Default Example 1

Transcription

Default Example 1

Job Number
Job Title
Software Consultants (Pty) Ltd
Client
Internet: http://www.prokon.com
Calcs by
E-Mail : [email protected]
Sheet
Checked by
Default Example 1
C13
General column design by PROKON. (GenCol Ver W2.5.07 - 21 Sep 2011)
Design code : ACI 318 - 2005
Input tables
General design parameters:
Code
X/Radius or
Bar dia. ( mm )
Y (mm )
Angle (?)
+
500
500
-250
-250
-250
250
-250
-500
-
125
c
55
+
-200
b
16
+
-50
b
16
+
300
b
16
+
450
b
16
+
-200
b
20
+
-50
b
20
+
300
b
20
+
450
b
20
+
-200
b
25
+
-50
b
25
+
300
b
25
+
450
b
25
Date
125
50
50
50
50
250
250
250
250
450
450
450
450
Job Number
Sheet
Job Title
Software Consultants (Pty) Ltd Client
Calcs by
Checked by
Date
Design loads:
Load
case
Ultimate limit state design loads
Designation
P (kN )
Mx top (kNm )
1
DL
2400
234
2
DL+LL
3200
431
My top (kNm )
Mx bot (kNm )
My bot (kNm )
-341
24
-13
Code specific parameters:
?d - see Clause 10.11.1
0
% of moments that are as a
result of sway - X direction
0
result of sway - Y direction
0
ds - see Clause 10.13.4.3
1
ACI 318 - 2005
Y
Given:
Lo = 7.000 m
fc' = 30 MPa
fy = 450 MPa
Ac = 298972 mm?
500
X
250
X
500
Y
250
0
-250
0
Assumptions:
(2) The specified design axial loads include
the self-weight of the column.
(3) The design axial loads are taken constant
over the height of the column.
Design approach:
The column is designed using an iterative procedure:
(1) An area of reinforcement is chosen.
(2) The column design charts are constructed.
(3) The corresponding slenderness moments are calculated.
(4) The design axis and design ultimate moment are determined .
(5) The design axial force and moment capacity is checked on
the relevant design chart.
(6) The safety factor is calculated for this load case.
(7) The procedure is repeated for each load case.
(8) The critical load case is identified as the case yielding the lowest
safety factor about the design axis
Through inspection:
Load case 2 (DL+LL) is critical.
Check column slenderness:
End fixity and bracing for bending about the Design axis:
At the top end: Condition 2 (partially fixed).
At the bottom end: Condition 2 (partially fixed).
The column is braced.
Effective length factor ? = 0.85
kLu = ? . Lo
Effective column height:
Table 3.19
Job Number
Job Title
Client
Calcs by
Sheet
Checked by
Date
Column slenderness about weakest axis:
k Lu/r =
=
k Lu
r
5.95
.14268
= 41.702
Minimum Moments for Design:
Check for mininum eccentricity:
Check that the eccentricity exceeds the minimum in the plane of bending:
10.12.3.2
eminx= 0.6 . 0.0254 + 0.03 . h
= 0.6 ×0.0254 + 0.03 ×.75
= 0.0377 m
10.12.3.2
eminy= 0.6 . 0.0254 + 0.03 . b
= 0.6 ×0.0254 + 0.03 ×.5
= 0.0302 m
Mminx= eminx. N
= .03024 ×3200
= 96.768 kNm
Check if the column is slender:
kLu x/r = 41.7 > 34 - 12
? M1/M2 = 34.6
10.12.2
10.12.2
Note that 34 - 12? M1/M2 is limited to a maximum value of 40.0
\ The column is slender.
Check maximum slenderness limit:
kLu /r = 41.7 < 100
\ Maximum Slenderness limit not exceeded.
10.11.5
Job Number
Job Title
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Checked by
Date
Initial moments:
The column is bent in double curvature about the X-X axis:
M1 = Smaller initial end moment = -24.0 kNm
M2 = Larger initial end moment = 431.0 kNm
The column is bent in double curvature about the Y-Y axis:
Moment Magnification:
For bending about the weakest axis:
The bending moments M1 and M2 about the X-X and Y-Y axes are transformed to
moments about the weakest axis and summed to give resultant moments about the
weakest axis. Moment magnification is now performed and the slenderness induced moments are transformed back to the X-X and Y-Y axes.
