Date July , 2015PCA Work shop
W. Mark McGinley Ph.D, PE
ASD Wall Design for a Single Story Masonry Building
Look at a Single Wythe single story CMU Building Example
Slide 2
3
Plan of Typical Big Box - single story Flexible diaphragm reinforced 8” CMU, Load bearing and non loadbearing walls.Flex Diaphragm- High Seismic
Single Story Building - Ex1
Item ValueRoof Live Load 20 psf
Roof Dead Load (including joist weight) 20 psf
8” CMU wall weight (grouted at 2’ OC) 60 psf
8” CMU wall weight (fully grouted) 80 psf
Weight of Glass doors and store front 10 psf
4” brick wall weight 40 psfDoor dead load 5 psf
Wind Zone 150 mph, Risk Category II
Wind Exposure CSoil Site Class D
Seismic S1 0.47Seismic Ss 1.50
East Wall Elevation
South Wall Elevation
West Wall Elevation
North Wall Elevation
4
Single Story Building Ex1 Elevations
5
Ex1 LoadsRoof Load: ΡDL = 11,900 lb ΡLL = 11,500 lb Ρuplift = 31,700 lb
6
Ex1 Loads
Using loads determine Critical P, M and V at support
and at mid-height of walls for all load
cases
7
Capacity of Wall – Single Wythe Reinforced
CMU’s
TMS 402 Design ProvisionsTwo Rational Design Methods Allowable Stress Design (ASD) – stresses under
service level loads must be less than allowable values.
Strength Design (SD) – Capacities must be greater than load effects under strength level loads (ultimate levels).
Both methods can be applied to unreinforced or reinforced masonry.
Today Address Reinforced ASD.Slide 8
9
ASD Load Combinations – IBC 2012 /ASCE 7-10
D +F D + H+ F+L D + H +F+(Lr or S or R) D +H+F+0.75(L) + 0.75(Lr or S or R) D +H+F+(0.6W or 0.7E) D +H+F+0.75(0.6W) + 0.75L+0.75(Lr or S or R) D +H+F+0.75(0.7E) + 0.75L+0.75(S) 0.6D +0.6W + H 0.6(D+F) +H + 0.7E No increase for E or W any more with Stress
Recalibration – Even with alternative load cases
ASD - Reinforced Masonry:TMS 402 Section 8.3 Masonry in flexural tension is cracked Reinforcing steel is needed to resist tension Linear elastic theory No minimum required steel area and As limited only
on special shear walls. Wire joint reinforcement can be used as flexural
reinforcement No unity interaction equation – Combined loads
must used interaction diagram
Slide 10
Allowable Stresses for Steel Reinforcement: TMS 402 Sec. 8.3.3.1 Tension
Grade 40 or 50 20,000 psi Grade 60 32,000 psi Wire joint reinforcement 30,000 psi
Compression Only reinforcement that is laterally tied (Section 8.3.3.3)
can be used to resist compression Allowable compressive stress = tensile values above
Slide 11
Allowable Axial Compressive Capacity For reinforced masonry, allowable compressive
capacity is expressed in terms of force rather than stress
Maximum compressive stress in masonry from axial load plus bending must not exceed 0.45f’m
Axial compressive stress alone must not exceed allowable axial stress from TMS 402 Section 8.2.4.1 (unreinforced Fa).
Slide 12
Allowable Axial Compressive Capacity: TMS 402 Section 8.3.4.2.1 TMS 402 Equations (8- 21) and (8 - 22) define axial
force limits (ASD). Slenderness reduction coefficients are identical to those used for unreinforced masonry.
Allowable capacity is sum of capacity of masonry plus compressive reinforcement
99for 70)65.025.0(
99 for 140
1)65.025.0(
2'
2'
rh
hrFAAfP
rh
rhFAAfP
sstnma
sstnma
Slide 13
TMS 402 Section 8.3.5 – Shear in Reinforced Masonry... In design we generally:
Check if shear can be resisted entirely by masonry. If not, increase cross – section or,
Add shear reinforcement. Check shear stress. If still no good, increase cross – section.
Next slide shows changes from past code edition!
