lecture 4 5 urm shear walls

Masonry Structures, slide 1

Classnotes for ROSE School Course in:Masonry Structures

Classnotes for ROSE School Course in:Masonry Structures

Notes Prepared by:Daniel P. Abrams

Willett Professor of Civil EngineeringUniversity of Illinois at Urbana-Champaign

October 7, 2004

Lessons 4 and 5: Lateral Strength and Behavior of URM Shear Wallsflexural strength, shear strength, stiffness, perforated shear walls


Existing URM Buildings


Damage to Parapets

1994 Northridge Earthquake, Filmore

1996 Urbana Summer


Damage Can Be Selective

1886 Charleston, South Carolina


Damage to Corners

1994 Northridge Earthquake, LA


Damage to In-Plane Walls

1994 Northridge Earthquake, Hollywood

URM cracked pier, Hollywood


Damage to Out-of-Plane Walls

1886 Charleston, South Carolina

1996 Yunnan Province Earthquake, Lijiang


Likely Consequences

St. Louis Firehouse

1999 Armenia, Colombia Earthquake


2001 Bhuj Earthquake


Lateral Strength of URM Shear Walls


URM Shear Walls

n

1=iiib

n

1=iib

n

1=iib

h H= M

H= V

P= P

Ref: BIA Tech. Note 24C The Contemporary Bearing Wall - Introduction to Shear Wall Design NCMA TEK 14-7 Concrete Masonry Shear Walls

P3

Pb

hi

H3

Hi

H1

Pi

P1

flexural tension crack

flexural compression cracks

VbMb

diagonaltension crack


URM Shear Walls Design Criteria

(a) allowable flexural tensile stress: -fa + fb < Ft

Ft given in UBC 2107.3.5 (Table 21 - I); Ft = 0 per MSJC Sec. 2.2.3.2

pg. cc-35 of MSJC Commentary reads: Note, no values for allowable tensile stress are given in the Code for in-plane bending because flexural tension in walls should be carried by reinforcement for in-plane bending.

where:Fa = allowable axial compressive stress (UBC 2107.3.2 or MSJC 2.2.3)Fb = allowable flexural compressive stress = 0.33 f´m (UBC 2107.3.3 or MSJC 2.2.3)

0.1F

f

F

f

b

b

a

a

(b) allowable axial and flexural compressive stress:

MSJC Sec. 2.2.3.1 and UBC 2107.3.4 unity formula:


Allowable Tensile Stresses, Ft

MSJC Table 2.2.3.2 and UBC Table 21-I

Direction of Tensionand

Type of Masonry

Mortar Type

Portland Cement/Limeor Mortar Cement Masonry Cement/Lime

M or S

M or S

N Ntension normalto bed joints solid units hollow units fully grouted units

tension parallelto bed joints solid units hollow units fully grouted units

40 25 68*

80 50 80*

30 19 58*

60 38 60*

24 15 41*

48 30 48*

15 9 26*

30 19 29*

* grouted masonry is addressed only by MSJC

all units are (psi)


URM Shear WallsDesign

Criteria(c) allowable shear stresses:

UBC Sec. 2107.3.7 shear stress, unreinforced masonry:clay units:

Fv = 0.3 (f’m)1/2 < 80 psi (7-44)concrete units:

with M or S mortar Fv = 34 psi with N mortar Fv = 23 psi

ev A

Vf Per UBC Sec. 2107.3.12 shear stress is average shear stress,

allowable shear stress may be increased by 0.2 fmd where fmd is compressive stress due to dead load



Criteria

Ib

VQfv Note: Per MSJC Sec. 2.2.5.1, shear stress is maximum stress,

(c) allowable shear stresses:

MSJC Sec. 2.2.5.2: shear stress, unreinforced masonry:Fv shall not exceed the lesser of:

(a) 1.5 (f’m)1/2

(b) 120 psi

(c) v + 0.45 Nv/An where v = 37 psi for running bond, w/o solid grout 37 psi for stack bond and solid grout

60 psi for running bond and solid grout

(d) 15 psi for masonry in other than running bond



Criteria(c) allowable shear stresses:

fvavg

for rectangular section

fvmax

netmaxv

maxvnet

avgv

A

V

2

3f

f3

2

A

Vf


URM Shear Walls

Possible shear cracking modes.

strong mortarweak units

through masonry units

Associated NCMA TEK Note

#66A: Design for Shear Resistance of Concrete Masonry Walls (1982)

low vertical compressive stress

sliding along bed joints

weak mortarstrong units

stair step through bed and head joints


Example: URM Shear WallsDetermine the maximum base shear per UBC and MSJC.

