2003 Ibc Diaphragm Design

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2003 IBC Chapter 16

Seismic Design

Diaphragms

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SCOPE

Diaphragm Design

•

• Diaphragm System ReviewDiaphragm System Review •

• Load CombinationsLoad Combinations •

• Vertical Distribution of Horizontal LoadsVertical Distribution of Horizontal Loads •

• Diaphragm LoadsDiaphragm Loads •

• Diaphragm DesignDiaphragm Design •

• Openings in DiaphragmsOpenings in Diaphragms

Wall Anchorage

•

• Wall SupportWall Support •

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Lateral Force

Resisting

Diaphragm

System

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Lateral Force Resisting System

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Diaphragm System

Diaphragm design depends on type of

diaphragm

Flexible Diaphragm

Computed maximum in

Computed maximum in--plane deflection of theplane deflection of the

diaphragm itself is more than 2 times the average diaphragm itself is more than 2 times the average drift of the adjoining vertical elements of the lateral drift of the adjoining vertical elements of the lateral force resisting system

force resisting system

Per Simplified Design Section 1617.5.3,

Per Simplified Design Section 1617.5.3, untoppeduntopped steel decking or wood panel diaphragms can be steel decking or wood panel diaphragms can be considered flexible

considered flexible

Rigid Diaphragm

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Diaphragm System

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Flexible Diaphragms

Load is transferred to lateral resisting elements Load is transferred to lateral resisting elements

based on tributary width based on tributary width

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Flexible Diaphragms

q = wL/2W

q = diaphragm shear

w = lateral load to diaphragm

L = length of diaphragm

W = depth of diaphragm

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Rigid Diaphragms

Rigid Diaphragm Analysis includes

torsional

moments with accidental

torsion

Rigid Diaphragms using Equivalent Lateral Force Rigid Diaphragms using Equivalent Lateral Force

Procedure in SDC C, D, E or F with Type 1

Procedure in SDC C, D, E or F with Type 1 torsionaltorsional irregularity per Table 9.5.2.3.2 must have the

irregularity per Table 9.5.2.3.2 must have the accidental

accidental torsionaltorsional moment,moment, MM_ta_ta, multiplied by A, multiplied by A_x_x,,

A

A_x_x need notneed not exceed 3.0 exceed 3.0 2 avg max x

_1.2;

;

A

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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

iy i iy r

R

x

R

x

ix i ix r

R

y

R

y

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Rigid Diaphragms

Distance between center of

mass and center of rigidity, e,

produces a

torsional

moment

under seismic lateral load

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Rigid Diaphragms

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Rigid Diaphragms

The lateral force is distributed to vertical lateral The lateral force is distributed to vertical lateral force resisting elements accounting for direct force resisting elements accounting for direct shear and

shear and torsionaltorsional shear using the equations:shear using the equations:

J

J_r_r = relative polar moment of= relative polar moment of intertiaintertia = = (R(R_ix_ixyy’’22_+R_+R iy iyxx’’22)) x py r iy py iy iy i y

F

e

J

x'

R

F

R

V

y px r ix px ix ix i x

F

e

J

y'

R

F

R

V

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Load Combinations

Section 1605

1605.2 Strength Design

1605.3 Allowable Stress Design

1605.3.3 Alternate Basic Load Combinations

•

• ASD Load CombinationASD Load Combination •

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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+f

₁₁

L+E

_m_m

0.9D+E

_m_m

E

E_m_m = Maximum effect of horizontal and vertical= Maximum effect of horizontal and vertical forces (1617.1)

forces (1617.1) f

f₁₁ = 1.0 for floors in places of public assembly,= 1.0 for floors in places of public assembly, live loads in excess of 100

live loads in excess of 100 psfpsf and parkingand parking garage live loads

garage live loads

= 0.5 for other live loads = 0.5 for other live loads

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Load Combinations

1617.1

E =

Q

_E_E

+ 0.2S

_DS_DS

D

E =

Q

_E_E

-

0.2S

_DS_DS

D

= Redundancy Coefficient (1617.2)

1.0 for design forces for diaphragms

and wall anchorage

Q

_E_E

= Effect of horizontal seismic forces

S

_DS_DS

= Design spectral response

acceleration at short periods

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Load Combinations

1617.1, Maximum Seismic Load Effect

E

_m_m

=

Q

_E_E

+0.2S

_DS_DS

D

E

_m_m

=

Q

_E_E

–

0.2S

_DS_DS

D

= System

Overstrength

Factor

(Table 1617.6.2)

An allowable stress increase of 1.7 (not to be combined with An allowable stress increase of 1.7 (not to be combined with

1/3 allowable stress increase due for wind or seismic loads) 1/3 allowable stress increase due for wind or seismic loads) is permitted for ASD designs

is permitted for ASD designs Term

Term QQ_E_E need not exceed force that can be transferred toneed not exceed force that can be transferred to the element by the other elements of the lateral force the element by the other elements of the lateral force resisting system

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Load Combinations

For designs utilizing ASCE 7, Equivalent

Lateral Force Procedure, the Special

Seismic Load is

E =

Q

_E_E

+0.2S

_DS_DS

D

E =

Q

_E_E

–

0.2S

_DS_DS

D

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

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Analysis Method

1.

