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Unformatted text preview: l.l7.4.3 l l BUILDING CODE REQUIREMENTS FOR MASONRY STRUCTURES C37 CHAPTER 3
STRENGTH DESIGN OF MASONRY 3.1 —— General 3.1.1 Scope This Chapter provides minimum requirements for
strength design of masonry. Masonry design by the
strength design method shall comply with the
requirements of Chapter 1, Section 3.1, and either
Section 3.2 or 3.3. 3.1.2 Required strength Required strength shall be determined in accordance
with the strength design load combinations of the legally
adopted building code. When the legally adopted
building code does not provide factored load
combinations, structures and members shall be designed
to resist the combination of loads speciﬁed in ASCE 7
for strength design. Members subject to compressive
axial load shall be designed for the factored moment
accompanying the factored axial load. The factored
moment, Mu, shall include the moment induced by
relative lateral displacement. 3.1.3 Design strength Masonry members shall be proportioned so that the
design strength equals or exceeds the required strength.
Design strength is the nominal strength multiplied by the strengthreduction factor, ¢, as specified in Section 3.1.4. 3.1.4 Strengthreduction factors
3.1.4.1 Combinations of flexure and axial load in reinforced masonry — The value of ¢ shall be taken
as 0.90 for reinforced masonry subjected to ﬂexure, axial
load, or combinations thereof. 3.1.4.2 Combinations of flexure and axial load in unreinforced masonry — The value of 415 shall be taken
as 0.60 for unreinforced masonry subjected to ﬂexure,
axial load, or combinations thereof. 3.1.4.3 Shear — The value of ¢ shall be taken
as 0.80 for masonry subjected to shear. 3.1.4.4 Anchor bolts —— For cases where the
nominal strength of an anchor bolt is controlled by
masonry breakout, by masonry crushing, or by anchor
bolt pryout, ¢ shall be taken as 0.50. For cases where the
nominal strength of an anchor bolt is controlled by anchor bolt steel, ¢ shall be taken as 0.90. For cases
Where the nominal strength of an anchor bolt is controlled by anchor pullout, ¢ shall be taken as 0.65. 3.1.4.5 Bearing — For cases
bearing on masonry, ¢ shall be taken as 0.60. involving 3.1.5 Deformation requirements
3.1.5.1 Deﬂection of unreinforced (plain) 1.17.2_4
masonry — Deﬂection calculations for unreinforced 
(plain) masonry members shall be based on uncracked section properties. 3.1.5.2 Deﬂection of reinforced masonry —
Deﬂection calculations for reinforced masonry members
shall consider the effects of cracking and reinforcement I
on member stiffness. The ﬂexural and shear stiffness
properties assumed for deﬂection calculations shall not
exceed onehalf of the gross section properties, unless a
cracked—section analysis is performed. 3.1.6 Anchor bolts embedded in grout m
3.1.6.1 Design requirements — Anchor bolts
shall be designed using either the provisions of 3.1.6.2
or, for headed and bentbar anchor bolts, by the
provisions of Section 3.1.6.3. 3.1.6.2 Nominal strengths determined by test
3.1.6.2.1 Anchor bolts shall be tested in
accordance with ASTM E488, except that a minimum of
ﬁve tests shall be performed. Loading conditions of the test
shall be representative of intended use of the anchor bolt. 3.1.6.2.2 Anchor bolt nominal strengths
used for design shall not exceed 65 percent of the
average failure load from the tests. 3.1.6.3 Nominal strengths determined by
calculation for headed and bent—bar anchor bolts —
Nominal strengths of headed and bent—bar anchor bolts
embedded in grout shall be determined in accordance with
the provisions of Sections 3.1.6.3.1 through 3.1.6.3.3. 3.1.6.3.1 Nominal tensile strength of
headed and bent—bar anchor bolts — The nominal axial
tensile strength of headed anchor bolts shall be computed
using the provisions of Sections 3.1.6.311. The nominal
