Columns Examples

8/13/2019 Columns Examples

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TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS

Columns

P o r t l a n d C e m e n t A s s o c i a t i o n

P a g e 2 o f 9

Use Fig. 1 to determine a preliminary sizefor the tied column at the 1stfloor level.Assuming a reinforcement ratio g =0.020, obtain Pu /Ag3.0 ksi (fc= 5 ksi).Since Pu = 1,499 kips, the required Ag =1,499/3.0 = 499.7 in.2Try a 22 x 22 in. column (Ag= 484 in.2)with a reinforcement ratio ggreater than0.020.Check if slenderness effects need to beconsidered.Since the column is part of a nonswayframe, slenderness effects can beneglected when the unsupported columnlength is less than or equal to 12h, whereh is the column dimension (Sect. 10.12.2).12h = 12 x 22 = 264 in. = 22 ft > 12 ftstory height, which is greater than theunsupported length of the column.Therefore, slenderness effects can beneglected.Use Fig. 1 to determine the required areaof longitudinal reinforcement.

For a 22 x 22 in. column at the 1st floorlevel:Pu /Ag= 1,499/484 = 3.10 ksiFrom Fig. 1, required g= 0.026, orAs= 0.026 x 22 x 22 = 12.58 in.2Try 8-No. 11 bars (As= 12.48 in.2)Check Eq. (10-2) of ACI 318-99:Pn(max)= 0.80[0.85fc (Ag Ast) + fy Ast]Pn(max)= 1,542 kips > 1,499 kips O.K.From Table 1, 5-No. 11 bars can beaccommodated on the face of a 22-in. widecolumn with normal lap splices and No. 4ties. In this case, only 3-No. 11 bars areprovided per face.Use 8-No. 11 bars (= 2.58 ).Determine required ties and spacing.According to Sect. 7.10.5.1, No. 4 ties arerequired when No. 11 longitudinal bars areused.


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P a g e 3 o f 9

According to Sect. 7.10.5.2, spacing ofties shall not exceed the least of:16 long. bar diameters = 16 x 1.4116 long. bar diameters = 22.6 in.48 tie bar diameters = 48 x 0.548 tie bar diameters= 24 in.Least column dimension = 22 in. (governs)Check clear spacing of longitudinal bars:

in.8542 2

4152spacelear=

=

Since the clear space between longitudinalbars > 6 in., cross-ties are required perSect. 7.10.5.3.Reinforcement details are shown below.See Sect. 7.8 for additional specialreinforcement details for columns.

22

22

8-No. 11

No. 4 ties @ 22


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P a g e 4 o f 9

Example 2In this example, a simplified interactiondiagram is constructed for an 18 x 18tied column reinforced with 8-No. 9 Grade60 bars (g= 8/182= 0.0247). Concretecompressive strength = 4 ksi.Use Fig. 3 to determine the 5 points onthe interaction diagram. Point 1: Pure compression

kips71))]50247 )5(86

)]5f50

2ccmax)

=

Point 2 (fs1= 0)Layer 1:

0ddC 1

12 = Layer 2:

4256500ddC 122 =Layer 3:

8456544d

dC 132 =

Since 1 C2 (d3 /d1) > 0.69, the steel inlayer 3 has yielded.Therefore, set 1 C2 (d3 /d1) = 0.69 toensure that the stress in the bars inlayer 3 is equal to 60 ksi.

d3

=

2.

44

181.

5

(typ.

)

d2

=

9.

00

d1

=

15.

56

18

No. 3 tie

3-No. 9

2-No. 9

3-No. 9


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P a g e 5 o f 9

kips44 )53090)]}9

)22)[(3)8659(0

ddC7 n1 1i2i

=

=

kipsft821281320

12]}4(9)(22

15.56))(9[(3 00565518

)8659(0

122h

ddC7

Cdh

in1 1i2i

211

=

=

Point 3 (fs1= -0.5fy)Layer 1:

344ddC 1

12 Layer 2:

23565004ddC 122 =Layer 3:

79565444d

dC 132 =

Use 0.69

kips14)31020

)]}9)324[(3

)8655(0ddC7 n1 1i2i

=

=


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P a g e 6 o f 9

kipsft461263470

12]}4(9)(32

15.56)(94[(3 34565518

)8655(0

122h

ddC7

Cdh

in1 1i2i

211

=

=

Point 4 (fs1= -fy)

Layer 1:699d

dC 112

Layer 2:02565009ddC 122 =

Layer 3:74565

449ddC 1

32 =Use 0.69

kips38780)]}9

)229[(3)8651(0

ddC7 n1 1i2i

=

=

kipsft801262360

12]}4(9)(22

15.56)(99[(3 69565518

)8651(0

122h

ddC7

Cdh

in 1 1i2i

211

=

=


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P a g e 7 o f 9

Point 5: Pure bendingUse iterative procedure to determineMn.Try c = 4.0 in.

0087 456503

cd03 11

=

=

kips800ksi0fseksi0

ksi51087009E

11s1

1

00384

903cd03 12

= =

kips200ksi0fseksi0

ksi08038009E

22

s2

2

0012 44403

cd03 13

= =

=

kips023ksi3012009

E

33

3

==

kips081855

ab5 c=

Total T = (-180) + (-120) = -300 kipsTotal C = 102 + 208 = 310 kipsSince T C, use c = 4.0 in.


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P a g e 8 o f 9

kipsft812652

18)80d2

hT 1s

=

0122

18)20d2

hT 2s

=

kipsft51242

1802d2

hC 3s

=

kipsft802542]808

M31 nsi=

=

kipsft5380n =

Compare simplified interaction diagram tointeraction diagram generated from thePCA computer program PCACOL.The comparison is shown on the next page.As can be seen from the figure, thecomparison between the exact (black line)and simplified (red line) interactiondiagrams is very good.


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P a g e 9 o f 9

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