SD-Lecture18-Diaphragm Theory-III.pdf

8/20/2019 SD-Lecture18-Diaphragm Theory-III.pdf

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Soil Dynamics

Lecture 18

Diaphragm Theory - Part III

In a Problem Format


2/47

Question #01.

A wood residence has a wood structural panel roof diaphragm (which is shown in

plan view below). Two retail stores are separated by a common wood shear wall,

parallel to the north-south lateral loading. The height of the building is 14 feet. The

roof diaphragm is adequately anchored to the shear wall. Assume ρ = 1.0.

Determine the roof diaphragm shear force along line 3.

A. 3,800 lb f .

B. 5,600 lb f .

C. 7,500 lb f .

D. 9,000 lb f .


3/47

Answer to #01.

C.

The shear wall along line 3 carries half of the diaphragm shear between lines 2 and

3.

( )( )150 1002

002

7 5 f

lb / ft ft wLV , lb= = =


4/47

Question #02.




roof diaphragm is adequately anchored to the shear wall. Assume ρ = 1.0).

Along line X between lines 1 and 2, determine the maximum diaphragm chord force

in the chord member.

A. 750 lbf .

B. 1,200 lbf .

C. 2,300 lbf .

D. 4,700 lbf .


5/47

Answer to #02.

B.

The maximum diaphragm chord force occurs at the midpoint between lines 1 and 2.

It is calculated as the bending moment of a simple beam under a distributed load,

per unit depth of diaphragm.

( )( )( )

22 150 50

8 81 17

00

4

f lb / ft ft M wL

C b b ft

, lb= = = =


6/47

Question #03.




roof diaphragm is adequately anchored to the shear wall. Assume ρ = 1.0).

Determine the diaphragm chord force at the intersection of lines Y and 1.

A. 0 lbf .

B. 290 lbf .

C. 600 lbf .

D. 1,200 lbf .


7/47

Answer to #03.

A.

Chords are the boundary members of a diaphragm, and are perpendicular to the

direction of the lateral load.

Chords are designed to carry moments and provide all the resistance to the flexural

stresses.

Chord forces at the chord ends are zero.


8/47

Question #04.

A wood frame shear wall in a seismic zone 2 is shown in elevation below. The shear

wall is part of a one-story structure with a flexible roof diaphragm. The shear wall

panels have the same thickness. What is the approximate lateral load carried by

panel 1?

A. 1,800 lbf.

B. 2,600 lbf.

C. 3,000 lbf.

D. 9,000 lbf.

drag supports


9/47

Answer to #04.

B.

( )( )

( )( )

10 ft + 20 ft + 5 ft = 35 ft

9 000 25735

257 10 2 570

f

f

f

b

, lbV lb / ft b ft

V b lb / f ,t lb ft

υ υυ υ

υ υυ υ

=

= = =

= = =


10/47

Question #05.

What is the purpose of providing drag struts (collectors) at the diaphragm

boundaries?

A. To resist moments.

B. To transmit unsupported horizontal diaphragm shear to the supporting shear

walls.

C. To distribute torsion.

D. None of the above.


11/47

Answer to #05.

B.

At points of discontinuity in the plan, or where the vertical resisting elements are not

provided because of windows or doors, the drag struts collect and drag the

horizontal diaphragm shear to the supporting vertical resisting elements.


12/47

Question #06.

The force in a drag strut is,

A. tension only.

B. compression only.

C. either tension or compression.

D. none of the above.


13/47

Answer to #06.

C.

Struts collect diaphragm load and carry it to a shear wall. Struts carry tension and

compression depending on the direction of seismic loading.


14/47

Question #07.

The wood structural panel roof diaphragm of a one-story building is shown below. Thenorth shear wall’s length is not the full width of the building. Determine the location of

the maximum strut force. Assume ρ = 1.0.

A. At point X.

B. At point Y.

C. At point Z.

D. At both points X and Z.


15/47

Answer to #07.

A.

The strut force changes along the length of the drag strut. The maximum force

occurs at the point where the drag strut attaches to the parallel wall.

For the north shear wall, considering the direction of seismic loading shown, this

location is at point X.


16/47

Question #08.

The plan view of a wood structural panel roof diaphragm is shown below. The left

wall has an opening at its center. For the loading shown, where in the drag strut does

the maximum force occur? Assume ρ = 1.0.

A. At point X.

B. At point Y.

C. At point Z.

D. At both points X and Z.


17/47

Answer to #08.

D.

The strut force changes along the length of the drag strut. The maximum force

occurs at the point where the drag strut attaches to the parallel wall. Points X and Z

frame into parallel walls. Therefore, the maximum drag strut forces occur at bothlocations.


18/47

Question #09.

Choose the correct statements for the strut force:

I. The strut force in compression will remain a compression force if the direction

of the lateral force is reversed.

II. The strut forces are zero at the outside ends of the walls.

III. The greater magnitude of the strut force or the chord force at corresponding

points along the wall determines the critical loading condition.

A. I and II.

B. I and III.

C. II and III.

D. I, II and III.


19/47

Answer to #09.

C.

