Calculation of Wind Drift in Staggered-Truss Buildings R. E. LEFFLER
Section 2 (by the floor slabs acting as diaphragms) as it accumulates down the building. However, the vertical reactions from the wind accumulate directly on each column in the section where they are first developed. An understanding of this load-transfer system is important in following the drift calculations and in checking the resistance to overturning as described in the following section. In general, only the calculations of overturning and wind drift are considered; all other conditions must be examined independently.
The important feature of a staggered-truss framing system is the story-high steel trusses that span the full width of the building. These trusses are used in a staggered arrangement, so that they occur at every other column row on each story. The floors, typically precast prestressed concrete, span from the bottom chord of one truss to the top chord of the adjacent truss, so that the span of the floor system is half the truss spacing. As the height of steel-framed staggered-truss buildings is extended beyond 20 stories, the need for calculating wind deflection (drift) becomes more important. However, because the staggered-truss system differs from conventional framing systems, the appropriate method for calculating drift may not have been apparent. Consequently, a method was developed for calculating drift in staggered-truss buildings and, through finiteelement computer analysis, the accuracy of the method was verified. Specifically, a NASTRAN model was made for the building described in this paper, and a complete analysis was made to calculate the deflections on each floor. The result showed that the difference in results between the handwork calculations and the NASTRAN analysis was generally less than 1.5%. A design example is presented to illustrate the hand-calculating method. Appendix B, Sheets 1 through 24, show the drift calculations for a typical interior bay from the building described in Ref. 1. (Appendix A presents the Nomenclature and Equations used in the analysis.) Although the drift of an end bay would be somewhat different, for most practical buildings the drift is considered to be governed by the behavior of the interior bays. Sheet 1 shows the building selected for the design example. 1 Sections 1 and 2 show the arrangement of the structural framing on adjacent column rows. As illustrated on Sheet 2, the horizontal wind load is transferred alternately between the trusses in Section 1 and the trusses in
OVERTURNING An important step in the early stages of building design is to check the resistance of the structure to overturning. The wind load acting on the vertical face of the building causes an overall bending moment on any horizontal cross section of the building. This moment, which reaches its maximum value at the base of the building, causes the building to tend to rotate about the leeward column and is called the overturning moment. The overturning moment causes compression in the leeward columns and tension (uplift) in the windward columns. The dead load generally causes compression in all columns, thus reducing the tension in the windward columns and providing resistance to overturning. Although the foundation weight could be used to provide resistance to overturning, it is usually considered desirable for all columns to be in a state of compression under the combined action of the wind load and dead load, with the compression load exceeding the tension load by a suitable margin in accord with building code requirements. In the staggered-truss system, the wind-induced tension in the columns is found by summing the verticaltruss reactions caused by the horizontal wind loads. From Sheet 2, R = 2W × (D/L)
(1)
where R is the vertical-truss reaction caused by the horizontal wind load W; D is the depth of the truss (story height); and L is the span of the truss. The factor of 2 arises
R. E. Leffler is Senior Research Engineer, U.S. Steel Corporation, Research Laboratory, Monroeville, Pennsylvania.
1 FIRST QUARTER / 1983
American Concrete Institute Code, 2 Sheet 7. The shear component of the displacement is found by first determining the angular shear distortion and then multiplying it by the span of the floor, Sheet 8. Because the floor slab acts as a beam with a depth of 2.5 times the span (depth = 60.0 ft; span = 24.0 ft), ordinary equations for bending deflection are not strictly correct. However, because the bending component of the displacement is only about 5% of the total, results obtained by using ordinary deflection equations are adequate. The total displacement is determined by summing the bending and shear displacement, as shown on Sheet 8. Because a unit load was assumed to act on the slab, the resulting displacement is the unit-deflection-force relationship for the slab.
because each truss resists the wind load of a two-bay width. Summing the vertical-truss reactions leads to the following two equations for the vertical-column forces at the base of the building caused by the wind load: R b1 = 2 (D/L) [W20 + W18 .... + W2 ] + F VKB R b2 = 2 (D/L) [W19 + W17 + .... + W3 ]
(2) (3)
R b1 and R b2 are the vertical-column forces at the base of the building for Sections 1 and 2, respectively, caused by the wind load; W is the total wind load transferred by the truss in the story denoted by the subscript; F VKB is the vertical component of the axial force in the wind brace caused by the wind load; and D and L are as defined in Eq. (1). The incremental wind loads, Wi , in Eqs. (2) and (3) are summed on the wind-load diagram1 on Sheet 3 and are shown as cumulative wind loads. The vertical-column reactions (tension on the windward column) are determined on Sheet 4. For Section 2, the final verticalcolumn reactions at the base are ±191.0 kips. However, the total wind load for the building in Section 1 (209.9 kips) is passed to the foundation by the wind braces. The vertical component of the axial force in the wind brace (267.4 kips) adds to the vertical-column force at the base in Section 1, resulting in final vertical-column reactions at the base equal to 477.3 kips as shown in the diagram on Sheet 4. Dead loads are taken from the design example,1 and the total dead load in the columns is determined. The ratio of the dead load to the vertical-column force caused by wind is 3.12 for Section 1 and 7.80 for Section 2.
