S.E. Exam Review:Masonry Design
Mark McGinley502-852-4068
[emailprotected]
Distributionofthewebinarmaterialsoutsideofyoursiteisprohibited.ReproductionofthematerialsandpictureswithoutawrittenpermissionofthecopyrightholderisaviolationoftheU.S.law.
1. Review Basics (Slides 5 – 29) -�Code documents, standards, terms, units, mortar, loads,
analysis methods, basic behavior, URM
2. Reinforced Masonry Design – Beams and Walls for Flexure (Slides 30 – 44)
3. Reinforced Masonry Walls Combined Axial and Flexure –Analysis and Design (Slides 45 – 57)
4. Shear, Shear Wall Analysis and Design, Seismic Details (Slides 58 – 91)
Table of Contents
2
1. Vertical Forces Exam – Friday Breadth�Masonry, 3 out of 40 questions: flexural members,
compression members, bearing walls, detailing
2. Vertical Forces Exam – Friday Depth� 4 - 1 hour problems, will include a masonry structure
3. Lateral Forces Exam – Saturday Breadth�Masonry, 3 out of 40 questions: flexural-compression
members, slender walls, ordinary or intermediate shear walls, special shear walls, anchorages, attachments
4. Vertical Forces Exam – Saturday Depth� 4 - 1 hour problems, may include a masonry structure
NCEES Guide
3
Masonry Experience – feedback from sites “via chat”
Most at site have:
A. Little or none
B. A short course and/or a little design
C. Design simple buildings/elements
D. Design masonry routinely
E. Design masonry in sleep – a masonry wiz, etc.
Any masonry design experience? – Courses or Design
4
Assume you have access to MSJC – recommend MDG 2013
NCEES uses 2013 Code (TMS402/ACI 530/ASCE 5-13) and Specification – VERT. & LATERAL
5
TMS 402 2011 vs TMS 402 2013 VERT. & LATERAL
6
Ch. 1 - General Requirements
Ch. 2AllowableStressDesign
Ch. 3 Strength Design
Ch. 4PrestressedMasonry
Ch. 5EmpiricalDesign
Ch. 6Veneer
Ch. 7Glass Block
2.1 - General ASD 2.2 - URM2.3 - RM
6.1 - General6.2 - Anchored6.3 - Adhered
3.1 - General SD 3.2 - URM3.3 - RM
MSJCTMS 602
Ch. 8 -AAC
2011 TMS 402 also included a new Appendix for Design of Masonry Infill
2011 TMS 402
2013 TMS 402 (Reorganized) – VERT. & LATERAL
7
Part 1: General
Chapter 1 –General
Requirements
Chapter 2 –Notations & Definitions
Chapter 3 –Quality &
Construction
Part 2: Design Requirements
Chapter 4: General Analysis & Design
Considerations
Chapter 5: Structural Elements
Chapter 6: Reinforcement,
Metal Accessories & Anchor Bolts
Chapter 7: Seismic Design
Requirements
Part 3: Engineered
Design Methods
Chapter 8: ASD
Chapter 9: SD
Chapter 10: Prestressed
Chapter 11: AAC
Part 4: Prescriptive
Design Methods
Chapter 12: Veneer
Chapter 13: Glass Unit Masonry
Chapter 14: Masonry Partition
Walls
Part 5: Appendices &
References
Appendix A –Empirical Design
of Masonry
Appendix B: Design of Masonry
infill
Appendix C: Limit Design of Masonry
References
2013 TMS 602 – VERT. & LATERAL
8
TMS 402 Code
Part 1General
Part 2Products
Part 3Execution
1.6 Qualityassurance
3.1 - Inspection3.2 - Preparation3.3 – Masonry erection3.4 – Reinforcement3.5 – Grout placement3.6 – Prestressing3.7 – Field quality control3.8 - Cleaning
2.1 - Mortar 2.2 - Grout2.3 – Masonry Units2.4 – Reinforcement2.5 – Accessories2.6 – Mixing2.7 - Fabrication
TMS 602Specification
Bond Patterns VERT. & LATERAL
9
Running Bond Stack Bond
1/3 Running Bond Flemish Bond
bedjoints
head joints
� Concrete masonry units (CMU) usually hollow & 8 x 16 x (8 or 10 or 12)� specified by ASTM C 90� minimum specified compressive