Struts carry tension and compression, depending on the direction of the seismic

loading. The outside ends of the walls can be considered to experience no strut forces.

The critical loading conditions depend on the greater magnitude of the strut and

chord forces at the corresponding points along the wall.


20/47

Question #10.

What is the purpose of providing chords at diaphragm boundaries?

A. To distribute torsion.

B. To distribute lateral load.

C. To resist moments.

D. None of the above.


21/47

Answer to #10.

C.

Chords are elements, such as walls or reinforcement at the diaphragm boundaries

that are perpendicular to the direction of the applied seismic loading. They resist

moments and are regarded as tension and compression members.


22/47

Question #11.

Which of the following materials would be suitable to use as diaphragm chords?

A. wood

B. steel

C. masonry

D. All of the above


23/47

Answer to #11.

C.

Chords are elements at the borders of the diaphragm along the walls perpendicular

to the direction of the applied seismic loading. Any continuous material that resists

all tensile and compressive forces can be used.

For example, masonry walls with adequate tensile reinforcement, the double top-

plate in wood stud walls, and steel are suitable.


24/47

Question #12.

The force in the chord member is obtained from which of the following formulas?

2

A. The force in the chord

B. The force in the chord8

C. The force in the chord 8

D. All of the above.

M

b

wLb

FL

b

=

=

=


25/47

Answer to #12.

D.

2

The chord force C is calculated as the bending moment of a simle beam due to a distributed

load per unit depth of the diaphragm,

and the bending moment of a simple beam is,8

Since the di

M w C

b

wL M

=

=

22

stributed load is the diaphragm force divided by the span,

the expressions for the chord force are,

8 8 8

F

w w L

F L M wL FL L

C b b b b

=

= = = =


26/47

Question #13.

For a flexible roof diaphragm adequately anchored to the shear walls, which of thefollowing statements identifies the correct chord forces at lines 1 and 2? Assume ρ = 1.0.

A. The maximum chord forces at lines 1 and 2 are equal.

B. The maximum chord force at line 2 is twice the chord force at line 1.

C. The chord force at line 1 is generally ignored.

D. The total chord force at lines 1 and 2 is 40w.

N


27/47

Answer to #13.

A.

The chords at lines 1 and 2 resist bending due to the diaphragm loads, when the

loading is in the direction shown.

At both lines, the maximum chord forces are equal and occur at mid-span where the

maximum moment develops.


28/47

Question #14.

The plan view of a residential one-story wood structure with a wood structural panel

roof diaphragm is shown below. The lateral force is in the north-south direction. The

calculated roof diaphragm shear capacity is 200 lb f /ft in the direction of the loading.

The roof diaphragm is adequately anchored to the shear walls. What will be the

maximum force in the chord members? Assume ρ = 1.0.

A. 7,500 lb f .

B. 11,000 lb f .

C. 12,700 lb f .

D. 14,300 lb f .


29/47

Answer to #14.

A.

( )( )The diaphragm shear is resisted by the parallel walls. The shear force on each

wall is, 200 50 10 000

Each parallel wall resists half of the diaphragm load. The total distributed l

f f V b lb / ft ft , lbυ υυ υ = = =

( )( )

oad is,

2 10 0002 133

150

The maximum chord force is calculated as the maximum bending stress due to the

distributed diaphragm load,

but

f

f

max

w

, lbV w lb / ft

L ft

M C M

b

= = =

=

( )( )( )( )

2

22

8133 150

therefore, 7 58 8

0050

f

max f

wL

lb / ft ft wLC

b ft , lb

=

= = =


30/47

Question #15.

The plan view of a one-story retail store is shown below. It is built with a woodstructure with a wood structural panel roof diaphragm. Determine the force in the

chord member at the intersection of lines X and 1. Assume ρ = 1.0.

A. 0 lb f .

B. 1,250 lb f .

C. 2,000 lb f .

D. 5,000 lb f .


31/47

Answer to #15.

A.

The chord forces at the chord ends are always zero. This can be demonstrated for the

current set of assumptions, but the results can be generalized to any chord member.

The chord force is the resistance to bending under the imposed diaphragm loads.

The diaphragm shear at the parallel walls is,

( )( )250 8010 000

2 2

10 000250

40

f

f

f

f

lb / ft ft wLV , lb

, lb / ft V lb / ft

b ft υ υυ υ

= = =

∴ = = =


32/47

As shown by the shear and bending moment diagrams above, the moments at the ends

of the chords are zero. Therefore, the chord forces at the ends are also zero.

C = M / b

Therefore, C = 0. The forces in the chord member at the intersection of lines X and 1 is

zero.


33/47

Question #16.

A one-story wood frame commercial building has a wood structural panel roofdiaphragm, and its south wall has a 40 foot opening. Determine the chord force at the

intersection of lines X and 1. Assume ρ = 1.0.

A. 2,350 lb f .

B. 4,700 lb f .

C. 9,400 lb f .

D. 11,750 lb f .


34/47

Answer to #16.

C.

