3. Estimated beam weight. The weight of the beam to be selected must be included in the calculation of the bending moment to be resisted, because the beam must support itself as well
4.10 Computer Example
4.10 Computer Example
Example 4.8
Repeat Example 4.4 using the Excel spreadsheet for Chapter 4.
S O L U T I O N
Use the worksheet called Beam Design. Enter material properties (fc, fy) and Mu (can be taken from the bottom part of the spreadsheet or just entered if you already know it). Inputρ = 0.009 (given in the example). The two tables with headings b and d give some choices for b and d based on theρ value you picked. Larger assumed values of ρ result in smaller values of b and d and vice versa. Select b = 18 in. and d = 31 in. (many other choices are also correct).
Add 2.5 in. or more to d to get h, and enter that value (used only to find beam weight below).
The spreadsheet recalculatesρ and Asfrom actual values of b and d chosen, so note thatρ is not the same as originally assumed (0.00895 instead of 0.009). This results in a slightly smaller calculated steel area than in Example 4.4. You can also enter the number of bars and size to get a value for As. This value must exceed the theoretical value or an error message will appear. You should check to see if this bar selection will fit within the width selected.
At the bottom of the spreadsheet, the design moment Mu can be obtained if the beam is simply supported and uniformly loaded with only dead and live loads. The beam self-weight is calculated based on the input values for b and h (Cells D23 and D25). You may have to iterate a few times before these values all agree. In this example, the dead load is 2 klf plus self-weight.
The input value for wDis 2.0 + 0.65 plf, with the second term being taken from the spreadsheet.
In working this problem the first time, you probably would not have these dimensions for b and h, hence the self-weight would not be correct. Iteration as done in Example 4.4 is also required with the spreadsheet, although it is much faster.
Design of singly reinforced rectangular beams
Instructions: Enter values only in cells that are highlighted in yellow. Other values calculated from those input values.
f 'c = fy =
Mu =
Mu
bd2 = = 17,215 in.3 b1 =
3 60 0.85 623.4
ksi ksi
ft-k
ffyr1 – rfy
1.7f'c Assume r = 0.009
These tables give some choices for b and d that you may round up to enter here.
R
R = d⏐R b d b d
1 25.82 25.82 14 35.07
1.2 22.86 27.44 15 33.88
1.4 20.63 28.88 16 32.80
1.5 19.70 29.56 17 31.82
1.6 18.87 30.20 18 30.93
1.7 18.13 30.82 19 30.10
1.8 17.45 31.41 20 29.34
1.9 16.83 31.98 21 28.63
2
<-- theoretical steel area
—
Calculation of Mu for simply supported beam with D and L uniformly distributed loads
2.65
Problem 4.1 The estimated service or working axial loads and bending moments for a particular column are as follows:
PD = 100 k, PL = 40 k, MD = 30 ft-k, and ML = 16 ft-k.
Compute the axial load and moment values that must be used in the design. (Ans. Pu = 184 k, Mu = 61.6 ft-k)
Problem 4.2 Determine the required design strength for a column for which PD = 120 k, PL = 40 k, and wind PW = 60 k compression or 80 k tension.
Problem 4.3 A reinforced concrete slab must support a dead working floor load of 80 psf, which includes the weight of the concrete slab and a live working load of 40 psf. Determine the factored uniform load for which the slab must be designed.
(Ans. wu = 160 psf)
Problem 4.4 Using the Chapter 4 spreadsheet, Load Combination worksheet, repeat the following problems:
(a) Problem 4.1 (b) Problem 4.2 (c) Problem 4.3
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For Problems 4.5 to 4.9, design rectangular sections for the beams, loads, and values given. Beam weights are not included in the given loads. Show sketches of beam cross sections, including bar sizes, arrangement, and spacing. Assume concrete weighs 150 lb/ft3. Use h = d + 2.5 in.
wD and wL
Problem
No. fy(psi) fc(psi) Spanl (ft) wDnot incl. beam wt (k/ft) wL (k/ft) ρ *
4.5 60,000 4000 30 2 1 0.18 fc/fy
4.6 60,000 4000 30 2 2 0.18 fc/fy
4.7 50,000 3000 18 3 4 12ρb
4.8 60,000 4000 32 2 1.8 12ρb
4.9 60,000 3000 25 1.8 1.5 t = 0.0075
*See Appendix A, Table A.7 forρ values that correspond to the tvalues listed.