axial tensile strength of bentbar anchor bolts shall be
computed using the provisions of Section 3.1.6.3.1.2. 3.1.6.3.1.1 Nominal axial tensile
strength of headed anchor bolts — The nominal axial tensile
strength, Ban, of headed anchor bolts embedded in grout shall
be determined by Eq. (3—1) (nominal axial tensile strength
governed by masonry breakout) or Eq. (32) (nominal axial
tensile strength governed by steel yielding). The nominal
axial tensile strength, BM, shall be the smaller of the values
obtained from Eqs. (31) and (32). Banb : 4Apt Bans : Abfy (31) (32) C38 1. 3.1.6.3.1.2 Nominal axial tensile
strength of bentbar anchor bolts — The nominal axial
tensile strength, Ban, for bentbar anchor bolts embedded
in grout shall be determined by Eq. (33) (nominal axial
tensile strength governed by masonry breakout), Eq. (3—
4) (nominal axial tensile strength governed by anchor
bolt pullout), or Eq. (3—5) (nominal axial tensile strength
governed by steel yielding). The nominal axial tensile
strength, Ban, shall be the smallest of the values obtained
from Eqs. (33), (34) and (35). B... =4Apn/E <3—3) B = 1.5 fjebdb + 300n(1,,+ e, + db)db (34) anp Ban. = Abfy (35) 3.1.6.3.2 Nominal shear strength of
headed and bentbar anchor bolts — The nominal shear
strength, an, of headed and bentbar anchor bolts shall
be determined by Eq. (36) (nominal shear strength
governed by masonry breakout), Eq. (3—7) (nominal
shear strength governed by masonry crushing), Eq. (38)
(nominal shear strength governed by anchor bolt pryout)
or Eq. (3—9) (nominal shear strength governed by steel
yielding). The nominal shear strength BM, shall be the
smallest of the values obtained from Eqs. (36), (37),
(38) and (3—9). anb : 4Apv BM 2 10504 f'm Ab (37)
3W = 208m = 8A.. 12'. (38)
am = 0.6Abfy (39) 3.1.6.3.3 Combined axial tension and
shear — Anchor bolts subjected to axial tension in
combination with shear shall satisfy Eq. (310).
b b
“f + V :1 (3—10)
¢ Ban ¢an TMS 40208/ACI 53008/ASCE 508 3.1.7 Nominal bearing strength The nominal bearing strength of masonry shall be
computed as 0.60 f ’m multiplied by the bearing area, Abr,
as deﬁned in Section 1.9.5. 3.1.8 Material properties
3.1.8.1 Compressive strength 3.1.8.1.1 Masonry compressive strength
—— The speciﬁed compressive strength of masonry, f ’m ,
shall equal or exceed 1,500 psi (10.34 MPa). The value
off 1,, used to determine nominal strength values in this
chapter shall not exceed 4,000 psi (27.58 MPa) for
concrete masonry and shall not exceed 6,000 psi (41.37
MPa) for clay masonry. 3.1.8.1.2 Grout compressive strength—
For concrete masonry, the speciﬁed compressive
strength of grout, f jg, shall equal or exceed the speciﬁed
compressive strength of masonry, ’m, but shall not
exceed 5,000 psi (34.47 MPa). For clay masonry, the
speciﬁed compressive strength of grout, f 'g, shall not
exceed 6,000 psi (41.37 MPa). 3.1.8.2 Masonry modulus of rupture— The
modulus of rupture, f,, for masonry elements subjected
to outofplane or inplane bending shall be in
accordance with the values in Table 3.1.8.2. For grouted
stack bond masonry, tension parallel to the bed joints
shall be assumed to be resisted only by the minimum
crosssectional area of continuous grout that is parallel to
the bed joints. 3.1.8.3 Reinforcement strength —— Masonry
design shall be based on a reinforcement strength equal
to the speciﬁed yield strength of reinforcement, j}, which
shall not exceed 60,000 psi (413.7 MPa). The actual
yield strength shall not exceed 1.3 multiplied by the
speciﬁed yield strength. The compressive resistance of
steel reinforcement shall be neglected unless lateral
reinforcement is provided in compliance with the
requirements of Section 1.14.1.3. BUILDING CODE REQUIREMENTS FOR MASONRY STRUCTURES Table 3.1.8.2 — Modulus of rupture, f,, psi (kPa) Mortar types Portland cement/lime or mortar Masonry cement or air
