example 3 One-way Slab Design
The shear width at 13.5 in. using a commercial
form for b 6.66 in. and 3 in. over 36 in. So 36
in - 13.5 in 22.5 in. increase in thickness
22.5/36(3 in.) 1.875 in. if it is on both sides
5in 3.75 in. 8.75 in. gt 6.66 in.
16
Example 3 One-way Slab Design
Compute the moments for the sections the
internal length is 24.5 ft. For the rib with a
positive moment
17
Example 3 One-way Slab Design
Determine the beff of the joist For the rib
positive
governs
18
Example 3 One-way Slab Design
Find (c/d) for external column, Mu 25.0 k-ft
(300 k-in) d 13.5 in.
19
Example 3 One-way Slab Design
The minimum k required for the t section
20
Example 3 One-way Slab Design
The minimum k required is 0.07843 Use 25
(As 0.62 in2) or 24 13 (As0.51 in2)
21
Example 3 One-way Slab Design
Look at the transverse moment along the joist
spacing is 30 in., w 0.197 k/ft2
Using an elastic analysis for concrete for the
positive moment
tension in concrete
22
Example 3 One-way Slab Design
The thickness of the slab is checked with an
elastic analysis So a 3 in. slab will work
23
Example 3 One-way Slab Design
The minimum reinforcement is governed by
temperature and shrinkage from ACI 7.12.2.1
Determine the spacing between the bars
24
Example 3 One-way Slab Design
Calculate the maximum allowable spacing (ACI
7.12.2.2) Use a wire mesh to get the necessary
steel, because 3 bar is 0.11 in2 and with a
maximum spacing of 9 in. the amount of steel will
still be too large. So use a wire mesh.
governs
25
Example Interaction Diagrams
The final design is given for joist 12 in deep
and slab thickness of 3 in.
26
Biaxial Loaded Short Columns
27
Biaxial Bending and Axial Load
Ref. PCA Notes on ACI 318-95
Unaxial bending about y-axis
28
Biaxial Bending and Axial Load
Ref. PCA Notes on ACI 318-95
The biaxial bending moments Mx Pey My Pex
29
Approximate Analysis Methods
Use Reciprocal Failure surface S2
(1/Pn,ex,ey) The ordinate 1/Pn on the surface S2
is approximated by ordinate 1/Pn on the plane S2
(1/Pn ex,ey) Plane S2 is defined by points A,B,
and C.
30
Approximate Analysis Methods
P0 Axial Load Strength under pure axial
compression (corresponds to point C ) Mnx Mny
0 P0x Axial Load Strength under uniaxial
eccentricity, ey (corresponds to point B ) Mnx
Pney P0y Axial Load Strength under uniaxial
eccentricity, ex (corresponds to point A ) Mny
Pnex
31
Approximate Analysis Methods
Design Pu Muy, Mux Pu, Puex, Puey
32
Approximate Analysis Methods
Pn Nominal axial load strength at
eccentricities, ex ey Limited to cases when
33
Biaxial Bending in Short Columns
Analysis Procedure Reciprocal Load Method
Breslers Formula
Steps
1) Calculate P0 2) Calculate P0y ( Pn for e ex,
ey 0 ) 3)Calculate P0x ( Pn for ex 0, e ey
) 4) Calculate Pn (from Breslers Formula )
34
Biaxial Bending in Short Columns
where, f 0.65
35
Biaxial Column Example
The section of a short tied column is 16 x 24 in.
and is reinforced with 8 10 bars as shown.
Determine the allowable ultimate load on the
section f Pn if its acts at ex 8 in. and ey
12 in. Use fc 5 ksi and fy 60 ksi.
36
Biaxial Column Example
Compute the P0 load, compression with no moments
37
Biaxial Column Example
Compute Pnx, by starting with ey term and assume
that compression controls. Check by Compute
the nominal load, Pnx and assume second
compression steel does not contribute
assume small
38
Biaxial Column Example
The components of the equilibrium equation
are Use similar triangles to find the
stress in the steel, fs
39
Biaxial Column Example
Compute the moment about the tension
steel where The resulting equation is
40
Biaxial Column Example
Combine the two equations and solve for Pn using
an iterative solution Set the two equation
equal to one another and sole for fs and the
definition
41
Biaxial Column Example
Combine the two equations and solve for c using
an iterative technique You are solving a cubic
equation
42
Biaxial Column Example
Check the assumption that Cs2 is close to zero
This value is small relative to the others
43
Biaxial Column Example
This Cs2 11 kips relatively small verses the
overall load, which is So Pnx 733.0 kips
44
Biaxial Column Example
Start with ex term and assume that compression
controls. Compute the nominal load, Pny and
assume second compression steel does not
contribute
assume small
45
Biaxial Column Example
The components of the equilibrium equation are
46
Biaxial Column Example
Compute the moment about the tension
steel where The resulting equation is
47
Biaxial Column Example
Combine the two equations and solve for Pn using
an iterative solution Set the two equation
equal to one another and sole for fs and the
definition
48
Biaxial Column Example
Combine the two equations and solve for c using
an iterative technique You are solving a cubic
equation
49
Biaxial Column Example
Check the assumption that Cs2 is close to zero
This value is negative so it does not contribute
50
Biaxial Column Example
This Cs2 - 2.1 kips relatively small verses the
overall load, which is So Pnx 684.6 kips
51
Biaxial Column Example
Compute the nominal load
52
Biaxial Column Example
Note the Pnx Pny include the corner steel bars
in both calculations a more conservative solution
would be to use 1/2 the steel in each direction
so As 2(1.27 in2) which would reduce Pu .
(Remember fs can not be greater than 60 ksi, so
that Pnx 620.3 k and Pny 578.4 k Pn 360.7
k and Pu 234.5 k )
53
Homework due (11/22/02)
Go back and prove the numbers for the last
assumption for biaxial loaded column Determine
the load capacity of the column, if the eccentric
loading is at ex 5.5 in. and ey 8 in. and fc
4 ksi and fy 60 ksi