1. Equivalent Lateral Force Procedure

Equivalent Lateral Force Procedure

ASCE 7

-

02 Section 9.5.5

2.

2. Simplified Analysis

Simplified Analysis

Permitted for:

Seismic Use Group I structures if

1.

1. Buildings of light framed construction notBuildings of light framed construction not exceeding 3 stories in height

exceeding 3 stories in height 2.

2. Buildings of any construction not exceeding 2Buildings of any construction not exceeding 2 stories with flexible construction at every level stories with flexible construction at every level

3.

3. Dynamic Analysis

Dynamic Analysis

ASCE 7

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Analysis Method

For structures designed using the

Simplified Analysis Procedures, the

requirements of Sections 1620.2

-

-1620.5 (IBC) must be met.

1620.5 (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

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Simplified Procedure

1617.5.1 Seismic Base Shear

(EQ. 16

-

56)

R = Response modification factor (Table 1617.6.2) R = Response modification factor (Table 1617.6.2)

W = Effective weight of structure W = Effective weight of structure

W

R

1.2S

V

DS

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Simplified Procedure

1617.5.2 Vertical Distribution of Horizontal

Forces

(EQ. 16

-

57)

w

_x_x

= Portion of effective weight of structure,

W, at Level x.

x DS x

w

R

1.2S

F

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Simplified Procedure

1620.2.5 Diaphragms

Designed to resist force:

F

_p_p

= 0.2I

_E_E

S

_DS_DS

w

_p_p

+

V

_px_px

(EQ. 16

-

60)

w

w_p_p = weight of diaphragm and other elements attached= weight of diaphragm and other elements attached to diaphragm

to diaphragm V

V_px_px = portion of seismic shear force required to be= portion of seismic shear force required to be transferred to lateral force resisting elements transferred to lateral force resisting elements

through diaphragm from other lateral force through diaphragm from other lateral force

resisting elements due to offsets or changes in resisting elements due to offsets or changes in stiffness of the lateral force resisting elements stiffness of the lateral force resisting elements

above or below the diaphragm above or below the diaphragm

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Simplified Procedure

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Simplified Procedure

1620.4.3 Diaphragms in SDC D

px n x i i n x i i px

w

F

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Simplified Procedure

Zw

₄₄

w

₄₄

Roof 4

Zw

₃₃

w

₃₃

3

3 Zw

Zw

₂₂

w

₂₂

2

2 Zw

Zw

₁₁

w

₁₁

Ground 1

Fpx

F

_i_i

=

F

_x_x

Weight,

w

_i_i

Level

R

1.2S

Z

DS

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Simplified Procedure

w

F

1 4 3 2 1 4 3 2 1 p1

w

Zw

F

1 1 4 3 2 1 4 3 2 1

_w

_Zw

)

w

(w

)

w

Z(w

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Simplified Procedure

Zw

₄₄

Zw

₄₄

w

₄₄

Roof 4

Zw

₃₃

Zw

₃₃

w

₃₃

3

3 Zw

Zw

₂₂

Zw

₂₂

w

₂₂

2

2 Zw

Zw

₁₁

Zw

₁₁

w

₁₁

Ground 1

F

_px_px

**

F

_i_i

=

F

_x_x

Weight,

w

_i_i

Level

R

1.2S

Z

DS

**

F

_px_px

max = 0.4S

_DS_DS

I

_E_E

w

_px_px

F

_px_px

min = 0.2S

_DS_DS

I

_E_E

w

_px_px

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Simplified Procedure

2.4

4.8

1.25

2

4

1.50

3

6

1.0

1.0 If R < (# below),

If R < (# below),

then

F

_px_px

max

controls

If R > (# below),

then

F

_px_px

min

controls

I

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30 30

Equivalent Lateral Force Procedure

ASCE 7 9.5.5.2 Seismic Base Shear

V =

C

_s_s

W

need not be

greater than

but not less than

and for SDC E

and F not less than

R/I

S

C

DS s

_T(R/I)

S

C

D1 s

I

0.044S

C

_s _DS

R/I

0.5S

C

1 s

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31 31

Equivalent Lateral Force Procedure

9.5.5.4 Vertical Distribution of Seismic Forces

C

C_vx_vx = vertical distribution factor= vertical distribution factor w

w_i_i andand ww_x_x = portion of total gravity load, W, assigned to= portion of total gravity load, W, assigned to Level i or x