strength (net area) of 1900 psi (Ave). New ASTM – 2000 psi (2011)
� net area is about 55% of gross area� nominal versus specified versus actual
dimensions� Type I and Type II designations no
longer exist
� Also, Concrete Brick – ASTM C 55
� Most masonry modules 8”
� See TEK NOTE 2-1A www.ncma.org (online resources)
Concrete Masonry Units VERT. & LATERAL
10
� ASTM C 62 or C 216 or C 652 (hollow)
� Usually solid, with small core holes for manufacturing purposes
� If cores occupy ≤ 25% of net area, units can be considered 100% solid
� Bia tech Note 10 B – see www.bia.org
Clay Masonry Units VERT. & LATERAL
11
�ASTM C 270 – mortar for unit masonry
� Three systems�Portland-cement-lime
mortar (PCL)�Masonry cement mortar�Mortar cement mortar� Two ways to spec –
proportion and property � Then have 4 types
Masonry Mortar VERT. & LATERAL
12
�Mortar Type (MaSoN wOrK)
�Going from Type K to Type M – more Portland cement; higher compressive and tensile bond strengths, stiffer.
� Types N and S are specified for modern masonry construction.
Masonry Mortar VERT. & LATERAL
13
�Reinforcing bars in grout; joint reinforcement (ties) embedded in mortar
�Usually center placement of reinforcement
�Protection – in code
�Hooks – in code
Reinforcement – Code Ch. 6 VERT. & LATERAL
14
�Concrete�Designer states assumed value of f’c�Compliance is verified by compression tests on cylinders
cast in the field and cured under ideal conditions
�Masonry�Designer states assumed value of f’m�Compliance is verified by “Unit Strength Method” or by
“Prism Test Method”
Role of f’m VERT. & LATERAL
15
�Unit strength method (Spec 1.4 B 2)� Compressive strengths from unit manufacturer� ASTM C 270 mortar� Grout meeting ASTM C 476 - min. f’m - 2,000 psi
�Prism test method (Spec 1.4 B 3)� Pro: can permit optimization of materials� Con: requires testing, qualified testing lab, and procedures in
case of non-complying results
Verify Compliance with Specified f’m VERT. & LATERAL
16
Concrete masonry units (Table 2)unit compressive strength ≥ 2,000Type S mortarf’m can be taken as 2,000 psi
Unit Strength Method VERT. & LATERAL
17
�
�
� or or
� 0.75 0.75 or or
� 0.6 or0.7
� 0.75 0.6 0.75 0.75 or or
� 0.75 0.7 0.75 0.75
� 0.6 0.6
� 0.6 0.7
� No increase for E or W any more with Stress Recalibration – even with alternative load cases
ASD Load Combinations – IBC 2015/ASCE 7-10 VERT. & LATERAL
18
�Unreinforced vs. Reinforced masonry
�Unreinforced masonry: masonry resists flexural tension, reinforcement is neglected
�Reinforced masonry: masonry in flexural tension neglected, reinforcement resists all tension
General Structural Analysis and Design VERT. & LATERAL
19
Design Methods
�ASD – applied stresses service loads ≤ allowed stresses� f ≤ F Ch 8
�Strength – Ch 9
� Factored load effects ≤ factored resistance�
General Structural Analysis and Design VERT. & LATERAL
20
�Specified masonry compressive strength, f’m�Compressive strength of masonry units�Mortar type
�Bond pattern
�Unit type – hollow or solid
�Extent of grouting
�Slenderness
� Type of stress – flexure, tension, compression, shear, etc.
Allowable Stresses (ASD) Depend On - VERT. & LATERAL
21
� Load distribution and deformation – elastic analysis based on uncracked sections, except beam defl. (Ieff was in Commentary now in Section 5.2 for beams)
�Member stresses and actions – calculated on minimum critical sections (reinforced – cracked). Section 4.3
�Member stiffness calculated based on average sections.
