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ALLOWABLE STRESS DESIGN TEK 14-7A

Structural (2004)
OF CONCRETE MASONRY
Keywords: allowable loads, allowable stress, allowable stress
design, axial strength, building code provisions, flexural
strength, reinforced concrete masonry, shear strength, struc- DESIGN LOADS
tural design, unreinforced concrete masonry
Utilizing allowable stress design, masonry elements are
sized and proportioned such that the anticipated service level
INTRODUCTION loads can be safely and economically resisted using the speci-
fied material strengths. The specified strength of masonry and
Loadbearing concrete masonry walls can be designed reinforcement are in turn reduced by appropriate safety factors.
using one of several methods in accordance with Building Minimum design loads for allowable stress design are included
Code Requirements for Masonry Structures (ref. 1): empirical in Minimum Design Loads for Buildings and Other Structures
design, strength design or allowable stress design (ASD). This (ref. 4) or obtained from the general building code such as the
TEK provides a basic overview of design criteria and require- International Building Code (IBC) (ref. 5). For load combina-
ments for concrete masonry walls designed using the ASD tions that include wind or earthquake loads, the code-pre-
provisions contained in Chapter 2 of Building Code Require- scribed allowable stresses (refs. 1, 5) are permitted to be
ments for Masonry Structures. For the empirical or strength increased by one-third. Note, however, that the one-third
design provisions, the reader is referred to TEK 14-8A Empiri- stress increase cannot be used when using the loads or load
cal Design of Concrete Masonry Walls and TEK 14-4A Strength combinations of Minimum Design Loads for Buildings and
Design of Concrete Masonry (refs. 2, 3), respectively. Other Structures (ref. 4).
This TEK is intended only to provide a general review of Using ASD, the calculated design stresses on a masonry
the pertinent ASD criteria. Tables, charts and additional aids member (indicated by lowercase f) are compared to code-
specific to the allowable stress design of concrete masonry prescribed maximum allowable stresses (indicated by a capital
elements can be found in other related TEK. F). The design is acceptable when the calculated applied
Allowable stress design is based on the following design stresses are less than or equal to the allowable stresses (f < F).
principles and assumptions.
· Within the range of allowable stresses, masonry elements
satisfy applicable conditions of equilibrium and compat- fb
ibility of strains.
· Plane sections before bending remain plane after bending.
Therefore, masonry strain is directly proportional to the
distance from the neutral axis.
· Stress is linearly proportional to strain within the allowable
stress range.
· For reinforced masonry design, all tensile stresses are
resisted by the steel reinforcement. The contribution of
the masonry to the tensile strength of the element is
ignored.
· The units, mortar, grout and reinforcement, if present, act
compositely to resist applied loads.
Based on this assumed design model, the internal distri-
bution of stresses and resulting equilibrium is illustrated in
Figure 1 for unreinforced masonry and Figure 3 for reinforced Wall width
masonry.
Figure 1—Unreinforced Masonry Stress Distribution

TEK 14-7A © 2004 National Concrete Masonry Association (replaces TEK 14-7)
UNREINFORCED MASONRY Mt P
fb = − Eqn. 1
2 I n An
For unreinforced masonry, the masonry assembly (units,
mortar and grout if used) is designed to carry all applied TEK 14-1 Section Properties of Concrete Masonry
stresses. The additional capacity from the inclusion of reinforc- Walls (ref. 6) provides typical values for the net moment of
ing steel, such as reinforcement added for the control of inertia, In, and cross-sectional area, An, for various wall
shrinkage cracking or prescriptively required by the code, is sections. If the value of the bending stress, fb, given by
neglected. Because the masonry is intended to resist both Equation 1 is positive, then the masonry section is controlled
tension and compression stresses resulting from applied loads, by tension and the limiting values of Table 1 must be satisfied.
the masonry must be designed to remain uncracked. Conversely, if fb as given by Equation 1 is negative, the
masonry section is in compression and the compressive
Unreinforced Out-of-Plane Flexure stress limitation of Equation 2 must be met.
