ASD Wall Design for a Single
Story Masonry Building
W. Mark McGinley Ph.D, PE
Date July , 2015
PCA Work shop
Look at a Single Wythe
single story CMU Building
Example
Slide 2
Single Story Building - Ex1
Plan of Typical Big Box - single
story Flexible diaphragm
reinforced 8” CMU, Load
bearing and non loadbearing
walls.
Flex Diaphragm- High Seismic
Item
Roof Live Load
Roof Dead Load (including
joist weight)
8” CMU wall weight (grouted
at 2’ OC)
8” CMU wall weight (fully
grouted)
Weight of Glass doors and
store front
4” brick wall weight
Door dead load
Wind Zone
Wind Exposure
Soil Site Class
Seismic S1
Seismic Ss
Value
20 psf
20 psf
60 psf
80 psf
10 psf
40 psf
5 psf
150 mph, Risk
Category II
C
D
0.47
1.50
3
Single Story Building Ex1 Elevations
South Wall Elevation
North Wall Elevation
West Wall Elevation
4
East Wall Elevation
Ex1 Loads
Roof Load:
ΡDL = 11,900 lb
ΡLL = 11,500 lb
Ρuplift = 31,700 lb
5
Ex1 Loads
Using loads
determine Critical P,
M and V at support
and at mid-height of
walls for all load
cases
6
Capacity of Wall – Single Wythe
Reinforced
CMU’s
7
TMS 402 Design Provisions
Two 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
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
9
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
Grade 60
Wire joint reinforcement
20,000 psi
32,000 psi
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
2

h
 h  
'
Pa  (0.25 f m An  0.65 Ast Fs ) 1  
 99
  for
r
  140r  
 70r 
Pa  (0.25 f An  0.65 Ast Fs ) 

 h 
'
m
2
h
for  99
r
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:
V
fv 
Anv

(8 - 24)
Allowable shear stresses
Fv  Fvm  Fvs   g
(8 - 25)
 g 10.75 for partially grouted shear walls, 1.0
otherwise.
Slide 15
Masonry Shear Stress:
TMS 402 Section 8.3.5

Allowable shear stresses limits:

M / Vdv ≤ 0.25
Fv   3 f m'  g


 M / Vdv  1
Fv   2 f m'  g



(8 - 26)
(8 - 27)
Can linear interpolate between limits
2

M
   g
Fv    5  2
3

Vd
v 
 
Slide 16
Masonry Shear Walls:
TMS 402 Section 8.3.5.1.3

Allowable Shear Stress Resisted by the Masonry

Special Reinforced Masonry Shear Walls
 
Fv  1


P
 M  '
4  1.75
f m  0.25 , (8 - 29)



4 
An
 Vd v 
All other masonry
 
Fv  1

P
 M  '
4  1.75
f m  0.25 , (8 - 28)



2 
An
 Vd v 
M/Vdv is positive and need not exceed 1.0.
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
 Av Fs d v
Fvs  0.5
 Anv s





(8 - 30)
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
ASD Reinforced MasonrySingly Reinforced - FLEXURE
fm ≤ Fb
C
kd
jd
M
V
fv bdx
T
n = Es/Em and from equil.
Ms= Asfsjd (at the limit) = AsFsjd
Mm = ½ bjkd2fm (at limit)= ½ bjkd2Fb
fs/n ≤ Fs/n
k  ( n )2  2( n )  n
j  1 k / 3
A
 s
bd
19
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
20
Allowable Stress Interaction
Diagrams
•
•
•
•
•
Assume single reinforced
Out-of-plane flexure
Grout and masonry the same
Solid grouted
Steel in center
CL
21
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.
22
beff
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 = bd2/6
Note 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
  h 2 
h
Pa  (0.25 f An  0.65 Ast Fs ) 1  
  for  99
r
  140r  
'
m
 70r 
Pa  (0.25 f m' An  0.65 Ast Fs ) 