Level i or x h

h_i_i andand hh_x_x = height from base to Level i or x= height from base to Level i or x k = 1.0 if period, T = 0.5s or less

k = 1.0 if period, T = 0.5s or less = 2.0 if T = 2.5s or more

= 2.0 if T = 2.5s or more

use linear interpolation for periods between 0.5 and 2.5 use linear interpolation for periods between 0.5 and 2.5

V

C

F

_x _vx _n 1 i k i i k x x vx

h

w

h

w

C

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Equivalent Lateral Force Procedure

9.5.2.6.2.7 Diaphragms

Must resist the larger of

1.

1. The portion of the design seismic force at

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.

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Equivalent Lateral Force Procedure

9.5.2.6.4.4 Diaphragms in SDC D

w

F

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34 34

Diaphragm Design Example

3 story CMU bearing special reinforced shear wall building with 3 story CMU bearing special reinforced shear wall building with

3 foot parapet 3 foot parapet

Level 2 and 3 concrete diaphragms on metal deck Level 2 and 3 concrete diaphragms on metal deck Roof steel roof deck diaphragm

Roof steel roof deck diaphragm S

S_DS_DS = 0.50= 0.50 R = 5.0 (Table 1617.6.2)R = 5.0 (Table 1617.6.2) SDC D

SDC D T = 0.4 secondsT = 0.4 seconds I = 1.0

I = 1.0 No plan irregularitiesNo plan irregularities Floor DL = 60

Floor DL = 60 psfpsf Wall RigiditiesWall Rigidities Roof DL = 15

Roof DL = 15 psfpsf R1 = .2R1 = .2 R3 = .1R3 = .1 Wall DL = 80

Wall DL = 80 psfpsf R2 = .1R2 = .1 R4 = .3R4 = .3 Analysis for Diaphragm Design

Analysis for Diaphragm Design

1. Cannot use Simplified Analysis per section 1616.1; we 1. Cannot use Simplified Analysis per section 1616.1; we

don't have light framed construction and we exceed 2 don't have light framed construction and we exceed 2 stories

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Diaphragm Design Example

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Diaphragm Design Example

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Diaphragm Design Example

2. Using the Equivalent Lateral Force Procedure from ASCE 7

2. Using the Equivalent Lateral Force Procedure from ASCE 7--02,02, find the base shear

find the base shear V =

V = CC_s_sWW ((EqEq. 9.5.5.2. 9.5.5.2--1)1)

Weight tributary to level 1 = 80psf*12'/2*(2*40'+2*60') = 96,000

Weight tributary to level 1 = 80psf*12'/2*(2*40'+2*60') = 96,000 poundspounds Weight tributary to level 2 and 3 = 80psf*12'*(2*40'+2*60') + 60

Weight tributary to level 2 and 3 = 80psf*12'*(2*40'+2*60') + 60psf*(40'*60')=psf*(40'*60')= 336,000 pounds

336,000 pounds

Weight tributary to level 4 = 80psf*(12'/2+3)*(2*40'+2*60') + 15

Weight tributary to level 4 = 80psf*(12'/2+3)*(2*40'+2*60') + 15psf*(40'*60') =psf*(40'*60') = 180,000 pounds

180,000 pounds

C

C_s_s = S= S_DS_DS/(R/I) = .50/(5/1) = 0.10/(R/I) = .50/(5/1) = 0.10 CHECK OTHER C

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Diaphragm Design Example

0 948,000 36 180,000 Roof 4 24 336,000 3 12 336,000 2 0 96,000 Ground 1 F_px F_x C_vx w_ih_i Height, h Weight, w Level

V =

C

_s_s

W

=0.10*948,000 = 94,800 pounds

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Diaphragm Design Example

3. Determine the vertical distribution of Seismic Forces 3. Determine the vertical distribution of Seismic Forces

F F_x_x == CC_vx_vxVV k = 1 (T<0.5) k = 1 (T<0.5) n 1 i k i i k x x vx

h

w

h

w

C

18576000 948,000 33070 0.3488 6480000 36 180,000 Roof 4 41153 0.4341 8064000 24 336,000 3 20577 0.2171 4032000 12 336,000 2 0 0 0 0 96,000 Ground 1 F_px F_x C_vx w_ih_i Height, h Weight, w Level