Allowable flexural tension values as prescribed in Build- f b ≤ F b= 1
3 f 'm Eqn. 2
ing Code Requirements for Masonry Structures (ref. 1), vary
with the direction of span, mortar type, bond pattern and Unreinforced Axial Compression and Flexure
percentage of grouting as shown in Table 1. For walls spanning While unreinforced masonry can be designed to resist
horizontally between supports, the code conservatively as- flexural tension stresses due to applied loads, unreinforced
sumes that masonry constructed in stack bond cannot reliably masonry may not be subjected to net axial tension, such as that
transfer flexural tension stresses across the head joints. As due to wind uplift on a roof connected to a masonry wall or due
such, the allowable flexural tension values parallel to the bed to the overturning effects of lateral loads. While compressive
joints (perpendicular to the head joints) for stack bond con- stresses from dead loads can be used to offset tensile stresses,
struction are assumed to be zero for design purposes. where the wall is subject to a net axial tension reinforcement
Because the compressive strength of masonry is much must be incorporated to resist the resulting tensile forces.
larger than its tensile strength, the capacity of unreinforced When masonry walls are subjected to compressive axial
masonry subjected to net flexural stresses is almost always loads only, the calculated compressive stress due to the
controlled by the flexural tension values of Table 1. For ma- applied load, fa, must not exceed the allowable compressive
sonry elements subjected to a bending moment, M, and a stress, Fa, as given by Equation 3 or 4, as appropriate.
compressive axial force, P, the resulting flexural bending stress For elements with h/r not greater than 99:
is determined using Equation 1.
⎡ ⎛ h ⎞2 ⎤
f a ≤ Fa = 4 f ' m ⎢1 − ⎜
1
⎟ ⎥ Eqn. 3
Table 1—Allowable Flexural Tension Stresses for ⎢⎣ ⎝ 140r ⎠ ⎥⎦
Unreinforced Concrete Masonry, psi (kPa) (ref. 1) For elements with h/r greater than 99:
2
⎛ 70r ⎞
Mortar types f a ≤ Fa = f 'm ⎜
1
4 ⎟ Eqn. 4
Direction of Portland cement/ Masonry cement or ⎝ h ⎠
flexural lime or mortar air entrained portland A further check for stability against an eccentrically ap-
tensile stress cement cement/lime plied axial load is included with Equation 5, whereby the axial
& masonry type M or S N M or S N compressive load, P, is limited to one-fourth the buckling load,
Pe. With Equation 5, the actual eccentricity of the applied load,
Normal to bed joints (walls spanning vertically): e, is used to determine Pe. Moments on the wall due to loads
Solid units 40 (276) 30(207) 24 (166) 15 (103) other than the eccentric load are not considered in Equation 5.
Hollow unitsa
Ungrouted 25 (172) 19 (131) 15 (103) 9 (62.1) P≤ 1
4 Pe = ( 4 )
h
1
π 2 Em I n
2
(1 − 0.577 e r )
3
Eqn. 5
Fully grouted 65 (448) 63 (434) 61 (420) 58 (400)
When unreinforced masonry elements are subjected to a
Parallel to bed joints in running bond (walls spanning combination of axial load and flexural bending, a unity equation
horizontally): is used to proportion the available allowable stresses to the
Solid units 80 (552) 60 (414) 48 (331) 30 (207) applied loads per Equation 6.
Hollow units fa f
Ungrouted & + b ≤1 Eqn. 6
Fa Fb
partially grouted 50 (345) 38 (262) 30 (207) 19 (131)
Fully grouted 80 (552) 60 (414) 48 (331) 30 (207)
Unreinforced Shear
a
For partially grouted masonry, allowable stresses are Shear stresses on unreinforced masonry elements are
determined by linear interpolation between fully grouted calculated based on the net cross-sectional properties of the
hollow units and ungrouted hollow units based on the masonry in the direction of the applied shear force using the
amount of grouting. following relation:
 VQ REINFORCED MASONRY
fv = Eqn. 7
I nb
The design of reinforced masonry in accordance with
Equation 7 is applicable to the determination of both in- Building Code Requirements for Masonry Structures (ref. 1)
plane and out-of-plane shear stresses. Because unreinforced neglects the tensile resistance provided by the masonry units,
masonry is designed to remain uncracked, it is not necessary mortar and grout in determining the strength of the masonry
to perform a cracked section analysis to determine the net cross- assemblage. Thus, for design purposes, the portion of ma-
sectional area of the masonry. sonry subjected to net tensile stresses is assumed to have
The theoretical distribution of shear stress, fv, along the cracked, transferring all tensile forces to the reinforcement.
