Lecture 6 7 Rm Shear Walls

Masonry Structures, slide 1

Reinforced MasonryWorking Stress Design of flexural members

bdAs=ρ

Ts = Asfs

Cm = fmb kd/2

M

b

d

t

Asgrout

unit

strain

Ref: NCMA TEK 14-2 Reinforced Concrete MasonryBIA Tech. Note 17 Reinforced Brick Masonry - Part IBIA Tech. Note 17A Reinforced Brick Masonry - Materials and Construction

n.a.kd

sε

mε fm

fs/n

stress


misi :fiber particularany at #5, Assumption from εε =

Assumptions1. plane sections remain plane after bending

(shear deformations are neglected, strain distribution is linear with depth)2. neglect all masonry in tension3. stress-strain relation for masonry is linear in compression4. stress-strain relation for steel is linear5. perfect bond between reinforcement and grout

(strain in grout is equal to strain in adjacent reinforcement)6. masonry units and grout have same properties


kk1nff

kddnf

kdf :ondistributi stress ofgeometry from ms

sm −=

−=

mimim

ssi

m

mi

s

si nffEEf

Ef

Ef :#4 and #3 sAssumption from ===


k

k1n2/k

kk1fnbdfbdfA

2bkdf

:T C m,equilibriu from

msssm

−=

−===

=

ρ

ρρ

0n2nk2k 2 =−+ ρρ


Allowable reinforcement tensile stress per MSJC Sec.2.3.2:Fs=20 ksi for Grades 40 or 50; Fs=24 ksi for Grade 60Fs=30 ksi for wire joint reinforcement

Allowable reinforcement tensile stress per UBC Sec.2107.2.11 :Fs= 0.5fy < 24 ksi for deformed bars; Fs= 0.5fy < 30 ksi for wire reinforcementFs= 0.4fy < 20 ksi for ties, anchors, and smooth bars

If fs=Fs then moment capacity will be limited by reinforcement.

from equilibrium:

s2

sss jfbdjdfAM ρ== where jd=d-kd/3 or j=1-k/3

∑ = 0C aboutM m


If fm= Fb then moment capacity will be limited by masonry.

UBC 2107.2.6 & MSJC Sec.2.3.3.2:Fb=0.33f’m

2mmm jkbdf5.0bkdjdf5.0M ==

from equilibrium: ∑ = 0T aboutM s



WSD: Balanced Condition

• For any section and materials, only one unique amount of balanced reinforcement exists.

• Although balanced condition is purely hypothetical case, it is useful because it alerts the engineer to whether the reinforcement or the masonry stress will govern the design. Balanced stresses are not a design objective.

Reinforced Masonry

Definition: The balanced condition occurs when the extreme fiber stress in the masonry is equal to the allowable compressive stress, Fb, and the tensile stress in the reinforcement is equal to the allowable tensile stress, Fs.

C = fmb kd/2

fs/n = Fs /n

d

kbd

fm = Fb

T = As fs


WSD: Balanced ConditionReinforced Masonry

bsb /F F+ n

n= k

dnF + F

dkF

sb

b

b =

from geometry:

fs/n = Fs /n

d

kbd

fm = Fb

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+

=bsbs

b F/F21

F/Fnn ρ

)(22

2d

b

ss

b

s

bbb

sbbb

FFn

nF

FFkF

FdbρkbF

+==

=

ρ

from equilibrium: C=T

nF + F

F= k s

b

bb


Example: Balanced ConditionDetermine the ratio of reinforcement that will result in a balanced condition per UBC.Given: f’m = 2000 psi and Grade 60 reinforcement

ksi 29,000 Eksi1500 = f'750 =Eentreinforcem60 Grade for ksi= 24 F psi 667= f' 0.33=F

smm

smb

=

0.48%= x24/0.667 2

1 24/0.667+ 3.19

3.19b ⎥⎦

⎤⎢⎣⎡

⎥⎦⎤

⎢⎣⎡=ρ

19.3= 1500

29,000= EE= n

m

s


Design Strategy for RM Flexural DesignProcedure for sizing section and reinforcement for given moment.

Calculate ρb knowing f’m and fydetermine Fb from f’mdetermine Fs from fydetermine Emfrom f’mdetermine n = Es/Em

Size section for some ρ < ρbdetermine k and jbd2 = M/ρjFsselect b and d using common units

Size reinforcementAs = M/Fsjdselect number and size of rebars

Check designMs = AsFsjd > Mfb = M/0.5jkbd2 < Fb

Note: Section must also be sized for shear.


Example: Reinforced MasonryDesign a beam section for a moment equal to 370 kip-in. Prisms have been tested and f’m is specified at 2000 psi. Use Grade 60 reinforcement and 8” CMU’s.

%48.0b example,previous From1. =ρ

0.4% isestimate good A govern. willsteel so b than lowerslightly be to Estimate .2

=ρ

ρρ

892.03/k1j323.0k0154.0k154.02k

154.0)3.19)(004.0(2n20n2nk2 2k

:k forSolve 3.

=−===−+

===−+

ρρρ

3in4321)ksi24)892.0)(004.0/()inkip370(2bdsFj/M2bd

:2bd for Solve.4

=−=

= ρ


Example: Reinforced Masonry5. Select dimensions of beam using 8” CMU’s:

b = 7.63”dreq’d = [4321 / 7.63]0.5 = 23.8”use four units and center bars in bottomunit, d = 27.8”

4 - 8” CMU’sd=27.8”

7.63”

)in (0.62 in 0.62 (27.8) (0.892) ksi)(24 / in)-kip(370 A

dj F/ M A :entreinforcem of amountEstimate 6.

