2003 Ibc Diaphragm Design

2003 IBC Chapter 16 Seismic Design Diaphragms

1

SCOPE Diaphragm Design • • • • • •

Diaphragm System Review Load Combinations Vertical Distribution of Horizontal Loads Diaphragm Loads Diaphragm Design Openings in Diaphragms

Wall Anchorage • Wall Support • Sub-diaphragms 2

Lateral Force Resisting Diaphragm System

3

Lateral Force Resisting System

4

Diaphragm System Diaphragm design depends on type of diaphragm

Flexible Diaphragm

Computed maximum in-plane deflection of the diaphragm itself is more than 2 times the average drift of the adjoining vertical elements of the lateral force resisting system Per Simplified Design Section 1617.5.3, untopped steel decking or wood panel diaphragms can be considered flexible

Rigid Diaphragm

5

Diaphragm System

6

Flexible Diaphragms Load is transferred to lateral resisting elements based on tributary width

7

Flexible Diaphragms

q = wL/2W q = diaphragm shear w = lateral load to diaphragm L = length of diaphragm W = depth of diaphragm

8

Rigid Diaphragms Rigid Diaphragm Analysis includes torsional moments with accidental torsion Rigid Diaphragms using Equivalent Lateral Force Procedure in SDC C, D, E or F with Type 1 torsional irregularity per Table 9.5.2.3.2 must have the accidental torsional moment, Mta, multiplied by Ax,

Ax

;max 1.2;avg

2

Ax need not exceed 3.0 9

Rigid Diaphragms Load to vertical lateral resisting elements is based on the rigidity of the elements R= 1/ Must locate center of gravity And center or rigidity

xr

Riy x i Riy

yr

Rix y i Rix

10

Rigid Diaphragms Distance between center of mass and center of rigidity, e, produces a torsional moment under seismic lateral load

11

Rigid Diaphragms

12

Rigid Diaphragms The lateral force is distributed to vertical lateral force resisting elements accounting for direct shear and torsional shear using the equations:

Vy Vx

Riy i

i

Riy

Fpy

Rix Fpx Rix

Riy x' Jr

Fpy e x

Rix y' Fpx e y Jr

Jr = relative polar moment of intertia = (Rixy’2+Riyx’2)

13

Load Combinations Section 1605 1605.2 Strength Design 1605.3 Allowable Stress Design 1605.3.3 Alternate Basic Load Combinations • ASD Load Combination • Increase in allowable stress permitted

14

Load Combinations 1605.3.2 Alternate Basic Load Combination D+L+S+E/1.4 0.9D+E/1.4 1605.4 Special Seismic Load Combinations 1.2D+f1L+Em 0.9D+Em

Em = Maximum effect of horizontal and vertical forces (1617.1) f1 = 1.0 for floors in places of public assembly, live loads in excess of 100 psf and parking garage live loads = 0.5 for other live loads 15

Load Combinations 1617.1 E = QE + 0.2SDSD E = QE - 0.2SDSD

= Redundancy Coefficient (1617.2) 1.0 for design forces for diaphragms and wall anchorage QE = Effect of horizontal seismic forces SDS = Design spectral response acceleration at short periods 16

Load Combinations 1617.1, Maximum Seismic Load Effect Em = QE +0.2SDSD Em = QE – 0.2SDSD = System Overstrength Factor (Table 1617.6.2) An allowable stress increase of 1.7 (not to be combined with 1/3 allowable stress increase due for wind or seismic loads) is permitted for ASD designs Term QE need not exceed force that can be transferred to the element by the other elements of the lateral force resisting system 17

Load Combinations For designs utilizing ASCE 7, Equivalent Lateral Force Procedure, the Special Seismic Load is E = QE +0.2SDSD E = QE – 0.2SDSD This E is then used in the load combinations from ASCE 7 (Same as Strength Design or basic ASD combinations from IBC) An allowable stress increase of 1.2 is permitted for ASD designs 18

Analysis Method

1. Equivalent Lateral Force Procedure ASCE 7-02 Section 9.5.5

2. Simplified Analysis Permitted for: Seismic Use Group I structures if 1. Buildings of light framed construction not exceeding 3 stories in height 2. Buildings of any construction not exceeding 2 stories with flexible construction at every level