One ans. Problem 4.5: 16 in.× 29 in. with 4 #10 bars.
One ans. Problem 4.7: 16 in.× 28 in. with 4 #11 bars.
One ans. Problem 4.9: 18 in.× 26 in. with 6 #8 bars.
For Problems 4.10 to 4.22, design rectangular sections for the beams, loads, andρ values shown. Beam weights are not included in the loads shown. Show sketches of cross sections, including bar sizes, arrangement, and spacing. Assume concrete weighs 150 lb/ft3, fy = 60,000 psi, and fc= 4000 psi, unless given otherwise.
Problem 4.10
PL = 30 k
wD= 3 k/ft
12 ft 12 ft
24 ft Use r = 0.18 f′c
fy
Problem 4.11 Repeat Problem 4.10, if wD = 2 k/ft and if PL = 20 k. (One ans. 14 in. × 28 in. with 3 #11 bars)
Problem 4.12
P = 20 kL P = 20 k
= 1.5 k/ft
L
wD
Use r =
10 ft 10 ft 10 ft
30 ft
0.18 f′c
fy
Problem 4.13 Repeat Problem 4.12 if wD = 2.0 k/ft and PL = 20 k. (One ans. 16 in. × 33 in. with 4 #11 bars)
Problem 4.14
PL = 36 k
wD= 2 k/ft
20 ft 30 ft Use = 12 10 ft
rb
r
Problem 4.15 Repeat Problem 4.14 if wD = 3 k/ft, PL = 40 k, fc= 3000psi, and ρ = 0.5ρb. (One ans.
18 in.× 37 in. with 5 #11 bars) Problem 4.16
14 ft
wD = 3 k/ft, wL = 2 k/ft
Use ρ = 0.18 f ' fy
c
Problem 4.17 Repeat Problem 4.16 if the beam span = 12 ft.
(One ans. 14 in.× 31 in. with 4 #10 bars in top) Problem 4.18
PL = 30 k
wD = 2 k/ft
Problem 4.19 Repeat Problem 4.18 if PL = 20 k, = 12 ft, andρ = 12ρb. (One ans. 20 in.× 26 in. with 7 #9 in top)
Problem 4.20
8 ft 8 ft
16 ft
PL = 30 k PL = 20 k wD = 2 k/ft
Use = ρ 12ρmax
Problem 4.21 Select reinforcing bars for the beam shown if Mu = 250 ft-k, fy = 60,000 psi, and fc= 4000 psi. (Hint:
Assume that the distance from the c.g. of the tensile steel to the c.g. of the compression block equals 0.9 times the effective depth, d, of the beam.) After a steel area is computed, check the assumed distance and revise the steel area if necessary. Is
t≥ 0.005? (Ans. As = 2.84 in.2,t = 0.00538 > 0.005)
15 in.
5 in. 5 in. 5 in.
6 in.
18 in.
As
Problem 4.22 Repeat Problem 4.21 for Mu = 150 ft-k.
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For Problems 4.23 and 4.24, design rectangular sections for the beams and loads shown. Beam weights are not included in the given loads. fy = 60,000 psi and fc= 4000 psi. Live loads are to be placed where they will cause the most severe conditions at the sections being considered. Select beam size for the largest moment (positive or negative), and then select the steel required for maximum positive and negative moment. Finally, sketch the beam and show approximate bar locations.