22dreq' s

sdreq's

==

=

s#5' use 2

psi ok667 psi488"827"6372840905050

1000xinkip370dkj50Μf

in. OKkip370in.kip374"8279050ksi24in620djFΑΜ

9050328401j2840 k0ρn2ρnk2 k

002920"827"637in620bdΑ ρ

esign:. Check d7

22

m

2sss

2

2s

<=).)(.)(.)(.)(.(

) − ( = . / =

−> − = ).)(.)( )( .( = =

. = /. − = . = = − +

. = ).( ).( / . = / =


Flexural Capacity of Partially Grouted MasonryCase A: neutral axis in flange * per MSJC Sec. 2.3.3.3

If kd < tf assumption is valid, determine moment capacity as for rectangular section. If kd > tf assumption is not valid, need to consider web portion.

t

tf

As per width b

b = 6t or 72” or s*

As

flangeb

d

kd

neutralaxis

If neutral axis is in flange, cracked section is the same as a solid rectangular section with width “b.” Therefore, depth to neutral axis from extreme compression fiber may be calculated using:

bdA0n2nk2k s2 ==−+ ρρρ


Shear Design of Reinforced Masonry

• Vm reduces• dowel action invoked

Once diagonal crack forms:• flexural stresses increase• fsa is related to Mb

d

s

Asfs

Cm

R Basic shear mechanisms:before cracking: Vext = Vint = Vm + Vd + Viy + Vs

Vd

Vm

Vext.

Viy

Vs

Presence of shear reinforcement will:• restrict crack growth• help dowel action

• resist tensile stress


Shear Design of Reinforced Masonry

after cracking: Vext = Vint = Vs = nAvfs where n is the number of transversebars across the diagonal crack. Assuming a 45 degree slope, n=d/s

Vs = (d/s)Avfs

s

s

s

sv

dFV

dfV

sA

== UBC Sec. 2107.2.17 (Eq. 7.38)MSJC Sec. 2.3.5.3 (Eq. 2-26)


Shear Design of Reinforced MasomryFlexural shear stress

fvb dx = dT = dM/jdfv = (dM/dx)/bjd

UBC Sec. 2107.2.17 (Eq. 7-38)bjdVfv =

bdVfv = MSJC Sec. 2.3.5.2.1 (Eq. 2-19)

dx

M M + dMC

jd

T

C + dC

T + dT

na

T T + dT

fvbdx


Shear Design of Reinforced MasonryAllowable shear stresses for flexural members per UBC and MSJC

UBC Sec. 2107.2.8.A and MSJC Sec. 2.3.5.2.2(a): members with no shear reinforcement

psi50 f'1.0 F mv <= UBC Eq. 7-17; MSJC Eq. 2-20

psi150 f' 3.0 F mv <=

UBC Sec. 2107.2.8.B and MSJC Sec. 2.3.5.2.3(a): members with shear reinforcement designed to take the entire shear

UBC Eq. 7-18; MSJC Eq. 2-23


Shear Design of Reinforced MasonryAllowable shear stresses for shear walls per UBC and MSJC

UBC Sec. 2107.2.9.i and MSJC Sec. 2.3.5.2.2(b): walls with in-plane flexural reinforcement and no shear reinforcement

psi)VdM4580('f)

VdM4(

31F 1

VdM for mv −<−=< UBC Eq. 7-19; MSJC Eq. 2-21

UBC Sec. 2107.2.9.ii and MSJC Sec. 2.3.5.2.3(b):walls with in-plane flexural reinforcement and shear reinforcement designed to take 100% of shear

)psiVdM45120(f')

VdM4(

21 F1

VdMfor mv −<−=< UBC Eq. 7-21; MSJC Eq. 2-24

UBC Eq. 7-20; MSJC Eq. 2-22psi35'f0.1F 1VdM for mv <=≥

UBC Eq. 7-22; MSJC Eq. 2-25hpsi75'f5.1F 1VdM for mv <=≥


Shear Design of Reinforced MasonryMoment-to-Shear Ratios

d

dh

VdVh

VdM

==

V

M

For a single-story cantilevered shear walls

h

V

M

M

For piers between openings

h d

d2h

Vd2/Vh

VdM

==


MSJC Sec. 2.3.5.3.1smax = d/2 or 48”

d/2

Vdesign

Shear Design of Reinforced MasonryAdditional MSJC Requirements

Vdesign

MSJC Sec. 2.3.3.4.2minimum reinforcement perpendicular to shear reinforcement = Av/3 smax = 8 ft

MSJC Sec. 2.3.5.5design for shear force at distance “d/2” out from support


Provide Reinforcement to Take 100% of Shear

s

svdFV

sA

=Determine Shear Stress

bdVor

bjdVfv =

consider as unreinforced

End

Resize Section is fv<Fv?

Determine Fv Assuming Shear Reinforcement to take 100% of Shear

Determine Fv Assuming No Shear Reinforcement

Determine Maximum Design Shear

is ft>Ft?

Shear Design of Reinforced MasonryShear Design Strategy for Reinforced Sections

Determine Flexural Tension Stressft= -P/A+Mc/I

Start

yes

no

is fv<Fv?

no

yes

no

yes


Determine the maximum lateral force, Hwind per UBC and MSJCExample: Design of RM Shear Wall

8” CMU wallType S - PCL mortarsolidly grouted f’m=3000 psi

#4 @ 32”

2 - #8’s each end of wall

6’-8”

6’-4”

8’-0

”

120 psi

Case A: neglect all reinforcementCase B: consider vertical reinf., neglect horizontal reinf.Case C: consider vertical and horizontal reinf.Case D: design horizontal reinforcement for max. shear


kips 1.47)1000

psi1.77)(0.80x63.7(FAV

psi 1.7733.1x)]120(2.034[33.1x]f2.0psi 34[F

vemax

deadav

===

=+=+=shear

Case A: neglect all reinforcementExample: Design of RM Shear Wall

32

g in 81396

8063.7S =×

=per UBC:flexure

kips 7.14.lbs 684,14H33.1x408139

H96120FSMf - ta ===

×+−=+

kips 2.10.lbs 174,10H 08139

H96120FS/Mf - ta ===×

+−=+

per MSJC:flexure

kips3.44)80x63.7)(psi 109(32btF

32V

psi 109psi 2.8233.1Fpsi 2.82f'1.5psi 114)120(45.060F

vmax

v

mv

===

=×=

=>=+=shear


133.16.0'8.0'M/Vd >==

Shear per UBC Sec.2107.2.9 or MSJC Sec.2.3.5.2

Case B: consider only vertical reinforcement Example: Design of RM Shear Wall

Flexure by UBC or MSJC: neglecting fa

lumping 2 - #8’s ave. d for 2 bars

Ms = AsFsjd = 2 x 0.79 in2 (1.33 x 24 ksi) (0.9 x 72.0”) = 3268 k-in Hwind = 34.0 kips