3. Dynamic Analysis ASCE 7-02 Sections 9.5.6, 9.5.7 or 9.5.8 19

Analysis Method For structures designed using the Simplified Analysis Procedures, the requirements of Sections 1620.21620.5 (IBC) must be met. Exception: Structures in SDC A For structures designed using the Equivalent Lateral Force Procedure, the requirements of 9.5.2.6 (ASCE 7) must be met 20

Simplified Procedure 1617.5.1 Seismic Base Shear

V

1.2SDS W R

(EQ. 16-56)

R = Response modification factor (Table 1617.6.2) W = Effective weight of structure

21

Simplified Procedure 1617.5.2 Vertical Distribution of Horizontal Forces

Fx

1.2SDS wx R

(EQ. 16-57)

wx = Portion of effective weight of structure, W, at Level x. 22

Simplified Procedure 1620.2.5 Diaphragms Designed to resist force:

Fp = 0.2IESDSwp + Vpx (EQ. 16-60) wp = weight of diaphragm and other elements attached to diaphragm Vpx = portion of seismic shear force required to be transferred to lateral force resisting elements through diaphragm from other lateral force resisting elements due to offsets or changes in stiffness of the lateral force resisting elements above or below the diaphragm 23

Simplified Procedure

24

Simplified Procedure 1620.4.3 Diaphragms in SDC D n

Fpx

i x n i x

Fi wi

wpx

25

Simplified Procedure Level Ground 1 2 3 Roof 4

Z

Weight, wi w1 w2 w3 w4

Fi = Fx Zw1 Zw2 Zw3 Zw4

Fpx

1.2SDS R 26

Simplified Procedure n

Fpx

i x n i x

Fp1

Fi wi

wpx

Zw1 Zw2 Zw3 Zw 4 w1 w1 w2 w3 w4 Z(w1 w2 w3 w 4 ) w1 (w1 w2 w3 w 4 )

Zw1 27

Simplified Procedure Level Ground 1 2 3 Roof 4

Z

1.2SDS R

Weight, wi w1 w2 w3 w4

Fi = Fx Zw1 Zw2 Zw3 Zw4

Fpx** Zw1 Zw2 Zw3 Zw4

** Fpx max = 0.4SDSIEwpx Fpx min = 0.2SDSIEwpx

28

Simplified Procedure If R > (# below), then Fpx min controls

If R < (# below), then Fpx max controls

6

3

1.25

4.8

2.4

1.50

4

2

IE

1.0

29

Equivalent Lateral Force Procedure ASCE 7 9.5.5.2 Seismic Base Shear V = CsW SD1 SDS need not be Cs Cs T(R/I) R/I greater than but not less than C s 0.044SDSI and for SDC E 0.5S1 and F not less than C s

R/I

30

Equivalent Lateral Force Procedure 9.5.5.4 Vertical Distribution of Seismic Forces

Fx

C vx V

C vx

k x x

wh n i 1

k i i

wh

Cvx = vertical distribution factor wi and wx = portion of total gravity load, W, assigned to Level i or x hi and hx = height from base to Level i or x k = 1.0 if period, T = 0.5s or less = 2.0 if T = 2.5s or more 31 use linear interpolation for periods between 0.5 and 2.5

Equivalent Lateral Force Procedure 9.5.2.6.2.7 Diaphragms Must resist the larger of 1. The portion of the design seismic force at the level of the diaphragm that depends on the diaphragm to transmit forces to the vertical elements of the lateral force resisting system 2. Fp = 0.2SDSIwp + Vpx 32

Equivalent Lateral Force Procedure 9.5.2.6.4.4 Diaphragms in SDC D n

Fi Fpx

i x n

w px wi

i x

33

Diaphragm Design Example 3 story CMU bearing special reinforced shear wall building with 3 foot parapet Level 2 and 3 concrete diaphragms on metal deck Roof steel roof deck diaphragm SDS = 0.50 R = 5.0 (Table 1617.6.2) SDC D T = 0.4 seconds I = 1.0 No plan irregularities Floor DL = 60 psf Roof DL = 15 psf Wall DL = 80 psf

Wall Rigidities R1 = .2 R3 = .1 R2 = .1 R4 = .3

Analysis for Diaphragm Design 1. Cannot use Simplified Analysis per section 1616.1; we don't have light framed construction and we exceed 2 stories

