The design dead and live loads for a column are shown in Table 7-2. The column tributary area is 600 ft2 at each level. Calculate the column factored axial force, Pu. Note that the code (ASCE), doesn't require that live loads from the roof and floor are not added together. Hence, the roof live load may be neglected for calculating the column axial force below the fifth floor. Table 7-2-Interior column loads Level Roof Fifth Fourth Third Second Average design load (psf) Dead Load 150 psf 200 200 200 250 Reduced Live Load 20 psf 20 20 20 U = 1.2D +1.6L 100 Story Dead Load 6oox DL book 2002 booxwoo - 600X150- 90,000 600x20 600X200 = 320,000 ・ho,000 Column load (kip) 600x250€ 150,000 Story Live Load book ALL 12000 12,000 12000 12p²² 60,000 Story Factored Load fo 127,200 211,200 4344.00 599,600 813,600 Sum of Factored Loads 121.20 211.20 434.40 597.60 873.60 8.11 Using the loads in Question 8.10, design a round column from the ground level to 3rd floor. The column should have a 2 diameter with spiral transverse reinforcement and eight vertical bers. Determine the required vertical bar sizes so that th column's capacity, Pn, is greater than the calculated factored force, Pu. Compassion capacity of columns with spiral reinforcement: fPn=0.851 [0.85 fc Ag+ (fy-0.85 fe') Ast] where: 1=0.75 fe'= 5.5 ksi, (concrete compressive strength) fy-60 ksi (steel yield strength) Ag = gross concrete area Ast area of vertical steel C₁ =1.50 in Iue-loft Ky 0.5 N= 8 Dour #14 Olay = 1.693 in Dstv # = 4 Osta=0.5 in G= 24/10 psi t₁ = 57000 Ay=px=314, 1592² Ast = N² px Day ²) - 18.009 1 ² Ast_min= 0.01 Ag = 3.142 in ² At max = 0.08 Ag= 25.133 Vert bars 5 5 (45)954/229233 psi #4 spral ~ 2010 A₁ = A$ ~ 9.005 in ² P₁ = 0.85 × (0.88 xf₂ ²x(Ag-Ast) +by 'Ast) = PP - 1591.471 kips Column Section Suxims: 21 in Ag = ph² = 314.1592²2 A5-p D²0-196² 63= 0.45 (Ag Ah) -1) (Pe/by) =0.016) S₂ = min as xpx (D₂ -Ostr) / 2.83 283521 SS Sutis use #14 bare (25x)
The design dead and live loads for a column are shown in Table 7-2. The column tributary area is 600 ft2 at each level. Calculate the column factored axial force, Pu. Note that the code (ASCE), doesn't require that live loads from the roof and floor are not added together. Hence, the roof live load may be neglected for calculating the column axial force below the fifth floor. Table 7-2-Interior column loads Level Roof Fifth Fourth Third Second Average design load (psf) Dead Load 150 psf 200 200 200 250 Reduced Live Load 20 psf 20 20 20 U = 1.2D +1.6L 100 Story Dead Load 6oox DL book 2002 booxwoo - 600X150- 90,000 600x20 600X200 = 320,000 ・ho,000 Column load (kip) 600x250€ 150,000 Story Live Load book ALL 12000 12,000 12000 12p²² 60,000 Story Factored Load fo 127,200 211,200 4344.00 599,600 813,600 Sum of Factored Loads 121.20 211.20 434.40 597.60 873.60 8.11 Using the loads in Question 8.10, design a round column from the ground level to 3rd floor. The column should have a 2 diameter with spiral transverse reinforcement and eight vertical bers. Determine the required vertical bar sizes so that th column's capacity, Pn, is greater than the calculated factored force, Pu. Compassion capacity of columns with spiral reinforcement: fPn=0.851 [0.85 fc Ag+ (fy-0.85 fe') Ast] where: 1=0.75 fe'= 5.5 ksi, (concrete compressive strength) fy-60 ksi (steel yield strength) Ag = gross concrete area Ast area of vertical steel C₁ =1.50 in Iue-loft Ky 0.5 N= 8 Dour #14 Olay = 1.693 in Dstv # = 4 Osta=0.5 in G= 24/10 psi t₁ = 57000 Ay=px=314, 1592² Ast = N² px Day ²) - 18.009 1 ² Ast_min= 0.01 Ag = 3.142 in ² At max = 0.08 Ag= 25.133 Vert bars 5 5 (45)954/229233 psi #4 spral ~ 2010 A₁ = A$ ~ 9.005 in ² P₁ = 0.85 × (0.88 xf₂ ²x(Ag-Ast) +by 'Ast) = PP - 1591.471 kips Column Section Suxims: 21 in Ag = ph² = 314.1592²2 A5-p D²0-196² 63= 0.45 (Ag Ah) -1) (Pe/by) =0.016) S₂ = min as xpx (D₂ -Ostr) / 2.83 283521 SS Sutis use #14 bare (25x)
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
Related questions
Question
![8.10 The design dead and live loads for a column are shown in Table 7-2. The column tributary area is 600 ft2 at each level.
Calculate the column factored axial force, Pu.
Note that the code (ASCE), doesn't require that live loads from the roof and floor are not added together.
Hence, the roof live load may be neglected for calculating the column axial force below the fifth floor.
