Assignment #3r
contact | homepage | spring 2008 index for ARCH 264/564 | current index for ARCH 264/564 | homework index | textIssued April 8, 2008.
Due: April 10, 2008 (not collected or graded)
For problems 1-3, use f'c = 4 ksi and fy = 60 ksi.
1. Reinforced concrete beam-columns: design of rectangular section with dimensions known
Find the required steel area and select 4 bars (preferable) for a 16" x 16" beam-column supporting a design load, Pu = 768 k; and a design moment, Mu = 136.53 ft-k. Assume 3" of concrete cover measured to the longitudinal steel centerline, and choose the closest value for γ. Check bar fit using "column" charts in Table A-4.11.
Use γ = 10/16 = 0.625 or approximately 0.6.
Pu / Ag = 768 / (16x16) = 3.0
Mu / (Ag x h) = 136.53 x 12 / (16x16x16) = 0.4
See Table A-9.1 to find ρg:
By finding the intersection of Pu / Ag and Mu / (Ag x h), we can find the reinforcement ratio, ρg = 0.06
As = ρg(Ag) = ρg(bxh) = 0.06(16 x 16) = 15.36 in2.
Use 4 No. 18 bars with As = 16 in2.
Check bar fit (Table A-4.11): With 2 bars per line, need 12"; have 16", so OK.
2. Reinforced concrete beam-columns: design of circular section with dimensions known
Find the required steel area and select 8 bars (preferable) for a 16"-diameter beam-column supporting a design load, Pu = 502.6 k; and a design moment, Mu = 107.23 ft-k. Assume 3" of concrete cover measured to the longitudinal steel centerline, and choose the closest value for γ. Check bar fit using "column" charts in Table A-4.11.
Use γ = 10/16 = 0.625 or approximately 0.6.
Pu / Ag = 502.6 / (π8x8) = 2.5
Mu / (Ag x h) = 107.23 x 12 / (π8x8x16) = 0.4
See Table A-9.2 to find ρg:
By finding the intersection of Pu / Ag and Mu / (Ag x h), we can find the reinforcement ratio, ρg = 0.05
As = ρg(Ag) = 0.05(π8x8) = 10.05 in2.
Use 8 No. 10 bars with As = 10.12 in2.
Check bar fit (Table A-4.11): With 8 bars per column, need 14"; have 16", so OK.
3. Reinforced concrete beam-columns: design of rectangular section with one dimension and reinforcement ratio known
Find the missing dimension, b, and the required steel area; and select 16 bars (preferable; may be split equally between 2 sizes) for a beam-column with dimension, h = 42", supporting a design load, Pu = 3,150 k; and a design moment, Mu = 1,764 ft-k. Use reinforcement ratio, ρg = 0.04. Assume 2-1/2" of concrete cover measured to the longitudinal steel centerline, and choose the closest value for γ. Check bar fit using "column" charts in Table A-4.11, based on a weighted average of the combined bar sizes used.
Use γ = 37/42 = 0.88 or approximately 0.9.
e = Mu / Pu = 1,764 x 12 / 3,150 = 6.72
e / h = 6.72 / 42 = 0.16
See Table A-9.1 with ρg = 0.04:
By finding the intersection of ρg and e/h, we can find Pu / Ag = 2.7
Solving for Ag, we get: Ag = 3,150 / 2.7 = 1,167 in2
We can now find the missing cross-section dimension, b = Ag / h = 1,167 / 42 = 27.8" or approximately 28"
As = ρg(bxh) = 0.04(28x42) = 47.04 in2.
Use 8 No. 18 bars + 8 No. 14 bars with As = 32 + 18 = 50 in2.
Check bar fit (Table A-4.11): With 5 bars per line (assume 3 No. 18 and 2 No. 14), we need approximately (3x30 + 2x24)/5 = 27.6" (weighted average based on 30" for No. 18 bars and 24" for No. 14 bars); have 28", so OK. [Note: this complex sort of approximate calculation will not be on exams.]
4. Reinforced concrete diagonal tension (shear)
For this problem only, assume f'c = 3 ksi and fy = 60 ksi. Use No. 3 stirrups.
A continuous beam with clear spans of 22 feet supports a uniformly-distributed load such that the maximum shear at the face of supports, Vu = 95 k. Consider one such "shear force triangle" defined by this maximum value of 95 k, and a length of one-half the span, or 132 inches. For cross-sectional dimensions of b = 14" and d = 27", find the following:
Minimum stirrup spacing.
Maximum stirrup spacing, and location (distance from beam centerline where this maximum spacing may begin)
Value of shear where no stirrups are needed and location (distance from beam centerline where "no stirrups" may begin)
Minimum spacing: Find Vu at distance, d, from face of supports. Using "similar triangles," Vu = 95(105/132) = 75.57 k.
Vc = 2(sq.rt.3000)(14x27) = 41,407.8# = 41.41 k.
Vs = (75.57 / 0.75) - 41.41 = 59.35 k.
spacing, s = 2(0.11)(60)(27) / 59.35 = 6".
Maximum spacing: Choose smaller of d/2 = 13.5" or 24" or 2(.11)(60,000) / (50 x 14) = 18.8". The maximum spacing is 13.5".
Vs = 2(0.11)(60)(27) / (13.5) = 26.4 k.
Vu = 0.75(41.41 + 26.4) = 50.86 k.
Use similar triangles to find location of maximum spacing: x = (50.86 / 95)(132) = 70.7".
Location where no stirrups are needed: No stirrups are needed when Vu = 0.5(0.75)(41.41) = 15.53 k
Use similar triangles to find location where stirrups are not needed: x = (15.53 / 95)(132) = 21.6".
5. Reinforced concrete diagonal tension (shear)
For this problem only, assume f'c = 5 ksi and fy = 60 ksi. Use No. 4 stirrups.
A simply-supported beam with a clear span of 30 feet supports a single concentrated design load at mid-span, Pu = 200 k. Neglecting the weight of the beam itself, find the spacing and distribution of web steel (stirrups). The cross-sectional dimensions are b = 16" and d = 29".
Minimum spacing: Vu = 200 / 2 = 100 k for the entire span of the beam. Since this is also the value of shear at a distance, d, from the face of supports, we use this value to determine the spacing.
Vc = 2(sq.rt.5000)(16x29) = 65,619# = 65.62 k.
Vs = (100 / 0.75) - 65.62 = 67.73 k.
The spacing, s = 2(0.20)(60)(29) / 67.735 = 10.28" or round down to 10".
Check maximum spacing: Maximum spacing is the smaller of d/2 = 14.5" or 24" or 2(.20)(60,000) / (50 x 16) = 30". The maximum spacing is 14.5", which doesn't apply since we have a "minimum" spacing of 10" throughout. Note that we check maximum spacing since it is possible that the maximum spacing would be smaller than the minimum spacing, in which case we would use the maximum spacing instead of the minimum spacing (think about it...).
With a clear span of 30 feet = 360", we would start the first No. 4 stirrup at s/2 = 5" from the face of support, then have 35 spaces @ 10", so that the last stirrup was also 5" from the face of the other support.
No web steel? There is no region where web steel is not needed because the shear force is constant for the entire span.
© 2008 Jonathan Ochshorn. First posted 07 April 2008. Last updated: 10 April 2008