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BRACING MEMBERS
SUMMARY.
Introduce the bracing member design concepts. Identify column bracing members requirements in terms of strength and stiffness. The assumptions and limitations of lateral bracing for unrestrained beams analysis are given Diaphragms design, limitations and applications are introduced.
OBJECTIVES.
Understand that the design requirements for column and portal frame bracing. Understand that the design requirements for unrestrained beams lateral bracing. Understand that the design requirements for diaphragms action use and its main approaches.
REFERENCES. [1] ENV 1993-1-3 Eurocode 3 Part 1-3 General rules - Supplementary rules for cold formed thin gauge
members and sheeting. [2] The Steel Construction Institute, Steel Designers Manual, 5th Edt., Blackwell, 1992. [3] Winter, G. Transactions of ASCE, 125, pp.807, 1960. [4] Kennedy, J. B., Neville, A. M., Basic statistical Methods for engineers and Scientists, 3rd., Harper
and Row, New York, pp. 114, 1986. [5] Canadian Sheet Steel Building Institute, Diaphragm Action of Cellular Steel floor and Roof Deck
Construction, CSSBI Suite 305, 201 Consumers Road, Willowdale, On, M2J 4GB, 1972. [6] McGuire, W., Steel Structures, Prentice-Hall, 1968.
1. INTRODUCTION. The bracing system:
Must provide both force and stiffness. Is used for beams, columns and frames. Reduces effective length of columns. Reduces effective length of beams. Provides overall stability to frames. May be discrete or continuous.
Consider:
If is equal to zero then Q is also equal to zero, but in a system that is less than ideal, 0, and the situation portrayed in (c) occurs:
Q = k for equilibrium:
L
PQQLPM A 0 or: LPk
L
Pk
Note if: kL > P there is no sideway kL < P there is sideway
The critical case occurs when kL = P
[1] k0 = Pcr /L; The maximum load in the member is: 2
2
02
2
L
EILk
L
EIPcr
-For kL/k0L > 1, the system is braced and can reach Pcr -Any k > kc, gives no benefit -For k < kc, the system is only partially braced and P/Pcr < 1
2. COLUMN BRACING. Consider a two-storey column braced at mid-height. The bracing member forces the column to buckle into two half waves, i.e.:
Free body diagram of the top half.
22
LkQLPcr
L
Pk cr2
0 [2]
For a three-storey column braced at each storey height:
Buckling Mode 1
330 LkQL
Pcr
L
Pk cr3
0 [3]
Buckling Mode 2
LkQLPcr 0
Not critical when compared to first mode
For a four-storey column buckling in a zigzag manner the deflection at mid-height is not necessarily :
Free body diagram of top quarter
12
1 2
PL
QQ and
12
1 2
PLk
22
1
1Pk [A]
Free body diagram of next quarter 0BM
02 211
21
PLQL
212 2 PLQ
2
212
L
Pk [B]
Equating [A] and [B]:
2
21
21
1 2
2
2
; 2
22121212121 22 ; 21414.1
LPk cr41.30 [4]
For a large number of storeys the deflections in opposite directions at interior points, remote from the ends, must be equal and the net shear on a panel height will be Q/2. Therefore from the free body diagram:
222 0 LkQL
P
l
Pk
40 [5]
Hence:
In general L
Pk
0
Thus the bracing stiffness k0, corresponding to zero initial deflection at the bracing points, can be determined. The bracing force required is Q = k0 .
In a real structure, however columns have initial crookedness. Forces in the bracing only occurs when the force in the column causes the brace to deform.
0BM
00 crPQL
00 1
L
PkPLk cr
cr
00 1kk [6]
There is still the need to know both 0 and
000
0 1
kkkQ [7]
For compression members, normal tolerances are 1/500 to 1/1000 of the length for plumbness
Say 0 = L/500 = 0.002L; let = 0 k = k0(1+1) = 2k0
Required stiffness l
Pk cr2 and Q = k = crcr
cr PPxl
P 004.0002.02
2k Q [8]
Note: Assumed 0 = L/500 and =0
The selection of braces based on equation [8] implies:
-Assume equation [8] applies to both columns and beams. -For beams deal with compressive force only. -Does not apply to plastically designed structures. -Pcr can be interpreted as the elastic or inelastic load. -Does not apply to locations where brace forces are included in the analysis i.e. P- effects.
Steps: 1- Establish brace locations. 2- Calculated Pcr = Compressive resistance for a column and Ac y for a beam. 3- Calculate or estimate .
4- Calculate L
Pk cr
0 for n equal spaces L
5- Select Ab i.e. the area of brace such that: L
P
L
EA cr
b
b 2 (units in
mm
N
mmmm
Nmm
12
2 )
6- Check that crP004.0 Q
First Example Provide braces at 1/3 points for a WWF 600x152, 250MPa yield stress, carrying a uniformly distributed load on a span of 18 metres. Only one beam is to be braced. The beam is designed to reach its capacity for an unbraced length of 6 metres. The brace length is 4.5 metres.