Weakest axis lies at an angle of -180.00? to the X-X axis
EsIs = Es. Ist
= 210 ×123.46
= 25.93×103 kNm?
Ec. Ig
+ Es. Ist
5
EI =
1 + ?d
25.924 ×6 086.7
+ 210 ×123.46
5
=
1+0
= 57.48×103 kNm?
Pc =
=
p2 . EI
k Lu2
p2 ×57484
5.95 2
= 16.03×103 kN
Pu = 3200.0 kN
f = 0.650
Where no end moment or pin exists, the minimum moment emin? N = 96.8 kNm is used.
Moments at Top of column:
10.12
Job Number
Job Title
Client
Calcs by
Sheet
Checked by
M1b
= cos(-180.00)*431.00 + sin(-180.00)*-341.00) = 96.77
M1s
= cos(-180.00)*0.00 + sin(-180.00)*-0.00) = 0.000
Date
Moments at Middle of column:
Mb
= cos(-180.00)*203.50 + sin(-180.00)*-164.00) = 96.77
Ms
= cos(-180.00)*0.00 + sin(-180.00)*0.00) = 0.000
Moments at Bottom of column:
M2b
= cos(-180.00)*96.77 + sin(-180.00)*13.00) = 96.77
M2s
= cos(-180.00)*-0.00 + sin(-180.00)*0.00) = 0.000
Cm = 0.6 -
0.4 . M1
M2
= 0.6 -
0.4 ×-24
431
= 0.6223
But
10.12.3.1
Cm > = 0.4
Cm is based on the assumption that the maximum moment occurs
at mid height - see commentary on clause 10.12.3.1 ACI code.
For checking top and bottom of the column, C m is taken as 1.
db =
=
Cm
Pu
10.75 . Pc
1
3200×103
10.75 ×1603×104
= 1.363
With the proviso that
(10-9)
db > = 1.0
1
db(top&bot)=
1=
Pu
0.75 . Pc
1
3200×103
10.75 ×1603×104
= 1.363
Job Number
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Calcs by
With the proviso that
Sheet
Checked by
Date
(10-9)
db > = 1.0
ds = 1.000
(10-9)
Additional moment at Top of column:
Madd = (db. Mb + ds. Ms) - (Mb + Ms)
= (1.3628 ×431 + 1 ×0 ) - (431 + 0 )
= 156.367 kNm
\ Maddx = Madd*cos(-180.00?) = -35.1 kNm
\ Maddy = Madd*sin(-180.00?) = 0.0 kNm
Additional moment at Middle of column:
= (1.3628 ×- 167.12 + 1 ×0 ) - (- 167.12 + 0 )
= -60.6311 kNm
Additional moment at Bottom of column:
= (1.3628 ×96.768 + 1 ×0 ) - (96.768 + 0 )
= 35.107 kNm
Design ultimate load and moment:
Design axial load:
Pu = 3200.0 kN
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
For bending about the X-X axis, the maximum design moment is the greatest of:
(a)
10.12.3
Mtopx = Mtop + Madd
= -24 + 35.112
= 11.112 kNm
(b)
Mbotx = Mbot - Madd
= 431 - 35.112
= 395.888 kNm
Moments about X-X axis( kNm)
+
Mxbot=24.0 kNm
Initial
Mx=466.1 kNm
Mxmin=96.8 kNm
Mxadd=-35.1 kNm
Mxadd=-35.1 kNm
Mxtop=431.0 kNm
=
Mxadd=-35.1 kNm
Additional
For bending about the Y-Y axis, the maximum design moment is the greatest of:
(a)
Mtopy = Mtop + Madd
= 13 + 0
= 13.000 kNm
(b)
Mboty = Mbot - Madd
= -341 - 0
= - 341.0000 kNm
Design
10.12.3
Job Number
Sheet
Job Title
Calcs by
Checked by
Moments about Y-Y axis( kNm)
Date
My=341.0 kNm
Mymin=96.8 kNm
Mytop=-341.0 kNm
+
=
Mybot=-13.0 kNm
Initial
Additional
Design
Design of column section for ULS:
The column is checked for applied moment about the design axis.