Slide 14
Masonry shear stress:TMS 402 Section 8.3.5 Shear stress is computed as:
Allowable shear stresses
10.75 for partially grouted shear walls, 1.0 otherwise.
24)-(8 nv
v AVf
Slide 15
25)-(8 gvsvmv FFF
g
Masonry Shear Stress:TMS 402 Section 8.3.5 Allowable shear stresses limits:
M / Vdv ≤ 0.25
M / Vdv 1
Can linear interpolate between limits
26)-(8 3 g'
mv fF
Slide 16
27)-(8 2 g'
mv fF
g2532
vv Vd
MF
Masonry Shear Walls:TMS 402 Section 8.3.5.1.3 Allowable Shear Stress Resisted by the Masonry
Special Reinforced Masonry Shear Walls
All other masonry
M/Vdv is positive and need not exceed 1.0.
29)-(8 ,25.075.1441 '
nm
vv A
PfVdMF
28)-(8 ,25.075.1421 '
nm
vv A
PfVdMF
Slide 17
Shear Design of Reinforced Masonry:TMS 402 Section 8.3.5 - (continued) If allowable shear stress in the masonry is
exceeded then: design shear reinforcement using Equation 8-30
and add Fvs to Fvm
Shear reinforcement is placed parallel the direction of the applied force at a maximum spacing of d/2 or 48 in.
One - third of Av is required perpendicular to the applied force at a spacing of no more than 8 ft. Slide 18
30)-(8 5.0
sAdFA
Fnv
vsvvs
19
C
MV
T
jdf bdxv
ASD Reinforced Masonry- Singly Reinforced - FLEXURE
n = Es/Em and from equil.
Ms= Asfsjd (at the limit) = AsFsjd
Mm = ½ bjkd2fm (at limit)= ½ bjkd2Fb
fm ≤ Fb
fs/n ≤ Fs/n
kd
bdA
kjnnnk
s
3/1
)(2)( 2
20
ASD Interaction Diagrams- Flexural Compression Members
To design reinforced walls under combined loading must construct interaction diagram
stress is proportional to strain ; assume plane sections remain plane ; vary stress ( stress ) gradient to maximum limits and position of neutral axis and back calculate combinations of P and M that would generate this stress distribution
21
Allowable Stress Interaction Diagrams
• Assume single reinforced • Out-of-plane flexure • Grout and masonry the same• Solid grouted• Steel in center
CL
22
ASD Interaction Diagrams Walls - Singly Reinforced allowable – stress interaction diagram Linear elastic theory – tension in masonry it is
ignored- Plane sections remain plane Limit combined compression stress to Fb = 0.45
F’m
P ≤ Pa
d usually = t/2 - ignore compression steel since not tied.
23
Allowable Stress Interaction Diagrams Walls - Singly Reinforced
Assume Stress gradient- Range A –All the Section in compression
Get equivalent force-couple about center line
Pa= 0.5(fm1+fm2)An
Ma= (fm1-fm2)/2 (S) S = bd2/6Note at limit – fm1 and fm2 ≤Fb (set fm1=Fb)Note much of this is from Masonry Course Notes By Dan Abrams
Also Pa must less than Eq 8-21or 8-22
b eff
99for 70)65.025.0(
99 for 140
1)65.025.0(
2'
2'
rh
hrFAAfP
rh
rhFAAfP
sstnma
sstnma
Distribution of concentrated loads, running bond: TMS 402 Section 5.1.3.1
Effective Length
1
hh
/ 2
Load
2
Load Load
12
LoadLoad LoadLoad
EffectiveLength
EffectiveLength
Slide 24
Distribution of concentrated loads, running bond: TMS 402 Section 5.1.3.1
Load
1
2
Load
EffectiveLength
Load
1
2
Load
EffectiveLength
Slide 25
26
Allowable Stress Interaction Diagrams Walls - Singly Reinforced
Assume Stress gradient- Range B –Not All Section in compression- but no tension in steel
Get equivalent force-couple about center line
Pb = Cm = 0.5(fm1)atb Mb= em x Cm
em =d-at/3 = t/2-at/3 =t(1/2-a/3)Note that at = kdThis is valid until steel goes into tensionSet fm1 = Fb at limit
b eff
27
Allowable Stress Interaction Diagrams Walls - Singly Reinforced
Assume Stress gradient- Range C –Section in compression- tension in steel
Get equivalent force-couple about center line
em =d-at/3 = t/2-at/3 =t(1/2-a/3) Cm = 0.5(fm1)atbPc = Cm –Ts & T= As x fs
From similar triangles of the stress diagram fs/n= ([d-at]/at)fm1
Mb= em x Cm –Ts(d-t/2) note that d=t/2 usually so second term goes to zero
At limit fs = Fs and fm1 = Fb one or the other governs Note that at = kd
b eff
at fs/n
Effective compression width per bar: TMS 402 Section 5.1.2 For running - bond masonry, or masonry with bond
beams spaced no more than 48 in. center – to – center, the width of compression area per bar for stress calculations shall not exceed the least of: Center - to - center bar spacing Six times the wall thickness (nominal) 72 in.
