Transcript Lean Construction

1
CE-401
Reinforced Concrete Design-II
By
Dr. Attaullah Shah
Swedish College of Engineering and Technology
Wah Cantt.
Course Outline:
− Analysis & design of axially loaded columns, Eccentrically
loaded columns by USD
− Analysis & design of strip footing for wall, spread footings
for columns and combined footings by USD.
− Design of retaining wall.
− Introduction to limit states.
− Detailing of reinforcement.
− Introduction to design of staircases and water tanks.
Columns subjected to eccentric loadings
Eccentric Compression
Interaction diagrams of combined bending and compression
Behavior under Combined Bending
and Axial Loads
Interaction Diagram Between Axial Load and Moment
( Failure Envelope )
Concrete crushes
before steel yields
Steel yields before concrete
crushes
Note:
Any combination of P and M outside the envelope will cause failure.
Behavior under Combined Bending and
Axial Loads
Axial Load and Moment Interaction Diagram – General
Behavior under Combined
Bending and Axial Loads
Resultant Forces action at Centroid
( h/2 in this case )
Pn  Cs1  Cc  Ts2



compression
is positive
Moment about geometric center
h
h

h a

M n  Cs1 *   d1   Cc *     Ts2 *  d 2  
2
2

 2 2

Columns in Pure Tension
Section is completely cracked (no concrete
axial capacity)
Uniform Strain
 y
N
Pn  tension     f y As
i 1
i
Columns
Strength Reduction Factor, f (ACI Code 9.3.2)
(a) Axial tension, and axial tension with flexure.
f = 0.9
(b) Axial compression and axial compression with
flexure.
Members with spiral reinforcement confirming
to 10.9.3
f  0.70
Other reinforced members
f  0.65
Columns
Except for low values of axial compression, f may be
increased as follows:
when
and
f y  60,000psi and reinforcement is symmetric
h  d   ds   0.70
h
ds = distance from extreme tension fiber to centroid of
tension reinforcement.
Then f may be increased linearly to 0.9 as fPn
decreases from 0.10fc Ag to zero.
Column
Columns
Commentary:
Other sections:
f may be increased linearly to 0.9 as the
strain s increase in the tension steel. fPb
Design for Combined Bending and
Axial Load (Short Column)
Design - select cross-section and reinforcement
to resist axial load and moment.
Design for Combined Bending and
Axial Load (Short Column)
Column Types
1) Spiral Column - more efficient for e/h < 0.1,
but forming and spiral expensive
2) Tied Column - Bars in four faces used when
e/h < 0.2 and for biaxial bending
General Procedure
The interaction diagram for a column is
constructed using a series of values for Pn and
Mn. The plot shows the outside envelope of the
problem.
General Procedure for
Construction of ID
− Compute P0 and determine maximum Pn in compression
− Select a “c” value (multiple values)
−Calculate the stress in the steel components.
−Calculate the forces in the steel and
concrete,Cc, Cs1 and Ts.
−Determine Pn value.
−Compute the Mn about the center.
−Compute moment arm,e = Mn / Pn.
General Procedure for
Construction of ID
−
−
−
−
Repeat with series of c values (10) to obtain a series of values.
Obtain the maximum tension value.
Plot Pn verse Mn.
Determine fPn and fMn.
−Find the maximum compression level.
−Find the f will vary linearly from 0.65 to 0.9
for the strain values
−The tension component will be f = 0.9
Example: Axial Load vs. Moment
Interaction Diagram
Consider an square column (20 in x 20 in.) with 8 #10
(r = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the
interaction diagram.
Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
Ast  8 1.27 in 2   10.16 in 2
Ag   20 in.  400 in 2
2
Ast 10.16 in 2
r

 0.0254
2
Ag
400 in
Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
P0  0.85 f c  Ag  Ast   f y Ast
 0.85  4 ksi   400 in  10.16 in
2
2

  60 ksi  10.16 in 2 
 1935 k
Pn  rP0
 0.8 1935 k   1548 k
[ Point 1 ]
Example: Axial Load vs. Moment
Interaction Diagram
Determine where the balance point, cb.
Example: Axial Load vs. Moment
Interaction Diagram
Determine where the balance point, cb. Using similar
triangles, where d = 20 in. – 2.5 in. = 17.5 in., one can
find cb
cb
17.5 in.

0.003 0.003  0.00207
0.003


 cb  
17.5 in.
 0.003  0.00207 
cb  10.36 in.
Example: Axial Load vs. Moment
Interaction Diagram
Determine the strain of the steel
 cb  2.5 in. 
 10.36 in.  2.5 in. 
 s1  
  cu  
  0.003
cb
10.36 in.




 0.00228
 cb  10 in. 
 10.36 in.  10 in. 
 s2  
  cu  
  0.003
cb
10.36 in.




 0.000104
Example: Axial Load vs. Moment
Interaction Diagram
Determine the stress in the steel
fs1  Es s1  29000 ksi  0.00228
 66 ksi  60 ksi compression
fs2  Es s1  29000 ksi  0.000104 
 3.02 ksi compression
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c
 0.85  4 ksi  20 in. 0.85 10.36 in.
 598.8 k
Cs1  As1  f s1  0.85 f c 
 3 1.27 in
2
  60 ksi  0.85  4 ksi  
 215.6 k
Cs2  2 1.27 in 2   3.02 ksi  0.85  4 ksi  
 0.97 k  neglect
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As fs  3 1.27 in
2
  60 ksi 
 228.6 k
Pn  Cc  Cs1  Cs2  Ts
 599.8 k  215.6 k  228.6 k
 585.8 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h
h a
h


M  Cc     Cs1   d1   Ts  d3  
2
2 2
2


 20 in. 0.85 10.85 in. 
 599.8 k 


2
 2

 20 in.