Through inspection: the critical section lies at the top end of the column.
The design axis for the critical load case 2 lies at an angle of 323.81? to the X-axis
The safety factor for the critical load case 2 is 0.88
For bending about the design axis:
Interaction Diagram
Moment max = 546.6kNm @ 1228kN
5000
4500
4000
3500
3200 kN
3000
Axial load (kN)
2500
2000
1500
-500
-1000
-1500
Bending moment (kNm)
Warning: The safety factor is < 1
Moment distribution along the height of the column for bending about the design axis:
The final design moments were calculated as the vector sum of the X- and Y- moments
of the critical load case. This also determined the design axis direction
At the top, Mx = 577.5 kNm
Near mid-height, Mx = 292.4 kNm
At the bottom, Mx = 96.8 kNm
600
578 kNm
500
400
300
200
100
0.00
-100
-200
-300
-400
-500
500
-600
1000
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
Stresses at the top end of the column for the critical load case 2
ACI 318 - 2005
D
Y
500
323.8?
X
250
X
500
250
Y
0
-250
0
D
Summary of design calculations:
Design table for critical load case:
Moments and Reinforcement for LC
Top
2:DL+LL
Middle
Bottom
Madd-x
(kNm) -35.1
-35.1
35.1
Madd-y
(kNm) 0.0
0.0
-0.0
Mx
(kNm) 466.1
213.9
59.1
My
(kNm) -341.0
-199.4
-13.0
M'
(kNm) 577.5
292.4
96.8
Design axis (?)
323.81
317.01
167.60
Safety factor
0.88
1.41
1.52
(mm?) 4024
4024
4024
1.33 %
1.33 %
2990
2990
Asc
Percentage
AsNom
1.33 %
(mm?) 2990
Critical load case: LC 2
Design results for all load cases:
Load case
Axis
N (kN )
Load case 1
DL
X-X
Y-Y
2400.0
M1 (kNm )
0.0
0.0
M2 (kNm )
234.0
0.0
Mi (kNm )
140.4
0.0
Madd (kNm )
-18.1
0.0
Design
Top
M (kNm )
252.1
0.0
M' (kNm )
Safety
factor
252.1
1.630
Job Number
Sheet
Job Title
Calcs by
Load case
Axis
N (kN )
Load case 2
DL +LL
X-X
Y-Y
3200.0
M1 (kNm )
-24.0
13.0
Checked by
M2 (kNm )
Mi (kNm )
431.0
-341.0
249.0
-199.4
Madd (kNm )
-35.1
0.0
Date
Design
Top
M (kNm )
466.1
341.0
M' (kNm )
Safety
factor
577.5
0.885
Job Number
Job Title
Client
Calcs by
Sheet
Checked by
Default Example 2
C13
Design code : BS8110 - 1997
Input tables
Code
X/Radius or
Bar dia. ( mm )
Y (mm )
Angle (?)
+
500
A
500
A
500
-90
+50
-90
-500
-
500
C
100
+
50
B
32
+
250
B
32
+
450
B
32
+
600
B
32
+
725
B
32
+
890
B
32
+
950
B
32
+
50
B
32
+
250
B
32
+
450
B
32
+
600
B
32
+
725
B
32
+
890
B
32
+
50
B
32
+
50
B
32
+
50
B
32
Date
550
50
50
50
50
110
275
500
1000
1000
1000
1000
940
775
525
275
775
Job Number
Sheet
Job Title
Calcs by
Checked by
Date
Design loads:
Load
case
Designation
P (kN )
Mx top (kNm )
1
DL
2400
234
2
DL+LL
3200
431
My top (kNm )
Mx bot (kNm )
-341
24
My bot (kNm )
BS8110 - 1997
-13
Y
1000
X
500
Y
1000
0
500
Assumptions:
(1) The general conditions of clause 3.8.1 are applicable.
X
0
Given:
Lo = 10.000 m
fcu = 30 MPa
fy = 450 MPa
Ac = 893945 mm?