Slide 28
29
Axial Load P
Moment M
Capacity envelop letting fm1 = Fb
Range B
Range A
Capacity envelop letting fm1 = Fb
Range CCapacity envelop letting fs = Fs
P cut off Eq 8-21 or 8-22
Ms Mm
Can get a three point interaction diagram easily Most walls have low axial loads
Allowable Stress Interaction Diagrams Walls - Singly Reinforced
NO GOOD ABOVE P CUT OFF
30
West Wall Design for Out-of-Plane Loads
• Guess at wall unit size –Usually 8” CMU.
• Guess at bar size location and spacing – can use max. Moment and assume steel stress governs
• Create interaction diagram for wall.
• Plot diagram and critical P &M values –
• If all load effects within capacity envelope wall is OK.
• No P-delta , or min. As, or max. As.
Ex 1 Out-of-Plane Wall Design
Slide 31
Flexural Design of the WallTrial reinforcing steel:D - .7E produces the maximum moment. Assume d =7.63/2 = 3.81 in.P = 3580 lb/ft M = 16,400 lb-in /ftAssume j = 0.9As = M / fsjd = 16,400 lb-in /(32,000 psi×0.9×3.81 in.) = 0.15 in.2/ftTry a No. 5 bar @ 16 in. on center.This provides As=0.232 in.2 /ft
Ex 1 Out-of-Plane Wall Design
Slide 32
21 Ft Wall w/ No. 5 @ 16in (Centered)total depth, t 7.63 Wall Height, h 21.00 feetf 'm 2,000.00 Radius of Gyration, r 2.20 in
E m 1,800,000.00 h/r 114.5
F b 900.00 Reduction Factor, R 0.37
E s 29,000,000.00 Allowable Axial Stress, F a 187 psi (MSJC8.3.4.2.1 )
F s 32,000.00 Net Area, A n 91.50 in2
d 3.81 Allowable Axial Compr, P a 17,086 lb
kbalanced 0.31
tensile reinforcement, A s 0.23 #5 @ 16 Centered
width, bef f 12.00
because compression reinforcement is not tied, it is not countedk kd f b C mas f s Axial Force Moment Axial Force
(psi) (lb) (psi) (lb) (lb-in) w/ stress axial limitPoints controlled by steel 0.01 0.04 20 5 -32,000 -7,435 -1 -7,435
0.05 0.19 105 119 -32,000 -7,321 429 -7,3210.10 0.38 221 504 -32,000 -6,936 1,841 -6,9360.15 0.57 351 1,202 -32,000 -6,238 4,335 -6,2380.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.22 0.84 560 2,817 -32,000 -4,623 9,936 -4,6230.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.30 1.14 851 5,838 -32,000 -1,602 20,014 -1,602
Points controlled by masonry 0.31 1.19 900 6,416 -32,000 -1,024 21,900 -1,0240.40 1.52 900 8,230 -21,750 3,173 27,182 3,1730.45 1.71 900 9,258 -17,722 5,138 29,996 5,1380.50 1.91 900 10,287 -14,500 6,916 32,679 6,916
Must change C mas to trapezoid 0.55 2.10 900 11,316 -11,864 8,557 35,230 8,557when kd>t 0.60 2.29 900 12,344 -9,667 10,097 37,651 10,097Moment needs to be adjusted 0.62 2.36 900 12,756 -8,887 10,690 38,583 10,690
0.65 2.48 900 13,373 -7,808 11,558 39,941 11,5580.68 2.59 900 13,990 -6,824 12,404 41,252 12,4040.70 2.67 900 14,402 -6,214 12,957 42,100 12,9570.80 3.05 900 16,459 -3,625 15,616 46,026 15,6160.90 3.43 900 18,517 -1,611 18,142 49,429 17,086
Pure compression 45,634 0 17,086
Range C
Range B
Ex 1 Out-of-Plane Wall Design