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Diaphragm Design Example

4. Determine forces to diaphragm at each level 4. Determine forces to diaphragm at each level

Fpx

Fpx (max) = 0.4SDSIwx(max) = 0.4SDSIwx Fpx

Fpx (min) = 0.2SDSIwx(min) = 0.2SDSIwx Fp3 = {(F3+F4)/(w3+w4)}w3 Fp3 = {(F3+F4)/(w3+w4)}w3 px n x i i n x i i px

w

F

18576000 948,000 33070 33070 0.3488 6480000 36 180,000 Roof 4 48331 41153 0.4341 8064000 24 336,000 3 37386 20577 0.2171 4032000 12 336,000 2 9600 0 0 0 0 96,000 Ground 1 F_px F_x C_vx w_ih_i Height, h Weight, w Level

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Diaphragm Design Example

5. Determine diaphragm shears to design diaphragm 5. Determine diaphragm shears to design diaphragm

Level 4 Level 4

Flexible Diaphragm Flexible Diaphragm

Diaphragm Shear due to Y direction load Diaphragm Shear due to Y direction load

q

q₁₁ = q= q₃₃ = (Fp/2)/b = (33070/2)/40 = 413= (Fp/2)/b = (33070/2)/40 = 413 plfplf (Ultimate)(Ultimate) ASD Load Combinations: E/1.4

ASD Load Combinations: E/1.4 q

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Diaphragm Design Example

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Diaphragm Design Example

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44 44

Diaphragm Design Example

Level 3 Rigid Diaphragm Level 3 Rigid Diaphragm

Center of Rigidity Center of Rigidity x x_r_r == RR_yi_yixx_i_i// RR_yi_yi = (R1(0)+R3(60))/(R1+R3)= (R1(0)+R3(60))/(R1+R3) = (0.20*(0)+0.10*(60))/(0.20+0.10) = (0.20*(0)+0.10*(60))/(0.20+0.10) x x_r_r = 20 feet= 20 feet y y_r_r == RR_xi_xiyy_i_i// RR_xi_xi = (R2(40)+R4(0))/(R2+R4)= (R2(40)+R4(0))/(R2+R4) = (0.10*(40)+0.30*(0))/(0.10+0.30) = = (0.10*(40)+0.30*(0))/(0.10+0.30) = y y_r_r = 10 feet= 10 feet

Center of Mass = center of diaphragm Center of Mass = center of diaphragm

e

e_x_x = 30= 30--20 = 10 feet plus 5% accidental torsion20 = 10 feet plus 5% accidental torsion e

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Diaphragm Design Example

Direct Shear Direct Shear

V

V_yit_yit = (= (RR_yi_yi// RR_yi_yi)F)F_py_py V

V_y1t_y1t = (0.20/(0.20+0.10))*48331 = 32221 pounds= (0.20/(0.20+0.10))*48331 = 32221 pounds V

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Diaphragm Design Example

Shear due to torsion Shear due to torsion

V V_yir_yir = (R= (R_yi_yix'/(x'/( RR_yi_yix'x'22₊₊ _R_R xi xiy'y'22))F))Fpypyeexx R R_yi_yix'x'22 _{= 0.20(20)}_{= 0.20(20)}22_+0.10((60_+0.10((60_-_-₂₀₎₂₀₎22 _{= 240}_{= 240} R R_xi_xiy'y'22 _{= 0.10(40}_{= 0.10(40}_-_-₁₀₎₁₀₎22_+0.30((10)_+0.30((10)22 _{= 120}_{= 120} V V_y1r_y1r = (0.20*20/(240+120))*48331*(13)= (0.20*20/(240+120))*48331*(13) = 6981 pounds = 6981 pounds V V_y3r_y3r = (0.10*(60= (0.10*(60--20)/(240+120))*48331*(13)20)/(240+120))*48331*(13) = 6981 pounds = 6981 pounds

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Diaphragm Design Example

Substituting

Substituting RR_xi_xiyy_i_i forfor RR_yi_yixx_i_i in above equation to findin above equation to find shear in wall 2 and 4 from

shear in wall 2 and 4 from torsionaltorsional shear due toshear due to load in Y

load in Y--directiondirection

V V_y2r_y2r = (0.10*30/(240+120))*48331*(13)= (0.10*30/(240+120))*48331*(13) = 5236 pounds = 5236 pounds V V_y4r_y4r = (0.30*10/(240+120))*48331*(13)= (0.30*10/(240+120))*48331*(13) = 5236 pounds = 5236 pounds X

X--direction load will likely control shear for wall lines 2 and 4direction load will likely control shear for wall lines 2 and 4 Do not decrease shear in wall due to negative

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Diaphragm Design Example

V

V₁₁ = V= V_y1t_y1t + V+ V_y1r_y1r = 32221 pounds= 32221 pounds q

q₁₁ = V= V₁₁/W = 32221/40 = 806/W = 32221/40 = 806 plfplf (Ultimate)(Ultimate) ASD Load Combinations: E/1.4

q₁₁ = 806/1.4 = 576= 806/1.4 = 576 plfplf

V

V₃₃ = V= V_y3t_y3t + V+ V_y3r_y3r = 16110+6981 = 23091 pounds= 16110+6981 = 23091 pounds q

q₃₃ = V= V₃₃/W = 23091/40 = 577/W = 23091/40 = 577 plfplf (Ultimate)(Ultimate) ASD Load Combinations: E/1.4

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Diaphragm Design Example

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Diaphragms with Openings

Analysis of diaphragms with large openings

assumes diaphragm behaves similar to

Vierendeel

Truss.