length of the shear wall for in-plane loads (Figure 2), or perpen- (While the determination of the capacity of a reinforced ma-
dicular to any wall for out-of-plane loads, has a parabolic shape sonry element conservatively assumes the portion of the
for a rectangular cross-section. The calculated shear stress due masonry subjected to net tensile stresses has cracked, this
to applied loads, fv, as given by Equation 7 is limited to the code- should be verified when establishing the stiffness and deflec-
prescribed allowable shear stress, Fv, from Table 2. While tion of a reinforced masonry element.)
Building Code Requirements for Masonry Structures (ref. 1)
designates the values in Table 2 as being applicable to in-plane Reinforcement
shear stresses only, no allowable shear stresses are provided The tensile stress in the reinforcement due to applied load,
for out-of-plane loads. In light of the absence of out-of-plane fs, is calculated as the product of the strain in the steel (which
allowable shear stress values, the Commentary on Building increases linearly in proportion to the distance from the neutral
Code Requirements for Masonry Structures suggests using axis) multiplied by its modulus of elasticity, Es. The modulus
the values in Table 2 for out-of-plane shear design. of elasticity, Es, of mild steel reinforcement is assumed to be
29,000,000 psi (200 GPa). The code-prescribed allowable steel
P stresses are as follows (ref. 1).
For Grade 60 reinforcement in tension:
M Fs = 24,000 psi (165.5 MPa)
Compressive For Grade 40 and 50 reinforcement in tension:
stress, fb Fs = 20,000 psi (137.9 MPa)
For wire reinforcement in tension:
Fs = 30,000 psi (206.9 MPa)
For all reinforcement in compression:
fv Fs = 24,000 psi (165.5 MPa) or 0.4fy, whichever is less
Unless ties or stirrups laterally confine the reinforcement
as prescribed by Building Code Requirements for Masonry
V Structures, the reinforcement is assumed to contribute no
compressive resistance to axially loaded elements. Additional
information on mild reinforcing steel can be found in TEK 12-
b 4C Steel Reinforcement for Concrete Masonry (ref. 7).
For design purposes, the effective width of the compres-
sion zone per bar is limited to the smallest of:
· six times the wall thickness,
· the center-to-center spacing of the reinforcement, or
· 72 in. (1,829 mm).
Figure 2—Unreinforced Masonry Shear Walls This requirement applies to masonry constructed in run-
ning bond and to masonry constructed in stack bond contain-
Table 2—Allowable Shear Stresses for ing bond beams spaced no further than 48 in. (1,219 mm) on
Unreinforced Masonry center. Where the center-to-center spacing of the reinforce-
ment does not control the effective width of the compression
Allowable shear stress, Fv, is the least of: zone, the resulting resisting moment or resisting shear is
1. 1.5 f ' m psi ( 0.125 f ' m MPa) proportioned over the width corresponding to the actual
reinforcement spacing.