psi 6.46psi35x33.1Fpsi35'f0.1F1VdMfor vmv ==<=>

governs kips 0.231000/)psi6.46)("72)(9.0)("63.7(bjdFVUBCfor vmax ===

governs kips 6.251000/)psi6.46)("72)("63.7(bdFVMSJCfor vmax ===


Flexure by UBC or MSJC: same as case B

Case C: consider all reinforcementExample: Design of RM Shear Wall

psi100 = psi75 x 1.33= F

psi75 > psi 82.2= 30001.5 = Fpsi75 f'1.5 = F1> VdM for

v

vmv ≤

Overall shear per UBC Sec. 2107.2.9.C or MSJC Sec. 2.3.5.2.3 (b)

okay psi100 psi 30.7 (72))9.0(7.63)(

1000)x kips(14.4 bjdV fv <===UBC

okay psi100 psi 27.7(7.63)(72)

1000)x kips(14.4 bdV fv <===MSJC

Vmax= Vs=(Av/s)Fsd = (0.20 in2/32”)(24 ksi x 1.33)(72”) = 14.4 kips governs

Shear per UBC Sec. 2107.2.17 or MSJC Sec.2.3.5.3


ontal in. horiz8 @ 4" use #3.9 0215.0 / 20.0 ) s in20.0 (Av rebars 4g #sinu

per in. in0215.0 ) 72 ksi)(24 x 33.1 kips/(4.49 d /F V/S A

sy govern kips oka34 kips 4.49 1000)/72)(9.0)(63.7 psi)(100 (bjd F V

2

2smaxv

vmax

===

===

>===

Case D: design horizontal reinforcement for maximum shear strengthExample: Design of RM Shear Wall

Summary: Hmax, kips

10.2*

27.0

15.2

34.0*

No steel

vertical steelno horizontal steel

vertical steel and#4 @ 32” horizontal steel

#4 @ 8” horizontal

14.7*

24.3

15.2

34.0*

CaseA

B

C

D

Consideration UBC MSJC

*flexure governs

CaseA

B

C

D

Consideration UBC MSJC

No steel 14.7* 10.2*vertical steelno horizontal steel 23.0 25.6

vertical steel and#4 @ 32” horizontal steel 14.4 14.4

#4 @ 8” horizontal 34.0* 34.0*


Flexural Bond StressM = TjdM + dM = (T + dT)jddM = dT jddT = dM/jd

allowable bond stress per UBC Sec.2107.2.2.4:60 psi for plain bars200 psi for deformed bars100 psi for deformed bars w/o inspection

dx

U

TT + dT

dx

M

C C + dC

T

jd

T + dT

M + dMdx

dx

U = bond force per unit length for group of barsU dx = dT = dM/jdU = (dM/dx)/jd = V/jdu = flexural bond stress =

where sum of perimeters of all bars in group

UBC Sec. 2107.2.16 Eq. 7-36

∑ o

U

=∑ o

jdVuοΣ

=


Development Length

82.Eq 2.8.1.2.Sec MSJCpsi 167uforFd0015.0l

9-Eq.7 3 2107.2.2.Sec.UBC psi 125u forfd002.04udf l

lduf 4d

ldufA

sbd

sbbs

d

dbs

2b

dbss

−==

===

=

=

ππ

π

db

ld

bduπ

ss fA


Embedment of Flexural ReinforcementUBC Sec. 2106.3.4 and MSJC Sec.2.1. 8.3

Rule #1: extend bars a distance of “d” or “12db” past the theoretical cutoff point

theoretical cutoff pointcapacity with bars “a”

Moment Diagram

moment capacitywith bars “a” and “b”

(#1) d or 12db

(#2)> ld

(#2)> ld

bars

“b”

bars

“a”

Example for shear wall:Rule #2: extend bars a distance of “ld” past the point of maximum stress


Combined Bending and Axial LoadsCode RequirementsUBC Sec. 2107.1.6.3use unity formula to check compressive stress: 0.1

Ff

Ff

b

b

a

a <+

UBC Sec. 2.14.2if h’/t >30 then analysis should consider effects of deflections on moments

UBC Sec. 2107.1.6.1P

fa = P/Ae

Note: unity formula is conservative -better approach is to use P-M interaction diagram.

MUBC Sec. 2107.2.15

As fs

jd

kd

)317.Eq( jkbd2Mf 2b −=

MSJC Sec. 2.3.3.2.2 fa + fb < 1/3 f’m provided that fa < FaIn lieu of approximate method, use an axial-force moment interaction diagram.


Axial Force-Moment Interaction DiagramsGeneral Assumptions

1. plane sections remain plane after bending• shear deformations neglected• strain distribution linear with depth

Strain Stressεm

εs

fm

Ts=Asfs

Cs

P

M

2. neglect all masonry in tension3. neglect steel in compression unless tied4. stress-strain relation for masonry is linear in compression5. stress-strain relation for steel is linear 6. perfect bond between reinforcement and grout

• strain in grout is equal to strain in adjacent reinforcement

7. grout properties same as masonry unit properties


Axial Force-Moment Interaction Diagram

Range “a”:large P, small M, e=M/P < t/6

Cm

fm2fm1

Out-of-Plane Bending of Reinforced Wall

Mb

Pa

unit

wid

th =

b

d = t/2

em

Pa = 0.5(fm1 + fm2)A

Ma= 0.5(fm1 - fm2)S where S = bt2/6


Axial Force-Moment Interaction DiagramOut-of-Plane Bending of Reinforced Wall

fm1

unit

wid

th =

b

Range “b”medium P, medium M, e > t/6, As in compression

d = t/2

tα

0.5 < α < 1.0 for section with reinforcement at center

em

Cm

3t

2tem

α−=

Pb

tb2

fCP 1mmb α==

Mb

mmb eCM =



unit

wid

th =

b

d = t/2Range “c” small P, large M, e > t/6, As in tension

3t

2tem

α−=

fm1em

CmTs

tα

α < 0.5 for section with reinforcement at center

)2td(TeCM smmc −+=

2td for f5.0f

ttd

nf

1m1ms =⎥⎦

⎤⎢⎣⎡ −

=⎥⎦⎤

⎢⎣⎡ −

=α

αα

α

Mc

Pc

tb2

fC 1mm α= sss fAT =smc TCP −=


fs < Fsfm1 = Fb

fs = Fsfm1 < Fb

fs < Fs?


e=0; M=0fm1= fm2=Fa

P=Fa A

Start

M = 0?

