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 DesignNCMA 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)

0.1Ff

Ff

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, FtMSJC Table 2.2.3.2 and UBC Table 21-I

Direction of Tensionand

Type of Masonry

Mortar TypePortland Cement/Lime

or Mortar Cement Masonry Cement/Lime

M or S M or SN Ntension normalto bed joints

solid unitshollow unitsfully grouted units

tension parallelto bed joints

solid unitshollow unitsfully 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 psiwith 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


URM Shear WallsDesign Criteria

IbVQfv =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 grout60 psi for running bond and solid grout

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


URM Shear WallsDesign Criteria(c) allowable shear stresses:

fvavg

for rectangular section

fvmax

netmaxv

maxvnet

avgv

AV

23f

f32

AVf

=

==


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 MH 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

33.1Ff

Ff

b

b

a

a <+ 99 rh' for f´ 0.23 ])

140rh( -[1 0.25f´ F m

2ma ≤==

33.1616

V0630.042650 b =+ govern) not (does lbs. 11,857 Vb =

ExampleMaximum 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 32 A F

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

ba

um===

==


ExampleMaximum 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 0630.042650

33.1Ff

Ff

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

UBCMSJC

7890 11,8571194

Vb maxSummary:

10,629 11,934794


URM Shear Walls

⎟⎠⎞

⎜⎝⎛

Le 2 -1 bL 3

4P= f m

Post-Cracked Behavior

h

av f=σ

toe

fm< Fa

3η

e

ηL/2

width = b

heel

H

P

PHhe;PeHh:mequilibriu

AP v

==

= σ

[1]bam F or Fstress edge ecompressivb

P2f ≤==η

[2]⎟⎠⎞

⎜⎝⎛ −=−= e

2L3 e

2L

3ηη

[3]22

LL1

e2Lb3

P2b

P2f

:equationsabove combining

m⎟⎠⎞

⎜⎝⎛ −

==η


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 ti

mes

cr

acki

ng lo

ad

MSJC/UBC assumed behavior


Perforated URM Shear Walls


Lateral Stiffness of Shear WallsCantilevered shear wall

h G1

AH

I3EHh=

:shearflexure

vm

3+∆∆

H

L

h bL= A12bL= 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

bLEHh 3 +

bLEHh4 = ∆

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ 3

Lh4

Lh

bEH=

2

m∆

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ 3

Lh4

Lh

bE= H/= stiffness lateral= k2

m∆


Lateral Stiffness of Shear WallsPier between openings

hG1

AH

I12EHh=

shearflexure

vm

3+∆

H

H

h

L

)EbL(0.4Hh 1.2 +

bLEHh=

m3

m

3∆

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ 3

Lh

Lh

bEH=

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

mbEk

Lh

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

3Lh4

Lh

1k2

cantilever

fixed pier

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

3Lh

Lh

1k2


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 Design

24D The Contemporary Bearing Wall- Example of Shear Wall Design

24I Earthquake Analysis of Engineered Brick Masonry Structures


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

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

3Lh4

Lh

bEk2

mi

*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

Σki = 0.2776 bEm


321 V V V H ++=equilibrium:

factorondistributik / kk / (k V

kkkk V

ii

iii

iiiiii

= ∑

)∑=

∑ / = = =

ΗΗ∆∆

shear force attracted to single pier:

Lateral-Force Distribution to PiersPerforated Shear Walls

pierfixedfixedfor

3Lh

Lh

Ebk2

i

i

i

i

miii −

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

overall story stiffness:

321i

i

kkkk stiffnessstory lateral Kwhere

kH

KHorK H

++ = ∑==

∑ = = = ∆∆

h3

L1

L2

H ∆

V1 V2

L3L2

V3

h 1 h 2


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 ==610420x30540x7.631397 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 1E 7.63

3Lh

Lh

bE= k2

mb ==

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

Ε164Ε121911121ki . = ). + . + .( = ∑

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


Perforated Shear WallsAxial Force due to Overturning

fmaxfai= ave. axial stress acrosspier “i”

c

y1 y2 y3

y1y2 y3

y

M

p1p2 p3

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

[5]iiaiii yAfyp M ∑=∑=

equilibrium of moments:

[6]c/yff imaxai =from similar triangles:

substituting in [5]:[7]

iii

max yAcyfM ∑ ⎟

⎠⎞

⎜⎝⎛=

[8]Ic

fM yAc

fM max2ii

max == ∑

[2]iaii Afp ∑= = ∑ 0

[3]axis neutral areas

pier of centroid thus,A i

=

≠∑ 0

[4]∑

∑=i

iiA

yAy


Perforated Shear WallsAxial Force due to Overturning

[10]

solving for fmax:

IMc= f max

substituting in [6]:

IyM=

cy

IMc= f ii

ai ⎟⎠⎞

⎜⎝⎛ [11]

[12]IyMA= 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 jkbd2M fncompressio <=

:forceaxialneglectingflexure: reinforced piers

flexure: unreinforced piers

1.0 Ff

Ffncompressio

bb

aa <+

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

diagramninteractiomoment -load (b)

controls) tension (only when bdF

PAs

s by modify (c)


ve

v

vnet

v

FAV= f

FbI

VQ= f

<

<

UBC

MSJC

shear: unreinforced piers

shear: reinforced piers

vv FbjdV= f < UBC

vv FbdV= f < MSJC

Perforated Shear WallsDesign Criteria for Piers between Openings

P

V

V

h

M

P

D+L D+L Pmax for small lateral loadUBC MSJC Sec. 2.1.1 EffectLoading 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 I12E

h1k

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

48 740,6712E

841k

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 I12E

h1k

mwebm

3b

+=

Stiffness of Pier “b”

7.63

”

40”

b

)E40)(0.4 7.63 (5/6

48 693,4012E

841k

mm

3b

××+

=

m

mm

b E 43.1

E472.0

E226.0

1k =+

=



)E)(0.4 A (5/6

h I12E

h1k

mwebm

3c

+=

32”

7.63

”

40”

Stiffness of Pier “c”

4g in 470,76I =

(same as Pier a)

b 1.43 Em 0.409 15.2 c 0.38 Em 0.109 4.0

Σk = 3.50 Em 1.000 37.2 k

k 2.37k4.7k 9.14k 9.14Vbase =++= pier ki 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

112 470,7612E

1121k

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

"8.101y 1 =

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

IyA = 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 : 0.75EExample: Perforated Shear Wall

* 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 Sh

ear (

kip)


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. DeflectionNorth-South Uni-directional Motion

0

510

15

2025

30

0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 Average First Floor Deflection (in)

Bas

e Sh

ear

(kip

)

PGA = 0.33 g

PGA = 0.75 gPGA = 0.98 gPGA = 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)

Base

She

ar (k

ip)

PGA = 0.30 g

PGA = 0.75 g

PGA = 1.09 gPGA = 1.40 g

slide from S. Sweeney


End of Lessons 4 & 5

lecture 4 5 urm shear walls

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