 h 
2
for
h
 99
r
23
Distribution of concentrated loads,
running bond: TMS 402 Section 5.1.3.1
Load
Load
Load
1
2
h
2
h/2
1
Effective Length
Effective Effective
Length Length
Slide 24
Distribution of concentrated loads,
running bond: TMS 402 Section 5.1.3.1
Load
1
Load
1
2
Effective
Length
2
Effective
Length
Slide 25
beff
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 = kd
This is valid until steel goes into tension
26
Set fm1 = Fb at limit
Allowable Stress Interaction
Diagrams Walls - Singly Reinforced
beff
Assume Stress gradient- Range C –
Section in compression- tension in steel
Get equivalent force-couple about center line
at
fs/n
em =d-at/3 = t/2-at/3 =t(1/2-a/3)
Cm = 0.5(fm1)atb
Pc = 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
27
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
Allowable Stress Interaction
Diagrams Walls - Singly Reinforced
NO GOOD ABOVE P CUT OFF
Axial Load P
Capacity envelop letting fm1 = Fb
P cut off Eq 8-21 or 8-22
Range A
Range B
Can get a three point interaction
diagram easily
Most walls have low axial loads
Capacity
envelop letting
fs = Fs
Ms Mm
Range C
Capacity envelop letting fm1 = Fb
Moment M
29
West Wall Design for Out-ofPlane 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.
30
Ex 1 Out-of-Plane Wall Design
Flexural Design of the Wall
Trial 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 /ft
Assume j = 0.9
As = M / fsjd =
16,400 lb-in /(32,000 psi×0.9×3.81 in.) = 0.15 in.2/ft
Try a No. 5 bar @ 16 in. on center.
This provides As=0.232 in.2 /ft
Slide 31
Ex 1 Out-of-Plane Wall Design
21 Ft Wall w/ No. 5 @ 16in (Centered)
total depth, t
f 'm
1,800,000.00
Fb
900.00
Reduction Factor, R
Es
29,000,000.00
Allowable Axial Stress, F a
Fs
32,000.00
Net Area, A n
d
3.81
Allowable Axial Compr, P a
k balanced
0.31
width, bef f
Range B
Wall Height, h
Radius of Gyration, r
Em
tensile reinforcement, A s
Range C
7.63
2,000.00
h/r
21.00 feet
2.20 in
114.5
0.37
187 psi (MSJC8.3.4.2.1 )
2
91.50 in
17,086 lb
0.23 #5 @ 16 Centered
12.00
because compression reinforcement is not tied, it is not counted
k
kd
fb
(psi)
Points controlled by steel
0.01
0.04
20
0.05
0.19
105
0.10
0.38
221
0.15
0.57
351
0.24
0.91
627
0.22
0.84
560
0.24
0.91
627
0.30
1.14
851
Points controlled by masonry
0.31
1.19
900
0.40
1.52
900
0.45
1.71
900
0.50
1.91
900
Must change C mas to trapezoid
0.55
2.10
900
when k d>t
0.60
2.29
900
Moment needs to be adjusted
0.62
2.36
900
0.65
2.48
900
0.68
2.59
900
0.70
2.67
900
0.80
3.05
900
0.90
3.43
900
Pure compression
C mas
(lb)
5
119
504
1,202
3,441
2,817
3,441
5,838
6,416
8,230
9,258
10,287
11,316
12,344
12,756
13,373
13,990
14,402
16,459
18,517
fs
Axial Force Moment
Axial Force
(psi)
(lb)
(lb-in)
w/ stress axial limit
-32,000
-7,435
-1
-7,435
-32,000
-7,321
429
-7,321
-32,000
-6,936
1,841
-6,936
-32,000
-6,238
4,335
-6,238
-32,000
-3,999
12,052
-3,999
-32,000
-4,623
9,936
-4,623
-32,000
-3,999
12,052
-3,999
-32,000
-1,602
20,014
-1,602
-32,000
-1,024
21,900
-1,024
-21,750
3,173
27,182
3,173
-17,722
5,138
29,996
5,138
-14,500
6,916
32,679
6,916
-11,864
8,557
35,230
8,557
-9,667
10,097
37,651
10,097
-8,887
10,690
38,583
10,690
-7,808
11,558
39,941
11,558
-6,824
12,404
41,252
12,404
-6,214
12,957
42,100
12,957
-3,625
15,616
46,026
15,616
-1,611
18,142
49,429
17,086
45,634
0
17,086
Slide 32
Ex 1 Out-of-Plane Wall Design
21 Ft Wall w/ No. 5 @ 16in (Centered)
total depth, t
f 'm
Em
7.63
2,000.00
Wall Height, h
Radius of Gyration, r
1,800,000.00
h/r
114.5
Fb
900.00
Reduction Factor, R
Es
29,000,000.00
Allowable Axial Stress, F a
Fs
32,000.00
Net Area, A n
d
3.81
Allowable Axial Compr, P a
k balanced
0.31
tensile reinforcement, A s
0.23 #5 @ 16 Centered
width, bef f
12.00
0.37
187 psi (MSJC8.3.4.2.1 )
2
91.50 in
17,086 lb
 70r 
Pa  (0.25 f m' An  0.65 Ast Fs ) 

