For the cross section shown below, the material and sectional properties are given as follows:   A) Determine the maximum steel area A_s that can lead to tension driven failure. One can use Table A.4 in the textbook to easily calculate design parameters. B) To ensure under-reinforcement, the steel area is to be decided as a value between 90% and 95% of estimated in P1(a). Determine the reinforcing bars (i.e., bar size, and the number of bars) to meet the required steel area. It is assumed that a single type of rebar is chosen (i.e., rebars of different sizes shall not be combined). C) Given the decided steel area, calculate the nominal and the design moment capacity. One cannot use the design aids.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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For the cross section shown below, the material and sectional properties are given as follows:

 

A) Determine the maximum steel area A_s that can lead to tension driven failure. One can use Table A.4 in the textbook to easily calculate design parameters.

B) To ensure under-reinforcement, the steel area is to be decided as a value between 90% and 95% of estimated in P1(a). Determine the reinforcing bars (i.e., bar size, and the number of bars) to meet the required steel area. It is assumed that a single type of rebar is chosen (i.e., rebars of different sizes shall not be combined).

C) Given the decided steel area, calculate the nominal and the design moment capacity. One cannot use the design aids.

1. For the cross section shown below, the material and sectional properties are given as follows:
b
T
d ¹2
in.
=
typ
T
fé = 4ksi, fy = 60 ksi, b = 30 in,
E = 29000000-psi, Ec
57000√f. Ey
d = 20 in,
= 0.002
concrete cover depth 1.5in.
(a) Determine the maximum steel area A, that can lead to tension driven failure. One can use
Table A.4 in the textbook to easily calculate design parameters.
(b) To ensure under-reinforcement, the steel area is to be decided as a value between 90% and
95% of A, estimated in P1(a). Determine the reinforcing bars (i.e., bar size, and the number of
bars) to meet the required steel area. It is assumed that a single type of rebar is chosen (i.e.,
rebars of different sizes shall not be combined).
(c) Given the decided steel area, calculate the nominal and the design moment capacity. One
cannot use the design aids.
Transcribed Image Text:1. For the cross section shown below, the material and sectional properties are given as follows: b T d ¹2 in. = typ T fé = 4ksi, fy = 60 ksi, b = 30 in, E = 29000000-psi, Ec 57000√f. Ey d = 20 in, = 0.002 concrete cover depth 1.5in. (a) Determine the maximum steel area A, that can lead to tension driven failure. One can use Table A.4 in the textbook to easily calculate design parameters. (b) To ensure under-reinforcement, the steel area is to be decided as a value between 90% and 95% of A, estimated in P1(a). Determine the reinforcing bars (i.e., bar size, and the number of bars) to meet the required steel area. It is assumed that a single type of rebar is chosen (i.e., rebars of different sizes shall not be combined). (c) Given the decided steel area, calculate the nominal and the design moment capacity. One cannot use the design aids.
TABLE A.4
Limiting steel reinforcement ratios for tension-controlled members
fy, psi
40,000
60,000
80,000
100,000
fé
Eu
fy Eu+Ey+0.003
Calculated using &, = 0.002
"p=0.85 p₁-
be
fc, psi
3000
4000
5000
6000
7000
8000
9000
10000
3000
4000
5000
6000
7000
8000
9000
10000
3000
4000
5000
6000
7000
8000
9000
10000
3000
4000
5000
6000
7000
8000
9000
10000
B₁
0.85
0.85
0.80
0.75
0.70
0.65
0.65
0.65
0.85
0.85
0.80
0.75
0.70
0.65
0.65
0.65
0.85
0.85
0.80
0.75
0.70
0.65
0.65
0.65
0.85
0.85
0.80
0.75
0.70
0.65
0.65
0.65
Pmax
a
0.0220
0.0294
0.0346
0.0389
0.0423
0.0449
0.0505
0.0562
0.0135
0.0181
0.0213
0.0239
0.0260
0.0276
0.0311
0.0345
0.0093
0.0124
0.0146
0.0164
0.0178
0.0189
0.0213
0.0237
0.0069
0.0092
0.0108
0.0121
0.0132
0.0140
0.0158
0.0175
Pmin =
200
fy
0.0050
0.0050
0.0050
0.0050
0.0050
0.0050
0.0050
0.0050
0.0033
0.0033
0.0033
0.0033
0.0033
0.0033
0.0033
0.0033
0.0025
0.0025
0.0025
0.0025
0.0025
0.0025
0.0025
0.0025
0.0020
0.0020
0.0020
0.0020
0.0020
0.0020
0.0020
0.0020
Pmin =
3√√fc
fy
0.0041
0.0047
0.0053
0.0058
0.0063
0.0067
0.0071
0.0075
0.0027
0.0032
0.0035
0.0039
0.0042
0.0045
0.0047
0.0050
0.0021
0.0024
0.0027
0.0029
0.0031
0.0034
0.0036
0.0038
0.0016
0.0019
0.0021
0.0023
0.0025
0.0027
0.0028
0.0030
Transcribed Image Text:TABLE A.4 Limiting steel reinforcement ratios for tension-controlled members fy, psi 40,000 60,000 80,000 100,000 fé Eu fy Eu+Ey+0.003 Calculated using &, = 0.002 "p=0.85 p₁- be fc, psi 3000 4000 5000 6000 7000 8000 9000 10000 3000 4000 5000 6000 7000 8000 9000 10000 3000 4000 5000 6000 7000 8000 9000 10000 3000 4000 5000 6000 7000 8000 9000 10000 B₁ 0.85 0.85 0.80 0.75 0.70 0.65 0.65 0.65 0.85 0.85 0.80 0.75 0.70 0.65 0.65 0.65 0.85 0.85 0.80 0.75 0.70 0.65 0.65 0.65 0.85 0.85 0.80 0.75 0.70 0.65 0.65 0.65 Pmax a 0.0220 0.0294 0.0346 0.0389 0.0423 0.0449 0.0505 0.0562 0.0135 0.0181 0.0213 0.0239 0.0260 0.0276 0.0311 0.0345 0.0093 0.0124 0.0146 0.0164 0.0178 0.0189 0.0213 0.0237 0.0069 0.0092 0.0108 0.0121 0.0132 0.0140 0.0158 0.0175 Pmin = 200 fy 0.0050 0.0050 0.0050 0.0050 0.0050 0.0050 0.0050 0.0050 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0033 0.0025 0.0025 0.0025 0.0025 0.0025 0.0025 0.0025 0.0025 0.0020 0.0020 0.0020 0.0020 0.0020 0.0020 0.0020 0.0020 Pmin = 3√√fc fy 0.0041 0.0047 0.0053 0.0058 0.0063 0.0067 0.0071 0.0075 0.0027 0.0032 0.0035 0.0039 0.0042 0.0045 0.0047 0.0050 0.0021 0.0024 0.0027 0.0029 0.0031 0.0034 0.0036 0.0038 0.0016 0.0019 0.0021 0.0023 0.0025 0.0027 0.0028 0.0030
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