Stop

Determine P and M per Range “b”

fm2 = 0?

Reduce fm2 from 2Fa-Fb by increment

Determine P & M per Range “a”

no yes is As in tension?

no yes

Reduce α from 1.0 by

increment

Range “b”Range “a”fm1= Fb= f’m/3

Reduce α from 0.5 by

increment

Range “c”

Determine P and M per Range “c”

no

no yes

tens

ion

cont

rolli

ng

com

pres

sion

con

trol

ling

yes



Axi

al F

orce Fb

Range “b”

fm2 = 2Fa - Fbfm1 = Fa

fm1 = Fb

Range “a”Fb

Fb

Fa

limit by unity form

ula

e1

fs= Fs Fs/n

fm

Moment tens

ion

cont

rols Range “c”balanced point

Fs/n

Fb

com

pres

sion

cont

rolsfs/n

Fb


Example: Interaction DiagramDetermine an axial force-moment interaction diagram for a fully grouted 8” block wall reinforced with #4 @ 16”. Prism compressive strength has been determined by test to be equal to 2500 psi. Reinforcement is Grade 60. Height of wall is 11.5 feet.

60Grade for ksi 24 Fs =

factor reduction withoutpsi625 f´0.25 F ma ==

psi 833 0.33f´ F mb ==

ksi 29,000 E UBC per ksi1875 f´750 E smm ===

15.5 /EE n ms ==32

g2

g in1166/63.7x"12S;in6.91"12x"63.7A : wallof foot per ====2

s in150.016/12x20.0ft/A ==

909.0j272.0k0509.0n0033.0)"81.3x16/(in20.0 2 ===== ρρ


Example: Interaction Diagram

Case fm1(psi)

α Cm(kips)

2 833 417 - 57.2 - - 57.2 24.13 833 0 - 38.1 1.27 - 38.1 48.4

5 833 - 0.50 19.1 2.54 - 19.1 48.5

fm2(psi)

em(in.)

Ts(kips)

P=Cm- Ts(kips)

M=Cm em(kip-in)

7 833 - 0.25 9.5 3.18 2.0 7.5 30.28 833 - 0.167 6.4 3.39 3.9 2.5 21.5

9 for P = 0: Mm= 0.5Fbjkbd2 = 0.5(833 psi)(0.909)(0.272)(12)(3.81)2 = 17.9

11 664* - 0.150 4.6 3.43 3.6 1.0 15.7

Range

4 833 - 0.75 28.6 1.91 - 28.6 54.5b

12 check for P = 0: Ms = AsFsjd = (0.15 in2)(24 ksi)(0.909)(3.81”) = 12.5

1 625 625 - 57.2 0 - 57.2 0a

Com

pres

sion

Con

trol

s

6 833 - 0.33 12.6 2.97 0.9 11.7 37.4

10 833 bal. - 0.175 6.7 3.37 3.6 3.1 22.5c

Ten

sion

C

ontr

ols

*masonry stress inferred from Fs and α: ⎟⎠⎞

⎜⎝⎛

−=

αα

5.0nFf s

1m


Example: Interaction Diagram

Axial Forcekips

50

40

30

20

10

10 20 30 40 50

12

Moment, kip-in

10

10 8333.6 k = AsFs

0.175t11

11 664 3.6 k = AsFs

0.15t

1 1 625

4172 833

2

03 833

3

4 833.75t

4

5 833

0.50t

5

6 833 0.9k0.33t

6

7 833 2.0 k

0.25t

7

8

8 8333.9 k > AsFs

0.167t

9


Flexural Capacity with Axial CompressionShort Cut Method

m

ssm

ms

EE= n where )

nf(

k-1k= f ;

kdf

kd-d/nf

=stress compatibility: [1]

d

Ts

Cm

jd

MP

Out-of-Plane Bending, Reinforcement at Center

kd

fs/n

fm

d

[2]bkdf0.5 = C mm

[3] bdf= fA= T ssss ρ


equilibrium:sm T -C= P [4]

Flexural Capacity with Axial CompressionShort Cut Method

[5]sm bdf -bkdf0.5 = P ρ

[6]ss fbd -bkd

k-1k)

nf(0.5 = P ρ⎟

⎠⎞

⎜⎝⎛

[7]ρn1

k1k5.0

bdfP 2

s−⎟

⎠⎞

⎜⎝⎛

−=

[8]s

ss bdFP setFf ,controlstensionif == α

[9]ρn1

k)(1k0.5a

2−⎟

⎠⎞

⎜⎝⎛

−=

[10]k1

kn21 2

−=+αρ

[11]0)(n2k)(n2k 2 =+−++ αραρ

[12]3/k1jwherebkjdf5.0jdCM 2mm −===


Strength Design of Reinforced MasonryUltimate Flexural Strength

As

t

b

stresses

Cm

Mn

Ts = Asfy

strains

cn.a.

d

ys ε>ε

muε

d

Note: rectangular stress block can representcompressive stress distribution if k2/k1 = 0.5

f’mfm

mεmuε

c

k3f’m

k2c

Cm ck2c

Cm = k1k3f’mbcklc

k3f’m

=


P0+P1

stress

k2c

k3f’m

k1c

summing moments about centroid:P1a = (Po + P1)g

= (Po + P1)(c/2 - k2c)

ca

PPP 0.5 - k

1o

12 +

=

Strength Design of Reinforced MasonryMeasuring k1k3 and k2

total compressive force:

bc'fPPkk

m

1o31

+=

Po + P1 = k3f’m k1cb

∆

a

Po in displacement controlP1 in force control

Po P1

increase P1 so that ∆ = 0

strainc

g


Sample experimentally determined constants k1k3, and k2

0

0.2

0.4

0.6

0.8

1

0 0.001 0.002 0.003 0.004 0.005 0.006Extreme Fiber Strain (in/in)

