Jonathan Ochshorn - ARCH 264/564 Structural Elements Homework Assignment

For problems 1-3, use f'_c = 5 ksi and f_y = 60 ksi. Assume a continuous reinforced concrete floor system consisting of T-beams with a clear span of 30-feet, and 1-way slabs spanning between the T-beams. The T-beams are spaced at 15-feet on center. The slab thickness, h = 8", and the portion of the T-beam below the slab has a height of 16" and a "web" width of 12". The slab cover, measured to the centerline of steel, is 1 inch; while the beam cover is 2-1/2". Use a live load = 100 psf, and a dead load consisting of slab and beam weight only (with concrete @ 150 pcf). See the typical plan and cross-section below.
section through T-beam

1. Slab design. Using "moment values" from Table A-8.7, and No. 3 bars, find the reinforcing bar spacing for both negative and positive moment. Use the "clear span" in computing moment values. Check if the spacing is within allowable limits. Also check that slab thickness is acceptable for deflection control (Table A-8.2). Do not interpolate in Table A-8.9; just use closest value for an approximate solution.

Design slab as 1-foot strip. Dead load per linear foot: 150(8/12) = 100#/ft. Live load per linear foot: 100#/ft.
sketch of slab showing weight

Design load, w_u = 1.2D + 1.6L = 1.2(100) + 1.6(100) = 280#/ft = 0.28 k/ft.

Negative moment.
Using moment values from Table A-8.7 (text), and using the clear span = 14 feet, M_u = (.28)(14²) / 11 = 4.989 ft-k = 59.87 in-k.

Then, using the design equation for reinforced concrete beams and slabs, i.e., M_u ≤ φbd²R, we get:

59.87 ≤ (0.9)(12)(7²)R; from which R = 0.1131.

From Table A-8.9, we find ρ = 0.00180 (approximately, without interpolating). This is no smaller than ρ_min. = 0.00180, so it is OK.

For slabs, we find the spacing directly, assuming a bar size: in this case, with No. 3 bars (A_s = 0.11 in²), we get: s = A_s / (ρ d) = 0.11 / (0.00180 x 7) = 8.73 inches. Round down to 8 inches.

Since s = 8" is not greater than either 18" or 3h = 24", it is acceptable. The answer is: Use No. 3 bars @ 8" o.c..

Positive moment.
Since we were already using ρ_min. for negative steel, and since the positive moment value is always smaller, we will be forced to use the same value of ρ_min. for positive steel. Therefore the answer will be the same: Use No. 3 bars @ 8" o.c.. However, if you wish to see the calculations, they begin as follows:

Using moment values from Table A-8.7 (text), and using the clear span = 14 feet, M_u = (.28)(14²) / 16 = 3.43 ft-k = 41.16 in-k.

Then, using the design equation for reinforced concrete beams and slabs, i.e., M_u ≤ φbd²R, we get:

41.16 ≤ (0.9)(12)(7²)R; from which R = 0.0778.

From Table A-8.9, the steel ratio corresponding to R = 0.0778 is too small; we must use the minimum steel ratio ρ_min. = 0.00180. From here on, the calculations follow those for negative moment.

2. T-beam design. Using "moment values" from Table A-8.7, select bars for both negative and positive moment. Use the "clear span" in computing moment values. Do not interpolate in Table A-8.9; just use closest value for an approximate solution. Check that steel ratio is within acceptable limits. Check bar fit (Table A-4.11). Check that T-beam thickness is acceptable for deflection control (Table A-8.2).

Dead load per linear foot: slab + "beam" = 150(8/12)(15) + 150(16/12)(12/12) = 1700#/ft. Live load per linear foot: 100(15) = 1500#/ft.
sketch of slab showing weight

Design load, w_u = 1.2D + 1.6L = 1.2(1700) + 1.6(1500) = 4440#/ft = 4.44 k/ft.

Negative moment.
Using moment values from Table A-8.7 (text), and using the clear span = 30 feet, M_u = (4.44)(30²) / 11 = 363.27 ft-k = 4359.27 in-k.

Then, using the design equation for reinforced concrete beams and slabs, i.e., M_u ≤ φbd²R, we get:

4359.27 ≤ (0.9)(12)(21.5²)R; from which R = 0.8732.

From Table A-8.9, we find ρ = 0.01700 (approximately, without interpolating: 0.01600 is also acceptable in this context). This is no smaller than ρ_min. = 0.00333, so it is OK.

Find required steel area, A_s = ρbd = 0.01700(12)(21.5) = 4.39 in².

From Table A-4.10, select 3 No. 11 bars with A_s = 4.68 in². Check bar fit (Table A-4.11): need 12", have 12" so OK.

Positive moment.
Using moment values from Table A-8.7 (text), and using the clear span = 30 feet, M_u = (4.44)(30²) / 16 = 249.75 ft-k = 2997 in-k.

Before going much further, we need to find the effective width, b, for a positive-moment T-beam. Based on the criteria listed in Table A-8.9, b is the smaller of:

1/4 beam span = 30 x 12 / 4 = 90"
centerline beam spacing = 15 x 12 = 180"
b_w + 16h = 12 + 16(8) = 140"

The effective width, b = 90 inches.

Then, using the design equation for reinforced concrete beams and slabs, i.e., M_u ≤ φbd²R, we get:

2997 ≤ (0.9)(90)(21.5²)R; from which R = R = 0.0800.

From Table A-8.9, we find ρ = 0.00167 (approximately, without interpolating). This is no smaller than ρ_min. for the T-beam, which is found by computing b / b_w = 90 / 12 = 7.5. From Table A-8.9 (right hand column), it can be seen that the proposed value for ρ = 0.00167 is not smaller than the minimum steel ratio corresponding to b / b_w = 7.5, so it is OK.

Find required steel area, A_s = ρbd = 0.00167(90)(21.5) = 3.23 in².

From Table A-4.10, select 3 No. 10 bars with A_s = 3.79 in². Check bar fit (Table A-4.11): need 12", have 12" so OK.

3. Sketch. Sketch the T-beam section for negative and positive moment. Include reinforcing bars and important dimensions.

4. Rectangular beam design. Using f'_c = 3 ksi and f_y = 60 ksi, and assuming a design moment, M_u = 235 ft-k, find the effective depth of a rectangular beam of width, b = 16", by assuming a steel ratio of 0.00667 (that is, approximately half the maximum steel ratio). Round the answer to the closest inch. It is not necessary to find the steel area or select bars.