34

Diaphragm Design Example

35

Diaphragm Design Example

36

Diaphragm Design Example 2. Using the Equivalent Lateral Force Procedure from ASCE 7-02, find the base shear V = CsW (Eq. 9.5.5.2-1) Weight tributary to level 1 = 80psf*12'/2*(2*40'+2*60') = 96,000 pounds Weight tributary to level 2 and 3 = 80psf*12'*(2*40'+2*60') + 60psf*(40'*60')= 336,000 pounds Weight tributary to level 4 = 80psf*(12'/2+3)*(2*40'+2*60') + 15psf*(40'*60') = 180,000 pounds

Cs = SDS/(R/I) = .50/(5/1) = 0.10 CHECK OTHER Cs EQUATIONS

37

Diaphragm Design Example Level Ground 1 2 3 Roof 4

Weight, w Height, h 96,000 0 336,000 12 336,000 24 180,000 36 948,000

wihi

Cvx

Fx

Fpx

0

V = CsW=0.10*948,000 = 94,800 pounds 38

Diaphragm Design Example 3. Determine the vertical distribution of Seismic Forces Fx = CvxV

C

w vx

xh

k x

n

w ih

k = 1 (T<0.5)

k i

i 1

Level

Weight, w

Height, h

w ih i

Cvx

Fx

Ground 1

96,000

0

0

0

0

2

336,000

12

4032000

0.2171

20577

3

336,000

24

8064000

0.4341

41153

Roof 4

180,000

36

6480000

0.3488

33070

948,000

Fpx

18576000 39

Diaphragm Design Example 4. Determine forces to diaphragm at each level n

Fpx (max) = 0.4SDSIwx Fpx (min) = 0.2SDSIwx

Fi Fpx

i x n

w px

Fp3 = {(F3+F4)/(w3+w4)}w3

wi i x

Level

Weight, w

Height, h

w ih i

Cvx

Fx

Fpx

Ground 1

96,000

0

0

0

0

9600

2

336,000

12

4032000

0.2171

20577

37386

3

336,000

24

8064000

0.4341

41153

48331

Roof 4

180,000

36

6480000

0.3488

33070

33070

948,000

18576000 40

Diaphragm Design Example 5. Determine diaphragm shears to design diaphragm Level 4 Flexible Diaphragm Diaphragm Shear due to Y direction load q1 = q3 = (Fp/2)/b = (33070/2)/40 = 413 plf (Ultimate) ASD Load Combinations: E/1.4 q1 = 413/1.4 = 295 plf 41

Diaphragm Design Example

42

Diaphragm Design Example

43

Diaphragm Design Example Level 3 Rigid Diaphragm Center of Rigidity xr = Ryixi/ Ryi = (R1(0)+R3(60))/(R1+R3) = (0.20*(0)+0.10*(60))/(0.20+0.10) xr = 20 feet yr = Rxiyi/ Rxi = (R2(40)+R4(0))/(R2+R4) = (0.10*(40)+0.30*(0))/(0.10+0.30) = yr = 10 feet Center of Mass = center of diaphragm ex = 30-20 = 10 feet plus 5% accidental torsion ex = 10 +0.5*60 = 13 feet

44

Diaphragm Design Example Direct Shear Vyit = (Ryi/ Ryi)Fpy Vy1t = (0.20/(0.20+0.10))*48331 = 32221 pounds Vy3t = (0.10/(0.20+0.10))*48331 = 16110 pounds

45

Diaphragm Design Example Shear due to torsion Vyir = (Ryix'/( Ryix'2+ Rxiy'2))Fpyex Ryix'2 = 0.20(20)2+0.10((60-20)2 = 240 Rxiy'2 = 0.10(40-10)2+0.30((10)2 = 120 Vy1r = (0.20*20/(240+120))*48331*(13) = 6981 pounds Vy3r = (0.10*(60-20)/(240+120))*48331*(13) = 6981 pounds

46

Diaphragm Design Example Substituting Rxiyi for Ryixi in above equation to find shear in wall 2 and 4 from torsional shear due to load in Y-direction Vy2r = (0.10*30/(240+120))*48331*(13) = 5236 pounds Vy4r = (0.30*10/(240+120))*48331*(13) = 5236 pounds X-direction load will likely control shear for wall lines 2 and 4 Do not decrease shear in wall due to negative torsional shear 47

