lecture 6 7 rm shear walls
DESCRIPTION
Dan Abrams + Magenes Course on MasonryTRANSCRIPT
Masonry Structures, slide 1
Reinforced MasonryWorking Stress Design of flexural members
bd
As
Ts = Asfs
Cm = fmb kd/2
M
b
d
t
Asgrout
unit
strain
Ref: NCMA TEK 14-2 Reinforced Concrete Masonry BIA Tech. Note 17 Reinforced Brick Masonry - Part I BIA Tech. Note 17A Reinforced Brick Masonry - Materials and Construction
n.a.
kd
s
m fm
fs/n
stress
Masonry Structures, slide 2
misi :fiber particularany at #5, Assumption from
Assumptions 1. plane sections remain plane after bending (shear deformations are neglected, strain distribution is linear with depth) 2. neglect all masonry in tension 3. stress-strain relation for masonry is linear in compression 4. stress-strain relation for steel is linear 5. perfect bond between reinforcement and grout (strain in grout is equal to strain in adjacent reinforcement) 6. masonry units and grout have same properties
Reinforced MasonryWorking Stress Design of flexural members
k
k1nff
kdd
nf
kd
f :ondistributi stress ofgeometry from ms
sm
mimim
ssi
m
mi
s
si nffE
Ef
E
f
E
f :#4 and #3 sAssumption from
Masonry Structures, slide 3
k
k1n2/k
k
k1fnbdfbdfA
2
bkdf
:T C m,equilibriu from
msssm
0n2nk2k 2
Reinforced MasonryWorking Stress Design of flexural members
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 where jd=d-kd/3 or j=1-k/3
0C aboutM m
Masonry Structures, slide 4
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
Reinforced MasonryWorking Stress Design of flexural members
Masonry Structures, slide 5
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
Masonry Structures, slide 6
WSD: Balanced Condition
Reinforced Masonry
bsb /F F+ n
n= k
dnF
+ F
dk
Fs
b
b
b
from geometry:
fs/n = Fs /n
d
kbd
fm = Fb
bsbs
b F/F2
1
F/Fn
n
)(22
2
d
b
ss
b
s
bbb
sbbb
FF
n
n
F
F
F
kF
FdbρkbF
from equilibrium: C=T
nF
+ F
F= k
sb
bb
Masonry Structures, slide 7
Example: Balanced Condition
Determine 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 =E
entreinforcem60 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
19.3= 1500
29,000=
E
E= n
m
s
Masonry Structures, slide 8
Design Strategy for RM Flexural DesignProcedure for sizing section and reinforcement for given moment.
Calculate b knowing f’m and fy
determine Fb from f’m
determine Fs from fy
determine Emfrom f’m determine n = Es/Em
Size section for some b determine k and j bd2 = M/jFs
select b and d using common units
Size reinforcement As = M/Fsjd select number and size of rebars
Check design Ms = AsFsjd > M fb = M/0.5jkbd2 < Fb
Note: Section must also be sized for shear.
Masonry Structures, slide 9
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.
0.4% isestimate good A
govern. willsteel so b than lowerslightly be to Estimate .2
892.03/k1j323.0k0154.0k154.02k
154.0)3.19)(004.0(2n20n2nk2 2k
:k forSolve 3.
