Design Example-columns-aci 318-05.pdf 216c5b

Design Example Objective In this example, a simplified interaction diagram is constructed for an 18 in x 18 in tied column reinforced with 8No.9 Grade 60 bars (ρg = 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"

1.5"

(8)-#9 bars h=18"

Figure 1 Square Column with Symmetrical reinforcement Solution •

Point 1: Pure Compression φPn(max) = 0.80φA g [0.85f c' + ρg (f y − 0.85f c' )] = 0.52 × 182[(0.85 × 4) + 0.0247(60 − (0.85 × 4))] = 808.3kips

1

•

Point 2 ( fs = 0 ) Layer 1: 1 − C2

d1 = 1 − 1(1) = 0 d1

Layer 2: 1 − C2

d2 ⎛ 9.00 ⎞ = 1 − 1⎜ ⎟ = 0.42 d1 ⎝ 15.56 ⎠

Layer 3: 1 − C3

d3 ⎛ 2.44 ⎞ = 1 − 1⎜ ⎟ = 0.84 d1 ⎝ 15.56 ⎠

Since 1 − C2

d3 > 0.69 , the steel in layer 3 has yielded. d1

Therefore, set 1 − C2

d3 = 0.69 to ensure that the stress in the bars in layer 3 is equal to 60 ksi. d1

n ⎡ ⎛ d ⎞⎤ φPn = φ ⎢C1d1b + 87∑ Asi ⎜1 − C 2 i ⎟ ⎥ d1 ⎠ ⎥⎦ i =1 ⎢⎣ ⎝ = 0.65{(2.89 ×15.56 ×18) + 87[(3 × 0) + (2 × 0.42) + (3 × 0.69)]}

= 0.65(809.4 + 253.2) = 690.9 kips n ⎡ ⎛ d ⎞⎛ h ⎞ ⎤ φM n = φ ⎢ 0.5C1d1b(h − C3d1 ) + 87∑ Asi ⎜1 − C2 i ⎟ ⎜ − di ⎟ /12 ⎥ d1 ⎠ ⎝ 2 ⎠ ⎦⎥ i =1 ⎝ ⎣⎢

= 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) /12 =168.6ft − kips

•

(

Point 3 fs = −0.5f y

)

Layer 1: 1 − C2

d1 = 1 − 1.35(1) = −0.35 d1

Layer 2: 1 − C2

d2 ⎛ 9.00 ⎞ = 1 − 1.35 ⎜ ⎟ = 0.22 d1 ⎝ 15.56 ⎠

Layer 3: 1 − C2

d3 ⎛ 2.44 ⎞ = 1 − 1.35 ⎜ ⎟ = 0.79 , use 0.69 d1 ⎝ 15.56 ⎠

2

n ⎡ ⎛ d ⎞⎤ φPn = φ ⎢C1d1b + 87∑ Asi ⎜1 − C2 i ⎟ ⎥ d1 ⎠ ⎦⎥ i =1 ⎝ ⎣⎢

= 0.65{(2.14 × 15.56 ×18) + 87[(3 × −(0.35)) + (2 × 0.22) + (3 × 0.69)]} = 0.65(599.4 + 127.0) = 474.9 kips n ⎡ ⎛ d ⎞⎛ h ⎞ ⎤ φM n = φ ⎢0.5C1d1b(h − C3d1 ) + 87∑ Asi ⎜1 − C2 i ⎟ ⎜ − di ⎟ /12 ⎥ d1 ⎠ ⎝ 2 ⎠ ⎥⎦ i =1 ⎢⎣ ⎝ = 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) /12 = 229.1ft − kips

•

(

Point 4 fs = −f y

)

Layer 1: 1 − C2

d1 = 1 − 1.69(1) = −0.69 d1

Layer 2: 1 − C2

d2 ⎛ 9.00 ⎞ = 1 − 1.69 ⎜ ⎟ = 0.02 d1 ⎝ 15.56 ⎠

Layer 3: 1 − C2

d3 ⎛ 2.44 ⎞ = 1 − 1.69 ⎜ ⎟ = 0.74 , use 0.69 d1 ⎝ 15.56 ⎠

n ⎡ ⎛ d ⎞⎤ φPn = φ ⎢C1d1b + 87∑ A si ⎜1 − C 2 i ⎟ ⎥ d1 ⎠ ⎥⎦ i =1 ⎢⎣ ⎝ = 0.65{(1.70 × 15.56 × 18) + 87[(3 × (−0.69)) + (2 × 0.02) + (3 × 0.69)]}