The chord force is the bending moment per unit depth of the diaphragm. The shear

load at the parallel walls is,

( )( )500 80

20 0002 2

The shear and bending moment diaphragms across the length of the

chord are as shown on the next slide,

f f



35/47

( )

( ) ( )

20 0005 000 at lines X and 1

40 10The moment at line 1 is calculated as the area under the shear diagram. The

chord force is the bending moment divided by the diaphragm depth.

155 000 30

f

f

f

, lb V V , lb

, M , lb ft

= ∴ =

= + ( )( )

9 400

000 30375 000

2

375 000

40

f f

f

f

lb ft , ft lb

, ft lb M

, l C b ft b

= −

∴

−

= = =


36/47

Question #17.

Which of the following statements regarding diaphragm chord members isincorrect ?

I. The maximum chord force occurs at the location of the maximum moment.

II. The maximum chord force occurs at the chord ends.

III. The maximum chord force is zero.

A. II

B. III

C. I and III

D. I, II and III


37/47

Answer to #17.

A.

The maximum chord force develops at mid-span (at the location of the maximum

moment), while the minimum chord force occurs at the chord ends and is zero.


38/47

Question #18.

The plan view of a wood-framed one-story retail store is shown below. The south wallhas a 35 foot shear wall panel. How long should this shear wall panel have to be in

order to develop an equal magnitude drag and chord force? Assume ρ = 1.0.

A. 0 ft

B. 40 ft

C. 60 ft

D. 75 ft


39/47

Answer to #18.

D.

Assume x equals the distance of an additional wall section measured from the east

wall. For the strut force on the south wall, E-W seismic loading applies.

( )( )

( )( )

600 5015 000

2 2

15 000150

100Therefore, the strut force on the south wall is,

150 150

For the chord force on the south wall, N-S seismic loading applies. Assum

f

f

f

f

f f

lb / ft ft wLV , lb

, lbV lb / ft

b ft

lb / ft x ft x lb

υ υυ υ

= = =

= = =

=

( )( )

e

x equals the distance of an additional wall section measured from the east wall.

200 10010 000

2 2

f

f



40/47

( ) ( )

( ) ( ) ( ) 2

U sing the shear and m om ent diagram s, the desi red shear is ,

10 00050 200 50

50

and the desired m om ent is ,

200 50 10 000 10 000 1002

T he cho rd force on the south w all i s

f

f

, lbV ft x x

ft

x M x , lb , x x

= − = −

= − + = −

22

,

10 000 100

200 250

M , x x

C x x b ft

−

= = = −


41/47

Here it is required to find how long the 35 foot south wall panel should be in

order to have equal magnitude strut and chord forces.

Therefore, the strut force should be set equal to the chord force,

150 2

extended length

200 2

Solve for (the shear wall additional length measured from the east wall),

25

the south wall 10 5 750 2

x x x

x

x feet

ft ft ft

= −

=

= − =


42/47

Question #19.

The plan view of a wood structural panel roof diaphragm is shown below. The wallsare of wood-frame construction. For the south wall, at what point are the magnitudes

of the drag strut and the chord force identical? Assume ρ = 1.0.

A. At point X.

B. At point Y.

C. At point Z.

D. None.


43/47

Answer to #19.

A.

( )( )300 609 000

2 2

f lb ft wL

V , lb= = =


44/47

( ) ( )

( )( )

( ) ( )

9 000 30135 000

2 2

3 000 10135 000 120 0002

6 000 20135 000 75 000

2

f

X f

f

Y f f

f

Z f f

, lb ft Vx M , ft lb

, lb ft M , ft lb , ft lb

, lb ft M , ft lb , ft lb

= = = −

= − − = −

= − − = −

135 000 ft lbM


45/47

135 0002 250

60

120 0002 000

60

75 0001 250

60

The strut shear stresses in the roof and the walls are,

9 000 9 00= 150 =

60

f X X f

f Y Y f

f Z

Z f

f

roof f wall

, ft lb M C , lb

b ft

, ft lb M C , lb

b ft

, ft lb M C , lb

b ft

, lb ,V V lb / ft and

b ft b

υ υ υ υ υ υ υ υ

−= = =

−= = =

−= = =

= = =

( ) ( )

( ) ( )

0225

40

At X, the drag strut force is, 150 30 225 10 0 2 250

At Y, the drag strut force is, 150 20 225 10 0 750

At Z, the

f

f

f f

X X f

f f Y Y f

lblb / ft

ft lb lb

D ft ft D , lb ft ft

lb lb D ft ft D lb

ft ft

=

+ − = ∴ =−

+ − = ∴ =−

( ) ( )drag strut force is, 150 10 225 10 0 750 f f Z Z f lb lb

D ft ft D lb ft ft

+ − = ∴ =

Q ti #20


46/47

Question #20.

A building’s drift depends on,

A. the story height.

B. the building height.

C. the shear load.

D. All of the above.

A t #20


47/47

Answer to #20.

D.

The drift ∆ is the displacement between adjacent stories due to applied forces. The

drift can be divided into different components (for example, shear drift and chord

drift) depending on the nature of the loading and the displacement that results.

The drift depends on variables, such as the building and story heights, shear load,

girder length and depth, column length and height, and frame length.

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SD-Lecture18-Diaphragm Theory-III.pdf