� For CMU – See Tek Note 14-1B Section Properties (www.ncma.org)
General Analysis Considerations VERT. & LATERAL
22
�Chord modulus of elasticity� 700 f’m for clay masonry� 900 f’m for concrete masonry
� Thermal expansion coefficients for clay and concrete masonry
�Moisture expansion coefficient for clay masonry
�Creep coefficients for clay and concrete masonry
Material Properties Code – 4.2 VERT. & LATERAL
23
�Masonry can have more than one wythe (thickness)
�Multiwythe walls may be designed for:�Composite action or noncomposite action
�Composite action requires that collar joints be:�Crossed by connecting headers, or filled with mortar or
grout and connected by ties
�Code 5.1.4.2 and 8.1.4.2 (ASD) limits shear stresses on collar joints or headers – 5 psi for mortar, 13 psi for grout, (Header strength)1/2
Composite vs. Noncomposite Construction VERT. & LATERAL
24
Assumed stress distribution in multiwythe composite walls
Stresses – Composite Action, Code Commentary Fig. CC-5.1-6 VERT. & LATERAL
25
collar joint filled
lateralload
lateralload
Com
p.
Tens
.
Horizontal Bendingtension parallel to bed joints
Vertical Bendingtension normal to bed joints
�Horizontal in-plane loads and gravity loads resisted to wythe applied to only
�Weak-axis bending moments are distributed to each wythein proportion to flexural stiffness
If Not a Composite Multiwythe Masonry Wall VERT. & LATERAL
26
Assumed stress distribution in multiwythe noncomposite walls
Stresses with Noncomposite Action, Code Commentary Fig. CC-5.1-8 VERT. & LATERAL
27
Com
p.
Tens
.
Tens
.C
omp.
collar joint open
lateralload
lateralload
Horizontal Bendingtension parallel to bed joints
Vertical Bendingtension normal to bed joints
Assumptions (Stresses on net section) – = , =
�Net flexural tension stress limited - Table 8.2.1.4
Ch. 8.2 in MSJC-ASD URM Masonry VERT. & LATERAL
28
�Compression stress limited , 1/3
0.25 ′ 1 for 99
0.25 ′ for 99 and
� P Pe 1 0.577
� Force unity equation – + 1
�Shear , = 1.5 , 120 psi, or 37 psi +0.45 , or
60 psi +0.45 , or 15 psi
Ch. 8.2 in MSJC-ASD URM Masonry VERT. & LATERAL
29
Assumptions
�Masonry in flexural tension is cracked
�Reinforcing steel is needed to resist tension
� Linear elastic theory
�No min. required steel area except columns
�Wire joint reinforcement can be used as flexural reinforcement
�No unity or interaction equation – use interaction curves
Ch. 8.3 in MSJC-ASD Reinforced Masonry VERT. & LATERAL
30
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 5.3.1.4) can be used to resist compression
�Allowable compressive stress = allowable tension stress if tied.
Allowable Stresses Steel VERT. & LATERAL
31
�ASD reinforced allowable compressive capacity is expressed in terms of force rather than stress
�Allowable capacity Σ(masonry + tied compressive reinforcement)
�Max. compressive stress in masonry from axial load & bending ≤ (0.45)f’m
�Axial compressive stress must not exceed allowable axial stress from Code 8.2.4.1
Allowable Axial Compression VERT. & LATERAL
32
Code equations (8-21) and (8-22) slenderness reduction factors are the same as unreinforced masonry.
0.25 ′ 0.65 1 for 99
0.25 ′ 0.65 for 99
Allowable Axial Compressive Capacity -VERT. & LATERAL
33
Code 5.3.1.4:
a) Longitudinal reinforcement – enclosed by lateral ties at least ¼ in dia.
b) Vertical spacing of ties ≤ 16 db, 48 dties, or least cross-sectional dimension of the member.
c) Lateral ties are required to enclose bar, max. 6 in along tie between bars, have splices and included angle of <135⁰. Can be in mortar.
d) ½ spacing at top and bottom.
e) terminated within 3” of beams
Axial Compression in Bars Can be Accounted for Only If Tied As: - VERT. & LATERAL
34
For running-bond masonry, or masonry with bond beams spaced no more than 48 in. center-to-center, the width of masonry in compression per bar for stress calculations less than or = to:
�Center-to-center bar spacing
�Six times the wall thickness (nominal)
� 72 in.