 h 
Fs = 32,000
because compression reinforcement is not tied, it is not counted
k
kd
fb
(psi)
Points controlled by steel
0.01
0.04
20
0.05
0.19
105
0.10
0.38
221
0.15
0.57
351
0.24
0.91
627
0.22
0.84
560
0.24
0.91
627
0.30
1.14
851
Points controlled by masonry
0.31
1.19
900
b
0.40
1.52
900
0.45
1.71
900
0.50
1.91
900
Must change C mas to trapezoid
0.55
2.10
900
when k d>t
0.60
2.29
900
Moment needs to be adjusted
0.62
2.36
900
0.65
2.48
900
0.68
2.59
900
0.70
2.67
900
0.80
3.05
900
0.90
3.43
900
Pure compression
fs/n [at/ ([d-at]) =f
32000/16.11 [.1(3.81/ ([3.81-.1(3.81])
21.00 feet
2.20 in
C mas
(lb)
5
119
504
1,202
3,441
2,817
3,441
5,838
6,416
8,230
9,258
10,287
11,316
m
12,344
12,756
13,373
13,990
14,402
16,459
18,517
 70(2.2) 
Pa  (0.25) 2000) 

 21x12 
2
for
h
 99
r
2
fs
Axial Force Moment
Axial Force
(psi)
(lb)
(lb-in)
w/ stress axial limit
-32,000
-7,435
-1
-7,435
-32,000
-7,321
429
-7,321
-32,000
-6,936
1,841
-6,936
-32,000
-6,238
4,335
-6,238
-32,000
-3,999
12,052
-3,999
-32,000
-4,623
9,936
-4,623
-32,000
-3,999
12,052
-3,999
-32,000
-1,602
20,014
-1,602
-32,000
-1,024
21,900
-1,024
-21,750
3,173
27,182
3,173
mas
-17,722
5,138
29,996
5,138
-14,500
6,916
32,679
6,916
-11,864
8,557
35,230
8,557
m
s
-9,667
10,097
37,651
10,097
-8,887
10,690
38,583
10,690
-7,808
11,558
39,941
11,558
-6,824
12,404
41,252
12,404
-6,214
12,957
42,100
12,957
-3,625
15,616
46,026
15,616
-1,611
18,142
49,429
17,086
45,634
0
17,086
P=221 – 0.23 x 32000
C
=0.5x0.1(3.81)(12)(221)
Mb= e x C –T (d-t/2)
=(3.81-( 0.1 x 3.81/3)x32000x0.23
Slide 33
Ex 1 Out-of-Plane Wall Design
21 Ft Wall w/ No. 5 @ 16in (Centered)
total depth, t
f 'm
Em
7.63
2,000.00
Wall Height, h
Radius of Gyration, r
1,800,000.00
h/r
114.5
Fb
900.00
Reduction Factor, R
Es
29,000,000.00
Allowable Axial Stress, F a
Fs
32,000.00
Net Area, A n
d
3.81
Allowable Axial Compr, P a
k balanced
0.31
tensile reinforcement, A s
width, bef f
21.00 feet
2.20 in
0.37
187 psi (MSJC8.3.4.2.1 )
2
91.50 in
17,086 lb
0.23 #5 @ 16 Centered
12.00
because compression reinforcement is not tied, it is not counted
k
kd
fb
(psi)
Points controlled by steel
0.01
0.04
20
0.05
0.19
105
0.10
0.38
221
0.15
0.57
351
0.24
0.91
627
0.22
0.84
560
0.24
0.91
627
0.30
1.14
851
Points controlled by masonry
0.31
1.19
900
0.40
1.52
900
0.45
1.71
900
0.50
1.91
900
Must change C mas to trapezoid
0.55
2.10
900
when
k
d>t
0.60
2.29
900
b
Moment needs to be adjusted
0.62
2.36
900
0.65
2.48
900
0.68
2.59
900