Slide 33
21 Ft Wall w/ No. 5 @ 16in (Centered)total depth, t 7.63 Wall Height, h 21.00 feetf 'm 2,000.00 Radius of Gyration, r 2.20 in
E m 1,800,000.00 h/r 114.5
F b 900.00 Reduction Factor, R 0.37
E s 29,000,000.00 Allowable Axial Stress, F a 187 psi (MSJC8.3.4.2.1 )
F s 32,000.00 Net Area, A n 91.50 in2
d 3.81 Allowable Axial Compr, P a 17,086 lb
kbalanced 0.31
tensile reinforcement, A s 0.23 #5 @ 16 Centered
width, bef f 12.00
because compression reinforcement is not tied, it is not countedk kd f b Cmas f s Axial Force Moment Axial Force
(psi) (lb) (psi) (lb) (lb-in) w/ stress axial limitPoints controlled by steel 0.01 0.04 20 5 -32,000 -7,435 -1 -7,435
0.05 0.19 105 119 -32,000 -7,321 429 -7,3210.10 0.38 221 504 -32,000 -6,936 1,841 -6,9360.15 0.57 351 1,202 -32,000 -6,238 4,335 -6,2380.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.22 0.84 560 2,817 -32,000 -4,623 9,936 -4,6230.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.30 1.14 851 5,838 -32,000 -1,602 20,014 -1,602
Points controlled by masonry 0.31 1.19 900 6,416 -32,000 -1,024 21,900 -1,0240.40 1.52 900 8,230 -21,750 3,173 27,182 3,1730.45 1.71 900 9,258 -17,722 5,138 29,996 5,1380.50 1.91 900 10,287 -14,500 6,916 32,679 6,916
Must change Cmas to trapezoid 0.55 2.10 900 11,316 -11,864 8,557 35,230 8,557when kd>t 0.60 2.29 900 12,344 -9,667 10,097 37,651 10,097Moment needs to be adjusted 0.62 2.36 900 12,756 -8,887 10,690 38,583 10,690
0.65 2.48 900 13,373 -7,808 11,558 39,941 11,5580.68 2.59 900 13,990 -6,824 12,404 41,252 12,4040.70 2.67 900 14,402 -6,214 12,957 42,100 12,9570.80 3.05 900 16,459 -3,625 15,616 46,026 15,6160.90 3.43 900 18,517 -1,611 18,142 49,429 17,086
Pure compression 45,634 0 17,086
1221
)2.2(70)2000)25.0(
99for 70)65.025.0(
2
2'
xP
rh
hrFAAfP
a
sstnma
Fs = 32,000
fs/n [at/ ([d-at]) =fb
32000/16.11 [.1(3.81/ ([3.81-.1(3.81])
Cmas=0.5x0.1(3.81)(12)(221)
P=221 – 0.23 x 32000
Mb= em x Cm –Ts(d-t/2)=(3.81-( 0.1 x 3.81/3)x32000x0.23
Ex 1 Out-of-Plane Wall Design
Slide 34
21 Ft Wall w/ No. 5 @ 16in (Centered)total depth, t 7.63 Wall Height, h 21.00 feetf 'm 2,000.00 Radius of Gyration, r 2.20 in
E m 1,800,000.00 h/r 114.5
F b 900.00 Reduction Factor, R 0.37
E s 29,000,000.00 Allowable Axial Stress, F a 187 psi (MSJC8.3.4.2.1 )
F s 32,000.00 Net Area, A n 91.50 in2
d 3.81 Allowable Axial Compr, P a 17,086 lb
kbalanced 0.31
tensile reinforcement, A s 0.23 #5 @ 16 Centered
width, bef f 12.00
because compression reinforcement is not tied, it is not countedk kd f b Cmas f s Axial Force Moment Axial Force
(psi) (lb) (psi) (lb) (lb-in) w/ stress axial limitPoints controlled by steel 0.01 0.04 20 5 -32,000 -7,435 -1 -7,435