Trusses—The deflection-force relationship for the trusses, determined by using the method of virtual work, 3 is shown on Sheets 9 through 13. In this method, the reactions and internal forces must be determined for both the “real load” and a dimensionless “virtual load.” Sheet 10 represents both loads, including their reactions and forces. The total deflection is the sum of that caused by bending effects in the top and bottom chords, Sheet 11, and that caused by axial effects, Sheet 12. The appropriate equations for calculating the deflection are shown on Sheets 11 and 12, and further explanation is presented as follows. A finite-element computer analysis used in verifying this calculation method indicated that the shear flow between the trusses and the floor slabs is such that the force in the truss chords can be considered to be zero except for the bottom chord of the second-story truss. Consequently, the wind load applied to the truss top chord and the reactions in the bottom chord, Sheet 10, are distributed so that the axial force in the chords is zero; the horizontal component of the diagonals balances the applied load (or reaction) at each truss panel point. With the virtual-work method, a unit virtual load is placed at the point where the deflection is to be determined. In the truss chords for which the axial force is zero, the horizontal displacement is the same everywhere along these chords, and the placement of the unit virtual load and its reactions is not important. In Ref. 1, the vertical shear in the center panel of the truss was distributed in accordance with the moment of inertia of the top and bottom chords. However, the computer analysis showed a nearly equal distribution of vertical shear between the top and bottom chords, which can be attributed to the flexibility of the truss verticals flanking the center panel. (The effect of this vertical shear distribution on the resulting truss displacement was found to be very small for the usual variation in moment of inertia.) Therefore, the vertical shear is equally divided between the top and bottom chords in the present example. Because of the symmetry of the truss, certain simplifying assumptions can be made in determining bending effects
WIND-LOAD DISTRIBUTION A pattern of shear flow through the floor slab caused by the wind loads is shown on Sheet 5. The wind shear shown is for one bay. The total wind shear in the trusses is twice this value because of the contribution from the adjacent bay. By using this pattern, the wind shear in the floor slabs and trusses throughout the building is tabulated on Sheet 6. COMPONENT DISPLACEMENT The deflection-force relationships for the structural components are determined first. These relationships are then used to tabulate the total drift. Equations used in the following calculations are shown in Appendix A. Floor Slab—The hollow, precast floor plank shown on Sheet 7, which is typical of those available, will be used in this example. Only the continuous thickness of the floor slab (the top 1¼-in. and the bottom 1¾-in.) is considered effective. Because the displacement of the floor slab is dominated by shear, with a small additional contribution from bending, fixed-boundary conditions are appropriate for an interior bay. Properties are calculated in accordance with the 2
ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
of the truss. These assumptions eliminate the need for an indeterminate analysis of the truss chords. First, an inflection point (point of zero moment) is assumed at the midpoint of the Vierendeel panel. Second, an investigation showed that the effect of the chords in the panels not adjacent to the Vierendeel panel was negligible. Therefore, truss-chord bending effects were only considered in the Vierendeel panel and those adjacent to it (Sheets 10, bottom, and 11). The calculation for the bending displacement and the axial displacement is shown on Sheets 11 and 12, respectively. The total displacement is the sum of the axial and bending displacements. The total displacement for a typical truss (except the second story) is 0.0079 in./kip as shown on Sheet 12. Because a 1-kip wind shear load was assumed, the resulting displacement is the unit-deflection-force ratio for the truss. At the second floor, the wind shear is transferred to the wind braces. Therefore, the axial force is not zero in the bottom chord of the second-story truss. The bottomchord axial forces are deduced from the axial-force diagram on Sheet 10 and are shown at the bottom of Sheet 12. The bottom-chord contribution to the displacement is calculated on Sheet 12 and is combined with the previously determined truss displacement to find the total unit-deflection-force ratio for the second-story truss, which is 0.001057 in./kip, as shown on Sheet 13. Wind Brace—Displacement of the wind braces results from the horizontal wind shear and the vertical-column reactions to the wind-induced moment. The method of virtual work was used to determine these displacements. Calculations are shown for one wind brace on Sheets 13 and 14. The horizontal and vertical loads result in both horizontal and vertical displacements of the knee brace. Because of the symmetry of the building, the total horizontal displacement of the other wind brace is identical to that of the one shown (0.4594 in.). The total vertical displacement of the other wind brace is equal to magnitude and opposite in sign to that of the one shown (0.0325 in.).
COLUMN-LENGTH CHANGE The vertical-truss reactions to the horizontal wind shears cause the columns to lengthen on the windward face of the building and shorten on the leeward face of the building. This column-length change causes the trusses to rotate and results in a component of drift (Sheet 17). Vertical-truss reactions are calculated for each truss and summed down the building (Sheet 18). The main reaction points for the trusses in this example are at the top chord, and the columns below the truss top chord contribute to its rotation and the associated drift. The per-story drift associated with column-length changes is determined for Sections 1 and 2 on Sheets 19 and 20, respectively, on the basis of the equations shown on Sheet 17. The drift caused by columnlength change in each story is determined on the basis of the truss in that story for Sections 1 and 2. FINAL DRIFT DETERMINATION The final drift determinations for Sections 1 and 2 are shown on Sheets 21 and 22. The structural drift and the drift caused by the column-length change are combined to find the total drift per story. The total per-story drift is summed, starting at the base of the building to find the total drift for each story. The per-story drift ratio is also determined by dividing the total drift per story by the corresponding story height. Except in the lower few stories, the drift ratio is 0.002 or less. Comparison of the total drift for each story in Section 1 with that in Section 2 indicates only a small variation between the two sections; the difference for stories 2 and 3 is about 1.6% and the difference for all the other stories is less than 1%. Consequently, final tabulation of drift for only one section would be an adequate procedure for most structures. CONTROL OF DRIFT In some instances the engineer may desire to reduce the drift ratio to 0.002, especially in the first and second stories. In the present example, this can be accomplished with a few changes in member sizes as will be described. For the first story, previous calculations indicated that the wind braces provide the dominant control of drift. Therefore, as shown on Sheet 23, an increase in the size of the wind brace with no change in column size reduces the first-story drift ratio to 0.002. To reduce the drift ratio for the second story to 0.002 or less, Sheet 24, the bottom chord of the truss is increased to the size of the top chord, and the verticals and diagonals are increased to the size of the largest vertical and diagonal. CONCLUSIONS
TABULATION OF STRUCTURAL DRIFT The drift of the structure resulting from the displacement of the trusses and floor slabs (structural drift) is tabulated on Sheets 15 (Section 1) and 16 (Section 2). First the appropriate wind shears, taken from Sheet 6, are entered in the table. Then the unit-deflection-force ratios, shown at the top of the sheet, are used to find the drift contribution of each element. Finally, the drift of the elements in each story is summed to find the total story drift. In a story where a truss occurs in the section being tabulated, the drift is that of the truss, and the floor slabs do not enter into the calculations. Where a truss does not occur in the story, the floor slabs above and below the story enter into the calculations along with the truss from the opposite section. For completeness, the drift associated with the wind braces is included in the tabulation.