cement entrained portland cement/lime 38(262) Direction of ﬂexural tensile stress
and masonry type Normal to bed joints in running or
stack bond Sohdunﬁs 100(689) 75(517) 60(413) Hollow unitsl 63(431)
163(1124) 48(331)
158(1089) 38(262)
153(1055) 23(158)
145(1000) Ungrouted Fully grouted Parallel to bed joints in running bond Solid units 200 (1379) 150 (1033) 120 (827) 75 (517) Hollow units 125(862)
200(1379) 95(655)
150(1033) 75(517)
120(827) 48(331)
75(517) Ungrouted and partially grouted Fully grouted Parallel to bed joints in stack bond 250(1734) ﬁOUBQ 2m(NM) ﬁOUBQ Continuous grout section parallel
to bed joints Other 0m) 0@ mm mm 1 For partially grouted masonry, modulus of rupture values shall be determined on the basis of linear interpolation between
ﬁllly grouted hollow units and ungrouted hollow units based on amount (percentage) of grouting. C 39 C39 040 3.2 ——Unreinforced (plain) masonry 3.2.1 Scope The requirements of Section 3.2 are in addition to the
requirements of Chapter 1 and Section 3.1 and govern masonry
design in which masonry is used to resist tensile forces. 3.2.1.1 Strength for resisting loads ~—
Unreinforced (plain) masonry members shall be designed
using the strength of masonry units, mortar, and grout in
resisting design loads. 3.2.1.2 Strength contribution from
reinforcement —— Stresses in reinforcement shall not be
considered effective in resisting design loads. 3.2.1.3 Design criteria — Unreinforced (plain)
masonry members shall be designed to remain uncracked. 3.2.2 Flexural and axial strength of unreinforced
(plain) masonry members 3.2.2.1 Design assumptions — The following
assumptions shall apply when determining the ﬂexural and
axial strength of unreinforced (plain) masonry members: (a) Strength design of members for factored ﬂexure and
axial load shall be in accordance with principles of
engineering mechanics. (b) Strain in masonry shall be directly proportional to the
distance from the neutral axis. (c) Flexural tension in masonry shall be assumed to be
directly proportional to strain. (d) Flexural compressive stress in combination with axial
compressive stress in masonry shall be assumed to be
directly proportional to strain. 3.2.2.2 Nominal strength — The nominal strength
of unreinforced (plain) masonry cross—sections for combined
ﬂexure and axial loads shall be determined so that: (a) the compressive stress does not exceed 0.80 f ’m. (b) the tensile stress does not exceed the modulus of
rupture determined from Section 3.1.8.2. 3.2.2.3 Nominal axial strength — The nominal
axial strength, P", shall not be taken greater than the
following: (a) For members having an h/r ratio not greater than 99: h 2
13,, = 0.80{0.80An fm [1—[140rj (b) For members having an h/r ratio greater than 99: 2
p, wits..." fig—r] ] (311) (3—12) TMS 40208lACI 53008lASCE 508 3.2.2.4 PDelta effects 3.2.2.4.1 Members shall be designed for the
factored axial load, Pu, and the moment magniﬁed for the
effects of member curvature, MC. 3.2.2.4.2 The magniﬁed moment, MC, shall
be determined either by a second—order analysis, or by a
ﬁrstorder analysis and Eqs. (313) and (314). Me = 6Mu (313)
6 — ——l——————— (314)
_ Pu
1"““——2
Anfif'm 3.2.2.4.3 It shall be permitted to take 6 = 1
for members in which h / r S 45 . 3.2.2.4.4 It shall be permitted to take 6 = 1
for members in which 45 < h/r s 60, provided the nominal strength deﬁned in Section 3.2.2.2 is reduced by
10 percent. 3.2.3 Axial tension — The tensile strength of
unreinforced masonry shall be neglected in design when
the masonry is subjected to axial tension forces. 3.2.4 Nominal shear strength — Nominal shear
strength, V", shall be the smallest of (a), (b) and the
applicable condition of (0) through (i): (a) 3.8A” M (b) 300A" (c) For running bond masonry not solidly grouted;
56A,, + 0.45Nu (d) For stack bond masonry with open end units and