Example:

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51 51

Diaphragms with Openings

Step 1

Step 1 –– Analyze Diaphragm as though no openings existedAnalyze Diaphragm as though no openings existed

Line 1 Line 1 V1 = wL/2 = 400(60)/2 = 12,000 #V1 = wL/2 = 400(60)/2 = 12,000 # q1 = V1/W = 12,000/40 = 300 q1 = V1/W = 12,000/40 = 300 plfplf Line 2 Line 2 V2 = w(L/2-V2 = w(L/2-x) = 400(60/2x) = 400(60/2--10) = 8000 #10) = 8000 # q2 = 8000/40 = 200 q2 = 8000/40 = 200 plfplf M2 = (wx/2)*(L M2 = (wx/2)*(L--x) = (400*10/2)*(60-x) = (400*10/2)*(60-10) = 100,000 #ft10) = 100,000 #ft T=C=M/d T=C=M/d F2@a = 100,000/40 = 2500 # C F2@a = 100,000/40 = 2500 # C F2@d = 2500 # T F2@d = 2500 # T Line 3 Line 3 V3 = 400(60/2-V3 = 400(60/2-15) = 6000 #; q3 = 6000/40 = 15015) = 6000 #; q3 = 6000/40 = 150 plfplf M3 = (400*15/2)*(60 M3 = (400*15/2)*(60--15) = 135,000 #ft15) = 135,000 #ft F3@a = 135,000/40 = 3375 # C; F3@d = 3365 # T F3@a = 135,000/40 = 3375 # C; F3@d = 3365 # T Line 4 Line 4 V4 = 400(60/2-V4 = 400(60/2-20) = 4000 #; q4 = 4000/40 = 100plf20) = 4000 #; q4 = 4000/40 = 100plf M4 = (400*20/2)*(60 M4 = (400*20/2)*(60--20) = 160,000 #ft20) = 160,000 #ft F4@a = 160,000/40 = 4000 # C. F4@d = 4000 # T F4@a = 160,000/40 = 4000 # C. F4@d = 4000 # T Line 5 V4 = 400(60/2 Line 5 V4 = 400(60/2--30) = 0 #; q4 = 030) = 0 #; q4 = 0 plfplf M4 = (400*30/2)*(60 M4 = (400*30/2)*(60--30) = 180,000 #ft30) = 180,000 #ft F4@a = 180,000/40 = 4500 # C. F4@d = 4500 # T F4@a = 180,000/40 = 4500 # C. F4@d = 4500 # T

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52 52

Diaphragms with

Openings

Step 2. Determine the shears and Step 2. Determine the shears and chord forces at the edges of the chord forces at the edges of the openings using free

openings using free--body diagramsbody diagrams for each segment

for each segment

Segment A Segment A V4(ab)=V4/2=4000/2= 2000 # V4(ab)=V4/2=4000/2= 2000 # q4(ab)=V4(ab)/15 q4(ab)=V4(ab)/15’’=2000/15= 133=2000/15= 133 plfplf V3(ab)=V4+400(5 V3(ab)=V4+400(5’’)=2000+2000)=2000+2000 = 4000 # = 4000 # q3(ab)=4000/15 q3(ab)=4000/15’’= 267= 267 plfplf F4@a F4@a MM_3b_3b=-=-3375(15)3375(15)--2000(5)2000(5)- -400(5 400(522_/2)+F4@b(15)_/2)+F4@b(15) F4@a=4375# C F4@a=4375# C F4@b F4@b FF_x_x=-=-4375+3375+F4@b4375+3375+F4@b F4@b=1000# T F4@b=1000# T

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Diaphragms with

Openings

Segment B Segment B V2(ab)= V3(ab)+400(5)=4000+2000 V2(ab)= V3(ab)+400(5)=4000+2000 = 6000 = 6000 q2(ab)=6000/15= 400 q2(ab)=6000/15= 400 plfplf F2@a