A method has been developed for the hand calculation of drift in a staggered-truss building. The accuracy of the method, which was verified through finite-element computer analysis of a typical building is excellent. A design 3 FIRST QUARTER / 1983
example was presented in detail to illustrate the application of the calculation method. Also, details of calculating resistance to the overturning moment from wind were reviewed.
Ii E ∆ AT
Fi Ui Li Ai
DISCLAIMER The material in this paper is intended for general information only. Any use of this material in relation to any specific application should be based on independent examination and verification of its unrestricted availability for such use, and a determination of suitability for the application by professionally qualified personnel. No license under any United States Steel Corporation patents or other proprietary interest is implied by the publication of this paper. Those making use of or relying upon the material assume all risks and liability arising from such use or reliance.
∆ TT ∆ BH/H
∑
F Hi U Hi ∆ BH/V
REFERENCES
1. 2. 3.
∑
Staggered Truss Framing Systems for High-Rise Buildings U.S. Steel Corp., ADUSS 27-5227-01, 1971. Building Code Requirements for Reinforced Concrete (ACI 318-77) American Concrete Institute, 1977. McGuire, W. and R. H. Gallagher Matrix Structural Analysis John Wiley and Sons, Inc., New York, New York, 1979.
F Vi = Axial force in the ith member caused by a real vertical force on the wind brace ∆ BV/H = Vertical wind-brace displacement caused by a horizontal load n F U L = ∑ Hi vi i Ai E i =1 U vi = Axial force in the ith member caused by a vertical virtual force on the wind brace ∆ BV/V = Vertical wind-brace displacement caused by a vertical load n F U L = ∑ vi vi i Ai E i =1 ∆ D/CLC = Displacement of a truss in a particular story (drift per story) caused by the length change in the total length of column segments supporting that truss D = 2[ ∑ ∆C ] T LT ∑∆C = Length change of supported-column segment D T = Truss depth (story height) L T = Truss length (span)
APPENDIX A EQUATIONS AND NOMENCLATURE ∆ ss = BV/AG c B = γ = V = A = Gc = ∆ sB = I Ec ∆ slab ∆ BT
= = = = =
Mi mi
= =
= Moment of inertia of the ith element = Modulus of elasticity = Displacement of truss caused by axial loads n FU L =∑ i i i i =1 Ai E = Real axial force acting on the ith member = Virtual axial force acting on the ith member = Length of the ith member = Cross-sectional area of the ith member = Total displacement of the truss = ∆ BT + ∆ AT = Horizontal wind-brace displacement caused by a horizontal load n FHi U Hi Li = Ai E i =1 = Axial force in the ith member caused by a real horizontal force on the wind brace = Axial force in the ith member caused by a horizontal virtual force on the wind brace = Horizontal wind-brace displacement caused by vertical load n FBiU Hi Li = Ai E i =1
Shear displacement of floor slab = B γ = Bay width or width of slab Angular shear distortion = V/AG c Wind load acting in shear on the floor slab Cross-sectional area of the floor slab Shear modulus of concrete Bending displacement of floor slab = VB 3 /12E c I Moment of inertia Modulus of elasticity of concrete Total displacement of floor slab = ∆ ss + ∆ sB Displacement of truss caused by bending n M i mi dx l EI i i =1 Bending moment caused by real load in the ith element Bending moment caused by virtual load in the ith element.
∑∫
APPENDIX B DRIFT CALCULATIONS See Calculation Sheets 1 through 24, following:
4 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. JOB NO.
1
OF
24
G ENERAL A RRANGEMENT
5 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
6 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
SHEET NO. JOB NO.
2
OF
24
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. JOB NO.
3
OF
24
W IND LOAD C UMULATIVE W IND LOAD , E/N S ECTION 1 S ECTION 2
INCREMENTAL W IND LOAD , Wi 4.16 k
12.48 k 20.80 29.12 37.44 45.76 54.08 62.40 70.72 78.00 84.24 90.48 96.72 102.96 109.20 114.92
B AY W IDTH = 24.0 T YPICAL
120.12 124.80 128.96 133.74*
T OTAL OF T RUSS S HEARS - 1 BAY T OTAL T RUSS S HEAR (2 B AYS)
726.44 k ×2
660.92 k ×2
1452.88 k
132184 . k
* 1 ST S TORY W IND LOAD IS R ESISTED B Y T HE W IND B RACE. IT IS NOT INCLUDED IN THE T RUSS S HEAR T OTACS
7 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. JOB NO.
4
O VERTURNING
S ECTION 1
S ECTION 2
V ERTICAL C OLUMN LOAD D UE TO W IND (IBAY) 2 ND S TORY 1452.88 k @
8.67 * = 209.9 k 60
1321.84 k @
8.67 = 191.0 k 60
1 ST S TORY N O C HANGE = 191.0 k
D EAD LOADS (S EE R EF. 1) R OOF = 0.065 k / FT2 @ 24 FT × 60 FT × ½ = 46.8 k W ALL = 0.050 k / FT2 @ 24 FT× 8.67 FT = 10.4 k /S TORY
FLOOR =0.093 k / FT2 @24 FT×27 FT = 60.3 k /FLOOR 0.073 k /FT2 @24 FT×3 FT = 5.2 65.5 k /FLOOR
D EAD LOAD IN 1 ST S TORY C OLUMNS
O VERTORNING C HECK
1 R OOF @ 46.8 =
46.8 k
S ECT. 1
DL 1489 = = 312 . ok WL 477
S ECT. 2
DL 1489 = = 7.80 ok WL 191
19 FLOORS @ 65.5 = 1244.5 19 W ALLS @ 10.4 = 197.6 T OTAL DL/1st S TORY = 1488.9 k
* N OTE : S UMMING
TRUSS SHEARS AND MULTIPLYING BY THE TRUSS DEPTH TO SPAN RATIO IS ALGEBRAICALLY EQUIVALENT TO
SUMMING INDIVIDUAL TRUSS REACTIONS CAUSED BY THE WIND SHEARS .
