Jonathan Ochshorn - ARCH 264/564 Structural Elements Homework Assignment

1. Reinforced concrete columns: analysis
What is the maximum design (factored) load, P_u, that can be supported on the following reinforced concrete columns? Use f'_c = 5 ksi; and f_y = 60 ksi. Check reinforcement ratio and bar fit:

Set the design load, P_u = αφP_n; that is: P_u = αφ[0.85f'_c(A_g - A_st) + f_yA_st]

P_u = (0.80)(0.65)[0.85(5)(16x20 - 16) + 60x16] = 1171 kips

Check reinforcement ratio, ρ_g = 16 / (16x20) = 0.05 which is between 0.01 and 0.08 so it is OK.

Check bar fit: from Table A-4.11, need 12", have 16", so OK.

Set the design load, P_u = αφP_n; that is: P_u = αφ[0.85f'_c(A_g - A_st) + f_yA_st]

P_u = (0.85)(0.70)[0.85(5)(π10² - 12.65) + 60x12.65] = 1214 kips

Check reinforcement ratio, ρ_g = 12.65 / (π10²) = 0.04 which is between 0.01 and 0.08 so it is OK.

Check bar fit: from Table A-4.11, need 16" diameter, have 20", so OK.

Note on steel area: since 10 No. 10 bars is not listed directly in Table A-4.10, one can add two columns together, e.g., the area for 2 bars + 8 bars. Because of rounding errors, your total area may vary slightly from the number used above.

2. Reinforced concrete columns: design
plan and section

Select steel bars for a 16" x 16" first-floor tied column. Assume that the dead load consists only of 8"-thick floor and roof slabs (weighing 150 pcf); and that the live load (with no live load reduction needed) is 100 psf on all floors, but not the roof. Use f'_c = 4 ksi; and f_y = 60 ksi. Check reinforcement ratio and bar fit.

First, find the governing load on the column (since only D + L are considered here, we just use 1.2D + 1.6L):

Tributary area per floor = 25x30 = 750 sq.ft. so:

D = 150(8/12)(750 x 4) = 300,000# = 300 k
L = 100(750 x 3) = 225,000# = 225 k
P_u = 1.2D + 1.6L = 1.2(300) + 1.6(225) = 720 k

P_u ≤ αφP_n; that is: P_u ≤ αφ[0.85f'_c(A_g - A_st) + f_yA_st]

720 ≤ (0.80)(0.65)[0.85(4)(16x16 - A_st) + 60A_st]

A_st ≥ 9.08 in². Select, for example, 6 No. 11 bars with A_st = 9.36 in².

Check reinforcement ratio, ρ_g = 9.36 / (16x16) = 0.037 which is between 0.01 and 0.08 so it is OK.

Check bar fit: from Table A-4.11, need 14", have 16", so OK.

column cross-section

3. Reinforced concrete columns: design
Assuming a reinforcement ratio of about 0.04 (i.e., 4%), how big would a round spiral column be that must support a design (factored) load, P_u = 1000 kips*. Round diameter to the closest even inch. Use f'_c = 4 ksi; and f_y = 60 ksi. Do not find steel area or select bars.

* Note that "design" or "factored" loads, P_u, already include the load factors. That is, they already are in the form: 1.2D + 1.6L.

P_u ≤ αφP_n; that is: P_u ≤ αφ[0.85f'_c(A_conc) + f_yA_st]

But since the reinforcement ratio is assumed to be 0.04, we can replace A_conc with 0.96A_g; and replace A_st with 0.04A_g. Then, we get:

1000 ≤ (0.85)(0.70)[0.85(4)(0.96A_g) + 60(0.04A_g)]

Solving for A_g, we get:

A_g ≥ 508.68 in² = π(r²). Solving for the radius, r, we get:

r = 12.72 inches, so the required diameter = 12.72 x 2 = 25.44 inches. The closest even inch would be diameter = 26 inches.