K1K

3 &

K2 K1K3

K2

Strength Design of Reinforced MasonryMeasured k1k3 and k2 values


Strength Design of Reinforced MasonryUltimate Flexural Strength

fs

fy

sε

)'ff

59.01(dfAMm

yysn ρ−=:thenandif 85.0k5.0

kk 3

1

2 ==

m31

y

yysm31

sm

'fkkdf

c

fbdfAbc'fkk0TC

ρ

ρ

=

==

=+

equilibrium

)'fkk

fk1(dfAM

)'fkk

dfkd(fAM

)ckd(fAM

m31

y2ysn

m31

y2ysn

2ysn

ρ

ρ

−=

−=

−=

summing moments about Cm


Balanced condition with single layer of reinforcement

ymu

mu

y

m31b f

'fkkεε

ερ+

=

yy

m1

syy

m1b

ss

yymu3

f000,87000,87

f'fk85.0

E/f003.0003.0

f'f)85.0(k

:psi000,000,29E Ef

003.085.0k if

+=

+=

====

ρ

εε

ys ε=ε

d

muε

strains

Strength Design of Reinforced Masonry

c

n.a.

stresses

k1c Cm

Ts= Asfy

Mn

dcordc

ymu

mu

ymu

mu

εεε

εεε

+==

+

strain compatibility

ybymu

mum31

sm

bdfdb'fkk

0TC

ρεε

ε=

+

=+

equilibrium


f’m

Grade 40 Grade 60

bρ tbρbρtbρ

1000 0.0124 0.0062 0.00360.0071

d2t steel of layerone forbt/Af000,87

000,87f

'fk85.0stb

yy

m1b ==

+= ρρ 85.0k1 =

Balanced condition with single layer of reinforcementStrength Design of Reinforced Masonry

2000 0.0247 0.0124 0.0143 0.00723000 0.0371 0.0186 0.0214 0.01074000 0.0495 0.0247 0.0285 0.01425000 0.0619 0.0309 0.0356 0.01786000 0.0742 0.0371 0.0428 0.0214


Balanced condition with multiple layers of reinforcementStrength Design of Reinforced Masonry

d592.0 c then ,003.0 if

00207.0 ksi29,000

ksi 60

dc ))(dd(

mu

y

ymu

musymu

4

imusi

==

==

+=+−<

ε

ε

εεεεεεε

60) (Grade

strain compatibility

sbal

sisbalm

sisim

siisbalsisi

ysissi

A for solve0fAbd'f428.0

0)TC(C

fATorC

fEf

=∑+−

=+∑+

=

<= εequilibrium

c

Asbal

b strains

ys4 εε =

muε

d4

d3

d2

d1 s1ε

s2ε

s3ε

0.85

c

stresses0.85f’m

Cs1Cm=0.85f’mb(0.85c)Cs2

Ts3

Ts4 = Asbal fy


Example: Flexural Strength of In-Plane WallMaximum steel is equal to one-half of that resulting in balanced conditions. f’m= 1500 psi Grade 60 reinforcement special inspection

Determine the maximum bar size that can be placed as shown.

Asbal ?

7.63”

5’-4

”

Ts3

Ts4 = Asbal fy

Pn = 0

Cs2

Cs1

Cm = 0.85f’mb(0.85c)

0.85f’m

60.0

”c

00207.0ys =ε=ε

2sε

3sε

1sε

n.a.

0.0034.

0”

44.0

”20

.0”


c = 0.003/0.00507 (60.0”) = 35.5”Cm = 0.85f’mb(0.85c) = -0.85(1500)(7.63”)(0.85 x 35.5) = -294 k

ysissi

1i

fEf

)003.0(cdc

≤=

−⎟⎠⎞

⎜⎝⎛ −

=

ε

ε

*bars larger than #9 are not recommended because of anchorage and detailing problems

without compression steel (neglect Cs1 andCs2 forces)Cm + Σ(Csi + Tsi) = -294 + Asbal (20.8+ 60.0) = 0 Asbal = 3.64 in2 Asmax = 1.82 in2

max. bar size is #ll (1.56 in2)*

Example: Flexural Strength of In-Plane WallDetermine the maximum bar size.

layer di εsi fsi

with compression steel (include Cs1 and Cs2 forces)Cm + Σ(Csi + Tsi) = -294 + Asbal (-60.0 - 38.0 + 20.8 + 60.0) = 0 Asbal = -17.1 in2

note: negative Asbal means that ΣC > ΣT , in such case no limit on tensile reinforcement

1 4.0” -0.00261 (C) -60.0

2 20.0” -0.00131 (C) -38.0

3 44.0” 0.00072 (T) 20.8

4 60.0” 0.00207 (T) 60.0


Determine flexural strength of wall.Example: Flexural Strength of In-Plane Wall

7.63”

5’-4

”f’m= 1500 psi Grade 60 reinforcement special inspection

#8 (typ)

44.0

”20

.0”4.0”

60.0

”

c

ys ε>ε

2sε

n.a.

0.0031sε

3sε

Ts4 = As fy= 0.79 in2 x 60 ksi = 47.4 k

0.85f’m

Ts3

Ts2

Cs1

Cm = 0.85f’mb(0.85c)


c27.8CfEf (-) )003.0(cdc

mysissii

i =<==−⎟⎠⎞

⎜⎝⎛ −

= εε strains ive compress

Example: Flexural Strength of In-Plane Wall

cd1 = 4.0”

f1 Csl1εd2 = 20.0”

f2 Ts2

d3 = 44.0”f3 Ts3

Cm )TC( +∑

-0.00240 -60 -47.4 0 0 0 0.00360 60.0 47.420.0 -165 -117.6

2ε 3ε

c85.0

inkip222,5)89.40.60)(4.47()89.40.44)(4.47()89.40.20)(4.47()89.400.4)(8.44(

)}2

d(fA{M isisin

−=−+−+−+−−=

−∑=

Determine flexural strength of wall.

15.0 -0.00220 -60 -47.4 0.00100 29.0 22.9 0.00580 60.0 47.4 -124 -54

11.0 -0.00191 -55 -43.7 0.00245 60.0 47.4 0.00900 60.0 47.4 -91 +7.5

close to zero, take c = 11.5”

11.5 -0.00196 -56 -44.8 0.00222 60.0 47.4 0.00848 60.0 47.4 -95 +2.3


Example: Flexural Strength of In-Plane Wall

7.63”

5’-4

”

#8 (typ)

Approximate flexural strength of wall.