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ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION DATE SUBJECT C ALCULATION OF SHEET NO. BY R EL CHKD. BY DATE W IND D RIFT IN JOB NO. S TAGGERED - T RUSS B UILDINGS
W IND LOAD D ISTRIBUTION A SSUME WIND TO BE DISTRIBUTED EQUALLY TO EACH COLUMN ROW. T HE WIND LOAD SHOWN IS FOR AN INDIVIDUAL BAY .
R OOF
20 TH F LOOR
19 TH F LOOR
18 TH F LOOR
9
5
OF
24
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
FIRST QUARTER / 1983 SHEET NO. 6 OF 24 JOB NO.
W IND LOAD AND S HEAR D ISTRIBUTION FOLLOWING THE PATTERN ESTABLISHED ON P AGE AND FLOOR WIND LOADS FOR THE BUILDING .
FLOOR R OOF 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
S TORY
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
5, TABULATE TRUSS
S ECTION 1 T RUSS† W IND LOAD @ FLOOR W IND S HEAR 2.08 k 4.16
4.16
2.08 k 8.32 16.64 24.96
k
20.80 37.44 54.08 4.16 3.64 3.12
70.72 84.24 96.72
3.12 2.86 2.60 2.34 2.08 2.39 1.35
S ECTION 2 FLOOR S LAB W IND LOAD W IND S HEAR @ FLOOR
109.20 120.12 128.96 133.74*
†
33.28 41.60 49.92 58.24 66.56 74.36 81.12 87.36 93.60 99.84 106.08 112.06 117.52 122.46 126.88 2.39*
T RUSS† W IND S HEAR
2.08 k 4.16 12.48 k
29.12 45.76 4.16 3.64 3.12
62.40 78.00 90.48
3.12 2.86 2.60 2.34 2.08 2.39 1.35
102.96 114.92 124.80
T RUSS W IND S HEAR FOR I B AY . T OTAL TRUSS SHEAR INCLUDES WIND LOAD FROM ADJACENT BAY. * W IND SHEAR IN 2 ND STORY TRUSS TRANSFERS TO THE WIND BRACE IN THE 1 ST STORY . 2 ND FLOOR SLAB TRANSFERS ONLY WIND LOAD FROM SECTION 2, 2 ND FLOOR TO SECTION 1.
10
ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION DATE SUBJECT C ALCULATION OF SHEET NO. BY R EL CHKD. BY DATE W IND D RIFT IN JOB NO. S TAGGERED - T RUSS B UILDINGS
7
S LAB D ISPLACEMENT - W IND LOAD R ELATIONSHIP
TYPICAL S ECTION THROUGH F LOOR S LAB
P LAN OF F LOOR S LAB
M ATERIAL P ROPERTIES U NIT W EIGHT OF C ONCRETE - W c = 145 # /FT3 C ONCRETE 28-D AY C OMPRESSIVE S TRENGTH - f'c = 4000 psi M ODULUS OF E LASTICITY - C ONCRETE - E C = W C 1.5 33 f ′ c = 145 × 33 4000 ≅ 3.6 × 10 3 k / IN 2
A SSUME P OISSON’ S R ATIO , v=0.2 S HEAR M ODULUS – C ONCRETE – G C =
E 3.6 × 10 3 k/IN 2 = = 1.5 × 103 k/IN 2 2(1 + v) 2(1 + 0.2)
M ODULUS OF E LASTICITY-S TEEL - E S = 29. × 10 3 k/IN 2 M ODULAR R ATIO OF E LASTICITY -n =
ES 29 × 10 3 k / IN 2 = ≅8 EC 3.6 × 10 3 k / IN 2
11
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24
DATE DATE
BY R EL CHKD. BY
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
FIRST QUARTER / 1983 SHEET NO. 8 OF 24 JOB NO.
S LAB D ISPLACEMENT - W IND LOAD R ELATIONSHIP Bv ( 24 × 12IN )10 . k = 8.89 × 10−5 IN ∆ ss = Bγ = = AGc 3IN (60 × 12IN)(1.5 × 103 k / IN 2 )
B ENDING D ISPLACEMENT
ICONC = ISTL = ITOT =
(3 IN ) (60 × 12 IN ) 3 = 93.3 × 10 6 IN . 4 2(10 IN 2 )(30×12 IN ) 2 8 = 20.7×10 6 =114.0×10 6 IN . 4 1
12
n
∆ SB =
10 . k (12 × 24IN ) 3 = 0.485 × 10 −5 IN 12(3.6 × 10 3 k / IN 2 )(114 × 10 6 IN 4 )
T OTAL D ISPLACEMENT ∆ SS
=
8.89×10 –5 IN
∆ SB
=
0.49×10 –5 IN
∆ SCAB =
9.38×10 –5 IN
D EFLECTION. FORCE R ELATIONSHIP - FLOOR S LAB = 0.0000938 IN./KIP
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BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED -T RUSS B UILDINGS
SHEET NO. JOB NO.
9
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24
T RUSS D ISPLACEMENT - W IND LOAD R ELATIONSHIP P ROPERTIES OF T RUSS
M EMBER
S ECTION
M EMBER P ROPERTIES A Ix Iy (IN. 2 ) (IN. 4 ) (IN. 4 )
T OP C HORD
W10×45
13.3
248.
53.4
B UTT. C HORD
W10×39
11.5
209.
45.0
D IAGONAL D1
2C8×11.5
][
6.76
65.2
4.84
D2
2C6×8.2
][
4.80
26.2
2.64
D3
2C3×4.1
][
2.42
3.32
0.854
V ERTICAL V1
2C6×10.5
[]
6.18
30.4
21.4
V2
2C5×6.7
[]
8.94
14.98
10.0
V3
2C3×4.1
[]
2.42
3.32
4.02
13 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. 10 JOB NO.
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24
T RUSS D ISPLACEMENT - W IND LOAD R ELATIONSHIP 1.0 k W IND LOAD AND U NIT V IRTUAL LOAD - T YPICAL T RUSS D ISTRIBUTE W IND LOAD AND R EACTIONS TO CAUSE ZERO AXIAL FORCE IN THE TOP AND B OTTOM CHORDS . U SE THE SAME FORCE DIAGRAM FOR THE 1.0 k W IND LOAD AND THE HORIZONTAL UNIT VIRTUAL LOAD .