Diaphragm Design Example V1 = Vy1t + Vy1r = 32221 pounds q1 = V1/W = 32221/40 = 806 plf (Ultimate) ASD Load Combinations: E/1.4 q1 = 806/1.4 = 576 plf V3 = Vy3t + Vy3r = 16110+6981 = 23091 pounds q3 = V3/W = 23091/40 = 577 plf (Ultimate) ASD Load Combinations: E/1.4 q3 = 577/1.4 = 412 plf 48

Diaphragm Design Example

49

Diaphragms with Openings Analysis of diaphragms with large openings assumes diaphragm behaves similar to Vierendeel Truss. Example:

50

Diaphragms with Openings Step 1 – Analyze Diaphragm as though no openings existed Line 1 V1 = wL/2 = 400(60)/2 = 12,000 # q1 = V1/W = 12,000/40 = 300 plf Line 2 V2 = w(L/2-x) = 400(60/2-10) = 8000 # q2 = 8000/40 = 200 plf M2 = (wx/2)*(L-x) = (400*10/2)*(60-10) = 100,000 #ft T=C=M/d F2@a = 100,000/40 = 2500 # C F2@d = 2500 # T Line 3 V3 = 400(60/2-15) = 6000 #; q3 = 6000/40 = 150 plf M3 = (400*15/2)*(60-15) = 135,000 #ft F3@a = 135,000/40 = 3375 # C; F3@d = 3365 # T Line 4 V4 = 400(60/2-20) = 4000 #; q4 = 4000/40 = 100plf M4 = (400*20/2)*(60-20) = 160,000 #ft F4@a = 160,000/40 = 4000 # C. F4@d = 4000 # T Line 5 V4 = 400(60/2-30) = 0 #; q4 = 0 plf M4 = (400*30/2)*(60-30) = 180,000 #ft F4@a = 180,000/40 = 4500 # C. F4@d = 4500 # T

51

Diaphragms with Openings Step 2. Determine the shears and chord forces at the edges of the openings using free-body diagrams for each segment Segment A V4(ab)=V4/2=4000/2= 2000 # q4(ab)=V4(ab)/15’=2000/15= 133 plf V3(ab)=V4+400(5’)=2000+2000 = 4000 # q3(ab)=4000/15’= 267 plf F4@a M3b=-3375(15)-2000(5)400(52/2)+F4@b(15) F4@a=4375# C F4@b Fx=-4375+3375+F4@b F4@b=1000# T

52

Diaphragms with Openings Segment B V2(ab)= V3(ab)+400(5)=4000+2000 = 6000 q2(ab)=6000/15= 400 plf F2@a M3b=-F2@a(15)+3375(15) +400(52/2)-6000(5) F4@a=1708# C F4@b Fx=-3375+1708+F4@b F4@b=1667# T

4375#

4000#

2000# 1000#

53

Diaphragms with Openings Segment C V4(cd)=V4/2=4000/2= 2000 # q4(cd)=V4(ab)/15’=2000/15= 133 plf V3(ab) Fy V3(ab)= 2000 # q3(cd)= 133 plf F4@c=V4(cd)(5’)/15’=2000(5)/15 = 667 # C F4@d=3375+667= 4042 # T

1708#

6000#

1667#

4375#

4000#

2000# 1000#

54

Diaphragms with Openings Segment D V2(cd) Fy V2(cd)= 2000 # q2(cd)= 133 plf F2@c=V4(cd)(5’)/15’=2000(5)/15 = 667 # T F2@d=3375-667= 2708 # T

1708#

6000#

4375#

4000#

2000# 1000#

1667#

667#

2000#

2000#

4042# 55

Diaphragms with Openings Step 3 The net change in the chord forces caused by the openings is determined by adding the results of step 2 to that of the diaphragm without openings, these net changes must be dissipated into the diaphragm