Now in Code 5.1.2
Amount of Masonry Effective Around Each Bar Is Limited by Code VERT. & LATERAL
35
�Span = clear span plus depth ≤ than distance between support centers
�Minimum bearing length = 4 in.
� Lateral support on beam compression face at a maximum spacing of 32 times the beam thickness (nominal) or 120b2/d (smaller of these).
�Must meet deflection limits of Code 5.2.1.4 – gives Ieff and lets you ignore deflection for Span ≤ 8d
Code Section 5.2 Beams Only - VERT
36
/ and from equil.(at the limit)
(at limit)
ASD Reinforced Masonry – Singly ReinforcedVERT. & LATERAL
37
21 /3
C
M
V
T
jdf bdxv
d
kd
/ /
Given a lintel over a door in a 8 CMU wall:
Max moment = 493.3 kip.in V = 8 kips
Assume by tests f’m = 2,000 psi
� Assume that the beam is sized for shear
� Try four courses h ~ 32
� ⁄ ,, . .
0.62
� Try 2 – #5 rebar As = 0.62 in2
�, ,
,16.11
Example Design Masonry for Flexure (ASD) - VERT
38
4-8” CMU’s d = 27.8”
7.63”
Guess j = 0.9
Check design.
. .0.00292
2 / 0.2635, 1 0.2635 3⁄ 0.912
=0.62 32,000 0.912 27.8 503 . 493.3
.. ,
. . . . .692.2psi 0.45 2,000 900psi
or
900 7.63 0.263 0.912 27.8
636.5 · 503 – steel stress governs
� Problem types – could ask you to calculate moment capacity, select the number of #5 bars needed to resist load, etc.
Lintel Design - VERT
39
Try a reinforced 8” CMU f’m = 1,500 psi, Grade 60 rebar
.6D +0.6W governs at mid-height – start with 1 ft design width of wall
0.6 33.33 13.5 /8
455.6lb · ft per ft of wall
Design Masonry Wall Flexure Out-of-Plane - LATERAL
40
W = 33.33 psf13.5 ft
Per foot of wall – assume 0.9 and 2⁄ 7.625 2⁄ 3.81“
⁄ 455.6 12 32,000 0.9 3.81⁄ 0.050in
Try #5 rebar at 56 OC ” 0.31in
ft⁄ ~0.066 in ft⁄ (based on 56/12)
Check section – effective width = 6t = 48, or s = 56 or 72
⁄ 0.31 48 3.81⁄ 0.001695,
290,00,000 900 1,500⁄ 21.48
Design Masonry Wall Flexure Out-of-Plane - LATERAL
41
2 ⁄ 0.236, 1 3⁄ 0.921
0.31 0.921 3.81 32,000 12⁄ 2,901lb · ft
0.45 0.45 1,500 675psi
1 2⁄ 0.921 0.236 48 3.81 675 12⁄ 4,259lb · ft
governs since 2,901 is less than 4,259 and is greater than the applied moment 455.6 56 12⁄ 2,126lb · ft
Design Masonry Wall Flexure Out-of-Plane LATERAL
42
Check depth of Neutral Axis:
0.236 3.81 0.90 the face shell of a 8 CMU is 1.25”
So compression stresses are in face shell and partial grouting is possible without recalculation.
Use #5 at 56” OC
Design Masonry Wall Flexure Out-of-Plane - LATERAL
43
t
tf
d
As
bkdb = 6t or 72” or s*
As per width b
flange
If Kd > face-shell for partial grouted section, you would need to sum the moment produced by each couple or just limit the moment to the flange stresses.