0.05 0.19 105 119 -32,000 -7,321 429 -7,3210.10 0.38 221 504 -32,000 -6,936 1,841 -6,9360.15 0.57 351 1,202 -32,000 -6,238 4,335 -6,2380.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.22 0.84 560 2,817 -32,000 -4,623 9,936 -4,6230.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.30 1.14 851 5,838 -32,000 -1,602 20,014 -1,602
Points controlled by masonry 0.31 1.19 900 6,416 -32,000 -1,024 21,900 -1,0240.40 1.52 900 8,230 -21,750 3,173 27,182 3,1730.45 1.71 900 9,258 -17,722 5,138 29,996 5,1380.50 1.91 900 10,287 -14,500 6,916 32,679 6,916
Must change Cmas to trapezoid 0.55 2.10 900 11,316 -11,864 8,557 35,230 8,557when kd>t 0.60 2.29 900 12,344 -9,667 10,097 37,651 10,097Moment needs to be adjusted 0.62 2.36 900 12,756 -8,887 10,690 38,583 10,690
0.65 2.48 900 13,373 -7,808 11,558 39,941 11,5580.68 2.59 900 13,990 -6,824 12,404 41,252 12,4040.70 2.67 900 14,402 -6,214 12,957 42,100 12,9570.80 3.05 900 16,459 -3,625 15,616 46,026 15,6160.90 3.43 900 18,517 -1,611 18,142 49,429 17,086
Pure compression 45,634 0 17,086
fb=Fb = 900
fs = [d-at/ ([at]) nfb
16.11 [(3.81 -0.4(3.81/ (0.4(3.81])(900 x 16.11
Cmas=0.5x0.4(3.81)(12)(900)
P=221 – 0.23 x 32000
Mb= em x Cm –Ts(d-t/2)=(3.81-( 0.1 x 3.81/3)x32000x0.23
Ex 1 Out-of-Plane Wall Design
Slide 35
21 Ft Wall w/ No. 5 @ 16in (Centered)total depth, t 7.63 Wall Height, h 21.00 feetf 'm 2,000.00 Radius of Gyration, r 2.20 in
E m 1,800,000.00 h/r 114.5
F b 900.00 Reduction Factor, R 0.37
E s 29,000,000.00 Allowable Axial Stress, F a 187 psi (MSJC8.3.4.2.1 )
F s 32,000.00 Net Area, A n 91.50 in2
d 3.81 Allowable Axial Compr, P a 17,086 lb
kbalanced 0.31
tensile reinforcement, A s 0.23 #5 @ 16 Centered
width, bef f 12.00
because compression reinforcement is not tied, it is not countedk kd f b C mas f s Axial Force Moment Axial Force
(psi) (lb) (psi) (lb) (lb-in) w/ stress axial limitPoints controlled by steel 0.01 0.04 20 5 -32,000 -7,435 -1 -7,435
0.05 0.19 105 119 -32,000 -7,321 429 -7,3210.10 0.38 221 504 -32,000 -6,936 1,841 -6,9360.15 0.57 351 1,202 -32,000 -6,238 4,335 -6,2380.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.22 0.84 560 2,817 -32,000 -4,623 9,936 -4,6230.24 0.91 627 3,441 -32,000 -3,999 12,052 -3,9990.30 1.14 851 5,838 -32,000 -1,602 20,014 -1,602
Points controlled by masonry 0.31 1.19 900 6,416 -32,000 -1,024 21,900 -1,0240.40 1.52 900 8,230 -21,750 3,173 27,182 3,1730.45 1.71 900 9,258 -17,722 5,138 29,996 5,1380.50 1.91 900 10,287 -14,500 6,916 32,679 6,916
Must change C mas to trapezoid 0.55 2.10 900 11,316 -11,864 8,557 35,230 8,557when kd>t 0.60 2.29 900 12,344 -9,667 10,097 37,651 10,097Moment needs to be adjusted 0.62 2.36 900 12,756 -8,887 10,690 38,583 10,690
0.65 2.48 900 13,373 -7,808 11,558 39,941 11,5580.68 2.59 900 13,990 -6,824 12,404 41,252 12,4040.70 2.67 900 14,402 -6,214 12,957 42,100 12,9570.80 3.05 900 16,459 -3,625 15,616 46,026 15,6160.90 3.43 900 18,517 -1,611 18,142 49,429 17,086
Pure compression 45,634 0 17,086
Ex 1 Out-of-Plane Wall Design
Slide 36