3in4321)ksi24)892.0)(004.0/()inkip370(2bd
sFj/M2bd
:2bd for Solve.4
Masonry Structures, slide 10
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 bottom unit, d = 27.8”
4 - 8” CMU’s
d=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
Masonry Structures, slide 11
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:
bd
A0n2nk2k s2
Masonry Structures, slide 12
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
Masonry Structures, slide 13
Shear Design of Reinforced Masonry
after cracking: Vext = Vint = Vs = nAvfs where n is the number of transverse
bars across the diagonal crack. Assuming a 45 degree slope, n=d/s
Vs = (d/s)Avfs
s
s
s
sv
dF
V
df
V
s
A UBC Sec. 2107.2.17 (Eq. 7.38)
MSJC Sec. 2.3.5.3 (Eq. 2-26)
Masonry Structures, slide 14
Shear Design of Reinforced MasomryFlexural shear stress
fvb dx = dT = dM/jd
fv = (dM/dx)/bjd
UBC Sec. 2107.2.17 (Eq. 7-38)bjd
Vfv
bd
Vfv 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
Masonry Structures, slide 15
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
Masonry Structures, slide 16
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)Vd
M4580('f)
Vd
M4(
3
1F 1
Vd
M 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
)psiVd
M45120(f')
Vd
M4(
2
1 F1
Vd
Mfor mv UBC Eq. 7-21; MSJC Eq. 2-24
UBC Eq. 7-20; MSJC Eq. 2-22psi35'f0.1F 1Vd
M for mv
UBC Eq. 7-22; MSJC Eq. 2-25hpsi75'f5.1F 1Vd
M for mv
Masonry Structures, slide 17
Shear Design of Reinforced MasonryMoment-to-Shear Ratios
d
d
h
Vd
Vh
Vd
M
V
M
For a single-story cantilevered shear walls
h
V
M
M
For piers between openings
hd
d2
h
Vd
2/Vh
Vd
M
Masonry Structures, slide 18
MSJC Sec. 2.3.5.3.1 smax = d/2 or 48”
d/2
Vdesign
Shear Design of Reinforced MasonryAdditional MSJC Requirements
Vdesign
MSJC Sec. 2.3.3.4.2 minimum reinforcement perpendicular to shear reinforcement = Av/3 smax = 8 ft
MSJC Sec. 2.3.5.5 design for shear force at distance “d/2” out from support
Masonry Structures, slide 19
Provide Reinforcement to Take 100% of Shear
s
sv
dF
V
s
A
Determine Shear Stress
bd
Vor
bjd
Vfv
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
Masonry Structures, slide 20
Determine the maximum lateral force, Hwind per UBC and MSJC
Example: 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
Masonry Structures, slide 21
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
H96120F
S
Mf - ta
kips 2.10.lbs 174,10H 08139
H96120FS/Mf - ta
per MSJC:flexure
kips3.44)80x63.7)(psi 109(3
2btF
3
2V
psi 109psi 2.8233.1F
psi 2.82f'1.5psi 114)120(45.060F
vmax
v
mv
shear
Masonry Structures, slide 22
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'f0.1F1Vd
Mfor 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
Masonry Structures, slide 23
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> Vd
M 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
bjd
V fv UBC
okay psi100 psi 27.7(7.63)(72)
1000)x kips(14.4
bd
V 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
Masonry Structures, slide 24
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*
Case
A
B
C
D
Consideration UBC MSJC
*flexure governs
Case
A
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*
Masonry Structures, slide 25
Flexural Bond Stress
M = 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
U
jd
Vu
Masonry Structures, slide 26
Development Length
82.Eq 2.8.1.2.Sec MSJCpsi 167uforFd0015.0l
9-Eq.7 3 2107.2.2.Sec.UBC psi 125u forfd002.04u
df l
lduf 4
d
ldufA
sbd
sbbs
d
dbs
2b
dbss
db
ld
bdu
ss fA
Masonry Structures, slide 27
Embedment of Flexural Reinforcement UBC 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
bar
s “b
”
bar
s “a
”
Example for shear wall:Rule #2: extend bars a distance of “ld” past the point of maximum stress
Masonry Structures, slide 28
Combined Bending and Axial LoadsCode RequirementsUBC Sec. 2107.1.6.3use unity formula to check compressive stress: 0.1
F
f
F
f
b
b
a
a
UBC Sec. 2.14.2 if h’/t >30 then analysis should consider effects of deflections on moments
UBC Sec. 2107.1.6.1
P
fa = P/Ae
Note: unity formula is conservative - better approach is to use P-M interaction diagram.
M
UBC Sec. 2107.2.15
As fs
jd
kd
)317.Eq( jkbd2
Mf
2b
MSJC Sec. 2.3.3.2.2 fa + fb < 1/3 f’m provided that fa < Fa
In lieu of approximate method, use an axial-force moment interaction diagram.