grouted solid; 56A,, +0.45Nu (e) For running bond masonry grouted solid; 90A,, + 0.45Nu (t) For stack bond other than open end units grouted
solid; 23A ll BUILDING CODE REQUIREMENTS FOR MASONRY STRUCTURES 3.3 — Reinforced masonry 3.3.1 Scope The requirements of this Section are in addition to the
requirements of Chapter 1 and Section 3.1 and govern
masonry design in which reinforcement is used to resist
tensile forces. 3.3.2 Design assumptions
The following assumptions apply to the design of
reinforced masonry: (a) There is strain continuity between the reinforcement,
grout, and masonry so that loads are resisted in a
composite manner. (b) The nominal strength of reinforced masonry cross
sections for combined flexure and axial load shall be
based on applicable conditions of equilibrium. (0) The maximum usable strain, emu, at the extreme
masonry compression ﬁber shall be assumed to be
0.0035 for clay masonry and 0.0025 for concrete
masonry. (d) Strain in reinforcement and masonry shall be assumed
to be directly proportional to the distance from the
neutral axis. (e) Compression and tension stress in reinforcement shall
be taken as E, multiplied by the steel strain, but not
greater than fy . (f) The tensile strength of masonry shall be neglected in
calculating ﬂexural strength but shall be considered in
calculating deﬂection. (g) The relationship between masonry compressive stress
and masonry strain shall be assumed to be deﬁned by
the following: Masonry stress of 0.80 f 1,, shall be assumed
uniformly distributed over an equivalent compression
stress block bounded by edges of the cross section and
a straight line located parallel to the neutral axis and
located at a distance a = 0.80 c from the ﬁber of
maximum compressive strain. The distance c from the
ﬁber of maximum strain to the neutral axis shall be
measured perpendicular to the neutral axis. 3.3.3 Reinforcement requirements and details
3.3.3.1 Reinforcing bar size limitations —
Reinforcing bars used in masonry shall not be larger than
No. 9 (M#29). The nominal bar diameter shall not exceed
oneeighth of the nominal member thickness and shall not
exceed onequarter of the least clear dimension of the cell,
course, or collar joint in which the bar is placed. The area of
reinforcing bars placed in a cell or in a course of hollow unit construction shall not exceed 4 percent of the cell area. C41 3.3.3.2 Standard hooks —— The equivalent
embedment length to develop standard hooks in tension,
1,. , shall be determined by Eq. (315): 1,:13d, (345) 3.3.3.3 Development —— The required tension or
compression reinforcement shall be developed in
accordance with the following provisions: The required development length of reinforcement
shall be determined by Eq. (316), but shall not be less
than 12 in. (305 mm). QBdfﬂJ lei—W K shall not exceed the smallest of the following: the
minimum masonry clear cover, the clear spacing between
adjacent reinforcement splices, and 5 d1, . y = 1.0 for No. 3 (M#lO) through No. 5 (M#16) bars;
7 = 1.3 for No. 6 (M#l9) through No. 7 (M#22) bars; @4@ and
7 = 1.5 for No. 8 (M#25) through No. 9 (M#29) bars. Development length of epoxycoated reinforcing bars
shall be taken as 150 percent of the length determined by
Eq. (3—16). 3.3.3.3.1 Bars spliced by noncontact lap
splices shall not be spaced farther apart than oneﬁfth the
required length of lap nor more than 8 in. (203 mm). 3.3.3.3.2 Shear reinforcement shall extend
the depth of the member less cover distances. 3.3.3.3.2.1 Except at wall intersections,
the end of a horizontal reinforcing bar needed to satisfy
shear strength requirements of Section 3.3.4.1.2 shall be
bent around the edge vertical reinforcing bar with a 180
degree hook. The ends of singleleg or Ustirrups shall be
anchored by one of the following means: (a) A standard hook plus an effective embedment of [4/2.