2. 120 psi (827 kPa)
3. a. For running bond not solidly grouted and for stack Reinforced Out-of-Plane Flexure
bond masonry with open end units and grouted solid, As with unreinforced masonry, the allowable compres-
37 psi + 0.45Nv/An (255 kPa + 0.45Nv/An) sive stress in masonry, Fb, due to flexure or due to a combina-
b. For masonry in running bond and solidly grouted, tion of flexure and axial load is limited by Equation 2. When axial
60 psi + 0.45Nv/An (414 kPa + 0.45Nv/An) loads are not present, or are conservatively neglected as may
c. For stack bond masonry with other than open end be appropriate in some cases, there are several circumstances
units grouted solid, 15 psi (103 kPa) to consider in determining the flexural capacity of reinforced
masonry walls. For a fully grouted d d
element, a cracked transformed sec- jd jd
tion approach is used wherein the kd T kd T
reinforcement area is transformed 1 kd 1 kd
3 3
to an equivalent area of concrete fb
masonry using the modular ratio. fb
Partially grouted walls are analyzed C C
in the same way, but with the addi-
tional consideration of the
ungrouted cores. For partially
grouted masonry there are two
types of behavior to consider.
1. The first case applies when the
neutral axis (the location of zero
stress) lies within the compres-
sion face shell as shown in Figure
3a. In this case, the wall is ana-
lyzed and designed using the pro-
Masonry cover Masonry cover
cedures for a fully grouted wall.
Bar diameter Bar diameter
2. The second type of analysis oc-
curs when the neutral axis lies Wall width Wall width
within the core area, rather than
the compression face shell, as a—Neutral axis within the face shell b—Neutral axis in core area
shown in Figure 3b. For this case,
Figure 3—Reinforced Masonry Stress Distribution
the portion of the ungrouted cells
must be deducted from the area of
masonry capable of carrying compression stresses. Tee Beam Analysis
The location of the neutral axis depends on the relative For partially grouted masonry walls where the neutral axis
moduli of elasticity of the masonry and steel, η, as well as the is located within the cores, the resisting flexural capacity, Mr,
reinforcement ratio, ρ, and the distance between the reinforce- is calculated using the neutral axis coefficient k given by
ment and the extreme compression fiber, d. Equation 14 and either Case A or Case B as follows:
When analyzing partially grouted walls, it is typically − Asη − t fs (b − bw )
assumed that the neutral axis lies within the compression face k=
shell, as the analysis is more straightforward. Based on this dbw
assumption, the resulting value of k and the location of the
neutral axis (kd) is calculated. If it is determined that the neutral
Eqn. 14
axis lies outside the compression face shell, the more rigorous
tee beam analysis is performed. Otherwise, the rectangular
beam analysis is carried out. A complete discussion and
(A) For cases where the masonry strength controls the design
derivation of this procedure is contained in Concrete Masonry
capacity:
Design Tables (ref. 8).
⎛1− k ⎞
f s = ηFb ⎜ ⎟ Eqn. 15
Rectangular Beam Analysis ⎝ k ⎠
For fully grouted masonry elements and for partially
grouted masonry walls with the neutral axis in the compression
If fs as determined using Equation 15 is greater than the
face shell, the resisting flexural capacity, Mr, is calculated as
allowable steel stress, Fs, then the steel controls the design
follows:
capacity per (B) below. Otherwise, the internal compression
η = Es / Em Eqn. 8 force, C, and tension force, T, are computed as follows:
ρ = As/bd Eqn. 9 C = 1/2 Fb bkd Eqn. 16

k = 2 ρη + ( ρη ) − ρη
2
Eqn. 10 T = As f s = AsηFb 1 − k ( k
) Eqn. 17
j = 1 - k/3 Eqn. 11
Mm = (B) For cases where the steel strength controls the design:
2
1
2 Fb kjbd Eqn. 12
M s = As Fs jd Eqn. 13 Fs ⎛ k ⎞
fb =
The resisting flexural capacity, Mr, is taken as the lesser of Mm
η ⎜⎝ 1 − k ⎟⎠ Eqn. 18

and Ms.