Masonry Structures, slide 29
Axial Force-Moment Interaction DiagramsGeneral Assumptions
1. plane sections remain plane after bending• shear deformations neglected• strain distribution linear with depth
Strain Stress
m
s
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
reinforcement7. grout properties same as masonry unit properties
Masonry Structures, slide 30
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
un
it w
idth
= b
d = t/2
em
Pa = 0.5(fm1 + fm2)A
Ma= 0.5(fm1 - fm2)S where S = bt2/6
Masonry Structures, slide 31
Axial Force-Moment Interaction DiagramOut-of-Plane Bending of Reinforced Wall
fm1
un
it w
idth
= b
Range “b”medium P, medium M, e > t/6, As in compression
d = t/2
t
0.5 < < 1.0 for section with reinforcement at center
em
Cm
3
t
2
tem
Pb
tb2
fCP 1m
mb
Mb
mmb eCM
Masonry Structures, slide 32
Axial Force-Moment Interaction DiagramOut-of-Plane Bending of Reinforced Wall
un
it w
idth
= b
d = t/2Range “c” small P, large M, e > t/6, As in tension
3
t
2
tem
fm1em
Cm
Ts
t
< 0.5 for section with reinforcement at center
)2
td(TeCM smmc
2
td for f
5.0f
t
td
n
f1m1m
s
Mc
Pc
tb2
fC 1m
m sss fAT smc TCP
Masonry Structures, slide 33
fs < Fs
fm1 = Fb
fs = Fs
fm1 < Fb
fs < Fs?
Axial Force-Moment Interaction DiagramOut-of-Plane Bending of Reinforced Wall
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 from 1.0 by increment
Range “b”Range “a”
fm1= Fb= f’m/3
Reduce from 0.5 by increment
Range “c”
Determine P and M per Range “c”
no
no yes
ten
sion
con
trol
lin
g
com
pre
ssio
n c
ontr
olli
ng
yes
Masonry Structures, slide 34
Axial Force-Moment Interaction DiagramOut-of-Plane Bending of Reinforced Wall
Axi
al F
orce Fb
Range “b”
fm2 = 2Fa - Fbfm1 = Fa
fm1 = Fb
Range “a”Fb
Fb
Fa
limit by unity form
ula
e1
f s = F s Fs/n
fm
Moment ten
sion
con
trol
s Range “c”balanced point
Fs/n
Fb
com
pre
ssio
nco
ntr
olsfs/n
Fb
Masonry Structures, slide 35
Example: Interaction Diagram
Determine 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
Masonry Structures, slide 36
Example: Interaction Diagram
Case fm1
(psi) 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
pre
ssio
n C
ontr
ols
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
5.0n
Ff s
1m
Masonry Structures, slide 37
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 6643.6 k = AsFs
0.15t
11 625
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
Masonry Structures, slide 38
Flexural Capacity with Axial CompressionShort Cut Method
m
ssm
ms
E
E= n where )
n
f(
k-1
k= f ;
kd
f
kd-d
/nf stress compatibility:
[1]
d
Ts
Cm
jd
M
P
Out-of-Plane Bending, Reinforcement at Center
kd
fs/n
fm
d
[2]bkdf0.5 = C m m
[3] bdf= fA= T ssss
Masonry Structures, slide 39
equilibrium:sm T -C= P [4]
Flexural Capacity with Axial CompressionShort Cut Method
[5]sm bdf -bkdf0.5 = P
[6]ss fbd -bkd
k-1
k)
n
f(0.5 = P
[7]ρn
1
k1
k5.0
bdf
P 2
s
[8]s
ss bdF
P setFf ,controlstensionif
[9]ρn
1
k)(1
k0.5a
2
[10]k1
k
n2
1 2
[11]0)(n2k)(n2k 2
[12]3/k1jwherebkjdf5.0jdCM 2mm
Masonry Structures, slide 40
Strength Design of Reinforced MasonryUltimate Flexural Strength
As
t
b
stresses
Cm
Mn
Ts = Asfy
strains
cn.a.