00398.05263.779.02"0.52

2)4460(d

AandAlumpingand,TandC neglecting 4s3s2ssl

=×

×==

+= ρ

answerof%86 inkip467,4

)50.1

60x00398.0x59.01)(0.52)(60)(in79.0(2

)'ff

59.01(dfAM

2

m

yysn

−=

−=

−= ρ


UBC Sec. 2108.2.4Limitations of Method:

Slender Wall Design

(a) for out-of-plane bending of solid, reinforced walls lightly stressed under gravity loads

(c) ρg= As/bt < 0.5 ρbal

Sec. 2108.1.3: Load factors

)W7.1L7.1D4.1(75.0U)ELD(4.1U

L7.1D4.1U

++=++=

+=

W3.1D9.0UE4.1D9.0U

±=±=

Ref: NCMA TEK 14-11A Strength Design of Tall Concrete Masonry Walls

30 t

h' ,0.20f'A

PP0.04f' m

g

fwm <<

+< that providing used be can still method when:Note

psi 6000f' where4.4.2.8 210Sec.) 198(f' 04.0 A

PP mm

g

fw <−≤+

(b) limited to:

(d) special inspection must be provided during construction(e) t > 6”


wuh2/8

transverse load

Required Flexural Strength: UBC Sec. 2108.2.4.4

)208()PP(2

eP8hwM

PPP

uufuwuf

2u

u

uwufu

−+++=

+=

∆

Slender Wall Design

e

h

Puf

t

wu

Puw

(Puw + Puf) u∆

h/2

h/2

P∆eccentric

load

Pufe/2

Pufe


Slender Wall Design

Design strength: Sec. 2108.2.4.422)-(8 MM nu φ<

Design Considerations

Assumptions for ultimate flexural strength (Sec. 2108.2.1.2)

1. equilibrium

Strength reduction factor:flexure φ = 0.8 Sec. 2108.1.4.2.1

2. strain compatibility3. εmu = 0.0034. fs = Esεs < fy5. neglect masonry strength in tension6. rectangular stress block, k1 = 0.85, k3 = 0.85


Slender Wall DesignEquivalent area of reinforcement, Aefor single wythe construction reinforced at center:

Eq. (8-24)y

ysuse f

)fAP(A

+=

flexural strength25)-(8 Eq.where 23)-(8 Eq.

b'f85.0)fAP(

a )2ad(fAM

m

ysuysen

+=−=

b

As

d

Pu = Cm - Asfy

Cm = Pu + Asfy= Asefy

Ts = Asfy

Ts = Asfy

Pu

Pu

c

d

a = 0.85c

0.85f’m

Cm =0.85f’mb(0.85c)

Cm


Slender Wall DesignLateral Deflections

M

∆

EIMh

485

EIh

8wh

485

EIwh

3845 2224

===∆

My

Mcr

Ms∆s

∆y

∆cr

Modulus of Rupture, fr Eqs. 8-31, 32, 33

allowednot psi125'f5.2

psi 125'f5.2 psi235f'4.0

m

mm

<

<<

fully grouted partially grouted

hollow unit

2-wythe brick

gm

2s

s IEhM

485

=∆for Ms < Mcr (8-28)

3)kd(b)kdd(nAIwhere

IEh)MM(

485

IEhM

485

32

secr

crm

2crs

gm

2cr

s

+−=

−+=∆for Ms > Mcr

(note “kd” may be replaced by “c” for simplicity)

Mcr = fr S

(8-29)


Slender Wall Design

uufuwuf2

uu )PP(

2eP

8hwM ∆+++=

Strength Criteria

Design Considerations

Serviceability Criteria

)278(h007.0 s −≤∆

crm

2cru

gm

2cr

u IEh)MM(

485

IEhM

485 −

+=∆


Example: Slender Wall Design

ftin 075.0)

"32"12( in20.0A

"31.72

"63.7"50.3e

22

s ==

=+=

Determine the maximum wind load, w, per UBC and MSJC

8” CMU, partially groutedf’m = 2000 psi, Grade 60

500 lbs/ft dead200 lbs/ft live

#4 @

32”20

’-0”

3’-0

”

3.5”

Pw

ok 0072.0)0143.0(21 ρ

21 000164.0

81.312075.0

bdA ρ bal

s ==<=×

==


U = 0.75 (1.4D + 1.7L + 1.7W)