0 10,000 20,000 30,000 40,000 50,000 60,000
-10,000
-5,000
0
5,000
10,000
15,000
20,000
ASD Interaction Diagram8 in Wall, 21 ft high, #5 @ 16 in. Centered
CapacityD + LrD + .6W: D-.7ED + 0.75 (.6)W + 0.75 Lr D - 0.75 (0.7E) 0.6 D + .6W:0.6 D +0.7 E:
M, lb-in per foot of length
P , l
b pe
r foo
t of l
engt
h
37
A
E
1
87
VD1N2
VD2N1VD1N1
VD1E1VD1E2
VD2W VD2E1
North Wall 2
North Wall 1
South Wall
East Wall 2
East Wall 1
West Wall Diaphragm 2
Diaphragm 1
VD2S
Plan of Typical Big Box - single story Flexible diaphragm
See MDG for Load determination and distribution to shear wall lines - Flex Diaphragm – SDC- D
Diaphragm2West Wall
North Wall 1
East Wall 1
East Wall 2West wall 2
Example Shear wall in a single story Building Shear Wall Ex2
38
To check wall segments under in -plane loads must first
• Distribute Load to Shear wall lines – Either by Trib. Width or Rigid Diaphragm analysis.
• Distribute Line load to each segment w.r.t. relative rigidity.
Look at Shear Wall Design
39
Shear Wall Loads DistributionSegments get load w.r.t. relative k
East Wall Elevation
South Wall Elevation
West Wall Elevation
North Wall Elevation
North Wall 2
234567
8
910 18
9 107654321
1 2 3 4 5 76 8 9
78
6 5 4 3 2 1
East Wall 2
1 2 160.8 kips Diaphragm
Shear due to seismic
40
Shear Wall Loads DistributionSegments get load w.r.t relative k
13'3'4
wwmc l
hlhtEk
• For Cantilevered Shear wall segments
13'3'
wwmc l
hlhtEk
• For Fixed-Fixed Shear wall segments
41
Shear Wall Load Distribution
Table18.1-2 DPC Box Building Shear Load on Wall Segments on the West Wall
DPC Box West Wall (Grid 1) Vd = 160.8 kips Segment h l Ri Vi from Diaphragm (lb) Vi wt (lb)
1 22 12 0.332 4.99 4.05 2 22 24 1.715 25.80 8.1 3 22 24 1.715 25.80 8.1 4 22 24 1.715 25.80 8.1 5 22 24 1.715 25.80 8.1 6 22 24 1.715 25.80 8.1 7 22 24 1.715 25.80 8.1 8 22 6.67 0.065 0.98 2.25 Sum 10.687 160.800
42
Shear Wall Load Distribution
Table18.1-2 DPC Box Building Shear Load on Wall Segments on the West Wall
DPC Box West Wall (Grid 1) Vd = 160.8 kips Segment h l Ri Vi from Diaphragm (lb) Vi wt (lb)
1 22 12 0.332 4.99 4.05 2 22 24 1.715 25.80 8.1 3 22 24 1.715 25.80 8.1 4 22 24 1.715 25.80 8.1 5 22 24 1.715 25.80 8.1 6 22 24 1.715 25.80 8.1 7 22 24 1.715 25.80 8.1 8 22 6.67 0.065 0.98 2.25 Sum 10.687 160.800
13
12223
1222410
iR
10.6871.715160.8iV
Segment 2 Designed in later Example
43
Design of Reinforced Masonry (ASD) In Plane Loading (shear Walls)
h
V
L
Axial Force
44
ASD Design of Reinforced Masonry -In Plane Loading (Shear Walls)
Still use interaction diagrams Axial Load is still dealt with as out of
plane (M=0) In plane load produces moment and thus
moment capacity is dealt with slightly differently
Initially assume fm = Fb and Neutral Axis then same as out-of-plane but area and S are based
on Length = d and t = b use OOP equations in Range A and B.