The effective embedment of a stirrup leg shall be taken
as the distance between the middepth of the member,
d/2, and the start of the hook (point of tangency). (b) For No. 5 (M #16) bars and smaller, bending around
longitudinal reinforcement through at least 135
degrees plus an embedment of [4/3. The M3
embedment of a stirrup leg shall be taken as the
distance between middepth of the member, d/2, and
the start of the hook (point of tangency). (0) Between the anchored ends, each bend in the
continuous portion of a transverse Ustirrup shall
enclose a longitudinal bar. c41 C42 C42 3.3.3.3.2.2 At wall intersections,
horizontal reinforcing bars needed to satisfy shear strength
requirements of Section 3.3.4.1.2 shall be bent around the
edge vertical reinforcing bar with a 90—degree standard hook
and shall extend horizontally into the intersecting wall a
minimum distance at least equal to the development length. 3.3.3.4 Splices — Reinforcement splices shall
comply with one of the following: (a) The minimum length of lap for bars shall be 12 in.
(305 mm) or the development length determined by
Eq. (316), whichever is greater. (b) A welded splice shall have the bars butted and welded
to develop at least 125 percent of the yield strength,
1; , of the bar in tension or compression, as required. (c) Mechanical splices shall have the bars connected to
develop at least 125 percent of the yield strength, fy,
of the bar in tension or compression, as required. 3.3.3.5 Maximum area of ﬂexural tensile
reinforcement
3.3.3.5.1 For masonry members where Mu/Vudvz 1, the crosssectional area of ﬂexural tensile
reinforcement shall not exceed the area required to
maintain axial equilibrium under the following conditions: (a) A strain gradient shall be assumed, corresponding to a
strain in the extreme tensile reinforcement equal to 1.5
multiplied by the yield strain and a maximum strain in
the masonry as given by 3.3.2(c). (b) The design assumptions of Section 3.3.2 shall apply. (c) The stress in the tension reinforcement shall be taken
as the product of the modulus of elasticity of the steel
and the strain in the reinforcement, and need not be
taken as greater than j}. ((1) Axial forces shall be taken from the loading
combination given by D + 0.75L + 0.525QE. (e) The effect of compression reinforcement, with or
without lateral restraining reinforcement, shall be
permitted to be included for purposes of calculating
maximum ﬂexural tensile reinforcement. 3.3.3.5.2 For intermediate reinforced
masonry shear walls subject to inplane loads where
Mu/Vudv 2 1, a strain gradient corresponding to a strain in
the extreme tensile reinforcement equal to 3 multiplied by
the yield strain and a maximum strain in the masonry as
given by 3.3.2(c) shall be used. For intermediate
reinforced masonry shear walls subject to outofplane
loads, the provisions of Section 3.3.3.5.1 shall apply. 3.3.3.5.3 For special reinforced masonry shear walls subject to inplane loads where Mu/Vud 2 l, a
strain gradient corresponding to a strain in the extreme
tensile reinforcement equal to 4 multiplied by the yield TMS 40208IACI 53008IASCE 508 strain and a maximum strain in the masonry as given by
3.3.2(c) shall be used. For special reinforced masonry
shear walls subject to outofplane loads, the provisions of
Section 3.3.3.5.1 shall apply. 3.3.3.5.4 For masonry members where
Mu/Vudv S 1 and when designed using R S 1.5, there is no
upper limit to the maximum ﬂexural tensile reinforcement.
For masonry members where Mu/Vudv S l and when
designed using R 2 1.5, the provisions of Section 3.3.3.5.1
shall apply. 3.3.3.6 Bundling of reinforcing bars —
Reinforcing bars shall not be bundled. 3.3.4 Design of beams, piers, and columns Member design forces shall be based on an analysis that
considers the relative stiffness of structural members. The
calculation of lateral stiffness shall include the contribution
of all beams, piers, and columns. The effects of cracking on
member stiffness shall be considered. 3.3.4.1 Nominal strength 3.3.4.1.1 Nominal axial and flexural
strength ~ The nominal axial strength, P” , and the nominal
ﬂexural strength, M", of a cross section shall be determined
in accordance with the design assumptions of Section 3.3.2
and the provisions of Section 3.3.4.1. Using the slendemess
dependent modiﬁcation factors of Eq. (3—17) [l—(h/140r)2)]
and Eq. (318) (70r/h)2, as appropriate, the nominal axial
strength shall be modiﬁed for the effects of slendemess. The
nominal ﬂexural strength at any section along a member shall
not be less than one—fourth of the maximum nominal ﬂexural
strength at the critical section. The nominal axial compressive strength shall not
exceed Eq. (317) or Eq. (3—18), as appropriate. (a) For members having an h/r ratio not greater than 99: h 2
= . . ' _ _  7
P" 080[0 80fm(An As,)+fyAS,][1 (14%] 1(3 1 ) (b) For members having an h/r ratio greater than 99: 2
P" = 0.8o[0.80f,;,(A,, —As,)+ fyAS, (3—18) 3.3.4.1.2 Nominal shear strength —
Nominal shear strength, V", shall be computed using Eq.