F2@a MM_3b_3b=-=-F2@a(15)+3375(15)F2@a(15)+3375(15) +400(5 +400(522_/2)_/2)_-_-_6000(5)_6000(5) F4@a=1708# C F4@a=1708# C F4@b F4@b FF_x_x=-=-3375+1708+F4@b3375+1708+F4@b F4@b=1667# T F4@b=1667# T 2000# 2000# 4375# 4375# 1000# 1000# 4000# 4000#

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Diaphragms with

Openings

Segment C Segment C V4(cd)=V4/2=4000/2= 2000 # V4(cd)=V4/2=4000/2= 2000 # q4(cd)=V4(ab)/15 q4(cd)=V4(ab)/15’’=2000/15= 133=2000/15= 133 plfplf V3(ab)

V3(ab) FF_y_y V3(ab)= 2000 #V3(ab)= 2000 # q3(cd)= 133 q3(cd)= 133 plfplf F4@c=V4(cd)(5 F4@c=V4(cd)(5’’)/15)/15’’=2000(5)/15=2000(5)/15 = 667 # C = 667 # C F4@d=3375+667= 4042 # T F4@d=3375+667= 4042 # T 2000# 2000# 4375# 4375# 1000# 1000# 4000# 4000# 1708# 1708# 6000# 6000# 1667# 1667#

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Diaphragms with

Openings

Segment D Segment D V2(cd) V2(cd) FF_y_y V2(cd)= 2000 #V2(cd)= 2000 # q2(cd)= 133 q2(cd)= 133 plfplf F2@c=V4(cd)(5 F2@c=V4(cd)(5’’)/15’)/15’=2000(5)/15=2000(5)/15 = 667 # T = 667 # T F2@d=3375 F2@d=3375--667= 2708 # T667= 2708 # T 2000# 2000# 4375# 4375# 1000# 1000# 4000# 4000# 1708# 1708# 6000# 6000# 1667# 1667# 2000# 2000# 667# 667# 4042# 4042# 2000# 2000#

(56)

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

(57)

Diaphragms with Openings

42 T 42 T 4042 T 4042 T 4000 T 4000 T F4@d F4@d 667 C 667 C 667 C 667 C 0 0 F4@c F4@c 1000 T 1000 T 1000 T 1000 T 0 0 F4@b F4@b 375 C 375 C 4375 C 4375 C 4000 C 4000 C F4@a F4@a 208 T 208 T 2708 T 2708 T 2500 T 2500 T F2@d F2@d 667 T 667 T 667 T 667 T 0 0 F2@c F2@c 1667 C 1667 C 1667 C 1667 C 0 0 F2@b F2@b 792 T 792 T 1708 C 1708 C 2500 C 2500 C F2@a F2@a Net Change Net Change Due to Openings Due to Openings With With Openings Openings Without Without Openings Openings Diaphragm Diaphragm Force Location Force Location Chord Force (lbs) Chord Force (lbs)

(58)

Diaphragms with Openings

(59)

Diaphragms with Openings

(60)

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

(61)

61 61

Diaphragms with Openings

--4.24.2 --4.24.2 0 0 V5 @ c to d V5 @ c to d +62.6 +62.6 +62.6 +62.6 0 0 V5 @ b to c V5 @ b to c --37.537.5 --37.537.5 0 0 V5 @ a to b V5 @ a to b +95.8 +95.8 --4.24.2 100 100 V4 @ c to d V4 @ c to d +162.5 +162.5 +62.5 +62.5 100 100 V4 @ b to c V4 @ b to c +62.5 +62.5 --37.537.5 100 100 V4 @ a to b V4 @ a to b +220.8 +220.8 +20.8 +20.8 200 200 V2 @ c to d V2 @ c to d +287.5 +287.5 +87.5 +87.5 200 200 V2 @ b to c V2 @ b to c +120.8 +120.8 --79.279.2 200 200 V2 @ a to b V2 @ a to b +320.8 +320.8 +20.8 +20.8 300 300 V1 @ c to d V1 @ c to d +387.5 +387.5 +87.5 +87.5 300 300 V1 @ b to c V1 @ b to c +220.8 +220.8 --79.279.2 300 300 V1 @ a to b V1 @ a to b Resultant Resultant Shear Shear Due to Due to Openings Openings Without Without Openings Openings Diaphragm Shear Diaphragm Shear and Location and Location Shear ( Shear (plfplf))

(62)

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

(63)

Diaphragms with Openings

(64)

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

(65)

Collector Elements

Collect force from diaphragms and transfer

them to shear walls (drag struts)

(66)

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

(67)

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

(68)

Diaphragm 1 Diaphragm 1 Diaphragm Diaphragm 2 2

Collector Element Example

(69)

Collector Element Example

Tearing at Tearing at Discontinuity Discontinuity F F_px_px F F_px_px

(70)

Collector Element Example

Tearing at

Discontinuity

(71)