0.70
2.67
900
b900
0.80
3.05
0.90
3.43
900
Pure compression
fb=Fb = 900
C mas
(lb)
5
119
504
1,202
3,441
2,817
3,441
5,838
6,416
8,230
9,258
10,287
11,316
12,344
12,756
13,373
13,990
14,402
m
16,459
18,517
fs
Axial Force Moment
Axial Force
(psi)
(lb)
(lb-in)
w/ stress axial limit
-32,000
-7,435
-1
-7,435
-32,000
-7,321
429
-7,321
-32,000
-6,936
1,841
-6,936
-32,000
-6,238
4,335
-6,238
-32,000
-3,999
12,052
-3,999
-32,000
-4,623
9,936
-4,623
-32,000
-3,999
12,052
-3,999
-32,000
-1,602
20,014
-1,602
-32,000
-1,024
21,900
-1,024
-21,750
3,173
27,182
3,173
-17,722
5,138
29,996
5,138
-14,500
6,916
32,679
6,916
-11,864
8,557
35,230
8,557
-9,667
10,097
37,651
10,097
mas 38,583
-8,887
10,690
10,690
-7,808
11,558
39,941
11,558
-6,824
12,404
41,252
12,404
-6,214
12,957
42,100
12,957
m
-3,625 s 15,616
46,026
15,616
-1,611
18,142
49,429
17,086
45,634
0
17,086
P=221 – 0.23 x 32000
fs = [d-at/ ([at]) nf
C =0.5x0.4(3.81)(12)(900)
16.11 [(3.81 -0.4(3.81/ (0.4(3.81])(900 x 16.11
M = e x C –T (d-t/2)
=(3.81-( 0.1 x 3.81/3)x32000x0.23
Slide 34
Ex 1 Out-of-Plane Wall Design
21 Ft Wall w/ No. 5 @ 16in (Centered)
total depth, t
f 'm
7.63
2,000.00
Wall Height, h
Radius of Gyration, r
Em
1,800,000.00
Fb
900.00
Reduction Factor, R
Es
29,000,000.00
Allowable Axial Stress, F a
Fs
32,000.00
Net Area, A n
d
3.81
Allowable Axial Compr, P a
k balanced
0.31
tensile reinforcement, A s
width, bef f
h/r
21.00 feet
2.20 in
114.5
0.37
187 psi (MSJC8.3.4.2.1 )
2
91.50 in
17,086 lb
0.23 #5 @ 16 Centered
12.00
because compression reinforcement is not tied, it is not counted
k
kd
fb
(psi)
Points controlled by steel
0.01
0.04
20
0.05
0.19
105
0.10
0.38
221
0.15
0.57
351
0.24
0.91
627
0.22
0.84
560
0.24
0.91
627
0.30
1.14
851
Points controlled by masonry
0.31
1.19
900
0.40
1.52
900
0.45
1.71
900
0.50
1.91
900
Must change C mas to trapezoid
0.55
2.10
900
when k d>t
0.60
2.29
900
Moment needs to be adjusted
0.62
2.36
900
0.65
2.48
900
0.68
2.59
900
0.70
2.67
900
0.80
3.05
900
0.90
3.43
900
Pure compression
C mas
(lb)
5
119
504
1,202
3,441
2,817
3,441
5,838
6,416
8,230
9,258
10,287
11,316
12,344
12,756
13,373
13,990
14,402
16,459
18,517
fs
Axial Force Moment
Axial Force
(psi)
(lb)
(lb-in)
w/ stress axial limit
-32,000
-7,435
-1
-7,435
-32,000
-7,321
429
-7,321
-32,000
-6,936
1,841
-6,936
-32,000
-6,238
4,335
-6,238
-32,000
-3,999
12,052
-3,999
-32,000
-4,623
9,936
-4,623
-32,000
-3,999
12,052
-3,999
-32,000
-1,602
20,014
-1,602
-32,000
-1,024
21,900