Level
Roof
Fifth
Fourth
Third
Second
Average design load
(psf)
Dead
Load
150 psf
200
200
200
250
Reduced
Live Load
20 psf
20
20
20
U = 1.2D +1.6L
100
Table 7-2- Interior column loads
Story Dead
Load
болхо
600x150- 90,000
600 X200-320,000
600x 200%
ho
booxwoo -
.
120, a
600×250€ 150,000
Column load (kip)
Story Live
Load
boox RUL
booxzo
= 12000
12,000
12000
12p00
bo poo
Story
Factored
Load boxu
127,200
211,200
434400
599,600
873,600
Sum of
Factored
Loads
121.20
211.20
434.40
597.60
873.60
8.11 Using the loads in Question 8.10, design a round column from the ground level to 3rd floor. The column should have a 20
diameter with spiral transverse reinforcement and eight vertical bars. Determine the required vertical bar sizes so that th
column's capacity, Pn, is greater than the calculated factored force, Pu.
Compassion capacity of columns with spiral reinforcement:
fPn=0.85 f [0.85 fc' Ag+ (fy-0.85 fc' ) Ast] where:
f = 0.75
fc'= 5.5 ksi, (concrete compressive strength)
fy = 60 ksi (steel yield strength)
Ag = gross concrete area
Ast = area of vertical steel
C₂ =1.50 in
lux = loft
Kx- 0.5
N= 8
Dow #=14
Olay = 1.693 in
Dstv # = 4
Distur=0.5 in
G₁ = 29/100psi
E₁ = 57000 € 0.5
Ag = pxh= 2 = 314.159.1²2
Ast = N² px Day ² ) - 18.009 2²
Ast - min = 0.01 Ay = 3.142 in ²
Alma 0.08 Aga
Ast =
25.13)
= 4221233 psi
Vert.
bars
5 (15) ⁰.5=
#4 spiral
A₁ = A$ ~ 9.005 1₂ ²
P₁ = 0.85 × (0.85 % f ₂²x (Ag * - Ast) + by Ast)
= t'Pr = 11591.871 kips/
20'0
Column Section
Sutins=
2.1 in
P.
Ag = ph² = 314.1592²
95²-p D²0-196²
5+ ૭૫5 (5}}} {{/
S₂ = min as x px (D₂ - Dstr) /
-0.016
(15)
= 2.83
28327~
SS Suthes
use #14 bars](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71a4784b-5a2b-4a09-bc37-6f446bca3f80%2Fe7271e94-8524-4a95-b4aa-da032f8facf9%2Fpr4fef_processed.png&w=3840&q=75)
Transcribed Image Text:8.10 The design dead and live loads for a column are shown in Table 7-2. The column tributary area is 600 ft2 at each level.
Calculate the column factored axial force, Pu.
Note that the code (ASCE), doesn't require that live loads from the roof and floor are not added together.
Hence, the roof live load may be neglected for calculating the column axial force below the fifth floor.
Level
Roof
Fifth
Fourth
Third
Second
Average design load
(psf)
Dead
Load
150 psf
200
200
200
250
Reduced
Live Load
20 psf
20
20
20
U = 1.2D +1.6L
100
Table 7-2- Interior column loads
Story Dead
Load
болхо
600x150- 90,000
600 X200-320,000
600x 200%
ho
booxwoo -
.
120, a
600×250€ 150,000
Column load (kip)
Story Live
Load
boox RUL
booxzo
= 12000
12,000
12000
12p00
bo poo
Story
Factored
Load boxu
127,200
211,200
434400
599,600
873,600
Sum of
Factored
Loads
121.20
211.20
434.40
597.60
873.60
8.11 Using the loads in Question 8.10, design a round column from the ground level to 3rd floor. The column should have a 20
diameter with spiral transverse reinforcement and eight vertical bars. Determine the required vertical bar sizes so that th
column's capacity, Pn, is greater than the calculated factored force, Pu.
Compassion capacity of columns with spiral reinforcement:
fPn=0.85 f [0.85 fc' Ag+ (fy-0.85 fc' ) Ast] where:
f = 0.75
fc'= 5.5 ksi, (concrete compressive strength)
fy = 60 ksi (steel yield strength)
Ag = gross concrete area
Ast = area of vertical steel
C₂ =1.50 in
lux = loft
Kx- 0.5
N= 8
Dow #=14
Olay = 1.693 in
Dstv # = 4
Distur=0.5 in
G₁ = 29/100psi
E₁ = 57000 € 0.5
Ag = pxh= 2 = 314.159.1²2
Ast = N² px Day ² ) - 18.009 2²
Ast - min = 0.01 Ay = 3.142 in ²
Alma 0.08 Aga
Ast =
25.13)
= 4221233 psi
Vert.
bars
5 (15) ⁰.5=
#4 spiral
A₁ = A$ ~ 9.005 1₂ ²
P₁ = 0.85 × (0.85 % f ₂²x (Ag * - Ast) + by Ast)
= t'Pr = 11591.871 kips/
20'0
Column Section
Sutins=
2.1 in
P.
Ag = ph² = 314.1592²
95²-p D²0-196²
5+ ૭૫5 (5}}} {{/
S₂ = min as x px (D₂ - Dstr) /
-0.016
(15)
= 2.83
28327~
SS Suthes
use #14 bars
Transcribed Image Text:8.12 Assume that the round cross section and spiral reinforcement are used throughout the entire column height.
Modify the column design in Question 8.11 for the smaller forces above the third floor (that is, revise its diameter
and/or vertical reinforcement).
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