 ⎛ F ⎛ k ⎞⎞ Fv = f m′ < 50 psi Eqn. 26
C= 1 f b bkd = 1 bkd ⎜⎜ s ⎜ ⎟ ⎟⎟
⎝ η ⎝1− k ⎠⎠
2 2 Eqn. 19
(metric: Fv = 0.083 f ' m < 345 kPa)
T = As Fs Eqn. 20 For shear walls:
where M/Vd < 1,
(C) The design capacity is then computed using the following
relationship: Fv = 1
3 [4 − (M / Vd )] f ' m < 80 - 45(M/Vd) Eqn. 27

⎛ 2t ⎞ ⎛ 2t ⎞ (metric: Fv = 0.028[4 − (M / Vd )] f ' m < 0.55 - 0.31(M/Vd))

M r = C ⎜ fs ⎟ − T ⎜ d − kd − fs ⎟ Eqn. 21
⎝ 3 ⎠ ⎝ 3 3 ⎠ where M/Vd > 1,
Vr = Fv bd Eqn. 22
Fv = f ' m < 35 psi Eqn. 28
Reinforced Axial Compression (metric: Fv = 0.083 f ' m < 241 kPa)
Axial loads acting through the axis of a member are When shear reinforcement is provided to resist the entire
distributed over the net cross-sectional area of masonry. The shear force, the minimum amount of shear reinforcement is
allowable axial compressive force is based on the compressive determined by Equation 29.
strength of masonry and the slenderness ratio of the element
in accordance with the following. Vs
Av = Eqn. 29
For elements with h/r not greater than 99, the allowable Fs d
compressive force, Pa, is: Where reinforcement is provided to resist the entire calcu-
⎡ ⎛ h ⎞ ⎤ 2 lated shear stress, fv, the allowable shear stress, Fv, is deter-
Pa = (0.25 f m′ An + 0.65 As Fs )⎢1 − ⎜ ⎟ ⎥ Eqn. 23 mined in accordance with the following:
⎢⎣ ⎝ 140r ⎠ ⎥⎦ For flexural members:
For elements with h/r greater than 99, the allowable com- Fv = 3 f m′ ≤ 150 psi (1034 kPa) Eqn. 30
pressive force, Pa, is:
(metric: Fv = 0.25 f m′ < 1,034 kPa)
2
⎛ 70r ⎞
Pa = (0.25 f m′ An + 0.65 As Fs )⎜ ⎟ Eqn. 24 For shear walls:
⎝ h ⎠ where M/Vd < 1,
Fv = 1
2 [4 − (M / Vd )] f ' m < 120 - 45(M/Vd) Eqn. 31
Reinforced Axial Compression and Flexure
Often, loading conditions result in both axial load and (metric: Fv = 0.042[4 − (M / Vd )] f ' m < 0.82 - 0.31(M/Vd))
flexure on a wall. Superimposing the stresses resulting from where M/Vd > 1,
axial compression and flexural compression produces the com-
Fv = 1.5 f ' m < 75 psi Eqn. 32
bined stress. Members are proportioned so that this maximum
combined stress does not exceed the allowable stress limitation (metric: Fv = 0.125 f ' m < 517 kPa)
imposed by Equation 2, Equation 5, Equation 6 and either
Equation 23 or 24, as appropriate. Several design approaches
are available for combined axial compression and flexure, most For Equations 27, 28, 31, and 32, the ratio of M/Vd is
commonly using either computer programs to perform the required to be taken as a positive value.
necessary iterative calculations or using interaction diagrams Shear reinforcement provided in accordance with Equa-
to graphically determine the required reinforcement for a given tions 29 through 32 must also comply with the following.
condition (refs. 8, 9, 10). · Shear reinforcement is oriented parallel to the direction of
the shear force.
Reinforced Shear · Shear reinforcement spacing must not exceed the lesser of
Shear acting on masonry flexural members and shear walls d/2 or 48 in. (1,219 mm).
is resisted either by the masonry (units, mortar and grout) or by · Reinforcement must also be provided perpendicular to the
shear reinforcement. For masonry members that are not sub- shear reinforcement. This prescriptive reinforcement must
jected to flexural tension, the allowable shear stresses provided have an area of at least one-third Av as given by Equation
in Table 2 for unreinforced masonry apply. For masonry ele- 28 and may not be spaced farther apart than 8 ft (2,438 mm).