Problem 4.23 (One ans. 12 in.× 28 in. with 3 #10 bars negative reinforcement and 3 #9 bars positive reinforcement)
9 ft
For Problems 4.25 and 4.26, design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft3, fy = 60,000 psi, and fc= 4000psi. Do not use the ACI Code’s mini-mum thickness for deflections (Table 4.1). Steel percentages are given in the figures. The only dead load is the weight of the slab.
Problem 4.25 (One ans. 7.5-in. slab with #8 @ 9 in. main reinf.)
Problem 4.27 Repeat Problem 4.25 using the ACI Code’s minimum thickness requirement for cases where deflections are not computed (Table 4.1). Do not use theρ given in Problem 4.26. (Ans. 14.5-in. slab with #6 @ 9 in. main reinf.) Problem 4.28 Using fc= 3000 psi, fy = 60,000 psi, and ρ corresponding tot = 0.005, determine the depth required for a simple beam to support itself for a 200-ft simple span.
Problem 4.29 Determine the depth required for a beam to support itself only for a 100-ft span. Neglect concrete cover in self-weight calculations. Given fc= 4000psi, fy = 60,000 psi, andρ ∼= 0.5ρb. (Ans. d = 32.5 in.)
Problem 4.30 Determine the stem thickness for maximum moment for the retaining wall shown in the accompanying illustration. Also, determine the steel area required at the bottom and mid-depth of the stem if fc= 4000 psi and fy = 60,000 psi.
Assume that #8 bars are to be used and that the stem thickness is constant for the 18-ft height. Also, assume that the clear cover required is 2 in. andρ = 0.5ρb.
18 ft
500 psf = asssumed lateral liquid pressure stem
Problem 4.31
(a) Design a 24-in.-wide precast concrete slab to support a 60-psf live load for a simple span of 15 ft. Assume minimum concrete cover required is 58 in. as per Section 7.7.3 of the code. Use welded wire fabric for reinforcing.
fy = 60,000 psi, fc= 3000psi, and ρ = 0.18fc/fy. (Ans. 4-in. slab with 4× 8 D12/D6)
(b) Can a 300-lb football tackle walk across the center of the span when the other live load is not present? Assume 100%
impact. (Ans. yes)
Problem 4.32 Prepare a flow chart for the design of tensilely reinforced rectangular beams.
Problem 4.33 Using the Chapter 4 spreadsheets, solve the following problems.
(a) Problem 4.6. (Ans. 16 in.× 33 in. with 5 #10 bars) (b) Problem 4.18. (Ans. 18 in.× 39 in. with 8 #10 bars)
Problems in SI Units
For Problems 4.34 to 4.39, design rectangular sections for the beams, loads, andρ values shown. Beam weights are not included in the loads given. Show sketches of cross sections including bar sizes, arrangements, and spacing. Assume con-crete weighs 23.5 kN/m3. fy = 420 MPa and fc= 28 MPa.
Problem 4.34
wD = 20 kN/m wL = 12 kN/m
10 m
ρb
ρ =12
Problem 4.35 (One ans. 450 mm× 890 mm with 6 #32 bars)
wD= 25 kN/m PL= 100 kN
12 m
6 m 6 m ρ =12ρb
Problem 4.36
wD = 26 kN/m wL = 20 kN/m
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Problems 111
Problem 4.37 Place live loads to cause maximum positive and negative moments.ρ = 0.18 fc/fy. (One ans.
450 mm× 900 mm with 6 #32 bars positive reinf.) wD= 30 kN/m, wL = 20 kN/m
3 m 12 m 3 m
Problem 4.38
Problem 4.39 Design the one-way slab shown in the accompanying figure to support a live load of 12 kN/m2. Do not use the ACI thickness limitation for deflections.
Assume concrete weighs 23.5 kN/m3. fc= 28 MPa and fy = 420 MPa. Use ρ = ρmax. (One ans. 240-mm slab with #25 @ 140-mm main steel)