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Read Example 9.11 page 339, Sect. 9.9. Practical Design ... · PDF filePractical Design...
Transcript of Read Example 9.11 page 339, Sect. 9.9. Practical Design ... · PDF filePractical Design...
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Reading Assignment
Read Example 9.11 page 339, Sect. 9.9. Practical Design Considerations Chapter 9 of text Chapter 10 of ACI
Design Example Given
4,00060,000
450385
0.03
c
y
u
u
g
f ksif psiP kipM ft kipρ
=
=
== −=
Find required b, and h (width and height of the cross section). Solution:
Select a tied column dimension h, h=20, use 3" cover, thus:
2 20 6 0.720
h dh
γ′− −
= = =
Use the design aid given in your book on page 792, Figure B.13 Eccentricity will be equal to:
385 12 10.26450
u
u
MeP
×= = =
10.26 0.5120
eh= =
with e/h = 0.51, from graph given on the next page read:
0.44
/ 0.44
n
c g
u
c g
Pf APf A
φ
=′
=′
Assume 0.65φ =
2
2
2
450 / 0.65 0.444
393
393393 19.6720
g
g
A
A in
bh in
b in
=×
=
=
= =
Use a column of 20x20. The area of steel will be: 20.03 20 20 12sA in= × × = Use 8 #11 bars 212.5sA in=
Note. For design must insure satisfying ACI code provisions: 1. Min cover consideration ACI 7.7. 2. Min bar spacing ACI 7.6. 3 Arrangement of steel to achieve approximate agreement with design aid assumptions.
4. Evaluation of capacity of actual section chosen after all details have been satisfied. 8 #11 bars
20”
20”
3” 3”
3
0.0625eh=
0.25eh=
0.5eh=
1.25eh=
4
Design Example Using the Design Aids Use of graphic design aid for a column with axial load and uniaxial bending. Consider that we wish to design a rectangular tied column to accept the following service dead and live loads and moments. Architectural considerations limit allowable column width b = 16 in and h = 20 in (tied column). For now neglect length effects and bending about weak axis.
4,00060,000
184213107124
c
y
D
L
D
L
f ksif psiP kipP kipM ft kipM ft kip
==
=== −= −
Solution Calculate design loads:
1.2 1.6 1.2(184) 1.6(213) 5611.2 1.6 1.2(107) 1.6(124) 327
u D L
u D L
P P P kipM M D ft kip
= + = + == + = + = −
Use a cover of 3.0 inches. The column parameters (assuming bending about the strong axis)
/ 561/ 0.65 0.674 320
/ / 327 12 / 0.65 0.2420 4 320
u
c g
u u
c g c g
Pf A
eP Mhf A hf A
φ
φ φ
= =′ ×
×= = =
′ ′ × ×
and
2 20 6 0.720
h dh
γ′− −
= = =
From the design aid (see next page) read: 0.031gρ = Area of steel will be:
20.031 20 16 9.92stA in= × × = Use 8 #10 bars with 210.12stA in= Check φ factor
5
0.0625eh=
0.25eh=
0.5eh=
1.25eh=
16”
20”
3” 3”