Collector Element Example

Design of Collector for Forces in Y

-

Direction

1. Determine diaphragm shear for large component along line 1. Determine diaphragm shear for large component along line

of collector of collector For given F

For given F_py1_py1=100 k, q=100 k, q_y1_y1 = 100k/2/200= 100k/2/200’’ = 250= 250 plfplf

2. Determine diaphragm shear for small component along line 2. Determine diaphragm shear for small component along line

of collector of collector For given F

For given F_py2_py2=300 k, q=300 k, q_y2_y2 = 30k/2/100= 30k/2/100’’ = 150= 150 plfplf 3. Determine load in collector

3. Determine load in collector Q

(72)

Collector Element Example

4. Collectors required to be designed per load combinations for

4. Collectors required to be designed per load combinations for specialspecial seismic loads (determined from ASCE 7 or IBC for procedure used seismic loads (determined from ASCE 7 or IBC for procedure used inin design)

design) For IBC,

For IBC, EE_m_m == QQ_E_E +0.2S+0.2S_DS_DSDD For given

For given = 2.5, the lateral load to the drag strut for design is= 2.5, the lateral load to the drag strut for design is EE_m_m = 2.5*40k = 100 k

= 2.5*40k = 100 k

5. If the drag strut carries dead loads, the additional seismic

5. If the drag strut carries dead loads, the additional seismic portion ofportion of the dead load must be added to the load combination

the dead load must be added to the load combination 1.2D+f

1.2D+f₁₁L+EL+E_m_m = (1.2+0.2S= (1.2+0.2S_DS_DS)D + 100k(E)D + 100k(E_m_m)) 0.9D+E

0.9D+E_m_m= (0.9-= (0.9-0.2S0.2S_DS_DS)D + 100k(E)D + 100k(E_m_m))

An allowable stress increase of 1.7 can be used with these load An allowable stress increase of 1.7 can be used with these load combinations

combinations

Note that this example meets the definition of Plan Structural I

Note that this example meets the definition of Plan Structural Irregularityrregularity #2 per Table 1616.5.1 and therefore per section 1620.4.1, the

#2 per Table 1616.5.1 and therefore per section 1620.4.1, the design forces shall be increased 25% for connections of diaphrag design forces shall be increased 25% for connections of diaphragmsms to vertical element and for collectors to vertical elements exce

to vertical element and for collectors to vertical elements except ifpt if using special seismic load combinations

(73)

Collector Element Example

Design of Collector for Forces in X

-

Direction

1. Determine diaphragm shear for small component along line 1. Determine diaphragm shear for small component along line

of collector (wall with slip connection) of collector (wall with slip connection) For given F

For given F_px2_px2=20 k, q=20 k, q_x2_x2 = 20k/2/100= 20k/2/100’’ = 100= 100 plfplf 2. Determine load in collector

2. Determine load in collector Q

(74)

Collector Element Example

3. Collectors required to be designed per load combinations 3. Collectors required to be designed per load combinations for special seismic loads (determined from ASCE 7 or for special seismic loads (determined from ASCE 7 or IBC for procedure used in design)

IBC for procedure used in design) For IBC,

For IBC, EE_m_m == QQ_E_E +0.2S+0.2S_DS_DSDD For given

For given = 2.5, the lateral load to the drag strut for= 2.5, the lateral load to the drag strut for design is

design is EE_m_m = 2.5*10k = 25 k= 2.5*10k = 25 k

4. Develop this drag force into the larger diaphragm. This 4. Develop this drag force into the larger diaphragm. This

example meets the definition of Plan Structural example meets the definition of Plan Structural

Irregularity #2 per Table 1616.5.1 but we are using Irregularity #2 per Table 1616.5.1 but we are using the special seismic load combinations

the special seismic load combinations

For given diaphragm capacity of 400plf, must extend drag For given diaphragm capacity of 400plf, must extend drag strut into diaphragm length, L = 25*1.33/.4/1.7 = 47.8 strut into diaphragm length, L = 25*1.33/.4/1.7 = 47.8’’ Say extend into main diaphragm 48

(75)

Collector Element Example

5. Design the larger diaphragm for

5. Design the larger diaphragm for FpxFpx of that diaphragm andof that diaphragm and add in the additional force caused by the drag strut

add in the additional force caused by the drag strut from the smaller section.

from the smaller section. For a given F

For a given F_px1_px1=90 k,=90 k, q

(76)

Bearing and Shear Wall Anchorage

Simplified Analysis

SDC B

Same force as used in wall design:

F

_p_p

= 0.40I

_E_E

S

_DS_DS

w

_w_w

w

(77)

