16c masonry wall design asd example

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


Assume Stress gradient- Range B –Not All Section in compression- but no tension in steel


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


Assume Stress gradient- Range C –Section in compression- tension in steel


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


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


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


Slide 33


E m 1,800,000.00 h/r 114.5



F s 32,000.00 Net Area, A n 91.50 in2


kbalanced 0.31


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


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


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


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


Slide 34


E m 1,800,000.00 h/r 114.5



F s 32,000.00 Net Area, A n 91.50 in2


kbalanced 0.31


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


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


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


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


Slide 35


E m 1,800,000.00 h/r 114.5



F s 32,000.00 Net Area, A n 91.50 in2


kbalanced 0.31


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


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


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



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


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


#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


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


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

16c masonry wall design asd example

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