d
ys
mu
d
Note: rectangular stress block can representcompressive stress distribution if k2/k1 = 0.5
f’mfm
mmu
c
k3f’m
k2c
Cm c
k2c
Cm = k1k3f’mbcklc
k3f’m
=
Masonry Structures, slide 41
P0+P1
stress
k2c
k3f’m
k1c
summing moments about centroid: P1a = (Po + P1)g = (Po + P1)(c/2 - k2c)
c
a
PP
P 0.5 - k
1o
12
Strength Design of Reinforced MasonryMeasuring k1k3 and k2
total compressive force:
bc'f
PPkk
m
1o31
Po + P1 = k3f’m k1cb
a
Po in displacement control
P1 in force control
Po P1
increase P1
so that = 0
strainc
g
Masonry Structures, slide 42
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 0.006
Extreme Fiber Strain (in/in)
K1
K3
& K
2
K1K3
K2
Strength Design of Reinforced MasonryMeasured k1k3 and k2 values
Masonry Structures, slide 43
Strength Design of Reinforced MasonryUltimate Flexural Strength
fs
fy
s
)'f
f59.01(dfAM
m
yysn :thenandif 85.0k5.0
k
k 3
1
2
m31
y
yysm31
sm
'fkk
df c
fbdfAbc'fkk
0TC
equilibrium
)'fkk
fk1(dfAM
)'fkk
dfkd(fAM
)ckd(fAM
m31
y2ysn
m31
y2ysn
2ysn
summing moments about Cm
Masonry Structures, slide 44
Balanced condition with single layer of reinforcement
ymu
mu
y
m31b f
'fkk
yy
m1
syy
m1b
ss
yymu3
f000,87
000,87
f
'fk85.0
E/f003.0
003.0
f
'f)85.0(k
:psi000,000,29E E
f003.085.0k if
ys
d
mu
strains
Strength Design of Reinforced Masonry
c
n.a.
stresses
k1c Cm
Ts= Asfy
Mn
dcord
c
ymu
mu
ymu
mu
strain compatibility
ybymu
mum31
sm
bdfdb'fkk
0TC
equilibrium
Masonry Structures, slide 45
f’m
Grade 40 Grade 60
b tbbtb
1000 0.0124 0.0062 0.00360.0071
d2t steel of layerone forbt/Af000,87
000,87
f
'fk85.0stb
yy
m1b
85.0k1
Balanced condition with single layer of reinforcement
Strength Design of Reinforced Masonry
2000 0.0247 0.0124 0.0143 0.0072
3000 0.0371 0.0186 0.0214 0.01074000 0.0495 0.0247 0.0285 0.0142
5000 0.0619 0.0309 0.0356 0.0178
6000 0.0742 0.0371 0.0428 0.0214
Masonry Structures, slide 46
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 ))(d
d(
mu
y
ymu
musymu
4
imusi
60) (Grade
strain compatibility
sbal
sisbalm
sisim
siisbalsisi
ysissi
A for solve
0fAbd'f428.0
0)TC(C
fATorC
fEf
equilibrium
c
Asbal
bstrains
ys4
mu
d4
d3
d2
d1 s1
s2
s3
0.85
c
stresses
0.85f’m
Cs1
Cm=0.85f’mb(0.85c)
Cs2
Ts3
Ts4 = Asbal fy
Masonry Structures, slide 47
Example: Flexural Strength of In-Plane Wall
Maximum 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
2s
3s
1s
n.a.
0.0034.
0”
44.0
”
20.0
”
Masonry Structures, slide 48
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(c
dc
*bars larger than #9 are not recommended because of anchorage and detailing problems
without compression steel (neglect Cs1 andCs2 forces)
Cm + (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 si fsi
with compression steel (include Cs1 and Cs2 forces)
Cm + (Csi + Tsi) = -294 + Asbal (-60.0 - 38.0 + 20.8 + 60.0) = 0 Asbal = -17.1 in2
note: negative Asbal means that C > T , 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
Masonry Structures, slide 49
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
2s
n.a.