Wall Anchorage

Simplified Analysis SDC C

Must meet requirements of SDC B and

For concrete or masonry walls

supported by flexible diaphragm

F

_p_p

= 0.8S

_DS_DS

I

_E_E

w

_w_w

Supported by rigid diaphragm

(78)

78 78

Wall Anchorage

Simplified Analysis SDC C Simplified Analysis SDC C

Supported by rigid diaphragm Supported by rigid diaphragm

With component amplification factor,

With component amplification factor, apap = 1.0 and component= 1.0 and component response modification factor,

response modification factor, RpRp = 2.5= 2.5 z = height to point of attachment

z = height to point of attachment h = average roof height

h = average roof height

F

_p_p

(max) = 1.6S

_DS_DS

I

_p_p

W

_p_p

F

_p_p

(min) = 0.3S

_DS_DS

I

_p_p

W

_p_p

h

z

2

1 /I

R

W

S

0.4a

F

p p p DS p p

(79)

Wall Anchorage

Simplified Analysis SDC C Simplified Analysis SDC C Additional Requirements Additional Requirements

Continuous ties or struts must be provided to Continuous ties or struts must be provided to transfer wall anchorage forces into diaphragm transfer wall anchorage forces into diaphragm Metal deck cannot be used as tie or strut

Metal deck cannot be used as tie or strut perpendicular to deck span

perpendicular to deck span

Wood sheathing cannot be used as tie or strut Wood sheathing cannot be used as tie or strut Steel elements used for wall anchorage shall Steel elements used for wall anchorage shall have the strength design forces (ultimate) have the strength design forces (ultimate) increased by 1.4

(80)

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

(81)

Wall Anchorage

Equivalent Lateral Force Procedure

SDC A

F

_p_p

= 0.133S

_DS_DS

w

_w_w

or minimum of

F

_p_p

= 0.05w

_w_w

Concrete or masonry walls

Minimum E = 280 lbs/liner foot of wall

(82)

Wall Anchorage

Equivalent Lateral Force Procedure SDC B

Must meet requirements of SDC A and

Concrete or masonry walls

F

_p_p

= 0.4S

_DS_DS

Iw

_c_c

or minimum of

F

_p_p

= 0.10w

_c_c

or minimum of

E = 400S

(83)

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

(84)

Wall Anchorage

Equivalent Lateral Force Procedure SDC C Equivalent Lateral Force Procedure SDC C

Must meet requirements of SDC B and Must meet requirements of SDC B and

For concrete or masonry walls supported by flexible For concrete or masonry walls supported by flexible

diaphragm diaphragm F

F_p_p = 0.8S= 0.8S_DS_DSIwIw_p_p

Supported by rigid diaphragm Supported by rigid diaphragm

F

F_p_p (max) = 1.6S(max) = 1.6S_DS_DSII_p_pWW_p_p F

F_p_p (min) = 0.3S(min) = 0.3S_DS_DSII_p_pWW_p_p

Same equations as Simplified Analysis SDC C requirement Same equations as Simplified Analysis SDC C requirement

except no 1.4 increase for anchor bolts and reinforcing except no 1.4 increase for anchor bolts and reinforcing

h

z

2

1 /I

R

W

S

0.4a

F

p p p DS p p

(85)

Wall Anchorage

Equivalent Lateral Force Procedure

SDC C

Additional Requirements

Same as those required for SDC C

and

(86)

Wall Anchorage

Equivalent Lateral Force Procedure

SDC D, E and F

No additional requirements regarding

wall anchorage forces

(87)

Wall Anchorage Example

(88)

Wall Anchorage Example

(89)

Wall Anchorage Example

(90)

Wall Anchorage Example

(91)

Wall Anchorage Example

(92)

Wall Anchorage Example

(93)

93 93

Sub

-

diaphragms

Continuous ties or struts must be provided Continuous ties or struts must be provided

between main diaphragm chords to transfer wall between main diaphragm chords to transfer wall anchorage forces to the diaphragm

anchorage forces to the diaphragm Chords may be added to form sub

Chords may be added to form sub--diaphragmsdiaphragms with maximum length to width ration of 2.5/1 with maximum length to width ration of 2.5/1 (may be less for wood diaphragms)

(may be less for wood diaphragms)

Wood diaphragm sheathing shall not be Wood diaphragm sheathing shall not be

considered effective as providing the ties or considered effective as providing the ties or struts

struts

In metal deck diaphragms, the metal deck shall In metal deck diaphragms, the metal deck shall not be considered effective as providing the ties not be considered effective as providing the ties or struts in the direction perpendicular to the

or struts in the direction perpendicular to the deck span

(94)

Sub

(95)

Sub

(96)

Sub

(97)

Sub

(98)

Sub

(99)

Seismic Design Diaphragms

QUESTIONS?

(100)