1

Design Example

Objective In this example, a simplified interaction diagram is constructed for an 18 in x 18 in tied column reinforced with 8-No.9 Grade 60 bars (g = 8/182 = 0.0247). Use the simplified equations to determine 5 control points on the interaction diagram.

Reference Iyad M. Alsamsam, Mahmoud E. Kamara, Simplified Design Reinforced Concrete Buildings of Moderate Size and Height, Third Edition, 2005, Portland Cement Association, Example 5.4.1.1, pp. 5-7

Code Building Code Requirements for Structural Concrete (ACI 318-05) and Commentary (ACI 318R-05)

Problem Use the simplified equations to establish the interaction diagram for the column section shown in Figure 1 by determining the following 5 points:

Point 1: Pure Compression

Point 2: Bar stress near tension face of member equal to zero, fs = 0

Point 3: Bar stress near tension face of member equal to 0.5fy (fs = 0.5fy)

Point 4: Bar stress near tension face of member equal to fy (fs = fy)

Point 5: At pure bending

Material Properties and Section Layout Concrete: fc' = 4000 psi

Steel: fy = 60,000 psi

b=18"

h=18"

(8)-#9 bars

1.5"

Figure 1 Square Column with Symmetrical reinforcement

Solution Point 1: Pure Compression

' 'n(max) g c g y cP 0.80 A [0.85f (f 0.85f )]

0.52 182[(0.85 4) 0.0247(60 (0.85 4))]808.3kips

= + = + =

2

Point 2 ( )sf 0= Layer 1:

12

1

d1 C 1 1(1) 0d

= =

Layer 2:

22

1

d 9.001 C 1 1 0.42d 15.56

= =

Layer 3:

33

1

d 2.441 C 1 1 0.84d 15.56

= =

Since 321

d1 C 0.69d

> , the steel in layer 3 has yielded.

Therefore, set 321

d1 C 0.69d

= to ensure that the stress in the bars in layer 3 is equal to 60 ksi.

ni

n 1 1 si 2i 1 1

dP C d b 87 A 1 Cd

0.65{(2.89 15.56 18) 87[(3 0) (2 0.42) (3 0.69)]}0.65(809.4 253.2)690.9kips

=

= + = + + + = +=

ni

n 1 1 3 1 si 2 ii 1 1

d hM 0.5C d b(h C d ) 87 A 1 C d /12d 2

0.65{(0.5 2.89 15.56 18) (18 0.85 15.56) 87[(3 0)(9 15.56) (2 0.42)(9 9)(3 0.69)(9 2.44)]}/12

0.65(1932.1 1181.4) /12168.6ft kips

=

= + = + + +

= +=

Point 3 ( )s yf 0.5f= Layer 1:

12

1

d1 C 1 1.35(1) 0.35d

= =

Layer 2:

22

1

d 9.001 C 1 1.35 0.22d 15.56

= =

Layer 3:

32

1

d 2.441 C 1 1.35 0.79d 15.56

= = , use 0.69

3

ni

n 1 1 si 2i 1 1

dP C d b 87 A 1 Cd

0.65{(2.14 15.56 18) 87[(3 (0.35)) (2 0.22) (3 0.69)]}0.65(599.4 127.0)474.9kips

=

= + = + + + = +=

( )

ni

n 1 1 3 1 si 2 ii 1 1

d hM 0.5C d b(h C d ) 87 A 1 C d /12d 2

0.65{(0.5 2.14 15.56 18) (18 0.63 15.56) 87[(3 0.35 )(9 15.56) (2 0.23)(9 9)(3 0.69)(9 2.44)]}/12

0.65(2456.6 1780.7) /12229.1ft k

=

= + = + + +

= +=

ips

Point 4 ( )s yf f= Layer 1:

12

1

d1 C 1 1.69(1) 0.69d

= =

Layer 2:

22

1

d 9.001 C 1 1.69 0.02d 15.56

= =

Layer 3:

32

1

d 2.441 C 1 1.69 0.74d 15.56

= = , use 0.69

ni

n 1 1 si 2i 1 1

dP C d b 87 A 1 Cd

0.65{(1.70 15.56 18) 87[(3 ( 0.69)) (2 0.02) (3 0.69)]}0.65(476.1 3.5)314.0kips

=

= + = + + + = +=

ni

n 1 1 3 1 si 2 ii 1 1

d hM 0.5C d b(h C d ) 87 A 1 C d /12d 2

0.65{(0.5 1.70 15.56 18) (18 0.50 15.56) 87[(3 (0.69))(9 15.56) (2 0.02)(9 9)(3 0.69)(9 2.44)]}/12

0.65(2433.1 2362.8) /12260ft k

=

= + = + + +

= +=

ips

Point 5: Pure bending Use iterative procedure to determine nM Try c 4.0in=

4

1s1

c d0.003c

4 15.560.0034

0.0087

= =

=

s1 s s1

s1

f E29000 ( 0.0087) 251.4ksi 60ksi,usef 60ksi

= = = > =

s1 s1 s1T A f 3 ( 60) 180kips= = = 2

s2c d0.003

c4 90.003

40.0038

= =

=

s2 s s2

s1

f E29000 ( 0.0038) 108.8ksi 60ksi,usef 60ksi

= = = > =

s2 s2 s2T A f 2 ( 60) 120kips= = = 3

s3c d0.003

c4 2.440.003

40.0012

= =

=

s3 s s3f E29000 (0.0012) 33.9ksi

= = =

s3 s3 s3C A f 3 33.9 102 kips= = = '

c cC 0.85f ab0.85 4 (0.85 4) 18208kips

== =

Total T ( 180) ( 120) 300 kips= + = Total C 102 208 310 kips= + = Since T C , use c 4.0in=

ns1 s1 1h 18M T d ( 180) 15.56 /122 2

98.4ft kips

= = =

ns2 s2 2h 18M T d ( 120) 9 /122 2

0

= = =

5

ns3 s3 3h 18M C d (102) 2.44 /122 2

55.8ft kips

= = =

( )( )

3

n c nsii 1

M 0.5C h a M

0.5 208 18 3.4 /12 154.2

280.7ft kips

== + = +

=

nM 0.9 280.7 253ft kips = = Compare simplified interaction diagram to interaction diagram generated from StructurePoints pcaColumn computer program. As can be seen from the figure, the comparison between the exact (black line) and simplified (red line) interaction diagrams is very good.

Point Parameter Calculations (simplified) pcaColumn

(exact) 1: Pure compression Pn (kips) 808.3 808.3

Pn (kips) 690.9 679.8 2: fs = 0 Mn (ft-kips) 168.6 164.99 Pn (kips) 474.9 463.8 3: fs = 0.5fy Mn (ft-kips) 229.1 225.43 Pn (kips) 314 307.3 4: fs = fy Mn (ft-kips) 260 256.3

5: Pure bending Mn (ft-kips) 253 247.72

6

Point 2

Point 3

Point 4

Point 1

Point 5

7

Design ExampleObjectiveReferenceCodeProblemMaterial Properties and Section LayoutSolution