V ERTICAL R EACTION = 10 . k×
C HORD R EACTIONS
8.67 60
D IVIDE SHEAR EQUALLY BETWEEN T OP AND B OTTOM C HORDS. N EGLECT OUTER TWO PANELS EACH END
k = 01445 .
T RUSS MEMBER FORCES ARE CALCULATED ON THE SKETCH STARTING AT THE REACTIONS BY USING THE METHOD OF JOINTS .
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SHEET NO. 11 JOB NO.
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24
T RUSS D ISPLACEMENT - W IND LOAD R ELATIONSHIP C HORD B ENDING (V IRTUAL W ORK ) ∆ BT = ∑ ∫ i =1,n
Mimi dx EIi
(2.60) 2 K - IN 2 (108IN + 36IN) 0.02233 ∆ BT = 2 = I 3(2.9 × 10 4 k / IN 2 )I 1 1 1 1 + + ∆ BT = 0.02238 = 0.02238 248 209 I TOP I BOTT
∆ BT = 1973 . × 10 −4 IN
15 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. 12 JOB NO.
T RUSS D ISPLACEMENT - W IND LOAD R ELATIONSHIP A XIAL E FFECTS (V IRTUAL W ORK ) FiUiLi i =1,n AiE
∆ AT = ∑
M EMBER
F
U
L
A
FUL A
D IAGONAL
0.2083 k 0.2083 0.2778 0.1445 0.1686 0.0963
0.2083 0.2083 0.2778 0.1445 0.1686 0.0963
149.93
6.76IN 2 4.80 2.42 6.18 3.94 2.42
0.962 k/IN 1.355 4.781 0.351 0.750 0.399
V ERTICAL
104
∑ E∆ A =
FUL FOR A
1
2
T RUSS
∑
FU L FOR FULL A T RUSS
C OMBINED D ISPLACEMENT ∆ AT = 17.196 k / IN / 29000 = 5.930 × 10 -4 IN. –3 ∆ TT=0.790×10 IN. ∆ BT = 1.973 × 10 -4 ∆ TR =0.000790 IN/KIP
2ND FLOOR T RUSS R EACTIONS @ E NDS OF B OTTOM C HORD - R EAL AND V IRTUAL LOAD A DD B OTTOM C HORD FORCES - O THER M EMBER FORCES DO NOT CHANGE.
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8.598 k/IN 17.196 k/IN
OF
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BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. 13 JOB NO.
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24
T RUSS D ISPLACEMENT D UE TO W IND S HEAR 2 ND FLOOR T RUSS
M EMBER
F
U
L
B OTT. C HD
0.50 k 0.35 0.20
0.50 0.35 0.20
108.IN.
FuL A
A 11.5 IN 2 11.5 11.5
2.348k/IN 1.150 . 3874 0.376 × 2 = 7.748k/IN.
E∆ A = 17.196k / IN. 24.944k / IN. 7.748
C OMBINED D ISPLACEMENT ∆ AXIAL = 24.944 / 29000 = 8.601 × 10 −4 IN −3 ∆ TOTAL = 1.057 × 10 IN. ∆ BENDING = (FROM S HT. 11) = 1.973 × 10 − 4 ∆ TR − 2ND = 0.001057 IN./KIP W IND B RACE FROM S HEET 3, 1 ST S TORY W IND LOAD 133.74 k /BAY × 2BAYS = 267.48 k D UE TO S YMMETRY - ½ P ER KNEE B RACE = 133.74 k FROM S HEET 4,2 ND S TORY C OL. LOAD (2 B AYS)= 209.9 k H ORIZONTAL LOAD
V ERTICAL LOAD
17 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - TRUSS B UILDINGS
SHEET NO. 14 JOB NO.
OF
24
W IND BRACE - DISPLACEMENTS V ERTICAL V IRTUAL LOAD
H ORIZONTAL V IRTUAL LOAD
H ORIZONTAL D ISPLACEMENT H ORIZ LOAD M EMB C OL W.B. ∆ BH / H =
F +267.48 K –299.05
∑
V ERT LOAD U +2 − 5
L 135 IN. 150.93
A 68.5 IN 2 . 8.82
M EMB C OL W.B.
FUL 12497 k / IN. = = 0.4309 IN. AE 29000 k / IN.2
∆ BH/V =
F +209.9 K 0
∑
U +2 − 5
L 135 IN. 150.93
A 68.5 IN 2 . 8.82
FOL 827.3k/IN. = = 0.0285 IN. AE 29000k/IN.2
∆ BH/TOT = 0.4309 + 0.0285 = 0.4594 IN.
V ERTICAL D ISPLACEMENT V ERT LOAD
H ORIZ LOAD M EMB C OL W.B. ∆ BV/ H =
F +267.48 K –299.05
∑
U +1 0
L 135 IN. 150.93
A 68.5 IN 2 . 8.82
M EMB C OL W.B.
FUL 527.2 k / IN. = = 0.0182IN. AE 29000 k / IN.2
∆ BV / V =
F +209.9 K 0
∑
U +1 0
L 135 IN. 150.93
A 68.5 IN 2 . 8.82
FUL 413.7 k / IN. = = 0.0143 IN. AE 29000 k / IN.2
∆ BV/TOT = 0.0182 + 0.0143 = 0.0325 IN.
18 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 15 JOB NO.
OF
24
S TRUCTURAL D RIFT - S ECTION 1 FOR S LAB A ND T RUSS S HEARS, S EE S HT. 6 S LAB D ISPLACEMENT T RUSS D ISP . EXCEPT 2 ND S TORY T RUSS D ISP . - 2 ND S TORY
S TORY 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
= S LAB S HEAR × 0.938×10 –4 IN ./KIP = T RUSS S HEAR × 0.790×10 –3 IN ./KIP = T RUSS S HEAR × 1.057×10 –3 IN ./KIP
S LAB A BOVE S HEAR, D RIFT, IN . KIPS 8.32 24.96 41.60 58.24 74.36 87.36 99.84 112.06 122.46 WIND B RACE -
0.0008 0.0023 0.0039 0.0055 0.0070 0.0082 0.0094 0.0105 0.0115 S EE S HT. 14
T RUSS
S EE S HT. 8 S EE S HT. 12 S EE S HT. 13
S HEAR,* KIPS
D RIFT, IN .