ments that are subjected to flexural tension, the applied shear
stress is calculated as follows: NOTATION:
An = net cross-sectional area of masonry, in.2 (mm2)
V As = effective cross-sectional area of reinforcement, in.2 (mm2)
fv = Eqn. 25
bd A v = cross-sectional area of shear reinforcement, in.2 (mm2)
Where reinforcement is not provided to resist the entire b = width of section, in. (mm)
calculated shear stress, fv, the allowable shear stress, Fv, is bw = for partially grouted walls, width of grouted cell plus each
determined in accordance with the following: web thickness within the compression zone, in. (mm)
For flexural members: C = resultant compressive force, lb (N)
d = distance from the extreme compression fiber to Pa = allowable compressive force in reinforced masonry
centroid of tension reinforcement, in. (mm) due to axial load, lb (N)
Em = modulus of elasticity of masonry, psi (MPa) Pe = Euler buckling load, lb (N)
Es = modulus of elasticity of reinforcement, psi (MPa) Q = first moment of inertia about the neutral axis, in.3 (mm3)
e = eccentricity of applied load, in. (mm) r = radius of gyration, in. (mm)
Fa = allowable compressive stress due to axial load only, s = spacing of shear reinforcement, in. (mm)
psi (MPa) T = resultant tensile force, lb (N)
Fb = allowable bending stress due to flexure only, psi (MPa) t = thickness of masonry element, in. (mm)
Fs = allowable tensile or compressive stress in reinforce- tfs = concrete masonry unit face shell thickness, in. (mm)
ment, psi (MPa) V = applied shear force, lb (N)
Fv = allowable shear stress, psi (MPa) Vr = shear capacity (resisting shear) of masonry, lb (N)
fa = calculated compressive stress due to axial load only, η = modular ratio
psi (MPa) ρ = reinforcement ratio
fb = calculated bending stress due to flexure only, psi (MPa)
f'm = specified compressive strength of masonry, psi (MPa) REFERENCES
fs = calculated tensile or compressive stress in reinforce- 1. Building Code Requirements for Masonry Structures,
ment, psi (MPa) ACI 530-02/ASCE 5-02/TMS 402-02. Reported by the Ma-
fv = calculated shear stress in masonry, psi (MPa) sonry Standards Joint Committee, 2002.
fy = specified yield strength of reinforcement, psi (MPa) 2. Empirical Design of Concrete Masonry Walls, TEK 14-8A.
h = effective height of masonry element, in. (mm) National Concrete Masonry Association, 2001.
In = moment of inertia of net cross-sectional area of a 3. Strength Design of Concrete Masonry, TEK 14-4A. Na-
masonry element, in.4 (mm4) tional Concrete Masonry Association, 2002.
j = ratio of distance between centroid of flexural compres- 4. Minimum Design Loads for Buildings and Other Structures,
sive forces and centroid of tensile forces to depth d ASCE 7-02. American Society of Civil Engineers, 2002.
k = ratio of distance between compression face of wall 5. 2000 International Building Code. Falls Church, VA:
and neutral axis to the effective depth d International Code Council, 2003.
M = maximum calculated bending moment at the section 6. Section Properties of Concrete Masonry Walls, TEK 14-
under consideration, in.-lb (N-mm) 1. National Concrete Masonry Association, 2003.
Mm = flexural capacity (resisting moment) when masonry 7. Steel Reinforcement for Concrete Masonry, TEK 12-4C.
controls, in.-lb/ft (N-m/m) National Concrete Masonry Association, 2002.
Mr = flexural capacity (resisting moment), in.-lb/ft (N-m/m) 8. Concrete Masonry Design Tables, TR 121. National Con-
Ms = flexural capacity (resisting moment) when reinforce- crete Masonry Association, 2000.
ment controls, in.-lb/ft (N-m/m) 9. Concrete Masonry Wall and Lintel Design Software,
Nv = compressive force acting normal to the shear surface, lb (N) CMS10. National Concrete Masonry Association, 2002.
P = applied axial load, lb (N) 10. Loadbearing Concrete Masonry Wall Design, TEK 14-
5A. National Concrete Masonry Association, 2000.

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NCMA does not assume responsibility for errors or omissions resulting from the use of this TEK.

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