0.003
1s
3s
Ts4 = As fy
= 0.79 in2 x 60 ksi = 47.4 k
0.85f’m
Ts3
Ts2
Cs1
Cm = 0.85f’mb(0.85c)
Masonry Structures, slide 50
c27.8CfEf (-) )003.0(c
dcmysissi
ii
strains ive compress
Example: Flexural Strength of In-Plane Wall
c d1 = 4.0” f1 Csl1
d2 = 20.0” f2 Ts2
d3 = 44.0” f3 Ts3
Cm )TC(
-0.00240 -60 -47.4 0 0 0 0.00360 60.0 47.4
20.0 -165 -117.6
2 3
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
Masonry Structures, slide 51
Example: Flexural Strength of In-Plane Wall
7.63”
5’-4
”
#8 (typ)
Approximate flexural strength of wall.
00398.05263.7
79.02"0.52
2
)4460(d
AandAlumpingand,TandC neglecting 4s3s2ssl
answerof%86 inkip467,4
)50.1
60x00398.0x59.01)(0.52)(60)(in79.0(2
)'f
f59.01(dfAM
2
m
yysn
Masonry Structures, slide 52
UBC Sec. 2108.2.4
Limitations of Method:
Slender Wall Design
(a) for out-of-plane bending of solid, reinforced walls lightly stressed under gravity loads
(c) g= As/bt < 0.5 bal
Sec. 2108.1.3: Load factors
)W7.1L7.1D4.1(75.0U
)ELD(4.1U
L7.1D4.1U
W3.1D9.0U
E4.1D9.0U
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”
Masonry Structures, slide 53
wuh2/8
transverse load
Required Flexural Strength: UBC Sec. 2108.2.4.4
)208()PP(2
eP
8
hwM
PPP
uufuwuf
2u
u
uwufu
Slender Wall Design
e
h
Puf
t
wu
Puw
(Puw + Puf) u
h/2
h/2
Peccentric
load
Pufe/2
Pufe
Masonry Structures, slide 54
Slender Wall Design
Design strength: Sec. 2108.2.4.4
22)-(8 MM nu
Design Considerations
Assumptions for ultimate flexural strength (Sec. 2108.2.1.2)
1. equilibrium
Strength reduction factor:flexure = 0.8 Sec. 2108.1.4.2.1
2. strain compatibility3. mu = 0.0034. fs = Ess < fy
5. neglect masonry strength in tension6. rectangular stress block, k1 = 0.85, k3 = 0.85
Masonry Structures, slide 55
Slender Wall DesignEquivalent area of reinforcement, Ae
for single wythe construction reinforced at center:
Eq. (8-24)y
ysuse f
)fAP(A
flexural strength
25)-(8 Eq.where 23)-(8 Eq. b'f85.0
)fAP(a )
2
ad(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
Masonry Structures, slide 56
Slender Wall DesignLateral Deflections
M
EI
Mh
48
5
EI
h
8
wh
48
5
EI
wh
384
5 2224
My
Mcr
Mss
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 IE
hM
48
5for Ms < Mcr
(8-28)
3
)kd(b)kdd(nAIwhere
IE
h)MM(
48
5
IE
hM
48
5
32
secr
crm
2crs
gm
2cr
s
for Ms > Mcr
(note “kd” may be replaced by “c” for simplicity)
Mcr = fr S
(8-29)
Masonry Structures, slide 57
Slender Wall Design
uufuwuf
2u
u )PP(2
eP
8
hwM
Strength Criteria
Design Considerations
Serviceability Criteria
)278(h007.0 s
crm
2cru
gm
2cr
u IE
h)MM(
48
5
IE
hM
48
5
Masonry Structures, slide 58
Example: Slender Wall Design
ft
in 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(2