-1,024
-21,750
3,173
27,182
3,173
-17,722
5,138
29,996
5,138
-14,500
6,916
32,679
6,916
-11,864
8,557
35,230
8,557
-9,667
10,097
37,651
10,097
-8,887
10,690
38,583
10,690
-7,808
11,558
39,941
11,558
-6,824
12,404
41,252
12,404
-6,214
12,957
42,100
12,957
-3,625
15,616
46,026
15,616
-1,611
18,142
49,429
17,086
45,634
0
17,086
Slide 35
Ex 1 Out-of-Plane Wall Design
ASD Interaction Diagram
8 in Wall, 21 ft high, #5 @ 16 in. Centered
20,000
P , lb per foot of length
15,000
10,000
5,000
0
0
10,000
20,000
30,000
40,000
50,000
60,000
Capacity
D + Lr
D + .6W:
D-.7E
D + 0.75 (.6)W + 0.75 Lr
D - 0.75 (0.7E)
0.6 D + .6W:
0.6 D +0.7 E:
-5,000
-10,000
M, lb-in per foot of length
Slide 36
Example Shear wall in a single story
Building Shear Wall Ex2
7
North Wall
8
VD1N2
2
1
East 2
Wall 2
West wall
North Wall 1
East Wall 2
Diaphragm
1
North Wall
A
1
Plan of Typical Big
Box - single story
Flexible diaphragm
VD1N1
VD1E2
VD1E1
East
Wall 1
East Wall 1
West
Wall
West Wall
Diaphragm
2
Diaphragm2
See MDG for Load
determination and
distribution to shear
wall lines - Flex
Diaphragm – SDCD
E
VD2S
South Wall
37
VD2W
VD2E1
VD2N1
Look at Shear Wall Design
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.
38
Shear Wall Loads Distribution
Segments get load w.r.t. relative k
East Wall 2
1
2
8
5
6
7
4
160.8 kips
Diaphragm
Shear
due to
seismic
1
2
3
West Wall Elevation
1
2
3
4
5
7
6
8
9
East Wall Elevation
1
2
3
4
5
7
6
8
9
10
South Wall Elevation
North Wall 2
10
9
8
7
6
5
4
3
2
North Wall Elevation
39
1
Shear Wall Loads Distribution
Segments get load w.r.t relative k
•
For Cantilevered Shear wall segments
•
  h' 
 h ' 
k c  Em t 4   3 
  lw 
 lw 
For Fixed-Fixed Shear wall segments
1
3
 h'   h' 
kc  Emt    3 
 lw   lw 
3
1
40
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)
Segment
h
l
1
22
12
2
22
24
3
22
24
4
22
24
5
22
24
6
22
24
7
22
24
8
22
6.67
Sum
Ri
0.332
1.715
1.715
1.715
1.715
1.715
1.715
0.065
10.687
Vd = 160.8 kips
Vi from Diaphragm (lb)
Vi wt (lb)
4.99
4.05
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
0.98
2.25
160.800
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)
Segment
h
l
1
22
12
2
22
24
3
22
24
4
22
24
5
22
24
6
22
24
7
22
24
8
22
6.67
Sum
  22 
 22 
Ri  10  4   3 
 12 
  12 
3
Ri
0.332
1.715
1.715
1.715
1.715
1.715
1.715
0.065
10.687
1
Vd = 160.8 kips
Vi from Diaphragm (lb)
Vi wt (lb)
4.99
4.05
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
25.80
8.1
0.98
2.25
160.800
 1.715 
Vi  160.8  