 215.6 k 
 2.5 in. 
 2

20 in. 

 228.6 k 17.5 in. 

2


 6682.2 k-in  556.9 k-ft
Example: Axial Load vs. Moment
Interaction Diagram
A single point from interaction diagram,
(585.6 k, 556.9 k-ft). The eccentricity of the point is
defined as
M 6682.2 k-in
e

 11.41 in.
P
585.8 k
[ Point 2 ]
Example: Axial Load vs. Moment
Interaction Diagram
Now select a series of additional points by selecting
values of c. Select c = 17.5 in. Determine the strain
of the steel. (c is at the location of the tension steel)
 c  2.5 in. 
 17.5 in.  2.5 in. 
 s1  
  cu  
  0.003
c
17.5 in.




 0.00257  f s1  74.5 ksi  60 ksi (compression)
 c  10 in. 
 17.5 in.  10 in. 
 s2  
  cu  
  0.003
c
17.5 in.




 0.00129  f s2  37.3 ksi (compression)
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c  0.85  4 ksi  20 in. 0.85 17.5 in.
 1012 k
Cs1  As1  fs1  0.85 f c   3 1.27 in 2   60 ksi  0.85  4 ksi  
 216 k
Cs2  2 1.27 in 2   37.3 ksi  0.85  4 ksi  
 86 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As f s  3 1.27 in
2
  0 ksi 
0k
Pn  1012 k  216 k  86 k
 1314 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h a
h

M  Cc     Cs1   d1 
2 2
2

 20 in. 0.85 17.5 in. 
 1012 k 


2
 2

 20 in.

 216 k 
 2.5 in. 
 2

 4213 k-in  351.1 k-ft
Example: Axial Load vs. Moment
Interaction Diagram
A single point from interaction diagram,
(1314 k, 351.1 k-ft). The eccentricity of the point is
defined as
M 4213 k-in
e

 3.2 in.
P
1314 k
[ Point 3 ]
Example: Axial Load vs. Moment
Interaction Diagram
Select c = 6 in. Determine the strain of the steel, c =6 in.
 c  2.5 in. 
 6 in.  2.5 in. 
 s1  
  cu  
  0.003
c
6 in.




 0.00175  fs1  50.75 ksi (compression)
 c  10 in. 
 6 in.  10 in. 
 s2  
  cu  
  0.003
c
6 in.




 0.002  fs2  58 ksi (tension)
 c  17.5 in. 
 6 in.  17.5 in. 
 s3  
  cu  
  0.003
c
6 in.




 0.00575  fs3  60 ksi (tension)
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c
 0.85  4 ksi  20 in. 0.85  6 in.
 346.8 k
Cs1  As1  fs1  0.85 f c 
 3 1.27 in 2   50.75 ksi  0.85  4 ksi  
 180.4 k  C 
Cs2  2 1.27 in 2   58 ksi 
 147.3 k  T 
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As fs  3 1.27 in
2
  60 ksi 
 228.6 k
Pn  346.8 k  180.4 k  147.3 k  228.6 k
 151.3 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h
h a
h


M  Cc     Cs1   d1   Ts  d3  
2
2 2
2


0.85  6 in. 

 346.8 k 10 in. 

2


 180.4 k 10 in.  2.5 in.
228.6 k 17.5 in.  10 in.
 5651 k-in  470.9 k-ft
Example: Axial Load Vs. Moment
Interaction Diagram
A single point from interaction diagram,
(151 k, 471 k-ft). The eccentricity of the point is
defined as
M 5651.2 k-in
e

 37.35 in.
P
151.3 k
[ Point 4 ]
Example: Axial Load vs. Moment
Interaction Diagram
Select point of straight tension. The maximum tension
in the column is
Pn  As f y  8 1.27 in
2
  60 ksi 
 610 k
[ Point 5 ]
Example: Axial Load vs. Moment
Interaction Diagram
Point
c (in)
Pn
Mn
1
-
1548 k
2
20
1515 k
253 k-ft
2 in
3
17.5
1314 k
351 k-ft
3.2 in
4
12.5
841 k
500 k-ft
7.13 in
5
10.36
585 k
556 k-ft
11.42 in
6
8.0
393 k
531 k-ft
16.20 in
7
6.0
151 k
471 k-ft
37.35 in
8
~4.5
0k
395 k-ft
infinity
9
0
-610 k
0
e
0 k-ft
0
Example: Axial Load vs. Moment
Interaction Diagram
Use a series of c
values to obtain the
Pn verses Mn.
Column Analysis
2000
1500
P (k)
1000
500
0
0
100
200
300
-500
-1000
M (k-ft)
400
500
600
Example: Axial Load vs. Moment
Interaction Diagram
Max. compression
Column Analysis
1200
1000
Location of the linearly
800
varying f.
fPn (k)
600
Cb
400
200
0
-200
0
100
200
300
-400
Max. tension
-600
-800
fMn (k-ft)
400
500
ACI Design Aids for Columns
Design Example 8.3
Bar splicing in Columns
Assignment No.1: (Total Marks100 each question carries 50 marks
− Design and Rectangular Column to carry dead load of 250K
live load of 350K dead load moment 150ft-K and live load
moment of 350ft-K Assume material properties.
− Determine the main steel required
− Determine the ties spacing
− Draw final neat to the scale sketch on graph paper
− Due Date: Sep,17 2012.