S LAB B ELOW S HEAR, D RIFT, IN . KIPS
8.32 24.96 41.60 58.24 74.88 91.52 108.16 124.80 141.44 156.00 168.48 180.96 193.44 205.92 218.40 229.84 240.24 249.60 257.92
0.0066 0.0197 0.0329 0.0460 0.0592 0.0723 0.0854 0.0986 0.1117 0.1232 0.1331 0.1430 0.1528 0.1627 0.1725 0.1816 0.1898 0.1972 0.2726
16.64 33.28 49.92 66.56 81.12 93.60 106.08 117.52 126.88 -
0.0016 0.0031 0.0047 0.0062 0.0076 0.0088 0.0100 0.0110 0.0119 -
D RIFT P ER S TORY , IN . 0.0066 0.0221 0.0329 0.0514 0.0592 0.0809 0.0854 0.1103 0.1117 0.1378 0.1331 0.1600 0.1528 0.1821 0.1723 0.2031 0.1898 0.2206 0.2726 0.4309
* T RUSS S HEAR FOR 2 B AYS
19 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. 16 JOB NO.
OF
S TRUCTURAL D RIFT - S ECTION 1 FOR S LAB A ND T RUSS S HEARS, S EE S HT. 6 S LAB D ISPLACEMENT T RUSS D ISP E XCEPT 2 ND S TORY T RUSS D ISP . - 2 ND S TORY
S TORY 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
S LAB A BOVE S HEAR, D RIFT, IN . KIPS 2.08 0.0002 16.64 0.0016 33.28 0.0031 49.92 0.0047 66.56 0.0062 81.12 0.0076 93.60 0.0088 106.08 0.0100 117.52 0.0110 126.88 0.0119 2.39 0.0002
= S LAB S HEAR × 0.938×10 –4 IN ./K = T RUSS S HEAR × 0.790×10 –3 IN ./K = T RUSS S HEAR × 1.057×10 –3 IN ./K
T RUSS
S EE S HT. 8 S EE S HT. 12 S EE S HT. 13
S LAB B ELOW S HEAR,* D RIFT, S HEAR, D RIFT, IN . IN . KIPS KIPS 8.32 0.0066 8.32 0.0008 24.96 0.0197 41.60 0.0329 24.96 0.0023 58.24 0.0460 74.88 0.0592 41.60 0.0039 91.52 0.0723 108.16 0.0854 58.24 0.0055 124.80 0.0986 141.44 0.1117 74.36 0.0070 156.00 0.1232 168.48 0.1331 87.36 0.0082 180.96 0.1430 193.44 0.1528 99.84 0.0094 205.92 0.1627 218.40 0.1725 112.06 0.0105 229.84 0.1816 240.24 0.1898 122.46 0.0115 249.60 0.1972 257.92 0.2726 –2.39 –0.0002 0.4309 D RIFT/W IND B RACE - S ECT. 1→
* T RUSS S HEAR FOR 2 B AYS.
20 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
D RIFT P ER S TORY , IN . 0.0076 0.0197 0.0368 0.0460 0.0662 0.0723 0.0956 0.0986 0.1249 0.1232 0.1489 0.1430 0.1710 0.1627 0.1930 0.1816 0.2123 0.1972 0.2843 0.4311
24
BY R EL CHKD. BY
DATE DATE
SUBJECT
C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 17 JOB NO.
OF
24
C OLUMN - LENGTH C HANGE
∆ C + / STORY =
∑∆ ∑∆
C+
=
∑ VERT R EACT × S TORY H EIGHT ∑ VERT R EACT × 104 * = C OL A REA × 29000 C OL A REA × 29000
∑ (∆
C + / STORY )i
WHERE
n IS THE NUMBER OF STORIES BELOW THE TRUSS REACTION POINTS.
i = 1,n
C−
∆ D/CLC
∑∆ = [∑ ∆
=−
∆ D /CLC = 2
C+
(LIKE COLUMNS SUPPORTING EACH END OF THE TRUSS ARE ASSUMED.)
C+
+
∑∆
[∑ ∆ ] 60'8'−−80
C-
] DL
T T
[∑ ∆ ] DL
=2
T
C+
T
C+
∑ V ERT R EACT
= T OTAL COLUMN LOAD IN A PARTICULAR STORY AS A RESULT OF WIND LOADS ACTING ON ALL THE TRUSSES SUPPORTED BY THAT COLUMN SEGMENT.
∆ CT / STORY ∑ ∆ c+
= =
P OSITIVE LENGTH CHANGE IN A PARTICULAR COLUMN SEGMENT CAUSED BY ∑V ERT. R EACT. T HE SUM OF THE P OSITIVE LENGTH CHANGES OF THE COLUMN SEGMENT-SUPPORTING A PART
∆ D / CLC
=
CULAR TRUSS
T HE DISPLACEMENT OF A TRUSS IN A PORT CULAR STORY ( DRIFT PER STORY) CAUSED BY THE SUM OF THE LENGTH CHANGES OF THE COLUMN SEGMENT SUPPORTING THAT TRUSS.
* 1 ST S TORY = 135. ∆ C + CALCULATED AS PART OF THE WIND BRACE.
21 FIRST QUARTER / 1983
BY R EL CH KD. BY
DATE DATE
SUBJEC T CALCULATION OF WIND D RIFT IN S TAGGERE D- T RUSS B UILDINGS
SHEET NO . 18 JOB NO.
OF
C OLUM N LENGT H C HANGE V ERTICA L C OLUM N R EACTION S D UE T O W IND S HEAR V ER T. R EACTIO N = T RUSS S HEA R ×
8'−8 60FT.
