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Lecture 33 Long Columns

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Title: Lecture 33 Long Columns


1
Lecture 33 - Long Columns
  • April 16, 2003
  • CVEN 444

2
Lecture Goals
  • Slender Column Design

3
Long Column
Eccentrically loaded pin-ended column.
Lateral deflection - increases moment
M P( e D )
4
Long Column
Eccentrically loaded pin-ended column.
Do first-order deflection due to Mo Da
second-order deflection due to Po
5
Long Column
Eccentrically loaded pin-ended column.
OA - curve for end moment OB - curve for maximum
column moment _at_ mid-height)
Axial capacity is reduced from A to B due to
increase in maximum moment due to Ds
(slenderness effects)
6
Long Columns
7
Long Column - Slenderness Ratio
Slenderness Ratio for columns
8
Long Column - Slenderness Ratio
Slenderness Ratio for columns
9
Long Column - Slenderness Ratio
Slenderness Ratio for columns in frames
10
Long Column - Slenderness Ratio
Slenderness Ratio for columns in frames
11
Long Column
Unsupported height of column from top of floor to
bottom of beams or slab in floor Radius of
gyration 0.3 overall depth of
rectangular columns 0.25 overall depth of
circular columns
lu
r
12
Long Column
double curvature
singular curvature
13
Long Columns
M1/M2 Ratio of moments at two column ends
where M2 gt M1 (-1.0 to 1.0 range) - single
curvature - double curvature
is typically conservative (non-sway frames)
Note Code (10.12.2) M1/M2 -0.5 non-sway frames
14
Long Column
15
Moment Magnification in Non-sway Frames
If the slenderness effects need to be considered.
The non-sway magnification factor, dns, will
cause an increase in the magnitude of the design
moment. where
16
Moment Magnification in Non-sway Frames
The components of the equation for an Euler
bucking load for pin-end column and the
stiffness, EI is taken as
17
Moment Magnification in Non-sway Frames
A coefficient factor relating the actual moment
diagram to the equivalent uniform moment diagram.
For members without transverse loads For other
conditions, such as members with transverse loads
between supports, Cm 1.0
18
Moment Magnification in Non-sway Frames
The minimum allowable value of M2 is The sway
frame uses a similar technique, see the text on
the components.
19
Design of Long Columns- Example
A rectangular braced column of a multistory frame
building has floor height lu 25 ft. It is
subjected to service dead-load moments M2 3500
k-in. on top and M12500 k-in. at the bottom.
The service live load moments are 80 of the
dead-load moments. The column carries a service
axial dead-load PD 200 k and a service axial
live-load PL 350 k. Design the cross section
size and reinforcement for this column. Given YA
1.3 and YB 0.9. Use a d2.5 in. cover with
an sustain load 50 and fc 7 ksi and fy
60 ksi.
20
Design of Long Columns- Example
Compute the factored loads and moments are 80 of
the dead loads
21
Design of Long Columns- Example
Compute the k value for the braced compression
members Therefore, use k 0.81
22
Design of Long Columns- Example
Check to see if slenderness is going to matter.
An initial estimate of the size of the column
will be an inch for every foot of height. So h
25 in. So slenderness must be considered.
Since frame has no side sway, M2 M2ns, ds 0
Minimum M2
23
Design of Long Columns- Example
Compute components of concrete The moment of
inertia is
24
Design of Long Columns- Example
Compute the stiffness The critical load is
25
Design of Long Columns- Example
Compute the coefficient The magnification
factor
26
Design of Long Columns- Example
The design moment is Therefore the design
conditions are
27
Design of Long Columns- Example
Assume that the r 2.0 or 0.020 Use 14 9
bars or 14 in2
28
Design of Long Columns- Example
The column is compression controlled so c/d gt
0.6. Check the values for c/d 0.6 Check the
strain in the tension steel and compression steel.
29
Design of Long Columns- Example
The tension steel
30
Design of Long Columns- Example
Combined forces
31
Design of Long Columns- Example
Combined force
32
Design of Long Columns- Example
Moment is
33
Design of Long Columns- Example
The eccentricity is Since the e 11.28 in. lt
13.62 in. The section is in the compression
controlled region f 0.65. You will want to
match up the eccentricity with the design.
34
Design of Long Columns- Example
Check the values for c/d 0.66 Check the
strain in the tension steel and compression steel.
35
Design of Long Columns- Example
The tension steel
36
Design of Long Columns- Example
Combined forces
37
Design of Long Columns- Example
Combined force
38
Design of Long Columns- Example
Moment is
39
Design of Long Columns- Example
The eccentricity is Since the e 11.28 in.
The reduction factor is equal to f 0.65.
Compute the design load and moment.
40
Design of Long Columns- Example
The design conditions are
41
Design of Long Columns- Example
Design the ties Provide 3 ties, spacing will be
the minimum of Therefore, provide 3 ties _at_
18 in. spacing.
42
Using Interaction Diagrams
  • Determine eccentricity.
  • Estimate column size required base on axial load.
  • Determine e/h and required fPn/Ag
  • Determine which chart to use.
  • Select steel sizes.
  • Design ties by ACI code
  • Design sketch

43
Homework(4/27/03)
Problem 9.13(a)
Try a 20 in square column and sustained load of
50
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