5000 lb. DL

H

H

9’-

4”9’

- 4”

6’ - 8”

8” CMU’s with face shell beddingblock strength = 2800 psiType N Portland cement lime mortar special inspection provided during construction

2net in 200 (80) 2.5 A

32net in 2667 6 / (80) 2.5 S

4.39r

'h 112" h’ 2.84" r

Net section with face shell bedding:80”

1.25

”


Example

bbb

2deada

bbb

b

V0.0630 /2667V 168 M/S f

weight)selfg (neglectin psi50 in 200 / lbs)(5000 2 f

V 168 12x ) 9.33x (1.5 V M

H 2 V

Forces and Stresses:

Maximum base shear capacity per UBC

govern) not (does lbs.7980 )in(200 1.33)x psi (33 V

2107.3.7)(Sec. psi 33 (50) 0.2 23 f 0.2 23 F2

max

mdv

shear stress

flexural tensile stress

(governs) lbs.1194 Vb 25.3 0.0630V 50 - F f f-

) 2107.3.5Sec. per(F psi 25.3 1.33x psi 19 F

btba

tt


flexural compressive stress

33.1F

f

F

f

b

b

a

a 99 r

h' for f´ 0.23 ])

140r

h( -[1 0.25f´ F m

2ma

33.1616

V0630.0

426

50 b govern) not (does lbs. 11,857 Vb

Example Maximum base shear capacity per UBC

Maximum base shear capacity per MSJCIn lieu of prism tests, a lower bound compressive strength of 1861 psi will be used based on the 2800 psi unit strength and Type N mortar per MSJC Spec. Table 2.

shear stress

govern) not (does lbs. 10,629

)in(200 1.33)x psi(60 3

2 A F

3

2 V

(governs) psi60 psi)(50 0.45 psi 37 F

or psi65 (1861)1.5 )(f´1.5 F

2netvmax

v

1/21/2mv

psi 616 (1850) 0.33 Fpsi 426 F

mortar NType and psi 2800 f´ for D-Table 21 per psi1850 f´

ba

um


Example Maximum base shear capacity per MSJC

flexural tensile stress

0= V0.0630 +50 -

0= f + f-

psi0 = F: 2.2.3.2Sec. MSJC per

b

ba

t

(governs) lbs. 794 Vb

flexural compressive stress

govern) not (does lbs.11,934 = V 33.1620

V 0630.0

426

50

33.1F

f

F

f

bb

b

b

a

a

psi620 (1861) 0.33 F

39.4 h/r for psi 426 f´ 0.23 ](h/140r) -[10.25f´ F

b

m2

ma

shear stress tension compressionaxial and flexural stress

UBC

MSJC

7890 11,8571194

Vb max Summary:

10,629 11,934794


URM Shear Walls

Le

2 -1 bL 3

4P= fm

Post-Cracked Behavior

h

av f

toe

fm< Fa

3

e

L/2

width = b

heel

H

P

P

Hhe;PeHh:mequilibriu

AP v

[1]bam F or Fstress edge ecompressivb

P2f

[2]

e

2

L3 e

2

L

3

[3]22

LL1

e2L

b3

P2

b

P2f

:equationsabove combining

m


Note: shear strength should be checked considering effects of flexural cracking

URM Shear WallsPost-Cracked Behavior

Lat

eral

Loa

d, H

Lateral Deflection at Top of Wall

first flexural cracking

resultant load, P, shifts toward toetoe crushing

2 to

3 t

imes

cr

ack

ing

load

MSJC/UBC assumed behavior


Perforated URM Shear Walls


Lateral Stiffness of Shear WallsCantilevered shear wall

h G

1

A

H

I3E

Hh=

:shearflexure

vm

3

H

L

h bL= A12

bL= I

g

3

g)E(0.4 A

Hh 1.2 +

I3E

Hh=

E0.4 = G A 5/6= A :sassumption common

mgm

3

mgv

m3

m

3

bLE

Hh 3 +

bLE

Hh4 =

3

L

h4

L

h

bE

H=

2

m

3

Lh

4Lh

bE= H/= stiffness lateral= k

2m


Lateral Stiffness of Shear WallsPier between openings

hG

1

A

H

I12E

Hh=

shear flexure

vm

3

H

H

h

L

)EbL(0.4

Hh 1.2 +

bLE

Hh=

m3

m

3

3

L

h

L

h

bE

H=

2

m

32

Lh

Lh

bE/H stiffnesslateralk m


Lateral Stiffness of Shear Walls

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.20 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

mbE

k

L

h

3Lh

4Lh

1k

2

cantilever

fixed pier

3Lh

Lh

1k

2


References

Associated NCMA TEK Note:

61A Concrete Masonry Load Bearing Walls - Lateral Load Distribution (1981)

Associated BIA Technical Note:

24C The Contemporary Bearing Wall - Introduction to Shear Wall Design24D The Contemporary Bearing Wall - Example of Shear Wall Design24I Earthquake Analysis of Engineered Brick Masonry Structures


Example: Lateral-Force DistributionDetermine the distribution of the lateral force, H, to walls A, B and C.