56

Diaphragms with Openings Chord Force (lbs) Diaphragm Force Location

Without Openings

With Openings

Net Change Due to Openings

F2@a

2500 C

1708 C

792 T

F2@b

0

1667 C

1667 C

F2@c

0

667 T

667 T

F2@d

2500 T

2708 T

208 T

F4@a

4000 C

4375 C

375 C

F4@b

0

1000 T

1000 T

F4@c

0

667 C

667 C

F4@d

4000 T

4042 T

42 T 57

Diaphragms with Openings

58

Diaphragms with Openings

59

Diaphragms with Openings Step 4 Determine resultant shears in diaphragm by combining the net shears due to openings to the shears for the diaphragm without openings

60

Diaphragms with Openings Shear (plf) Diaphragm Shear and Location

Without Openings

Due to Openings

Resultant Shear

V1 @ a to b

300

-79.2

+220.8

V1 @ b to c

300

+87.5

+387.5

V1 @ c to d

300

+20.8

+320.8

V2 @ a to b

200

-79.2

+120.8

V2 @ b to c

200

+87.5

+287.5

V2 @ c to d

200

+20.8

+220.8

V4 @ a to b

100

-37.5

+62.5

V4 @ b to c

100

+62.5

+162.5

V4 @ c to d

100

-4.2

+95.8

V5 @ a to b

0

-37.5

-37.5

V5 @ b to c

0

+62.6

+62.6

V5 @ c to d

0

-4.2

-4.2

61

Diaphragms with Openings Step 5 Determine forces in the framing members in the direction perpendicular to the applied load by adding the shear forces at the edge of the opening

62

Diaphragms with Openings

63

Other Considerations Must verify plan and vertical irregularities (Tables 1616.5.1.1 and 1616.5.5.2) Verify diaphragm requirements specific to material being used

64

Collector Elements Collect force from diaphragms and transfer them to shear walls (drag struts)

65

Collector Elements SDC B and C Must have design strength to resist the special load combinations of 1605.4 Exception: Structures that use light framed shear walls entirely Note: Collector need not be designed for a force that exceeds the force that can be transferred to it from other members 66

Collector Elements SDC D, E and F Must have design strength to resist the special seismic load combinations Must resist forces in accordance with diaphragm forces required for SDC D Exception: Structures that use light framed shear walls entirely Note: Collector need not be designed for a force that exceeds the force that can be transferred to it from other members

67

Collector Element Example

Diaphragm 1

Diaphragm 2

68

Collector Element Example

Fpx

Tearing at Discontinuity

Fpx

69

Collector Element Example

Tearing at Discontinuity

70

Collector Element Example Design of Collector for Forces in Y-Direction 1. Determine diaphragm shear for large component along line of collector For given Fpy1=100 k, qy1 = 100k/2/200’ = 250 plf 2. Determine diaphragm shear for small component along line of collector For given Fpy2=300 k, qy2 = 30k/2/100’ = 150 plf 3. Determine load in collector QE = (250+150)*100’ = 40,000 # = 40 k

71

Collector Element Example 4. Collectors required to be designed per load combinations for special seismic loads (determined from ASCE 7 or IBC for procedure used in design) For IBC, Em = QE +0.2SDSD For given = 2.5, the lateral load to the drag strut for design is Em = 2.5*40k = 100 k 5. If the drag strut carries dead loads, the additional seismic portion of the dead load must be added to the load combination 1.2D+f1L+Em = (1.2+0.2SDS)D + 100k(Em) 0.9D+Em = (0.9-0.2SDS)D + 100k(Em) An allowable stress increase of 1.7 can be used with these load combinations Note that this example meets the definition of Plan Structural Irregularity #2 per Table 1616.5.1 and therefore per section 1620.4.1, the design forces shall be increased 25% for connections of diaphragms to vertical element and for collectors to vertical elements except if using special seismic load combinations 72

Collector Element Example Design of Collector for Forces in X-Direction 1. Determine diaphragm shear for small component along line of collector (wall with slip connection) For given Fpx2=20 k, qx2 = 20k/2/100’ = 100 plf 2. Determine load in collector QE = (100)*100’ = 10,000 # = 10 k

73

Collector Element Example 3. Collectors required to be designed per load combinations for special seismic loads (determined from ASCE 7 or IBC for procedure used in design) For IBC, Em = QE +0.2SDSD For given = 2.5, the lateral load to the drag strut for design is Em = 2.5*10k = 25 k 4. Develop this drag force into the larger diaphragm. This example meets the definition of Plan Structural Irregularity #2 per Table 1616.5.1 but we are using the special seismic load combinations For given diaphragm capacity of 400plf, must extend drag strut into diaphragm length, L = 25*1.33/.4/1.7 = 47.8’ Say extend into main diaphragm 48’ 74