)2ad(fAM)PP(

2eP

8hwM ysenuufuw

uf2

uu −=<+++= φφ∆

Example: Slender Wall DesignFlexural Strength per UBC

Mu=φMn = φAsefy(d - a/2)=0.8(0.103in2)(60ksi)(3.81in - 0.302in/2)=18.1 kip-in

.lbs1654P.lbs874)'13xpsf64x4.1(75.0P

.lbs 780)200x7.1500x4.1(75.0PPPP

u

uw

ufuwufu

=

==

=+=+=

2

y

ysuse in103.0

ksi 60)ksi60x075.0kips 65.1(

f)fAP(

A =+

=+

=

section rrectangula as treat shell,face withinaxis neutral"2.1"355.085.0

"302.0c

"302.0)"12xksi0.2x85.0(

)60(103.0b'f85.0

)fAP(a

m

ysu

<==

==+

=


crm

2cru

gm

2cr

u IEh)MM(

485

IEhM

485 −

+=∆

3.191500

000,29EEnksi1500'f750E

m

smm =====

Example: Slender Wall Design

( ) 43

223

2secr in9.23

3)"355.0(12)355.081.3(in103.03.19

3bc)cd(nAI =+−=+−=

to avoid iteration, assume Mmax = Mu

"955.0"837.0"118.0)9.23)(1500(

)12x20)(1.131.18(485

)444)(1500()12x20)(1.13(

485 22

u =+=−

+=∆

Flexural Strength per UBC

psf 17.87.1x75.0

ww us ==

wu=22.7 psf

"12/.ink1.18)12955.0)(654.1(

12)2/"31.7(780.0

8)'20(wM2

uu −=++=

for simplicity, use gross sectioneven though partially grouted

.ink1.13)6

63.7x"12(ksi112.0M psi112)'f(5.2f SfM2

cr5.0

mrgrcr −=====


crm

2crs

gm

2cr

s

s

IEh)MM(

485

IEhM

485

h007.0

−+=

<

∆

∆

Example: Slender Wall DesignCheck Service Load Deflections per UBC

.)in.lb(1532239,1315322

)"31.7(70012x8

)20)(psf8.17(

)PP(2/eP8hwM

ss

2

sowo

2s

s

−+=++⎥⎥⎦

⎤

⎢⎢⎣

⎡=

+++=

∆∆

∆

( )ok"68.1)12x'20007.0h007.0"19.0)9.23)(500,1(

)12x20)(1.13532.1239.13(485"118.0

s

2s

s

==<=

−++=

∆

∆∆


ss

s

2s

soswsos

2s

s

1532.inlb2562w600

)lbs1532()"66.3(lbs70012x8

)'20(w

)PP(2

eP8hwM

∆

∆

∆

+−+=

++=

+++=

Example: Slender Wall DesignMaximum Wind Load per MSJC

Determine Icr considering axial compression

0)(n2k)(n2k 2 =+−++ αραρ

1.25

”

#4 @ 32”

d = 3.82”

00104.0ksi 3282312

k 532.1 00164.0"82.3"12

in 075.0 3.192

=××

===×

=="."bdF

Pns

αρ

3.47"jd 908.03k1 j

ok thickness, shellface 1.05"kd 0.275 k 00104.0k0104.0k 2

==−=

<===−+

4223

2s

3

cr in 7.15)"05.1"82.3)(in 075.0(3.193

)"05.1("12)kdd(nA3

)kd(bI =−+=−+=


psf) in is w( 55.247.18win-kip 64.13532.1.in-kip56.2w600.0

in-kip 64.13"47.3]532.1ksi) 32075.0[(jd)PP(FAM

sss

ss

oswssss

∆∆

−=

=++

=+×=++=

Note: same wind load as by UBC slender wall design procedure. Should also check compressive stress with an axial force-moment interaction diagram

Example: Slender Wall DesignMaximum Wind Load per MSJC

21.4w 251.034.3 390.0652.0w 153.0"118.0

)in ksi)(15.7 1500()1220)(1.13532.1.in-kip56.2w600.0(

485"118.0

IEh)MM(

485

IEhM

485

ss

sss

4

2ss

s

crm

2crs

gm

2cr

s

−=

−+++=

×−+++=

−+=

∆∆∆

∆∆

∆

psf 17.8ws =)21.4w 251.0( 55.247.18w sss −−= ∆


Strength Design of RM Shear Walls

UBC Sec. 2108.1.1: Strength procedure may be used as an alternative to Sec. 2107 for design of reinforced hollow-unit masonry walls.

UBC Requirements

state limitflexure for0.80 statelimitshearfor60.0

==

φφB. shear

0P85.0P25.0PA'f1.0P65.0

n

bnemn

==

>>=

φφφφφ

fororfor

2. Design strengthA. axial load and flexure (see next slide)

UBC Sec. 2108.1.2: Special inspection must be provided during construction. Prismsshould be tested or unit strength method should be used.

1. Required strengthA. earthquake loading: U = 1.4 (D+L+E) (12-1)

U = 0.9D + - 1.4E (12-2)

UBC Sec.2108.1.3: Shear wall design procedure

B. gravity loading: U = 1.4D + 1.7E (12-3)C. wind loading: U = 0.75(1.4D + 1.7 L + 1.7W) (12-4)

U = 0.9D + - 1.3W (12-5)D. earth pressure: U = 1.4D + 1.7L + 1.7H (12-6)


Strength Design of RM Shear WallsDefinition of Balanced Axial Load, Pb

d)

Ef

e(

e85.0awhere ba'f85.0P:wallsgroutedsolidlyfor

CPsoTCassume

s

ymu

mubbmb

mbsisi

+==

== ΣΣ

b

Lw

Cm = 0.85f’mbab

d

0.85f’m

Cs1

Cs2

Pb

Ts3

Ts4 = Asbalfy

c 85.0ab =

muε

c

n.a.

0.00207== ys εε



3. Design assumptions (same as for Slender Wall Design Procedure, UBC Sec. 2108.2.1.2)

1. equilibrium

UBC Requirements

2. strain compatibility3. εmu = 0.0034. fs = Esεs < fy5. neglect masonry tensile strength6. use rectangular stress block, k1 = 0.85, k3 = 0.857. 1500 psi < f’m < 4000 psi


2. for flexural failure modeMn > = 1.8 Mcr for fully grouted wallMn > = 3.0 Mcr for partially grouted wall


3. anchor all continuous reinforcement

M

Mcr

Mn > Mcrductile

Mn < Mcrnonductile

∆

4. Reinforcement per UBC Sec. 2108.2.5.2

0"4'spacing0.0007and

0.002

hv

hv

−≤

≥

≥+

ρρρρ

1. minimum reinforcement

UBC Requirements

5. Axial strength (no flexure)Po = 0.85 f’m(Ac-As) + fyAs Pu < = φ (0.80)Po

4. As vertical > 1/2 As horizontal

5. maximum spacing of horizontal reinforcement within plastic hinge region = 3t or 24”



2. for walls limited by shear strength:

37)-(8where 'fACV

VVV

mmvdm

smn

=

+=

K)-21 (Table 0.1VdM for 2.1

25.0VdM for 4.2C and d

≥=

≤=

6. Shear Strength UBC Sec. 2108.2.5.5

) 0.1VdM for A250'fA0.4

25.0VdM for A380'fA0.6V

eme

emen

J-21 (Table≥≤=

≤≤=

1. maximum nominal shear:

UBC Requirements

Amv = net area of masonry wall section bounded by wall thickness and length of section in direction of shear

Amv

Lw

Vu

t



yhorizontalsw

yhorizontals

ws

yplanevertical

horizontalsmvs

planeverticalhorizontalsn

ynmvs

f)A(h

Lf)ht

A(tLV

f)AA

(AV

A/A

fAV

==

=

=

−=

ρ

ρ

where

38)(8

UBC Requirements

As fyh

Lw Lw

As fy

Lw

As fy

h

t6. Shear Strength UBC Sec. 2108.2.5.5.2 (continued)

Avertical plane

As horizontal



6. Shear Strength UBC Sec. 2108.2.5.5: continued

3. for walls limited by flexural strength:

UBC Requirements

39)(8 −== ynmsn fAVV ρ

within hinge region, distance of Lw above base:

(Vu determined at Lw/2 from base)

smn VVV +=

above hinge region:


1. Provide boundary members when the extreme fiber strain exceeds 0.0015.


Boundary Members: Sec.2108.2.5.6

UBC Requirements

Section at Base of Wall

> 3twall

centroid#3 @ 8” min.