Design Masonry Wall Flexure Out-of-Plane – Partial Grouting - LATERAL
44
t
tf
d
kdb = 6t or 72” or s*
As
bb’
tf
Asf Asf = As - Asf
flange web
� 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
Allowable Stress Interaction Diagrams – Flexural-Compression Members - VERT. & LATERAL
45
�Assume single reinforced
�Out-of-plane flexure
�Grout and masonry the same
�Solid grouted
�Steel in center
Allowable Stress Interaction Diagrams VERT. & LATERAL
46
CL
M
P
�Allowable – stress interaction diagram
� Linear elastic theory – tension in masonry it is ignored, plane sections remain plane
� Limit combined compression stress to 0.45
�
� d usually = t/2 – no compression steel since not tied, ignore in compression
Allowable Stress Interaction Diagrams Walls – Singly Reinforced VERT. & LATERAL
47
Assume stress gradient range A:
All sections in compression
Get equivalent force-couple about center line
0.5
/2 , /6
Note at limit – and (set )
Note much of this is from Masonry Course notes by Dan Abrams
Allowable Stress Interaction Diagrams Walls – Singly Reinforced - VERT. & LATERAL
48
Pa
Mb
d = t/2
fm2fm1
em
Cm
b eff
Assume stress gradient range B:
Not all section in compression, but no tension in steel
Get equivalent force-couple about center line
0.5
2⁄ 3⁄
Note that
This is valid until steel goes into tension
Set at limit
Allowable Stress Interaction Diagrams Walls – Singly Reinforced VERT. & LATERAL
49
Pb
Mb
d = t/2
fm1
em
Cm
b eff
αt
Assume stress gradient range C:
Section in compression, tension in steel
Get equivalent force-couple about center line
2⁄ 3⁄
0.5
and
From similar triangles on stress diagram
⁄ /
2⁄ ; note that 2⁄usually, so second term goes to zero
At limit and or and and the other governs – balance point
when both occur.
Note that
Allowable Stress Interaction Diagrams Walls – Singly Reinforced VERT. & LATERAL
50
Pc
Mc
d = t/2
fm1
em
Cm
b eff
αt fs/n
Ts
Allowable Stress Interaction Diagrams Walls – Singly Reinforced VERT. & LATERAL
51
Axial Load P
Moment M
Capacity envelop letting fm1 = Fb
Range B
Range A
Capacity envelope letting fm1 = Fb
Range CCapacity envelope letting fs = Fs
P cut off Eq 8-21 or 8-22(slide32)
Ms Mm
Can get a three point interaction diagram easily
Most walls have low axial loads
No Good Above P Cut Off
Construct the interaction diagram for a solidly grouted 8” CMU wall, f’m = 1,500 psi, with height 16.67 ft and grade 60 #5 rebar at 16” OC. Also, see if the wall is adequate for the loads below. Assume pinned top and bottom of the wall.
ASD Interaction Diagram Walls – Singly Reinforced Example - VERT. & LATERAL
52
P(kip)
M(k*in)
D + 0.75L + 0.75(0.6W) at mid-height 2.072 9.204
D + L at top 2 5.50.6D + 0.6W at mid-height 0.642667 13.33
ASD Interaction Diagram - VERT. & LATERAL
53
Spreadsheet for calculating allowable-stress M-N diagram for solid masonry wall –center rebar
16.67 ft. wall w/ No. 5 at 16 in. (centered)total depth, t 7.625 in. Wall Height, h 16.67 feet
f'm, 1,500 psi Radius of Gyration, r 2.20 in.
Em 1,350,000 psi h/r 90.9
Fb 675.00 psi Reduction Factor, R 0.578
Es 29,000,000 psi Allowable Axial Stress, Fa 217 psi
Fs 32,000 psi Net Area, An 121.7 in.2
d 3.81 in. Allowable Axial Compr, Pa
26384 lb/ft (MSJC 8.3.4.2.1)
Kbalanced 0.311828 Note: axial force is per foot of wall
tensile reinf., As/beff
0.31 #5 @ 16 centered
width, beff 16 in.