(319) and either Eq. (320) or Eq. (321), as appropriate. V" = Vnm + Vm (319)
where V" shall not exceed the following: (a) Where Mu/Vu a", S 0.25:
V, 3 6A,, J7; (3—20) BUILDING CODE REQUIREMENTS FOR MASONRY STRUCTURES (b) Where Mu/Vu d, 2 1.00 V,smydff (c) The maximum value of V,, for Mu/Vu dv between 0.25
and 1.0 shall be permitted to be linearly interpolated. 3.3.4.1.2.1 Nominal masonry shear
strength — Shear strength provided by the masonry, Vnm ,
shall be computed using Eq. (322): M .
Vnm = [4.0—1.75{V 6: HA" ‘lfm +0.25Pu (3—22) M,/( V" d.) need not be taken greater than 1.0. 3.3.4.1.2.2 The value of Mu/(Vu dv) shall
be taken as a positive number. 3.3.4.1.2.3 Nominal shear strength
provided by reinforcement —— Nominal shear strength
provided by shear reinforcement, V”, shall be computed as follows: (321) AV S (323) tgs=05£ Jfgdv 3.3.4.2 Beams ~ Design of beams shall meet
the requirements of Section 1.13 and the additional
requirements of Section 3.3.4.2. 3.3.4.2.1 Members designed primarily to
resist ﬂexure shall comply with the requirements of
Section 3.3.4.2. The factored axial compressive force on a
beam shall not exceed 0.05 A, f ’m . 3.3.4.2.2 Longitudinal reinforcement 3.3.4.2.2.1 The variation in longitudinal
reinforcing bars in a beam shall not be greater than one bar
size. Not more than two bar sizes shall be used in a beam. 3.3.4.222 The nominal ﬂexural
strength of a beam shall not be less than 1.3 multiplied by
the nominal cracking moment of the beam, M6,. The
modulus of rupture, f,, for this calculation shall be
determined in accordance with Section 3.1.8.2. 3.3.4.2.2.3 The requirements of Section
3.3.4.222 need not be applied if at every section the area
of tensile reinforcement provided is at least onethird
greater than that required by analysis. 3.3.4.2.3 Transverse reinforcement — Transverse reinforcement shall be provided where V" exceeds ¢ Vnm.
The factored shear, V”, shall include the effects of lateral
load. When transverse reinforcement is required, the
following provisions shall apply: (a) Transverse reinforcement shall be a single bar with a
l80—degree hook at each end. 043 (b) Transverse reinforcement shall be hooked around the
longitudinal reinforcement. (c) The minimum area of transverse reinforcement shall
be 0.0007 bdv. (d) The ﬁrst transverse bar shall not be located more than one
fourth of the beam depth, dv , from the end of the beam. (e) The maximum spacing shall not exceed onehalf the
depth of the beam nor 48 in. (1219 mm). 3.3.4.2.4 Construction — Beams shall be
grouted solid. 3.3.4.2.5 Dimensional limits — The nominal
depth of a beam shall not be less than 8 in. (203 mm). 3.3.4.3 Piers
3.3.4.3.1 The factored axial compression
force on piers shall not exceed 0.3 A” f 'm . 3.3.4.3.2 Longitudinal reinforcement v A
pier subjected to inplane stress reversals shall be reinforced
symmetrically about the neutral axis of the pier. Longitudinal
reinforcement of piers shall comply with the following: (a) At least, one bar shall be provided in each end cell. (b) The minimum area of longitudinal reinforcement shall
be 0.0007 bd. 3.3.4.3.3 Dimensional limits w Dimensions
shall be in accordance with the following: (a) The nominal thickness of a pier shall not exceed 16 in.