3Lh

4Lh

bEk

2m

i

*based on cantilever action

type of masonry and wall thickness is the same for each wall

iii k/k*k h/LLwall

A 10’ 1.50 0.0556 bEm 0.20

10’-0”

h=15’

A

H 18’-0”

B

B 18’ 0.83 0.2077 bEm 0.75

C

6’-0”

C 6’ 2.50 0.0143 bEm 0.05

ki = 0.2776 bEm


321 V V V H equilibrium:

factorondistributik / k

k / (k V

kkkk V

ii

iii

iiiiii

shear force attracted to single pier:

Lateral-Force Distribution to PiersPerforated Shear Walls

pierfixedfixedfor

3Lh

Lh

Ebk

2

i

i

i

i

miii

overall story stiffness:

321i

i

kkkk

stiffnessstory lateral Kwhere

k

H

K

HorK H

h3

L1

L2

H

V1V2

L3L2

V3

h1 h2


A

H

56”

a

40”11

2”

24”

64”

24”

7.63”

Va

Example: Lateral Force Distribution to PiersDetermine the distribution of story shear, H, to each pier.

Section A-A

Elevation

b

40” 32”

Vb

A

8”groutedconcrete block

c

Vc


piers a and c

4322gross in84,273/127.63(40.0)11.2)305(203.81)366(11.2I

"2.11671

7501y 610420x30540x7.63

1397 7.63/2x 366 x48 7.63

671 7501

40”

7.63”

Example: Lateral Force Distribution to Piers

y

48”

7.63

”

2webv in 254 )(40.0)(7.63" (5/6) A (5/6) A

E12.1

)254)(E4.0(64

)12E(84,273)64(

1

GAh

12EIh

1= k= k

3

v

3ca

pier b

64”

7.63

”

E91.1 3]+ [1 1

E 7.63

3Lh

Lh

bE= k

2m

b

Ε164Ε121911121ki

46.0 16.4/91.1 DF27.0 16.4/12.1 DF DF bca


Perforated Shear WallsAxial Force due to Overturning

fmax

fai= ave. axial

stress acrosspier “i”

c

y1 y2y3

y1y2 y3

y

M

p1

p2 p3

[1]forcegravity zero forp i 0equilibrium of pier axial forces:

[5]iiaiii yAfyp M

equilibrium of moments:

[6]c/yff imaxai from similar triangles:

substituting in [5]:[7]

iii

max yAc

yfM

[8]Ic

fM yA

c

fM max2

iimax

[2]iaii Afp 0

[3]

axis neutral areas

pier of centroid thus,A i

0

[4]

i

ii

A

yAy


Perforated Shear WallsAxial Force due to Overturning

[10]

solving for fmax:

I

Mc= fmax

substituting in [6]:

I

yM=

c

y

I

Mc= f ii

ai

[11]

[12]I

yMA= fA= p ii

iii

[13]

distribution factor for overturning moment

2ii

iii

yA

yA M= p


Perforated Shear WallsDesign Criteria for Piers between Openings

b2b F jkbd

2M fncompressio

:forceaxialneglecting

flexure: reinforced piers

flexure: unreinforced piers

1.0 F

f

F

fncompressio

b

b

a

a P

P = Pdead + Plive + Plateral

V

V

h

M

P

M=Vih/2

tba < F f +f-tension

ss

s FjdA

M ftension

only) (UBC equation unity (a)

:forceaxialgconsiderin

diagram ninteractiomoment -load (b)

controls) tension (only when bdF

PA

ss by modify (c)


ve

v

vnet

v

FA

V= f

FbI

VQ= f

UBC

MSJC

shear: unreinforced piers

shear: reinforced piers

vv Fbjd

V= f UBC

vv Fbd

V= f MSJC

Perforated Shear WallsDesign Criteria for Piers between Openings

P

V

V

h

M

P

D+L D+L Pmax for small lateral load

UBC MSJC Sec. 2.1.1 Effect

Loading Combinations

M=Vih/2

0.75(D+L+W/E) D+L+W/E Pmax and Mmax for large lateral0.9D-0.75E 0.9D+E Pmin for smallest moment capacity D+W


Example: Perforated Shear WallCheck stress per the UBC for the structure shown below. Design pier reinforcement if necessary.