S ECTIO N 1
S ECTIO N 2
S TORY
T RUSS* S HEA R, k
V ERT R EAC T, k
∑ V ERT R EAC T, k
20 18 16 14 12 10 8 6 4 2
8.32 41.60 74.88 108.16 141.44 168.48 193.44 218.40 240.24 257.92
1.202 6.009 10.816 15.623 20.430 24.336 27.941 31.547 34.701 37.255
1.202 7.211 18.027 33.650 54.080 78.416 106.357 137.904 172.605 209.860
S TORY
T RUSS* S HEA R, k
V ERT R EAC T, k
∑ V ERT R EAC T, k
19 17 15 13 11 9 7 5 3
24.96 58.24 91.52 124.80 156.00 180.96 205.92 229.84 249.60
3.605 8.412 13.220 18.027 22.533 26.139 29.744 33.199 36.053
3.605 12.017 25.237 43.264 65.797 91.936 121.680 154.879 190.932
* INCLUDE S T RUSS S HEA R FROM TH E A DJACEN T BAY .
22 ENG INEER ING JOURNA L / AMER ICAN INSTITUTE O F STEE L CONSTRUCT ION
24
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 19 JOB NO.
OF
24
C OLUMN LENGTH C HANGE S ECTION 1 (FOR DEFINITION OF S YMBOLS, S EE S HEET 17.)
S TORY
C OLUMN S ECTION
C OLUMN A REA , IN 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
W14×233 W14×233 W14×211 W14×211 W14×193 W14×193 W14×159 W14×159 W14×145 W14×145 W14×120 W14×120 W14×90 W14×90 W14×74 W14×74 W14×53 W14×53 W14×43 W14×43
68.5 68.5 62.0 62.0 56.8 56.8 46.7 46.7 42.7 42.7 35.3 35.3 26.5 26.5 21.8 21.8 15.6 15.6 12.6 12.6
∑ V ERT R EACT, K 209.86 209.86 172.61 172.61 137.90 137.90 106.36 106.36 78.42 78.42 54.08 54.08 33.65 33.65 18.03 18.03 7.21 7.21 1.20 1.20
∆C+ S TORY , IN .
∑ ∆ C +, IN . .
0.0325* 0.0110 0.0100 0.0100 0.0087 0.0087 0.0082 0.0082 0.0066 0.0066 0.0055 0.0055 0.0046 0.0046 0.0030 0.0030 0.0017 0.0017 0.0003 0.0003
0.0325 0.0435 0.0535 0.0635 0.0722 0.0809 0.0891 0.0973 0.1039 0.1105 0.1160 0.1215 0.1261 0.1307 0.1337 0.1367 0.1384 0.1401 0.1404 0.1407
P ER S TORY D RIFT (∆ D/CLC ), IN . 0.0285* 0.0126 0.0183 0.0234 0.0281 0.0319 0.0351 0.0378 0.0395 0.0405 0.0406
* W IND B RALE, S EE S HEET 14.
23 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED - T RUSS B UILDINGS
SHEET NO. 20 JOB NO.
OF
24
C OLUMN LENGTH C HANGE S ECTION 2 (FOR DEFINITION OF S YMBOLS, S EE S HT. 17)
S TORY
∑ V ERT R EACT, K
C OLUMN* A REA , IN 2
∆C+ S TORY , IN .
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
190.93 190.93 190.93 154.88 154.88 121.68 121.68 91.94 91.94 65.80 65.80 43.26 43.26 25.24 25.24 12.02 12.02 3.61 3.61 -
68.5 68.5 62.0 62.0 56.8 56.8 46.7 46.7 42.7 42.7 35.3 35.3 26.5 26.5 21.8 21.8 15.6 15.6 12.6 12.6
0.0130 0.0100 0.0110 0.0090 0.0098 0.0077 0.0093 0.0071 0.0077 0.0055 0.0067 0.0044 0.0059 0.0034 0.0042 0.0020 0.0028 0.0008 0.0010 -
∑∆ C +, IN .
P ER S TORY D RIFT (∆ D /CLC ), IN .
D RIFT/S TORY , IN .** IN . S ECT. 1 S ECT.2
0.0130 0.0230 0.0340 0.0430 0.0528 0.0605 0.0698 0.0769 0.0846 0.0901 0.0968 0.1012 0.1071 0.1105 0.1147 0.1167 0.1195 0.1203 0.1213 0.1213
0.0098 0.0153 0.0202 0.0244 0.0280 0.0309 0.0331 0.0345 0.0350 -
0.0285 0.0126 0.0098 0.0183 0.0153 0.0234 0.0202 0.0281 0.0244 0.0319 0.0280 0.0351 0.0309 0.0378 0.0331 0.0395 0.0345 0.0405 0.0350 0.0406
* FOR C OLUMN S ECTION, S EE S HT. 19 ** B ECAUSE THE FLOORS FUNCTION AS VERY EFFICIENT DIAPHRAMS, THE DRIFT CAUSED BY THE LENGTH CHANGE OF THE COLUMNS IS THE SAME FOR S ECTION 1 AND S ECTION 2. IT IS D ETERMINED IN EACH STORY BY THE TRUSS IN THAT STORY .
24 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 21 JOB NO.
OF
24
FINAL D RIFT D ETERMINATION S ECTION 1 S TRUCT
C OL LGTH
T OTAL D RIFT
T OTAL
S TORY
D RIFT
C HG
P ER S TORY
D RIFT
D RIFT R ATIO P ER S TORY
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0.4309 0.2726 0.2206 0.1898 0.2031 0.1723 0.1821 0.1528 0.1600 0.1331 0.1378 0.1117 0.1103 0.0854 0.0809 0.0592 0.0514 0.0329 0.0221 0.0066
0.0285 0.0126 0.0098 0.0183 0.0153 0.0234 0.0202 0.0281 0.0244 0.0319 0.0280 0.0351 0.0309 0.0378 0.0331 0.395 0.0345 0.0405 0.0350 0.0406
0.4594 0.2852 0.2304 0.2081 0.2184 0.1957 0.2023 0.1809 0.1844 0.1650 0.1658 0.1468 0.1412 0.1232 0.1140 0.0987 0.0859 0.0734 0.0571 0.0472
0.4594 0.7446 0.9750 1.1831 1.4015 1.5972 1.7995 1.9804 2.1648 2.3298 2.4956 2.6424 2.7836 2.9068 3.0208 3.1195 3.2054 3.2788 3.3359 3.3831
0.0034 0.0027 0.0022 0.0020 0.0021 0.0019 0.0019 0.0017 0.0018 0.0016 0.0016 0.0014 0.0014 0.0012 0.0011 0.0009 0.0008 0.0007 0.0005 0.0005
H EIGHT = 164′–8+11′–3 = 175′–11 = 2111 IN . 3.3831 = 0.0016 O VERALL D RIFT R ATIO = 2111. A LL D IMENSIONS ARE IN INCNES.