ASD Interaction Diagram - VERT. & LATERAL
54
Note: moment equation not valid after k > 2
kKd�Dt)
fb(psi)
Cmas(lb)
fs(psi)
Axial Force(lb)/ft
Moment(lb-in)/ft
Axial Forcew/ force limit/ft
RANGE C Points controlled by steel 0.01 0.04 15 5 -32,000 -7,437 -6 -7,4370.05 0.19 78 119 -32,000 -7,350 317 -7,350
0.1 0.38 166 504 -32,000 -7,062 1,376 -7,0620.15 0.57 263 1,202 -32,000 -6,539 3,246 -6,5390.24 0.91 470 3,441 -32,000 -4,859 9,034 -4,8590.22 0.84 420 2,817 -32,000 -5,327 7,447 -5,3270.24 0.91 470 3,441 -32,000 -4,859 9,034 -4,859
0.3 1.14 638 5,838 -32,000 -3,062 15,006 -3,0620.311828 1.19 675 6,416 -32,000 -2,628 16,420 -2,628
0.4 1.52 675 8,230 -21,750 1,115 20,383 1,1150.5 1.91 675 10,287 -14,500 4,344 24,507 4,3440.6 2.29 675 12,344 -9,667 7,011 28,237 7,0110.7 2.67 675 14,402 -6,214 9,357 31,574 9,357
Points controlled by masonry 0.8 3.05 675 16,459 -3,625 11,502 34,519 11,502
0.9 3.43 675 18,517 -1,611 13,513 37,072 13,5131.1 4.19 675 22,631 0 16,974 41,000 16,9741.3 4.95 675 26,746 0 20,060 43,359 20,0601.5 5.72 675 30,861 0 23,146 44,151 23,1461.7 6.48 675 34,976 0 26,232 43,374 26,232
RANGE B 2 7.62 675 41,148 0 30,861 39,271 30,861RANGE A Pure compression 675 82,350 0 82,350 0 82,350
Axial Force Limits 26,384 0 26,38426,384 44,151 26,384
⁄ /32,000 21.48⁄ 0.1 3.81 3.81 0.1 3.81⁄
K balanced
Fs/n kdfb
d
ASD Interaction Diagram Walls – Singly Reinforced Example - VERT. & LATERAL
55
Possible Answers:
A. Nothing
B. The entire capacity curve would shift up and to the right.
C. The moment capacity governed by steel stress would increase, but this would not increase the wall capacity.
D. The lower section of the curve would shift to the right (increase M).
What would happen to previous problem if I increased the steel size? - VERT. & LATERAL
56
a) Given a non load bearing wall with out-of-plane loading (wall size and f’m). Size rebar placed in center of wall. Assume steel governs, j = 0.9, and set Ms = Mmax.Applied. Find As. Check Mm. Iterate if needed.
b) Given a wall configuration – size of units, rebar location and size, etc. Find max moment capacity. Get smaller of Mm or Ms.
c) Given a wall configuration – size of units, rebar location and size, etc. Find axial load capacity. Eq. 8-21 or 8-22 (slide 32)
Possible Wall Breadth Exam Questions - VERT. & LATERAL
57
No shear reinforcing masonry resists all shear.
applied shear stress
varies with type of element
ASD – Reinforced Masonry – Shear - VERT. & LATERAL
58
dx
C C + dC
M M + dM
V V + dV
T T + dT
µdb dx = bond force
jdfvbdx
�Shear stress is computed as:
(8-24)
�Allowable shear stresses
(8-25)
0.75 for partially grouted shear walls, 1.0 otherwise.
Reinforced Masonry Shear Stresses - VERT. & LATERAL
59
�Allowable shear stress limits:� ⁄ 0.25
3 (8-26)� ⁄ 1
2 (8-27)�Can linear interpolate between limits
5 2
Shear Stress Cutoffs - VERT. & LATERAL
60
�Allowable shear stress resisted by the masonry�Special reinforced masonry shear walls
4 1.75 0.25 (8-28)
�All other masonry
4 1.75 0.25 (8-29)
⁄ is positive and need not exceed 1.0.
Shear Stresses - VERT. & LATERAL
61
� If allowable shear stress in the masonry is exceeded, then:�Design shear reinforcement using Equation 8-30 and add
to
0.5 (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.