Design approach:
The column is designed using the following procedure:
Through inspection:
le = ? . Lo
max_sl=
=
le
h
8.5
.94351
= 9.009
Table 3.19
Job Number
Job Title
Client
Calcs by
Sheet
Checked by
Date
Where h is an equivalent column depth derived from the radius of gyration*square root of 12
Use emin = 20mm
3.8.2.4
Mmin = emin. N
= .02 ×3200
= 64.000 kNm
le/h = 9.0 < 15
\ The column is short.
3.8.1.3
Initial moments:
The initial moment near mid-height of the column :
3.8.3.2
Mi = 0.4 . M1 + 0.6 . M2
= 0.4 ×-24 + 0.6 ×431
= 249.000 kNm
Mi2 = 0.4 . M2
= 0.4 ×431
= 172.400 kNm
\ Mi ? 0.4M2 = 249.0 kNm
The initial moment near mid-height of the column :
Mi = 0.4 . M1 + 0.6 . M2
= 0.4 ×13 + 0.6 ×-341
= - 199.4000 kNm
Mi2 = 0.4 . M2
= 0.4 ×-341
= - 136.4000 kNm
3.8.3.2
Job Number
Sheet
Job Title
Calcs by
Checked by
Date
\ Mi ? 0.4M2 = -199.4 kNm
Design axial load:
Pu = 3200.0 kN
Mx=431.0 kNm
Mxmin=64.0 kNm
Mxtop=431.0 kNm
+
=
Mxbot=24.0 kNm
Initial
Additional
Design
My=341.0 kNm
Mymin=64.0 kNm
Mytop=-341.0 kNm
+
=
Mybot=-13.0 kNm
Initial
Additional
Design
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
Axial load (kN)
3000
2500
2000
1500
1000
0.00
-500
550 kNm 500
-1000
-2000
-3000
-4000
-1000
-1500
-2000
-2500
-3000
3200 kN
Moment max = 2939 kNm @ 5216kN
Interaction Diagram
16E3
15E3
14E3
13E3
12E3
11E3
10E3
9000
8000
7000
6000
5000
4000
3000
2000
1000
Job Number
Sheet
Job Title
Calcs by
Checked by
Date
BS8110 - 1997
Y
D
1000
321.6?
X
X
500
0
500
1000
D
0
Y
Moments and Reinforcement for LC
Top
2:DL+LL
Middle
Bottom
Madd-x
(kNm) 0.0
0.0
-0.0
Madd-y
(kNm) 0.0
0.0
-0.0
Mx
(kNm) 431.0
249.0
24.0
My
(kNm) -341.0
-199.4
-13.0
M'
(kNm) 549.6
319.0
64.0
Design axis (?)
321.65
321.31
151.56
Safety factor
3.66
4.29
4.80
12868
12868
1.42 %
1.42 %
3576
3576
Asc
(mm?) 12868
Percentage
AsNom
1.42 %
(mm?) 3576
Load case
Axis
N (kN )
Load case 1
DL
X-X
Y-Y
2400.0
M1 (kNm )
0.0
0.0
M2 (kNm )
234.0
0.0
Mi (kNm )
140.4
0.0
Madd (kNm )
0.0
0.0
Design
Top
M (kNm )
234.0
0.0
M' (kNm )
Safety
factor
234.0
5.748
Job Number
Sheet
Job Title
Client
Calcs by
Load case
Axis
N (kN )
Load case 2
DL +LL
X-X
Y-Y
3200.0
M1 (kNm )
-24.0
13.0
Checked by
M2 (kNm )
Mi (kNm )
431.0
-341.0
249.0
-199.4
Madd (kNm )
0.0
0.0
Date
Design
Top
M (kNm )
431.0
341.0
M' (kNm )
Safety
factor
549.6
3.662
Job Number
Job Title
Calcs by
Sheet
Checked by
Default Example 2
C13
Design code : ACI 318 - 2005
Input tables
Code
X/Radius or
Bar dia. ( mm )
Y (mm )
Angle (?)