Evaluate the 6m unbraced length beam capacity. First the beam class has to be determined.
-Flange 7.8250
2359
235996
252
300
yfxt
c i.e.: Class 1
-Web 8.69250
23572
23572728.68
8
550
yft
c i.e.: Class 1
It can reach: Mpl = Z fy = 920 x 103 x 250 x 10-6 = 1230 kNm
Mcr = 5,0
2
2
2
2
z
t
z
wz
EI
GIL
I
I
L
EI
=
= kNmxxxx
xxxx
x
xxxx2167
1011310200
10322010776000
10113
109300
6000
10113102005,0
632
332
6
9
2
632
LT =cr
yy
M
fW= 75.0
2167
1230
2167
10250920000 6
xx
2)2.0(15,0 LTLTLTLT a = 92.075.0)2.075.0(49.015,0 2
LT
LT = 5,022
1
LTLTLT = 69.0
75.092.092.0
15,022
Mb.Rd = LT w Wpl.y fy/m1 = 0.69x 920 x 103 x 250 x 10-6 / 1.1 =771.5kNm
The unfactored compressive force is
1475.9kN25600
105.7711.1 3
xx
td
MF
f
rmc
For braces at 1/3 points: 3
mmNxx
L
Pk cr /738
6000
109.14753 3
0
20 2.33200000
450073822mm
xx
E
LkA b
b
As a compression member mmrrL 5.22200
4500200 min
Try L 150x150x10: rmin = 29.7mm; A = 2930mm2
17.71kN9.14753004.0 004.0 Q xxPcr
9.88250200000
1 yf
E 71.19.887.29
4500
1
xr
L
i
cr
2)2.0(15,0 a = 44.471.1)2.071.1(34.015,0 2
= 5,022
1
= 12.0
71.144.444.4
15,022
Nb.Rd = A fy/m1 = 0.12 x 2930 x 250 x 10-3 / 1.1 =78 kN >> Q = 17.71 kN
Notes:
1. mmx
xx
AE
PL14.0
2000002930
10450071.17 3
; assumed = 0.002L = 0.002x6000 = 12 mm.
The force of 17.71kN is generated in 0.14 mm not 12 mm
2. kNL
PkQ cr 914.012
1000
7380002
This is considerably less than the 17.71 kN assumed and in turn generates less .
3- If mm61000
10 was used rather than 1/500 then Q < 738 (6.14)/1000 =4.5 kN
mmx
xx
AE
PL03.0
2000002930
1045005.4 3
5.4 then Q3 < 738 (6.03)/1000 =4.45 kN
ok.
4. The brace have to designed for:
0.01 force in the compression flange if in an elastic design or 0.025 force in the compression flange if in a plastic design
0.001 kN4.1325600
105.77101.001.0 3
xx
td
MF
f
rc
This is 0.75 Q1 and 3Q3
5- What if braces can be provided in 2 directions so that brace can be assumed to act in tension?
Then as Ab = 47mm2 with a slenderness limit of 300, as a tension member:
mmrrL 15300
4500300 min
Try L 90x90x7: rmin = 17.7mm; A = 1220mm2 Npl.Rd = A fy/m1 = 1220 x 250 x 10-3 / 1.1 =272.3 kN O.K
17.71kN9.14753004.0 004.0 Q xxPcr
mmx
xx
AE
PL33.0
2000001220
10450071.17 3' << 12mm assumed.
kNQ 1.933.0121000
738'
mmx
xx
AE
PL16.0
2000001220
1045001.9 3'' kNQ 916.012
1000
738''
mmx
xx
AE
PL16.0
2000001220
1045009 3''' kNQ 914.012
1000
738'''
This is approximately 2/3 of the required by the design standard.
-Examination of Equations [6] and [7]
000 11
L
Pkk cr
[6]
000 L
PkQ cr
[7]
0
0
1k
k [6a]
0
00
1k
Q [7a]
If 10k
k
Select 210
andk
k
- As stiffness of brace, k/k0, decreases the deflections increase. - As stiffness of brace decreases the force Q that must be developed increases. - If the brace is too soft it will need a large force therefore the design should be based on both
strength and stiffness requirements.
3. BRACING OF PARALLELL MEMBERS. - Assume that bracing is of constant size.
- The extreme value of out-of-straightness for m members is mi , [5]. This states that the
extreme value for m members is less than for one: - Members are equally spaced. - Example m =3.