If Shear Reinforcement Present VERT. & LATERAL
62
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 - LATERAL
63
Shear Wall in a Single Story Building – Shear Wall Example 1 - LATERAL
64
Lateral Loads Breadth or Depth
Plan of typical big box single story flexible diaphragm
See MDG for load determination and distribution to shear wall lines – Flex Diaphragm – SDC-D
E
Diaphragm 2West Wall
West Wall 2
South Wall
North Wall
VD1E2 VD1E1
East Wall 2
VD1N1 VD2N1
VD1N2
East Wall 1
A1
VD2W VD2E1
VD2S
Shear Wall Loads Distribution - LATERAL
65
Segments get load w.r.t. relative k
160.8 kips
Diaphragm shear due to seismic
Segments get load w.r.t. relative k.
� For cantilevered shear wall segments
4 3
� For fixed-fixed shear wall segments
3
Shear Wall Loads Distribution - LATERAL
66
Shear Wall Load Distribution - LATERAL
67
Table 18.1-2 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
Shear Wall Load Distribution - LATERAL
68
Table 18.1-2 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
10 4 3160.8
1.71510.687
Segment 2 designed in later example
Design of Reinforced Masonry (ASD) in Plane Loading (Shear Walls) - LATERAL
69
h
V
L
Axial Force
�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
ASD Design of Reinforced Masonry – In Plane Loading (Shear Walls) - LATERAL
70
� Initially assume 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 as before until rebars start to go into tension. Note that
� Determine from similar triangles & get
� Check extreme and
� (or when )
� M capacity ∑ ∑ 2⁄ 2⁄ 3⁄
P-M Diagrams ASD-In Plane - LATERAL
71
Reinforced Masonry Shear Walls – ASD - LATERAL
72
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 - LATERAL
73
(P = 0) Can use the singly reinforced equations
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 = FsAs Ti Ti Ti Ti Ti Ti Ti T1
Tension Compression
fsi/N fsi/N fsi/N fsi/N fsi/N fsi/Nfsi/N
d* – location centroid of all bars in tension f*si/N
� To locate neutral axis, guess how many bars on tension side – As*
� Find d* (centroid of tension bars) and ∗ / ∗
�Get ∗ ∗ 2 ∗ ⁄ ∗
�Unless tied, ignore compression in steel.
Moment Only ASD in Plane - LATERAL
74
�Check k*d* to ensure assume tension bars correct – iterate if not
�Determine fsi from similar triangles and then Ti = (fsi xAi)
�M capacity ∑about ∑ ∗ ∗ 3⁄
Moment Only ASD in Plane LATERAL
75
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
1,500psi
Shear Wall Example 2 - LATERAL
76
VD
VP22’-0”
25’-4”
12’-8”
24’-0”
West wall seismic load condition
25,800lb acting 22 ft above foundation
8,100lb acting 12.7 ft above foundation
8 in. CMU grouted solid (maximum possible dead load)
80 lb ft⁄ 253ft 24ft 48,600lb
Vertical seismic: 0.2 0.2 1.11 48,600 10.800lb
ASD Load Combination: 0.6 0.7
0.6 48,600 0.7 0.2 1.11 48,600 21,600lb
0.6 0 0.7 25,800lb 22ft 8,100lb 12.7ft
469,000lb · ft 5,630,000lbin.
0.6 0 0.7 25,800lb 8,100lb 23,700lb
Shear Wall Example 2 - LATERAL
77
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)
Shear Wall Example 2 - LATERAL
78
#5 bar (typ)
24”4” 8” 8” 24”
For the 24 ft long wall panel between control joints subjected to in-plane loading, the flexural depth, d*, is the wall length less the distance to the centroid of the vertical steel at the ends of the wall.
∗ ℓ 12in 24ft 12 in ft⁄ 12in 276in
We are using three #5 bars, but if needed, an estimate of can be determined by assuming 0.9 and applied moment, M.
∗, ,
, . 0.71in
, ,,
21.48 21.5
Shear Wall Example 2 - LATERAL
79
Try three No. 5 bars, 3 0.31 0.93 . Calculate j and k:
∗∗
. .