= 0.65(476.1 + 3.5) = 314.0 kips n ⎡ ⎛ d ⎞⎛ h ⎞ ⎤ φM n = φ ⎢ 0.5C1d1b(h − C3d1 ) + 87 ∑ A si ⎜ 1 − C 2 i ⎟ ⎜ − d i ⎟ /12 ⎥ d1 ⎠ ⎝ 2 ⎠ ⎦⎥ i =1 ⎝ ⎣⎢

= 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) /12 = 260ft − kips

•

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

3

⎛ c − d1 ⎞ εs1 = 0.003 ⎜ ⎟ ⎝ c ⎠ ⎛ 4 − 15.56 ⎞ = 0.003 ⎜ ⎟ 4 ⎝ ⎠ = −0.0087

fs1 = E s εs1 = 29000 × (−0.0087) = −251.4 ksi > − 60 ksi, use fs1 = −60 ksi Ts1 = As1fs1 = 3 × (−60) = −180 kips ⎛ c − d2 ⎞ εs2 = 0.003 ⎜ ⎟ ⎝ c ⎠ ⎛ 4−9 ⎞ = 0.003 ⎜ ⎟ ⎝ 4 ⎠ = −0.0038

fs2 = Es εs2 = 29000 × (−0.0038) = −108.8ksi > − 60 ksi, use f s1 = −60 ksi Ts2 = As2fs2 = 2 × (−60) = −120 kips

⎛ c − d3 ⎞ εs3 = 0.003 ⎜ ⎟ ⎝ c ⎠ ⎛ 4 − 2.44 ⎞ = 0.003 ⎜ ⎟ 4 ⎝ ⎠ = 0.0012 fs3 = Es εs3 = 29000 × (0.0012) = 33.9 ksi Cs3 = As3fs3 = 3 × 33.9 = 102 kips Cc = 0.85f c' ab = 0.85 × 4 × (0.85 × 4) × 18 = 208kips

Total T = (−180) + (−120) = −300 kips Total C = 102 + 208 = 310 kips Since

T ≈ C , use c = 4.0in

⎛h ⎞ ⎛ 18 ⎞ M ns1 = Ts1 ⎜ − d1 ⎟ = (−180) ⎜ − 15.56 ⎟ /12 2 2 ⎝ ⎠ ⎝ ⎠ = 98.4ft − kips ⎛h ⎞ ⎛ 18 ⎞ M ns2 = Ts2 ⎜ − d 2 ⎟ = (−120) ⎜ − 9 ⎟ /12 ⎝2 ⎠ ⎝ 2 ⎠ =0

4

⎛h ⎞ ⎛ 18 ⎞ M ns3 = Cs3 ⎜ − d 3 ⎟ = (102) ⎜ − 2.44 ⎟ /12 ⎝2 ⎠ ⎝ 2 ⎠ = 55.8ft − kips 3

M n = 0.5Cc ( h − a ) + ∑ M nsi i =1

= ⎡⎣0.5 × 208 × (18 − 3.4 ) ⎤⎦ /12 + 154.2 = 280.7 ft − kips φM n = 0.9 × 280.7 = 253ft − kips

Compare simplified interaction diagram to interaction diagram generated from StructurePoint’s 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

1: Pure compression 2: fs = 0

Parameter

Calculations (simplified) 808.3 690.9 168.6 474.9 229.1 314

pcaColumn (exact) 808.3 679.8 164.99 463.8 225.43 307.3

4: fs = −fy

φPn (kips) φPn (kips) φMn (ft-kips) φPn (kips) φMn (ft-kips) φPn (kips) φMn (ft-kips)

260

256.3

5: Pure bending

φMn (ft-kips)

253

247.72

3: fs = −0.5fy

5

Point 1 Point 2

Point 3

Point 4

Point 5

6

7