0.0015

t

εmu > 0.0015

2. The minimum length of boundary members shall be 3t.3. Boundary members shall be confined with a minimum of #3 bars @ 8” spacing, or equivalent confinement to develop an ultimate compressive masonry strain equal to 0.006.


Example: Strength DesignDetermine the maximum wind force, H, and design horizontal reinforcement to develop the wall flexural strength.

Consider: zero vertical load and Pdead = 40 kips and Plive = 30 kips

7.63

”

4 - #8’s

5’-4”H

10’-

8”

8” concrete block, fully groutedGrade 60 reinforcement , f’m= 1500 psi

in.kip439,4)222,5(85.0MφMin. kip222,5M

nu

n

−===

−=from previous example:

ok 8.1

Min-kip 807)6

64ksi)(7.63 155.0(M

psi 15515004.0f

psi 235f'4.0f SfM

30-8Eq.permomentcrackingcheck

n2

cr

r

mrgrcr

<=×=

==

<==

kips 7.263.1

HH kips 7.34"128

MH :existsstate limitflexure if uuu ====


U = 1.3WExample: Strength DesignShear Reinforcement (neglecting vertical force)

ksi)60(A)5.0(fA )h

L( fA V V horizsyhorizsw

ynmvsn ==== ρ

shear design within Lw (5’-4”) of base:

shear design for top 5’-4” of wall:

ksi))(60 (0.5)(A kips 22.7)f)(Ah

L( 150064.00)x 1.2(7.63

fA f'AC V V V

horizsyhorizsw

ynmmmvdsmn

+=+=

+=+= ρ

ksi)](60 )(A 0.80[(0.5) V kips 34.7H V horiznuu ==== φ

222providedhorizs

max2

horizs

in1.45 in1.60 )in8(0.20 A

courses 8 bottom for 8" @ s#4'use 24" s in1.45 A

>==

==

222providedhoriz s

max2

horizs

horizsn

uu

in 0.69 in0.80 )in4(0.20 A

courses 8 top for 16" @ s#4'use 48" s in 0.69 A

ksi)])(60 (0.5)(A 0.80[22.7 V kips 34.7 H V

>==

==

+==

==

φ


Confinement requirements for vertical reinforcement per Sec. 2108.2.5.6

Example:Shear Wall Strength DesignConfinement Reinforcement (neglecting vertical force)

7.63

”

5’-4”

#8

Mu = 4,439 kip-in.

11.5”

5.75”

ε = 0.0015

3t > 5.7”

ε = 0.003

#3 @ 8”bottom 8 courses

Strain Diagram per Previous Example


Case 1: Pu = 0.75(1.4 x 40 + 1.7 x 30) = 80.3 kips perhaps maximum flexuralcapacity and critical for shear design

Example: Strength DesignFlexural Strength considering Vertical Loads

kips306)0.602.200.36-0.60 ( in79.0294 P

kips5.73P25.0 kips 294")2.30")(63.7 ksi)(5.1(85.0 P

"2.30"6000207.0003.0

003.085.0d

Ef85.0a

ba.85f'0 P

2b

bb

s

ymu

mub

bmb

−=++−+−=

===

=+

=+

=

=

:entreinforcem gconsiderin

ε

ε

capacity reduction factors

Case 2: Pu = 0.9(40) = 36.0 kips perhaps minimum flexural capacity and lowest Hu

0.65 kips73.5 P0.25 kips 80.3 P bu ==>= φCase 1:

Case 2: 0.75 )0.20 73.536.0( 0.65 kips73.5 P0.25 kips 36.0 P bu =+==<= φ


c 8.27 C f E f )003.0( c

dcmysissi

ii =<=−⎟

⎠⎞

⎜⎝⎛ −

= εε

)"0.28(T )"0.12(T )"0.12(T -)(28.0"C )2

0.85c -(32.0"C M s4s3s2slmcl +++=∑


20.0 -0.00240 -60 -47.4 0 0 0 0.00360 60.0 47.4 -165 -118 6,983

d1 = 4.0” d2 = 20.0” d3 = 44.0”ε1 f1 Cs1 ε2 f2 Ts2 ε3 f3 Cs3 Cm Pn Mn

(kips) (kip-in)(kips)(kips)(kips)(kips)(ksi) (ksi) (ksi)

15.0 -0.00220 -60 -47.4 0.00100 29.0 22.9 0.00580 60.0 47.4 -124 -54 6,126

16.8 -0.00229 -60 -47.4 0.00057 16.6 13.1 0.00185 53.8 42.5 -139 -83 6,463

13.1 -0.00208 -60 -47.4 0.00104 30.0 23.7 0.00708 60.0 47.4 -108 -37 5,794



Axi

al C

ompr

essi

ve F

orce

, kip

s 140

120

100

80

60

40

20

5500 6000 6500 7000Moment, Mn kip-in.

Case 2

6450

kip

-in.

80.3 kips

5820

kip

-in.

Case 1

36.0 kips

Hu = 34.1 kips (Case 2) ~ 34.7 kips (w/o vertical force). Use same shear design as for first part of problem.

Mu = 4,365 kip-in. (Case 2) ~ 4,439 kip-in. (w/o vertical force). Use same boundary members as for first part of problem.

Case 2: in.-kip4,365 0.75(5820) M M nu === φ

kips 26.2 1.3H H u ==

kips 34.1 12)x (10.67

4,365 Hu ==

Case 1: in.-kip 4,192 0.65(6450) M M nu === φ

governs kips 25.7 1.7)x 0.75 (

H H u ==

kips 32.7 12)x (10.67

4,192 Hu ==

Lecture 6 7 Rm Shear Walls

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Transcript of Lecture 6 7 Rm Shear Walls