1 ρ
2
1 000164.0
81.312
075.0
bd
A ρ bal
s
Masonry Structures, slide 59
U = 0.75 (1.4D + 1.7L + 1.7W)
)2
ad(fAM)PP(
2
eP
8
hwM ysenuufuw
uf2
uu
Example: Slender Wall DesignFlexural Strength per UBC
Mu=Mn = Asefy(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(103.0
b'f85.0
)fAP(a
m
ysu
Masonry Structures, slide 60
crm
2cru
gm
2cr
u IE
h)MM(
48
5
IE
hM
48
5
3.191500
000,29
E
Enksi1500'f750E
m
smm
Example: Slender Wall Design
43
223
2secr in9.23
3
)"355.0(12)355.081.3(in103.03.19
3
bc)cd(nAI
to avoid iteration, assume Mmax = Mu
"955.0"837.0"118.0)9.23)(1500(
)12x20)(1.131.18(
48
5
)444)(1500(
)12x20)(1.13(
48
5 22
u
Flexural Strength per UBC
psf 17.87.1x75.0
ww u
s wu=22.7 psf
"12/.ink1.18)12
955.0)(654.1(
12
)2/"31.7(780.0
8
)'20(wM
2u
u
for simplicity, use gross sectioneven though partially grouted
.ink1.13)6
63.7x"12(ksi112.0M psi112)'f(5.2f SfM
2
cr5.0
mrgrcr
Masonry Structures, slide 61
crm
2crs
gm
2cr
s
s
IE
h)MM(
48
5
IE
hM
48
5
h007.0
Example: Slender Wall DesignCheck Service Load Deflections per UBC
.)in.lb(1532239,1315322
)"31.7(70012x
8
)20)(psf8.17(
)PP(2/eP8
hwM
ss
2
sowo
2s
s
ok"68.1)12x'20007.0h007.0"19.0
)9.23)(500,1(
)12x20)(1.13532.1239.13(
48
5"118.0
s
2s
s
Masonry Structures, slide 62
ss
s
2s
soswsos
2s
s
1532.inlb2562w600
)lbs1532()"66.3(lbs70012x8
)'20(w
)PP(2
eP
8
hwM
Example: Slender Wall DesignMaximum Wind Load per MSJC
Determine Icr considering axial compression
0)(n2k)(n2k 2
1.25
”
#4 @ 32”
d = 3.82”
00104.0ksi 3282312
k 532.1 00164.0
"82.3"12
in 075.0 3.19
2
"."bdF
Pn
s
3.47"jd 908.03
k1 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(nA
3
)kd(bI
Masonry Structures, slide 63
psf) in is w( 55.247.18w
in-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 251.0
34.3 390.0652.0w 153.0"118.0
)in ksi)(15.7 1500(
)1220)(1.13532.1.in-kip56.2w600.0(
48
5"118.0
IE
h)MM(
48
5
IE
hM
48
5
ss
sss
4
2ss
s
crm
2crs
gm
2cr
s
psf 17.8w s )21.4w 251.0( 55.247.18w sss
Masonry Structures, slide 64
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
state limit shear for60.0
B. shear
0P85.0
P25.0PA'f1.0P65.0
n
bnemn
for
orfor2. Design strength
A. axial load and flexure (see next slide)
UBC Sec. 2108.1.2: Special inspection must be provided during construction. Prisms
should be tested or unit strength method should be used.
1. Required strength A. 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)
Masonry Structures, slide 65
Strength Design of RM Shear WallsDefinition of Balanced Axial Load, Pb
d
)E
fe(
e85.0awhere ba'f85.0P:wallsgroutedsolidlyfor
CPsoTCassume
s
ymu
mubbmb
mbsisi
b
Lw
Cm = 0.85f’mbab
d
0.85f’m
Cs1
Cs2
Pb
Ts3
Ts4 = Asbalfy
c 85.0ab
mu
c
n.a.