Collector Element Example 5. Design the larger diaphragm for Fpx of that diaphragm and add in the additional force caused by the drag strut from the smaller section. For a given Fpx1=90 k, qx1 = qx3 = 90k/2/200’+10/2/48’ = 329 plf

75

Bearing and Shear Wall Anchorage Simplified Analysis SDC B Same force as used in wall design: Fp = 0.40IESDSww ww = weight of wall

76

Wall Anchorage Simplified Analysis SDC C Must meet requirements of SDC B and For concrete or masonry walls supported by flexible diaphragm Fp = 0.8SDSIEww Supported by rigid diaphragm

77

Wall Anchorage Simplified Analysis SDC C Supported by rigid diaphragm

Fp

0.4apSDSWp R p /Ip

z 1 2 h

With component amplification factor, ap = 1.0 and component response modification factor, Rp = 2.5 z = height to point of attachment h = average roof height

Fp (max) = 1.6SDSIpWp Fp (min) = 0.3SDSIpWp

78

Wall Anchorage Simplified Analysis SDC C Additional Requirements Continuous ties or struts must be provided to transfer wall anchorage forces into diaphragm Metal deck cannot be used as tie or strut perpendicular to deck span Wood sheathing cannot be used as tie or strut Steel elements used for wall anchorage shall have the strength design forces (ultimate) increased by 1.4

79

Wall Anchorage

Simplified Analysis SDC D Must meet requirements of SDC C and Concrete and masonry walls anchored to flexible diaphragms must also Be designed for the forced induced by the eccentricity of wall anchorage connections by elements that are not perpendicular to the wall Be designed for additional forces collected by pilasters in the wall 80

Wall Anchorage Equivalent Lateral Force Procedure SDC A Fp = 0.133SDSww or minimum of Fp = 0.05ww Concrete or masonry walls Minimum E = 280 lbs/liner foot of wall 81

Wall Anchorage Equivalent Lateral Force Procedure SDC B Must meet requirements of SDC A and Concrete or masonry walls Fp = 0.4SDSIwc or minimum of Fp = 0.10wc or minimum of E = 400SDSI lbs/liner foot of wall 82

Wall Anchorage Equivalent Lateral Force Procedure SDC B Additional Requirements Connections must have sufficient ductility, rotational capacity or strength to resist shrinkage, thermal changes, and differential foundation settlement when combined with seismic forces Walls must be designed for bending if anchorage spacing exceeds 4 feet 83

Wall Anchorage Equivalent Lateral Force Procedure SDC C Must meet requirements of SDC B and For concrete or masonry walls supported by flexible diaphragm Fp = 0.8SDSIwp Supported by rigid diaphragm

Fp

0.4apSDSWp R p /Ip

z 1 2 h

Fp (max) = 1.6SDSIpWp Fp (min) = 0.3SDSIpWp Same equations as Simplified Analysis SDC C requirement except no 1.4 increase for anchor bolts and reinforcing

84

Wall Anchorage Equivalent Lateral Force Procedure SDC C Additional Requirements Same as those required for SDC C and SDC D in Simplified Analysis

85

Wall Anchorage Equivalent Lateral Force Procedure SDC D, E and F No additional requirements regarding wall anchorage forces

86

Wall Anchorage Example

87

Wall Anchorage Example

88

Wall Anchorage Example

89

Wall Anchorage Example

90

Wall Anchorage Example

91

Wall Anchorage Example

92

Sub-diaphragms Continuous ties or struts must be provided between main diaphragm chords to transfer wall anchorage forces to the diaphragm Chords may be added to form sub-diaphragms with maximum length to width ration of 2.5/1 (may be less for wood diaphragms) Wood diaphragm sheathing shall not be considered effective as providing the ties or struts In metal deck diaphragms, the metal deck shall not be considered effective as providing the ties or struts in the direction perpendicular to the 93 deck span

Sub-diaphragms

94

Sub-diaphragm Example

95

Sub-diaphragm Example

96

Sub-diaphragm Example

97

Sub-diaphragm Example

98

Seismic Design Diaphragms

QUESTIONS?

99