Gravity LoadsLevel Dead Live3 50 kip 80 kip

Special inspection is providedf’m = 2500 psifully grouted but unreinforcedGrade 60 reinforcementType N mortar with Portland Cement

Earthquake Loads

14.9 kip

7.4 kip

10’-

0”10

’-0”

9’-8

”

14.9 kip14.9 kip

18’-8”

2 60 kip 80 kip1 60 kip 80 kip

total 170 kip 240 kip


Example: Perforated Shear Wall18’-8”Pier Dimensions

9’-4

” 8” groutedconcrete block

3’-4

”

40” 32”

3’-4” 5’-4” 3’-4” 3’-4”3’-4”2’

-8”

4’-0

”2’

-8”

7.63”

a b c


Example: Perforated Shear Wall

12.82" 549 / /2)](40 7.63 2/ (7.63) [32 y 22

)E)(0.4 A (5/6

h

I12Eh

1k

mwebm

3a

Stiffness of Pier “a”

7.63”

32”

7.63

”

40”y

a

423

23g

in470,76 12.82) 0 -7.63(40)(2 /12 7.63(40)

7.63/2) -2.827.63(32)(1 (32)/12(7.63) I

)E40)(0.4 7.63 (5/6

48

740,6712E84

1k

mm

3a

m

mm

a E 69.1

E472.0

E120.0

1k



43g in 40,693 12 / (40) 7.63I

)E)(0.4 A (5/6

h

I12Eh

1k

mwebm

3b

Stiffness of Pier “b”

7.63

”

40”

b

)E40)(0.4 7.63 (5/6

48

693,4012E84

1k

mm

3b

m

mm

b E 43.1

E472.0

E226.0

1k



)E)(0.4 A (5/6

h

I12Eh

1k

mwebm

3c

32”

7.63

”

40”

Stiffness of Pier “c”

4g in 470,76I

(same as Pier a)

b 1.43 Em0.409 15.2

c 0.38 Em 0.109 4.0

k = 3.50 Em 1.000 37.2 k

k 2.37k4.7k 9.14k 9.14Vbase pier k i DF i V i

a 1.69 Em 0.483 18.0

Distribution of Story Shear to Piers

c

)E40)(0.4 7.63 (5/6

112

470,7612E112

1k

mm

3c

m

mm

c E 38.0

E101.1

E526.1

1k


7.63”40

.0”

12.82”

Distribute Overturning Moments to PiersExample: Perforated Shear Wall

"6.1141403/807,160y

pier Ai yi Aiyi

a 549 12.8” 7038

a

"8.101y1

y

124.0”

"38.9y 2

b

b 305 124.0” 37,820

"58.96y 3

c

211.2”

c 549 211.2” 115,949

Ai =1403 Aiyi=160,807


total story moment = M1 (@top of window opening, first story) = 14.9k x 23.0’ + 14.9k x 13.0’ + 7.4k x 3.0’ = 558k-ft

Distribute Overturning Moments to PiersExample: Perforated Shear Wall

11,029IyA 1403A 2iii

b 305 -9.38 27 41 68 -2.9 -1.8

c 549 -96.58” 5120 76 5196 -53.0 -32.1

pier

(in2)

Ai y

(in)

2

ii yA

(1000 in4) (1000 in4)

IyA 2ii

(1000 in4)

1n

iii M

I

yA = P

(kips)

iiyA

(1000 in3)

I

a 549 101.8” 5689 76 5765 55.9 33.9


*based on tributary wall length: pier a: (32” + 40” + 32”)/288 = 0.361(assuming that floor loads are pier b: (32” + 40” + 20”)/288 = 0.319applied uniformly to all walls) pier c: (20” + 40” + 32”)/288 = 0.319

Summary of Pier ForcesExample: Perforated Shear Wall

pier % gravity* Pd Pl Peq Veq Meq=Veq(h/2)(kips) (kips) (kips) (kips) (kip-in)

a 0.361 61.4 86.6 33.9 18.0 432

b 0.319 54.2 76.6 -1.8 15.2 365

c 0.319 54.2 76.6 -32.1 4.0 224


Loading CombinationsExample: Perforated Shear Wall

*UBC 2107.1.7 for Seismic Zones 3 and 4

axial compressive force, P moment, M shear, V dcase 1 case 2 case 3

pier D+L 0.75(D+L+E) 0.9D-0.75E 0.75Meq

0.75Veqx1.5*(kips) (kips) (kips) (kip-in) (kips) (in.)

a 148.0 136.4 29.8 327 20.3 36

b 130.8 99.5 47.4 274 17.1 36

c 130.8 122.2 24.7 168 4.5 36


*Fa = 0.25f’m[1-(h/140r)2] 112"4"9'h 2.2"7.630.2890.289tr Note that conservative assumption is used for Fa calculation, r is the lowest and h is the full height.