(406 mm). (b) The distance between lateral supports of a pier shall
not exceed 25 multiplied by the nominal thickness of
a pier except as provided for in Section 3.3.4.3.3(c). (c) When the distance between lateral supports of a pier
exceeds 25 multiplied by the nominal thickness of the
pier, design shall be based on the provisions of
Section 3.3.5. (d) The nominal length of a pier shall not be less than
three multiplied by its nominal thickness nor greater
than six multiplied by its nominal thickness. The clear
height of a pier shall not exceed five multiplied by its
nominal length. Exception: When the factored axial force at the
location of maximum moment is less than 0.05 f ’,,, Ag,
the length of a pier shall be permitted to be equal to
the thickness of the pier. 3.3.4.4 Columns — Design of columns shall
meet the requirements of Section 1.14 and the additional
requirements of Section 3.3.4.4. 3.3.4.4.] Construction — Columns shall be
solid grouted. 1.9.6 C44 3.3.4.4.2 Dimensional limits —— Dimensions
shall be in accordance with the following: (a) The distance between lateral supports of a colurrm
shall not exceed 30 multiplied by its nominal width. (b) The nominal depth of a column shall not be less than
8 in. (203 mm) and not be greater than three
multiplied by its nominal width. 3.3.5 Wall design for outofplane loads
3.3.5.1 Scope —— The requirements of Section
3.3.5 are for the design of walls for outofplane loads. 3.3.5.2 Moment and deﬂection calculations —
Moment and deﬂection calculations in Sections 3.3.5.3
and 3.3.5.4 are based on simple support conditions top and
bottom. For other support and fixity conditions, moments
and deﬂections shall be calculated using established
principles of mechanics. 3.3.5.3 Walls with factored axial stress of
0.20 f ’m or less — The procedures set forth in this Section
shall be used when the factored axial load stress at the
location of maximum moment satisﬁes the requirement computed by Eq. (324). P
[ “ J3 0.20f,;,
A8 When the slenderness ratio, h/t, exceeds 30, the
factored axial stress shall not exceed 0.05f’m. (3—24) Factored moment and axial force shall be determined
at the midheight of the wall and shall be used for design.
The factored moment, Mu, at the midheight of the wall
shall be computed using Eq. (325). wu h 2 eu
Mu = 8 + Puf 7 + Pu 6“ (325)
Where
Pu = PW + Pu] (3—26) The deﬂection due to factored loads (6“) shall be
obtained using Eq. (331) and (332) and replacing Mm
with Mu and dwith 6“. The design strength for outofplane wall loading shall
be in accordance with Eq. (327). Mu s ¢Mn (327) The nominal moment shall be calculated using Eqs.