Adjust aL as before until rebars start to go into tension. Note that aL = kd
Determine fsi from similar triangles & get Ti
Check extreme fsi ≤ Fs & fm ≤ Fb
Cm = aL x b x ½ Fb (or fm when fsn = Fs)
M capacity (Σabout center)= Σ (Ti x (di -L/2) + Cm x(L/2 – aL/3))
45
P-M Diagrams ASD-In Plane
46
Reinforced Masonry Shear Walls - ASD
h
V
LP
V= base shear
M = overturning moment
Axial Force ~ 0
Flexure Only P = 0 on diagram
h
V
L
P- self weight only ignore
V= base shear
M = over turning moment
Multiple rebar locations
Reinforced Masonry Shear Walls - ASD
47
48
L V= base shear
M = over turning moment
fm
k*d*
Fsc/NTi
Fs1/Nfsi/N
fsn/N<= Fs/N
di – location to centroid of each bar
CmTsn= Fs As Ti Ti Ti Ti Ti Ti Ti T1
Tension Compression
fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N
Reinforced Masonry Shear Walls – ASD (P =0) Can use the singly reinforced equations
d*– location centroid of all bars in tension f*si/N
49
Moment only ASD In Plane
To locate Neutral Axis – Guess how many bars on tension side – As*
Find d* (centroid of tension bars) and *=As*/bd*
Get k* = ((*n)2 + 2*n)1/2 – n*
Unless tied ignore compression in steel.
50
Moment only ASD In Plane
Check k* d* to ensure assume tension bars correct – iterate if not
Determine fsi from similar triangles and Ti
M capacity (Σabout C)= Σ (Ti x (di – k*d*/3))
51
Shear Wall Example 3
Geometry: typical wall element:25 ft – 4 in. total height3 ft – 4 in. parapet24 ft length between control joints8 in. CMU grouted solid: 80 psf dead
f’m = 1500 psi
VD
VP
22’-0”
25’-4”
12’-8”
24’-0”
52
Shear Wall Example 3
West Wall seismic load condition Vdiaphragm = 25,800 lb acting 22 ft above foundation
Vpier = 8,100 lb acting 12.7 ft above foundation
8 in. CMU grouted solid (maximum possible dead load) Pbase = 80 lb/ft2 x 25.3 ft x 24 ft = 48,6 00 lb
Vertical Seismic: Vpier = 0.2 SDSD = (-0.2 (1.11)(48,600 )=10,800 lb
ASD Load Combination 0.6 D + 0.7 E P = 0.6 x 48,600 + 0.7 x -0.2 (1.11) (48,600 ) = 21,600 lb
M = 0.6 x 0 + 0.7 x (25,800 lb x 22 ft + 8,100 lb x 12.7 ft) = 469,000 lb-ft = 5,630,000 lb in.