0
3 c
M
300303
2 k
L
PQ cr
2002032
2 k
L
PQQ cr
1001021
2 k
L
PQQ cr
Also: k
Q
EA
LQ
E
b 111
k
k
Q
k
Q
EA
LQ
E
b 2121212
k
QQQ
EA
LQ
E
b 321323
3210321
0003 QQQQQQQ
k
kkQ
Where: k00 =Q0 and 10 k
k
01 0321 QQQQ 210210
0012 QQQQQk
kkQQ
01 0321 QQQQ
And: 1010
0021 QQQk
kkQQ 01 021 QQQ
The three simultaneous equations are;
01 0321 QQQQ [A]
01 0321 QQQQ [B]
01 021 QQQ [C]
Where: k
k0 ; L
PkQ 0
000
and
mi0
The previous conclusions induce a selection of: 210 k
k
Second Example Will the L 150 x 150 x 10 mm selected for the first example provide adequate bracing for three parallel WWF 600x152, 250MPa yield stress, when spanning of 18 metres. The brace length is 4.5 metres.
1- The unfactored compressive force is equal to 1475.9 kN. 2- The braces at the 1/3 points i.e.: = 3
3- Therefore mmNxx
L
Pk cr /738
6000
109.14753 3
0
4- mmx
mi 93.6
3
60000020.00
Q0 = k00 = 6.93 x 738 = 5114.3 N
5- mmNx
L
EAk
B
B /2.1302224500
2000002930
6- 25.176738
2.130222
0
K
K ; 0057.00
k
k
7- The simultaneous equations are:
0 5114.310057.00057.00057.0 321 QQQ [A]
0 5114.310057.00057.0 321 QQQ [B]
0 5114.310057.0 21 QQ [C]
kN 15.7N 15763 1 Q ; kN 10.6N 10559 2 Q ; kN .35N 5295 3 Q
121.02.130
7.1511
k
Q; 202.0
2.130
6.10121.02
12 k
Q; 243.0
2.130
3.5202.03
23 k
Q
The capacity of L 150 x 150 x 10 mm, calculated in the first example, was: Nb.Rd = 78 kN >> Q = 15.7 kN
Summary 243.03
202.02
121.01
mmmi 9.60
Note: (i) One brace is in tension (ii) It is tacitly assumed that the braces area installed to limit imax to 0.002 of 6000mm. (iii) If braces are installed with imax = 36 mm there is a change in the forces previously calculated. (iv) If braces are installed with initial forces present, these forces should be considered. Third Example
Part floor framing plan
Column C1extends 3700 mm above and below this level. It is prevented from buckling about its weak axis at this level only by the flexural stiffness about the weak axis of beams AC-1 and CE-1. The diaphragm action of the floor prevents relative movement of A1, B1, D1 and E1. Column loads above and below are 1560 kN and 1780 kN. Assume 0 is L/1000 = 2x3700/1000 = 7.4 mm. Question: Do beams provide adequate support so that column can be designed for kL = 3700mm?
1- Take kNPc 16702
17801560
; 903.0
3700
167020
x
L
Pk c
kNQ )4.7(903.0 ; mmkNQ
k /4.7
1903.0
EI
aLPa
3
2 ; aLa
EIPk
2
3
kNmmNxxxxxx
aLa
EI
aLa
EIktotal 05.1/1051273778
73502750
103.252000003
73001500
103.212000003332
6
2
6
22
221
1
16.1903.0
051.1
0k
kno good,
too large, see previous graph, a stiffer system is needed, greater than double of the initially adopted.
2- Try a WWF 300x95.3 & WWF 300x115 i.e.: comparable W and Z (Iz = 72x106 mm4, 85.5x106 mm4) ktotal = 2.629 + 0.922 = 3.55 kN/mm
Q = 3.55 3.55 = 0.903 ( + 7.4) = 2.52mm Q = 8.96kN
3- An alternative solution takes into account that most of the restraint comes from AC-1 beam. Therefore as good solution will be to increase its size and leave CE-1 as it is i.e.:
ktotal = 2.629 + 0.273 = 2.902 kN/mm 2.902 = 0.903 ( + 7.4) = 3.34mm Q = 9.7kN
4- the bending moments developed in the members should also be checked, i.e. for AC-1:
kNmM dy 1.135.1902.2
629.2x7.9
Mb.Rd = Wpl.y fy/m1 =726x103 x 250x10-6 / 1.1 =165kNm
..108.0165
1.130
,,
KOM
M
M
M
ryb
dy
rdzb
dz
4. DIAGONAL BRACING FOR COLUMNS. The general procedure is composed of four steps:
1- Select bracing, 2- Compute deflections; 3- Determine the P- effects; 4- Iterate till the desired solution convergence.
Fourth Example The specified (1/50) win forces acting on a braced bay and the resulting forces are show in the illustration below. Select diagonal bracing to resist the imposed loads including the P- effects.