25 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 22 JOB NO.
FINAL D RIFT D ETERMINATION S ECTION 2
S TORY
S TRUCT D RIFT
C OL LGTH C HANGE
T OTAL D RIFT P ER S TORY
T OTAL D RIFT
D RIFT R ATIO P ER S TORY
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0.4311 0.2843 0.1972 0.2123 0.1816 0.1930 0.1627 0.1710 0.1430 0.1489 0.1232 0.1249 0.0986 0.0956 0.0723 0.0662 0.0460 0.0368 0.0197 0.0076
0.0285 0.0126 0.0098 0.0183 0.0153 0.0234 0.0202 0.0281 0.0244 0.0319 0.0280 0.0351 0.0309 0.0378 0.0331 0.0395 0.0345 0.0405 0.0350 0.0406
0.4596 0.2969 0.2070 0.2306 0.1969 0.2164 0.1829 0.1991 0.1674 0.1808 0.1512 0.1600 0.1295 0.1334 0.1054 0.1057 0.0805 0.0773 0.0547 0.0482
0.4596 0.7565 0.9635 1.1941 1.3910 1.6074 1.7903 1.9894 2.1568 2.3376 2.4888 2.6488 2.7783 2.9117 3.0171 3.1228 3.2033 3.2806 3.3353 3.3835
0.0034 0.0029 0.0020 0.0022 0.0019 0.0021 0.0018 0.0019 0.0016 0.0017 0.0015 0.0015 0.0012 0.0013 0.0010 0.0010 0.0008 0.0007 0.0005 0.0005
H EIGHT = 164′–8+11′–3 = 175′–11 = 2111. IN . O VERALL D RIFT R ATIO =
3.3835 = 0.0016 2111.
A LL D IMENSIONS ARE IN INCHES.
26 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
OF
24
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 23 JOB NO.
OF
24
C ONTROL OF D RIFT 1 ST S TORY
KNEE B RACE –2C9×15
INCREACE–2C10×30
R EFER TO S HEET 14. C ONSIDER ONLY H ORIZONTAL D ISPLACEMENT OF THE KNEE B RACE CAUSED BY H ORIZONTAL LOAD .
H ORIZ. LOAD
M EMB C OL K.B R .
F +267.48 k –299.05
U +2 – 5
L 135 IN . 150.93
∑
S H/ H =
∑ AE
FUL
∆1ST STORY DRIFT
FUL A 1054.3 S721.4
A 68.5 IN . 2 17.64 FUL A
=
6775.7
6775.7 = 0.2336 IN. 29000 = −0.4309(S HT 14) IN. . − 01973 + 0.2336 =
R EVISED D RIFT
S ECTION 1
S ECTION 2
O RIGINAL D RIFT = 0.4594 IN . ∆1 ST S TORY D RIFT = –0.1973 IMPROVED D RIFT = 0.2621 IN . 0.2621 D RIFT R ATIO = = 0.0019 135
O RIGINAL D RIFT = 0.4596 IN . ∆1 ST S TORY D RIFT= –0.1973 IMPROVED D RIFT = 0.2623 IN . 0.2623 D RIFT R ATIO = = 0.0019 135
27 FIRST QUARTER / 1983
BY R EL CHKD. BY
DATE DATE
SUBJECT C ALCULATION OF W IND D RIFT IN S TAGGERED. T RUSS B UILDINGS
SHEET NO. 24 JOB NO.
C ONTROL OF D RIFT 2 ND S TORY – T RUSS 2 ND S TORY D RIFT R ATIO A PPROACHES 0.003 T O A TTAIN A D RIFT R ATIO OF 0.002, USE 2 C 8×11.5 FOR ALL D IAGS 2C6×10.5 FOR ALL V ERTS W10×45 FOR B OTT. C HD N OTE: N O NEW S HAPES A DDED TO THE P ROJECT . A XIAL E FFECTS (V IRTUAL W ORK )
M EMBER T OP C HORD D IAGONAL
V ERTICAL
B ETT. C HORD
F O 0.2083 k 0.2083 0.2778 0.1445 0.1686 0.0963 0.5000 0.3500 0.2000
U
L
A
0.2083 0.2083 0.2778 0.1445 0.1686 0.0963 0.5000 0.3500 0.2000
149.93 IN
6.76 IN 2
104
6.18
108
13.3
∑
E∆ A = ∑
FUL A
FUL A 0.962 K /IN . 0.962 1.712 0.351 0.478 0.156 2.030 0.995 0.325
( 1 2 TRUSS)
7.971 K /IN .
FUL (FULL T RUSS) A
15.942 K /IN .
C OMBINED D ISPLACEMENT ∆ AXIAL ∆ BENDING
= 0.550 × 10 -3 IN ∆ = 0.730 × 10 −3 IN −4 −3 TR = 2 × 0.902 × 10 (S HT. 11) = 0.180 × 10 = 15.942 / 29000
∆ TR − 2ND FL = 0.730 × 10−3 IN / k
S ECTION 1
S ECTION 2
TRUSS = 257.92 × 0.730 × 10 −3 = 0.1883 IN. C OL L.C HG = 0.0126
TRUSS = 01883 . S LAB = 0.0117
0.2009 IN.
D RIFT R ATIO =
COL L.C HG
0.2009 = 0.0019 104
D RIFT R ATIO =
0.2000 0.2126IN. 0.0126
0.2126 = 0.0020 104
28 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCTION
OF
24