(328) and (329) if the reinforcing steel is placed in the
center of the wall. Mn = (Asfy +Pu(d—%] (328) TMS 40208lACI 53008/ASCE 508 Pu + As
a = ( 0.80 f?) (3'29) The nominal shear strength shall be determined by
Section 3.3.4.1.2. 3.3.5.4 Deflections — The horizontal midheight
deﬂection, 8, , under service lateral and service axial loads
(without load factors) shall be limited by the relation: 53 S 0.007 h (330) Pdelta effects shall be included in deﬂection
calculation. The midheight deﬂection shall be computed
using either Eq. (331) or Eq. (332), as applicable. (a) Where Mm < MC, 5M h2
as = (331)
48Em1g
(b) Where MC, < M5,, < Mn
2 _ 2
65 = SMcrh + 5 (Mser MCI‘) h 48Emlg 485mg, The cracking moment of the wall shall be computed using
the modulus of rupture, fr, taken from Table 3.1.8.2. 3.3.6 Wall design for inplane loads 3.3.6.1 Scope — The requirements of Section
3.3.6 are for the design of walls to resist inplane loads. 3.3.6.2 Reinforcement — Reinforcement shall
be provided perpendicular to the shear reinforcement and
shall be at least equal to one—third Av. The reinforcement
shall be uniformly distributed and shall not exceed a
spacing of 8 ft (2.44 m). 3.3.6.3 Flexural and axial strength —— The
nominal ﬂexural and axial strength shall be determined in
accordance with Section 3.3.4.1.1. 3.3.6.4 Shear strength — The nominal shear strength shall be computed in accordance with Section
3.3.4.1.2. ‘ 3.3.6.5 The maximum reinforcement
requirements of Section 3.3.3.5 shall not apply if a shear
wall is designed to satisfy the requirements of 3.3.6.5.1
through 3.3.6.5.5. 3.3.6.5.1 Special boundary elements need not
be provided in shear walls meeting the following conditions: 1. PuSOJOAgf ’m for geometrically symmetrical
wall sections Pu S 0.05 Agf C" for geometrically unsymmetrical
wall sections; and either BUILDING CODE REQUlREMENTS FOR MASONRY STRUCTURES 01‘ 3. V, sum/f," and Mu
S 3.0
Vulw
3.3.6.5.2 The need for special boundary
elements at the edges of shear walls shall be evaluated in
accordance with Section 3.3.6.5.3 or 3.3.6.54. The
requirements of Section 3.3.6.5.5 shall also be satisﬁed. 3.3.6.5.3 This Section applies to walls
bending in single curvature in which the ﬂexural limit state
response is governed by yielding at the base of the wall.
Walls not satisfying those requirements shall be designed
in accordance with Section 3.3.6.5.4 (a) Special boundary elements shall be provided over
portions of compression zones where: c 2 _J__
600 (Cdan, /hw) and c is calculated for the PM given by ASCE 7
Strength Design Load Combination 5
(1.2D + 1.0E + L + 0.25) or the corresponding
strength design load combination of the legally
adopted building code, and the corresponding nominal
moment strength, Mn , at the base critical section. The
load factor on L in Combination 5 is reducible to 0.5,
as per exceptions to Section 2.3.2 of ASCE 7. (b) Where special boundary elements are required by
Section 3.3.6.5.3 (a), the special boundary element
reinforcement shall extend vertically from the critical section a distance not less than the larger of [w or
Mu/4 Vu. C45 3.3.6.5.4 Shear walls not designed by
Section 3.3.6.5.3 shall have special boundary elements at
boundaries and edges around openings in shear walls
where the maximum extreme ﬁber compressive stress,
corresponding to factored forces including earthquake
effect, exceeds 0.2 f ’m. The special boundary element
shall be permitted to be discontinued where the calculated
compressive stress is less than 0.15 f ’m. Stresses shall be
calculated for the factored forces using a linearly elastic
model and gross section properties. For walls with ﬂanges,
an effective ﬂange width as deﬁned in Section 1.9.4.2.3 shall be used. 3.3.6.5.5 Where special boundary elements
are required by Section 3.3.6.5.3 or 3.3.6.5.4, requirements
(a) through (d) in this section shall be satisﬁed and tests shall
be performed to verify the strain capacity of the element: (a) The special boundary element shall extend
horizontally from the extreme compression ﬁber a
distance not less than the larger of (c  0.11,,) and c/2. (b) In ﬂanged sections, the special boundary element shall
include the effective ﬂange width in compression and
shall extend at least 12 in. (305 mm) into the web. (0) Special boundary element transverse reinforcement at
the wall base shall extend into the support a minimum
of the development length of the largest longitudinal
reinforcement in the boundary element unless the
special boundary element terminates on a footing or
mat, where special boundary element transverse
reinforcement shall extend at least 12 in. (305 mm)
into the footing or mat. (d) Horizontal shear reinforcement in the wall web shall
be anchored to develop the speciﬁed yield strength, fy ,
within the conﬁned core of the boundary element. ...
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This note was uploaded on 04/18/2009 for the course CE 478 taught by Professor Brandow during the Fall '08 term at USC.
 Fall '08
 BRANDOW

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