 10.687 
42
Segment 2 Designed in later Example
Design of Reinforced Masonry (ASD)
In Plane Loading (shear Walls)
Axial Force
V
h
L
43
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
44
P-M Diagrams ASD-In Plane

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
45
x(L/2 – aL/3))
Reinforced Masonry Shear Walls - ASD
Axial Force ~ 0
V
h
L
V= base shear
P M = overturning moment
46
Reinforced Masonry Shear Walls - ASD
V

Flexure Only P = 0 on diagram
Multiple rebar locations
h
L
M = over turning moment
V= base shear
P- self weight only ignore
47
Reinforced Masonry Shear Walls – ASD
(P =0) Can use the singly reinforced equations
di – location to centroid of each bar
Tension
fsn/N<= Fs/N
Compression
d*– location centroid of all bars in tension
f*si/N
k*d*
fsi/N
fsi/N
fsi/N
fsi/N
fsi/N
fsi/N
fsi/N
Fs1/N
fsi/N
Fsc/N
Tsn= Fs As
Ti
Ti
Ti
Ti
Ti
L
Ti
Ti
Ti
T1
fm
Cm
V= base shear
M = over turning moment
48
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.
49
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))
50
Shear Wall Example 3
Geometry:
typical wall element:
25 ft – 4 in. total height
3 ft – 4 in. parapet
24 ft length between
control joints
8 in. CMU grouted solid:
80 psf dead
VD
25’-4”
VP
22’-0”
12’-8”
f’m = 1500 psi
24’-0”
51
Shear Wall Example 3
West Wall seismic load condition
Vdiaphragm= 25,800 lb acting 22 ft above foundation
Vpier = 8,100lb acting 12.7 ft above foundation
8 in. CMU grouted solid (maximum possible dead load)
Pbase = 80lb/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 ) = 21,600lb
M = 0.6 x 0 + 0.7 x (25,800 lb x 22ft + 8,100lb x 12.7ft)
= 469,000lb-ft = 5,630,000 lb in.
V = 0.6 x 0 + 0.7 x (25,800 lb + 8,100 lb) = 23,700 lb
52
Shear Wall Example 3
#5 bar (typ)
4”
8”
8”
24”
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)
53
Shear Wall Example 3
For 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.
d    12 in  (24 ft  12 in / ft )  12 in  276 in
We are using 3 #5 Bars but if needed an estimate of As can be
Assumedetermined
that j = 0.9 The
area jof=reinforcing
steel canmoment
be calculated
by required
assuming
.9 and applied
, Mas:
5,630,000 lb  in
M
M
5
,
630
,
000
lb 2 in
M  As Fs jd
 As req 'd 

 0.944 in 2 2
Fs jd 24,000 lb in  0.9  276
Asreq 

in 0.71in
Fs jd 32,000 psi  0.9  276in
Es
29000000
n

 21.48  21.5
Em 900(1500)
54
Shear Wall Example 3
Try 3 No. 5 bars, As = 3 × 0.31 in.2 = 0.93 in.2 Calculate j and k:
As
0.93 in 2
 
 0.000442
bd 7.63 in  276 in
n  21.5  0.000442  0.00950
k  2n  n   n  2  0.00950  (0.00950) 2  0.00950  0.129
2
k
0.129
j  1  1
 0.957
3
3
You need to get the stress at the centroid based on the extreme bar fs=Fs
276-kd
x 32,000  31,476 psi Should get third bar stress then S Moments but M ≈
280  kd
M  0.93 in 2  31,476 lb in 2  0.957  276 in  7,732,000 lb  in  5,630,000 lb  in ok
Check Masonry Compression Stresses
Fb  0.45 f' m  0.45  1500 psi  675 psi
fb 
M
5,630,000

 157 psi  0675 psi
2
2
0.5 jkbd
0.5(0.957)0.129(7.63)(276)
55
Shear Wall Example 3
Check Shear Stress
Assume no shear reinforcing and
thus
Fv  Fvm  Fvs  Fvm
 

P
 M  '
1
Fvm 
4

1
.
75
f

0
.
25
  m
2 
An
 Vd 

 5,630,000 
0
1



4  1.75
 1500  0.25

2 
An
 23,700  ( 276 ) 

 
 48.3 psi  Fv  (Conservatively)  2 f m'  2 1500  77.5 psi OK
56
Shear Wall Example 3
Check Shear Stress
Conservatively assume just face
shell bedded areas resist shear
V
23,700lb
fv 

 33.9 psi  48.3 psi OK
Anv
280(1.25)2
57
Shear Wall Design
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)
4”
8”
8”
24”
24”
Check Prescriptive Seismic Reinforcing
58
Requirements for Detailed Plain SWs
and SDC C: TMS 402 Section 7.3.2.3
roof connectors
@ 48 in. max oc
roof
diaphragm
#4 bar (min) within
16 in. of top of parapet
Top of Parapet
#4 bars around
openings
#4 bar (min)
within 8 in. of
corners &
ends of walls
24 in. or 40 db
past opening
#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
#4 bars @ 10 ft oc or W1.7 joint
reinforcement @ 16 in. oc
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
roof connectors
@ 48 in. max oc
roof
diaphragm
#4 bar (min) within
16 in. of top of parapet
Top of Parapet
#4 bars around 24 in. or 40 db
openings
past opening
#4 bar (min)
within 8 in. of
corners &
ends of walls
#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
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