0.000442 ∗ 21.5 0.000442
∗ 2 ∗ ∗ ∗ 2 0.00950 0.00950 0.00950 0.129
1 1 . 0.957 and ∗ ∗ 35.6in. Don’t need to check since other bars not lapped
You need to get the stress at the centroid based on the extreme bar
∗ ∗∗ ∗
32,000 30,970psi. Should get third bar stress then ∑moments but
0.93in 30,970 lb in⁄ 0.957 276in 7,608,000lb · in 5,630,000lb · in OK
Check masonry compression stresses
0.45 0.45 1,500psi 675psi
. ∗ ∗, ,
. . . .157psi 675psi
Shear Wall Example 2 - LATERAL
80
Check shear stress.
Assume no shear reinforcing, and thus:
4 1.75 0.25
4 1.75 , ,,
1,500 0.25
48.3psi J 0.75 48.3 36.2 conservatively 2
2 1,500 77.5psi OK
Shear Wall Example 2 - LATERAL
81
Check shear stress.
Conservatively assume just face shell bedded areas resist shear.
, .
33.9psi 36.2psi OK
Shear Wall Example 2 - LATERAL
82
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.
Shear Wall Example 2 - LATERAL
83
#5 bar (typ)
24”4” 8” 8” 24”
a) Given a diaphragm shear line load, determine the critical shear and overturning moment on a shear wall segment. SW Ex1.
b) Given a shear wall segment size and rebar config., find max. diaphragm shear at top of wall. SW Ex2 – just back calculate V after setting applied stresses = allowable stress values. Look at both shear and flexure, take lowest resulting V.
c) Given a SW segment loading, wall size, and rebar location, select size of bars needed. SW Ex 2. Flexure only – assume governs. Find , check .
Possible Breadth Exam Problems - LATERAL
84
�Define Seismic Design Category ASCE 7
�SDC determines�Required types of shear walls�Prescriptive reinforcement for other masonry elements
(non-participating walls must be isolated)� Type of design allowed for lat. force resisting system –
note, for special shear walls, 1.5 .
Seismic Detailing Code – Ch. 7 - LATERAL
85
Minimum Reinf., SW Types, etc. – Cumulative -LATERAL
86
SW Type Minimum Reinforcement SDC
Empirically Designed None – drift limits and connection force A
Ordinary Plain None – same as A A, B
Detailed Plain
Vertical reinforcement = 0.2 in2 at corners, within 16 in. of openings, within 8 in. of movement joints, maximum spacing 10 ft; horizontal reinforcement
W1.7 @ 16 in. or #4 in bond beams @ 10 ft
A, B
Ordinary Reinforced Same as above A, B, C
Intermediate Reinforced Same as above, but vertical reinforcement @ 4 ft A, B, C
Special Reinforced
Same as above, but horizontal reinforcement @ 4 ft, and U = 0.002 – no stack bond any
Minimum reinforcement for detailed plain shear walls and SDC C - LATERAL
87
#4 bar (min) within 8 in. of corners & ends of walls
roofdiaphragm
roof connectorsAs per IBC or ASCE 7IBC – 4 ft usual
#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 2 leg W1.7 joint reinforcement @ 16 in. oc
#4 bars @ 10 ft oc & within16 in. of openings
24 in. or 40 dbpast opening
#4 bars around openings
�Seismic Design Category D�Masonry – part of lateral force-resisting system must be
reinforced so that 0.002, and and 0.0007
� Type N mortar & masonry cement mortars are prohibited in the lateral force-resisting system, except for fully grouted.
�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
MSJC 7.4 - LATERAL
88
Minimum Reinforcement for Special Reinforced Shear Walls – Running Bond - LATERAL
89
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 (min) @ smallest of 4 ft, L/3, or H/3 (int. SW just 4 ft)
#4 bar (min) within 8 in. of corners & ends of walls
24 in. or 40 db past opening
#4 bars around openings
#4 bars min @ smallest of4 ft, L/3, or H/3 (int. SW just 4 ft)Hook to vert.
Asvert > 1/3 AsvAsvert + Asv ≥ .002AgAsvert or Asv ≥ .007Ag
a) Select type of shear wall for a given SDC
b) Define prescriptive detailing requirements needed for a specific shear wall type and SDC.
Possible Breadth Exam Problems - LATERAL
90
Thank you for your attention!
Any questions?
Conclusion
91