V = 0.6 x 0 + 0.7 x (25,800 lb + 8,100 lb) = 23,700 lb
53
Shear Wall Example 3
#5 bar (typ)
24”4” 8” 8” 24”
Assume the rebar in the wall are as shown – Axial load is negligible – ignore- To simplify assume that only three end bars are effective (only lap these to foundation)
54
Shear Wall Example 3For the 24 ft long wall panel between control joints subjected to in-plane loading, the flexural depth is the wall length less the distance to the centroid of the vertical steel at the ends of the wall. ininftinftind 27612)/1224(12 Assume that j = 0.9 The required area of reinforcing steel can be calculated as:
22' 944.0
2769.0000,24000,630,5 in
ininlbinlb
jdFMAjdFAMs
dreqsss
5.2148.21)1500(900
29000000
m
s
EEn
We are using 3 #5 Bars but if needed an estimate of As can be determined by assuming j = .9 and applied moment , M
271.02769.0000,32
000,630,5 ininpsi
inlbjdF
MAs
sreq
55
Shear Wall Example 3Try 3 No. 5 bars, As = 3 × 0.31 in.2 = 0.93 in.2
psi 675 psi 1500 0.45 mf' 0.45 F
000,630,5000,732,7276957.0476,3193.0
476,31000,32280276
957.03129.01
31
129.000950.0)00950.0(00950.022
00950.0000442.05.21000442.027663.7
93.0
b
22
22
2
okinlbinlbininlbinM
psix kd
-kd
kj
nnnk
ninin
inbdAs
psipsijkbdMfb 0675157
)276)(63.7(129.0)957.0(5.0000,630,5
5.0 22
Calculate j and k:
You need to get the stress at the centroid based on the extreme bar fs=Fs
Check Masonry Compression Stresses
Should get third bar stress then S Moments but M ≈
56
Shear Wall Example 3
F vs vmvmv FFF
Check Shear StressAssume no shear reinforcing and thus
OK psi 5.77150022vely)Conservati(348
0250150027670023
0006305751421
250751421
'mv
n
n
'mvm
fFpsi.
A .
)(,,,.
AP .f
VdM.F
57
Shear Wall Example 3
Check Shear Stress
OK psi 48.3 psi 33.9 2)25.1(280 23,700lb
nvv A
Vf
Conservatively assume just face shell bedded areas resist shear
58
So the Final Design – can use the # 5 at the ends of the wall – ignoring any bars that will
likely be there for out-of-plane loading
#5 bar (typ)
24”4” 8” 8” 24”
Shear Wall Design
Check Prescriptive Seismic Reinforcing
Requirements for Detailed Plain SWs and SDC C: TMS 402 Section 7.3.2.3
#4 bar (min) within 8 in. of corners & ends of walls
roofdiaphragm
roof connectors@ 48 in. max oc
#4 bar (min) within16 in. of top of parapet
Top of Parapet
#4 bar (min) @ diaphragms continuous through control joint#4 bar (min) within 8 in. of all control joints
control joint
#4 bars @ 10 ft oc or W1.7 joint reinforcement @ 16 in. oc
#4 bars @ 10 ft oc
24 in. or 40 db past opening
#4 bars around openings
Slide 59
Seismic Design: TMS 402 Chapter 7
Seismic Design Category D Masonry that is part of the lateral force – resisting
system must be reinforced so that v + h 0.002, and v and h 0.0007
Type N mortar and masonry cement mortars are prohibited in the seismic force – resisting system
Shear walls must meet minimum prescriptive requirements for reinforcement and connections (special reinforced)
Other walls must meet minimum prescriptive requirements for horizontal and vertical reinforcement
Slide 60
Requirements for Special Reinforced Shear Walls: TMS 402 Section 7.3.2.6
roofdiaphragm
roof connectors@ 48 in. max oc
#4 bar (min) within16 in. of top of parapet
Top of Parapet
#4 bar (min) @ diaphragms continuous through control joint#4 bar (min) within 8 in. of all control joints
control joint
#4 bars @ 4 ft oc#4 bars @ 4 ft oc
#4 bar (min) within 8 in. of corners & ends of walls
24 in. or 40 db past opening
#4 bars around openings
Slide 61
Seismic Design Categories E and F: TMS 402 Section 7.4.5 Additional reinforcement requirements for masonry
not laid in running bond and used in nonparticipating elements Horizontal Reinforcement of at least 0.0015 Ag
Horizontal Reinforcement must be no more than 24 in. oc.
Must be fully grouted and constructed of hollow open-end units or two wythes of solid units
Slide 62