0.00207 ys
Masonry Structures, slide 66
Strength Design of RM Shear Walls
3. Design assumptions (same as for Slender Wall Design Procedure, UBC Sec. 2108.2.1.2)
1. equilibrium
UBC Requirements
2. strain compatibility3. mu= 0.0034. fs = Ess < fy 5. neglect masonry tensile strength6. use rectangular stress block, k1 = 0.85, k3 = 0.857. 1500 psi < f’m < 4000 psi
Masonry Structures, slide 67
2. for flexural failure modeMn > = 1.8 Mcr for fully grouted wallMn > = 3.0 Mcr for partially grouted wall
Strength Design of RM Shear Walls
3. anchor all continuous reinforcement
M
Mcr
Mn > Mcr ductile
Mn < Mcr
nonductile
4. Reinforcement per UBC Sec. 2108.2.5.2
0"4'spacing
0.0007and
0.002
hv
hv
1. minimum reinforcement
UBC Requirements
5. Axial strength (no flexure)Po = 0.85 f’m(Ac-As) + fyAs Pu < = (0.80)Po
4. As vertical > 1/2 As horizontal
5. maximum spacing of horizontal reinforcement within plastic hinge region = 3t or 24”
Masonry Structures, slide 68
Strength Design of RM Shear Walls
2. for walls limited by shear strength:
37)-(8where 'fACV
VVV
mmvdm
smn
K)-21 (Table 0.1Vd
M for 2.1
25.0Vd
M for 4.2C and d
6. Shear Strength UBC Sec. 2108.2.5.5
) 0.1Vd
M for A250'fA0.4
25.0Vd
M 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
Masonry Structures, slide 69
Strength Design of RM Shear Walls
yhorizontalsw
yhorizontals
ws
yplanevertical
horizontalsmvs
planeverticalhorizontalsn
ynmvs
f)A(h
Lf)
ht
A(tLV
f)A
A(AV
A/A
fAV
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
Masonry Structures, slide 70
Strength Design of RM Shear Walls
6. Shear Strength UBC Sec. 2108.2.5.5: continued
3. for walls limited by flexural strength:
UBC Requirements
39)(8 ynmsn fAVV
within hinge region, distance of Lw above base:
(Vu determined at Lw/2 from base)
smn VVV
above hinge region:
Masonry Structures, slide 71
1. Provide boundary members when the extreme fiber strain exceeds 0.0015.
Strength Design of RM Shear Walls
Boundary Members: Sec.2108.2.5.6
UBC Requirements
Section at Base of Wall
> 3twall
centroid#3 @ 8” min.
0.0015
t
mu > 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.
Masonry Structures, slide 72
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(85.0MφM
in. 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
30-8 Eq. per moment cracking check
n2
cr
r
mrgrcr
kips 7.263.1
HH kips 7.34
"128
MH :existsstate limitflexure if uu
u
Masonry Structures, slide 73
U = 1.3W
Example: Strength DesignShear Reinforcement (neglecting vertical force)
ksi)60(A)5.0(fA )h
L( fA V V horizsyhorizs
wynmvsn
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
ksi)](60 )(A 0.80[(0.5) V kips 34.7H V horiznuu
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
Masonry Structures, slide 74
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”
0.0015
3t > 5.7”
0.003
#3 @ 8”bottom 8 courses
Strain Diagram per Previous Example
Masonry Structures, slide 75
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
E
f85.0a
ba.85f'0 P
2b
bb
s
ymu
mub
bmb
:entreinforcem gconsiderin
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 Case 1:
Case 2: 0.75 )0.20 73.5
36.0( 0.65 kips73.5 P0.25 kips 36.0 P bu
Masonry Structures, slide 76
c 8.27 C f E f )003.0( c
dcmysissi
ii
)"0.28(T )"0.12(T )"0.12(T -)(28.0"C )2
0.85c -(32.0"C M s4s3s2slmcl
Example: Strength DesignFlexural Strength considering Vertical Loads
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”
f Cs f Ts
f Cs 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
Masonry Structures, slide 77
Example: Strength DesignFlexural Strength considering Vertical Loads
Axi
al C
omp
ress
ive
For
ce, k
ips 140
120
100
80
60
40
20
5500 6000 6500 7000
Moment, 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
kips 26.2 1.3
H H u
kips 34.1 12)x (10.67
4,365 Hu
Case 1: in.-kip 4,192 0.65(6450) M M nu
governs kips 25.7 1.7)x 0.75 (
H H u
kips 32.7 12)x (10.67
4,192 Hu