Axial and Flexural Stresses, Load Case 1 = D + LExample: Perforated Shear Wall

pier PD+L fa Fa* fa/Fa

(kips) (psi) (psi)

a

y

a 148.0 270 543 0.497 < 1.0 ok

b

b 130.8 430 543 0.792< 1.0 ok

y

c

c 130.8 239 543 0.440< 1.0 ok


Example: Perforated Shear WallAxial and Flexural Stresses, Load Case 2: 0.75 (D + L + E)

* minimum Sg is taken to give maximum fb for either direction of building sway** Fb= 0.33f’m = 833 psi

pier 0.75(PD+L+EQ) fa=P/A Fa fa/Fa 0.75Me Sg fb fb/Fb** fa/Fa+fb /Fb

(kips) (psi) (psi) (kip-in) (in3) (psi)

a

y

a 136.4 249 543 0.459 327 2813* 116 0.139 0.598 < 1.0 ok

b

b 99.5 326 543 0.600 274 2035 135 0.162 0.762 < 1.0 ok

y

c

c 122.2 223 543 0.411 168 2813* 60 0.072 0.483 < 1.0 ok


minimum axial compression: check tensile stress with Ft = 30 UBC Sec 2107.3.5

Example: Perforated Shear WallAxial and Flexural Stresses, Load Case 3: 0.9D -

0.75Peq

* minimum Sg is taken to give maximum fb for either direction of building sway**tensile stresses

pier (0.9PD-0.75PEQ) fa=P/A 0.75Meq Sg fb - fa+fb

(kips) (psi) (kip-in) (in3) (psi) (psi)**

a

y

a 29.8 54 327 2813* 116 62 > 30 psi provide reinf.

b

b 47.4 155 274 2035 135 -20 < 30 psi ok

y

c

c 24.7 45 168 2813* 60 15 < 30 psi ok


Pier Shear Stress, Load Case 4 : 0.75E


* from Case 3 0.9Pd-0.75Peq

** UBC 2107.3.7

pier V=0.75Veq x 1.5 fv = V/Aweb fao = P/A* Fv = 23 + 0.2fao** (kips) (psi) (psi) (psi)

a

y

a 20.3 67 54 34 < 67 provide shear reinf.

b

b 17.1 56 155 54 < 56 provide shear reinf.

y

c

c 4.5 15 45 32 > 15 ok


Case Study: Large-Scale Test


Georgia Tech Large-Scale Test

24’

photo from Roberto Leon


Final Crack Pattern

Load Direction

slide from Roberto Leon


Final Crack Pattern

Load Direction



Results- Global Behavior

Wall 1 Force-Displacement Response

-80

-60

-40

-20

0

20

40

60

80

-0.30 -0.20 -0.10 0.00 0.10 0.20 0.30

Roof Displacement (in)

Bas

e S

hea

r (k

ip)


Overturning Effect (Vertical Stress)

Base strains recorded during loading in the push and pull direction



USA CERL Shaking Table Tests

12’

photos from S. Sweeney


Damage on North Wall

Permanent offsets of 0.25” – 0.35” due to rocking of pier.

Final Cracking Pattern

slide from S. Sweeney


Peak Force vs. Deflection

North-South Uni-directional Motion

0

5

10

15

20

25

30

0 0.025 0.05 0.075 0.1 0.125 0.15 0.175

Average First Floor Deflection (in)

Bas

e S

hea

r (k

ip)

PGA = 0.33 g

PGA = 0.75 g

PGA = 0.98 g

PGA = 1.08 g

East-West Uni-directional Motion

0

5

10

15

20

25

0 0.1 0.2 0.3 0.4 0.5 0.6

Average First Floor Deflection (in)

Bas

e S

hea

r (k

ip)

PGA = 0.30 g

PGA = 0.75 g

PGA = 1.09 g

PGA = 1.40 g

slide from S. Sweeney


End of Lessons 4 & 5

lecture 4 5 urm shear walls

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