Additional information 1- Dead plus wind load combination case has been established to be critical. Dead loads
contributing to P- effects are: Level 2: 6248kN, Level3: 6248kN, and Level4: 2384kN.
2- Storey drifts due to: bracing force B aAE
BlB
2
in the storey;
Compressive force C aAE
ChC
2
in the leeward column;
Tensile force T aAE
TChT
2
in all the storeys above.
3- Area of column is equal to 9280mm2 in all the levels
Select bracing
Level 1-2 kNxQT qf 5870.3915.1
mmLr 1.333009931300min
Try 2 150x90x10 long legs back to back with a 10mm gusset plate: Ag= 4640mm2 and ry= 35.8mm
Adopting one bolt hole for 22mm diameter bolt the net area can be evaluated: An = 4640 – 2(10x26) = 4120 mm2
kNxxAf
NM
yRdpl 5.1054
1.1
102504640 3
0.
; kN
xxfAN
M
unetRdu 6.1186
25.1
1040041209.09.0
3
2.
kNNkNN duRd 5875.1054 O.K. Levels 2-3 and 3-4 Use 2 150x90x10 long legs back to back with a 10mm gusset plate, i.e.: Ag= 4640mm2 and ry= 35.8mm to provide minimum slenderness ratio. Lateral deflections (specified wind) Level 1-2
Bracing mmxx
xx
aAE
BlB 62.4
20000046409000
993110391 232
Column mmxx
xx
aAE
ChC 31.0
20000092809000
420010293 232
=4.93mm Level 2-3
Bracing mmxx
xx
aAE
BlB 70.2
20000046409000
980910234 232
Column Level 1-2 mmx
C 29.04200
390031.0
Level 2-3 mmxx
xxC 12.0
20000092809000
390010127 23
2
Level 2-3 mmxx
xxT 13.0
20000092809000
420010127 23
1
=3.24mm Level 3-4
Bracing mmxx
xx
aAE
BlB 00.1
20000046409000
9809106.86 232
Column Level 1-2 mmx
C 29.04200
390031.0
Level 2-3 mmC 12.0
Level 3-4 mmC 12.0
Level 2-3 mmT 13.0
Level 3-4 mmx
T 12.04200
390013.0
=1.784mm
P- effects (D + Q)
In any storey
Non-iterative method
QV
V
VV '
''
max
1
The deflected shape and factored P- effects are:
Equivalent Equivalent Shears, V’ Lateral Force, H ― ― → ――→
1.09 D Q 1.09 D Q
← ― ―
― ― → ――→
7.17 D Q 6.08 D Q
← ― ―
― ― → ――→
17.47 D Q 11.39 D Q
← ― ― ←――
17.47 D Q
For Level 3-4 (2384 D x 1.78 Q)/3900 = 1.09 D Q
For Level 2-3 ((2384 + 6248) D x 3.24 Q)/3900 = 7.17 D Q
For Level 3-4 ((2384 + 6248 + 6248) D x 4.93 Q)/3900 = 17.47 D Q
For level 1-2 kNxx
V
V
VV
Q
4.36
531
07.341
07.345.13.147.17
1'
''
max
The force in the bracing including the P- effects;
kNNkNa
LN uRdPd 5.10541.626
9000
99314.5674.36531, O.K.
For Levels 2-3 and 3-4 the diagonals selected for maximum slenderness ratio will also be ok.
2384 D
6248 D
1.78 Q
3.24 Q
4.93 Q
6248 D
5. DIAPHRAGMS.
The diaphragm action:
- Transfer in-plane lateral loads applied at roof and floor levels to stiffer members; - Need strength and stiffness;
- Must look at flow of forces (how forces are carried from point to point) intersheet connections and connections at boundaries are important.
Consider:
The shear strength of diaphragm depends on: - Seam capacity. - Edge capacity.
- Purlin capacity (a controlled by fasteners). overall buckling strength.
Stiffness of diaphragm depends on: -Panel warping;
- Movement at seams; - Edge panel slip.
The shear stiffness
G and a
;
a
bt
FG
bt
F or
b
FaGtG '
“spring stiffness” is equal to a
bGF '
Wall Cladding
Can cladding take horizontal force (in-plane) from girts to reduce effective length of columns? Must establish force in girt as a function of . Cladding must provide this force and stiffness. Roof Diaphragm
The diaphragm acts as a deep beam to carry the shear to the ends where it must be transferred to the ground.
The force in the flanges is equal to D
wL
D
M
8
2
6
4
102384
5
x
vLF
EI
wLVft
Where F is the flexibility factor equal to a function of: -Deck span/average length supplied; -Deck gauge;
- -Side lap connection; - -Transverse welds.
And I is the moment of inertia of the flanges (which must be connected to the web).