1

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”