+
500
A
500
A
500
-90
+50
-90
-500
-
500
C
100
+
50
B
32
+
250
B
32
+
450
B
32
+
600
B
32
+
725
B
32
+
890
B
32
+
950
B
32
+
50
B
32
+
250
B
32
+
450
B
32
+
600
B
32
+
725
B
32
+
890
B
32
+
50
B
32
+
50
B
32
+
50
B
32
Date
550
50
50
50
50
110
275
500
1000
1000
1000
1000
940
775
525
275
775
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
Design loads:
Load
case
Designation
P (kN )
Mx top (kNm )
1
DL
2400
234
2
DL+LL
3200
431
My top (kNm )
Mx bot (kNm )
My bot (kNm )
-341
24
-13
Code specific parameters:
?d - see Clause 10.11.1
0
result of sway - X direction
0
result of sway - Y direction
0
ds - see Clause 10.13.4.3
1
ACI 318 - 2005
Y
1000
Design approach:
The column is designed using the following procedure:
Through inspection:
kLu = ? . Lo
X
500
Y
1000
0
500
Assumptions:
X
0
Given:
Lo = 10.000 m
fc' = 30 MPa
fy = 450 MPa
Ac = 893945 mm?
Job Number
Job Title
Calcs by
Sheet
Checked by
Date
k Lu/r =
=
k Lu
r
8.5
.27237
= 31.208
10.12.3.2
eminx= 0.6 . 0.0254 + 0.03 . h
= 0.6 ×0.0254 + 0.03 ×1
= 0.0452 m
10.12.3.2
eminy= 0.6 . 0.0254 + 0.03 . b
= 0.6 ×0.0254 + 0.03 ×1.05
= 0.0467 m
Mminx= eminx. N
= .04524 ×3200
= 144.768 kNm
kLu x/r = 31.2 < 34 - 12
? M1/M2 = 34.6
Note that 34 - 12? M1/M2 is limited to a maximum value of 40.0
\ The column is short.
Initial moments:
10.12.2
10.12.2
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
Design axial load:
Pu = 3200.0 kN
Mx=431.0 kNm
Mxmin=144.8 kNm
Mxtop=431.0 kNm
+
=
Mxbot=24.0 kNm
Initial
Additional
Design
My=341.0 kNm
Mymin=144.8 kNm
Mytop=-341.0 kNm
+
=
Mybot=-13.0 kNm
Initial
Additional
Design
Job Number
Sheet
Job Title
Calcs by
Checked by
Date
4000
3500
3000
2500
2000
1500
1000
0.00
-500
-1000
550 kNm 500
-1000
-2000
-3000
-4000
-5000
-1500
-2000
-2500
-3000
-3500
3200 kN
Moment max = 3685 kNm @ 7532kN
Interaction Diagram
Axial load (kN)
16E3
15E3
14E3
13E3
12E3
11E3
10E3
9000
8000
7000
6000
5000
4000
3000
2000
1000
Job Number
Sheet
Job Title
Client
Calcs by
Checked by
Date
ACI 318 - 2005
Y
D
1000
321.6?
X
X
500
0
500
1000
D
0
Y
Moments and Reinforcement for LC2:DL+LL
Middle
Bottom
Madd-x
(kNm) 0.0
Top
0.0
-0.0
Madd-y
(kNm) 0.0
0.0
-0.0
Mx
(kNm) 431.0
249.0
24.0
My
(kNm) -341.0
-199.4
-13.0
M'
(kNm) 549.6
319.0
144.8
Design axis (?)
321.65
321.31
151.56
Safety factor
4.48
4.59
4.59
12868
12868
1.42 %
1.42 %
8939
8939
Asc
(mm?) 12868
Percentage
AsNom
1.42 %
(mm?) 8939
Load case
Axis
N (kN)
Load case 1
DL
X-X
Y-Y
2400.0
M1 (kNm) M2 (kNm) Mi (kNm) Madd (kNm) Design
0.0
0.0
234.0
0.0
140.4
0.0
0.0
0.0
Top
M (kNm)
234.0
0.0
M' (kNm)
Safety
factor
234.0
6.123
Job Number
Sheet
Job Title
Calcs by
Load case
Axis
N (kN)
Load case 2
DL+LL
X-X
Y-Y
3200.0
Checked by
Date
M1 (kNm) M2 (kNm) Mi (kNm) Madd (kNm) Design
-24.0
13.0
431.0
-341.0
249.0
-199.4
0.0
0.0
Top
M (kNm)
431.0
341.0
M' (kNm)
Safety
factor
549.6
4.476

Default Example 1

Transcription

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