Masonry Design

4% of exam

TMS allowable-stress design of reinforced and unreinforced CMU and brick: flexure, shear, axial and slender-wall capacity, reinforcement, and brick-veneer anchorage.

3 concepts

Exam areas

E. Masonry

TMS 402 Masonry: ASD Flexure & Axial

Allowable-stress flexure and axial design of concrete masonry per TMS 402-16: Fb, the slenderness-reduced axial capacity, the cracked-transformed flexure check, and the combined unity equation.

Masonry is the one structural material the NCEES PE Civil Reference Handbook leaves almost entirely to the code: you will not find an allowable stress, a slenderness factor, or a flexure procedure anywhere in it. Everything you design on exam day comes from the building code masonry standard

\textbf{TMS 402-16}

— and NCEES supplies that exact edition, so cite it and use its allowable-stress design (ASD) provisions, not strength design, unless a problem says otherwise. Points are lost here in three predictable ways: confusing the reinforced flexural compressive allowable

0.45 f'_m

with the

\tfrac{1}{3}f'_m

allowable for unreinforced flexure or the

\tfrac{1}{4}f'_m

coefficient of the allowable axial stress, forgetting the slenderness reduction on axial capacity, and treating masonry like reinforced concrete (wrong stress-block factor, wrong

n

, wrong allowable steel stress). This concept builds the working fluency to size a wall or pilaster for axial load, for out-of-plane bending, and for the two acting together.

Allowable stresses: where the numbers come from

ASD masonry caps service-load stresses at fractions of the specified compressive strength

f'_m

. For reinforced masonry in bending, TMS 402-16 §8.3.4.2.2 (Eq. 8-21) sets the allowable flexural compressive stress at

F_b = 0.45 f'_m

. Do not confuse this with

\tfrac{1}{3}f'_m

, the flexural compressive allowable for

\textit{unreinforced}

masonry, or with

\tfrac{1}{4}f'_m

, the leading coefficient of the allowable axial stress

F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]

(§8.2.4.1), the same one-quarter as the

0.25 f'_m

term in

P_a

. The allowable tensile stress in Grade 60 reinforcement is

F_s = 32{,}000\,\text{psi}

(32 ksi). The modular ratio uses

E_m = 900 f'_m

for concrete masonry and

E_s = 29{,}000\,\text{ksi}

, so

n = E_s/E_m

— for

f'_m = 2000\,\text{psi}

n \approx 16.1

, far larger than concrete's

n \approx 8

F_b = 0.45 f'_m\ (\text{reinforced flexure}), \qquad F_s = 32\,\text{ksi}\ (\text{Gr.\ 60}), \qquad n = \dfrac{E_s}{E_m} = \dfrac{29{,}000\,\text{ksi}}{900 f'_m}

Axial capacity and the slenderness reduction

A masonry wall or column carries axial load through both the net masonry area

A_n

and any longitudinal steel

A_{st}

, but slender members buckle, so TMS 402-16 §8.3.4 multiplies the squash capacity by a slenderness factor based on

h/r

, where

r

is the radius of gyration of the net section. For

h/r \le 99

the factor is

\left[1-(h/140r)^2\right]

; for

h/r > 99

it switches to the Euler-like

(70r/h)^2

. The allowable axial force is then

P_a

below. For a solid rectangular wall of actual thickness

t

r = t/\sqrt{12}

, so an 8-in CMU wall (

t = 7.625\,\text{in}

) has

r \approx 2.20\,\text{in}

and crosses

h/r = 99

at roughly an 18-ft height.

P_a = \left(0.25\,f'_m A_n + 0.65\,A_{st} F_s\right)\left[1-\left(\dfrac{h}{140\,r}\right)^2\right] \quad (h/r \le 99)

Unreinforced flexure: tension governs

When a wall has no flexural reinforcement, masonry resists bending elastically over its net (or transformed) section and the limit is almost always the allowable flexural

\textit{tension}

, not compression. TMS 402-16 §8.2.4 tabulates allowable flexural tension by mortar type, grouting, and direction relative to the bed joints — for example a fully grouted hollow CMU with Type S Portland-cement-lime mortar allows on the order of

163\,\text{psi}

normal to bed joints. The allowable moment is simply

M = F_t\,S

with

S

the net section modulus. The

\textit{unreinforced}

flexural-compression allowable is

\tfrac{1}{3}f'_m

(§8.2.4.1); at

f'_m = 2000\,\text{psi}

that is about

667\,\text{psi}

, several times larger than the

\sim 163\,\text{psi}

tension values, so you check the tension face first and only verify compression if the result is close. Note this is the unreinforced value — once a flexural bar is added, the reinforced-flexure allowable becomes

F_b = 0.45 f'_m

f_b = \dfrac{M}{S} \le F_t, \qquad S = \dfrac{b\,t^2}{6}\ (\text{solid/fully grouted strip})

Reinforced flexure: the cracked transformed section

Add a flexural bar and the masonry is assumed cracked in tension; the analysis is the classic working-stress cracked-transformed section, identical in form to reinforced-concrete WSD but with masonry's

n

. Define the reinforcement ratio

\rho = A_s/(bd)

, then the neutral-axis depth ratio is

k = \sqrt{2\rho n + (\rho n)^2} - \rho n

and the internal lever-arm ratio is

j = 1 - k/3

. The allowable moment is the smaller of the steel-controlled and masonry-controlled values,

M_s = A_s F_s\,j\,d

and

M_m = \tfrac{1}{2}F_b\,k\,j\,b\,d^2

with

F_b = 0.45 f'_m

. The bar is normally centered in the wall, so

d = t/2

for a single curtain.

k = \sqrt{2\rho n + (\rho n)^2} - \rho n, \quad j = 1-\dfrac{k}{3}, \quad M_{\text{allow}} = \min\!\left(A_s F_s\,j\,d,\ \tfrac{1}{2}F_b\,k\,j\,b\,d^2\right)

Combined axial plus flexure: the unity equation

Walls and pilasters rarely see pure axial load or pure bending — wind plus a roof reaction produces both. TMS 402-16 §8.3.4.2.2 checks the demands with a linear interaction (unity) equation on the stresses,

f_a/F_a + f_b/F_b \le 1.0

, where

f_a = P/A_n

is the axial stress,

F_b = 0.45 f'_m

is the allowable flexural compressive stress, and

f_b

is the computed flexural compressive stress. The allowable axial stress is

F_a = \tfrac{1}{4}f'_m\left[1-(h/140r)^2\right]

for

h/r \le 99

(§8.2.4.1), the same one-quarter that leads

P_a

; it is

\textit{not}

the bending allowable. A second, independent check guards against buckling instability under eccentric axial load: the applied axial force must not exceed one-quarter of the eccentric Euler load,

P \le \tfrac{1}{4}P_e

. Both must pass.

\dfrac{f_a}{F_a} + \dfrac{f_b}{F_b} \le 1.0, \quad F_a = \tfrac{1}{4}f'_m\!\left[1-\left(\tfrac{h}{140r}\right)^2\right], \quad F_b = 0.45 f'_m \qquad\text{and}\qquad P \le \tfrac{1}{4}P_e

Exam strategy

Identify the mode first: unreinforced (check

f_b = M/S \le F_t

from the §8.2.4 table) versus reinforced (cracked

k

–

j

). Keep

f'_m

\textbf{psi}

everywhere and remember

F_s = 32\,\text{ksi}

and

E_m = 900 f'_m

. Memorize which coefficient goes where: reinforced flexure uses

F_b = 0.45 f'_m

(§8.3.4.2.2);

\tfrac{1}{3}f'_m

is the

\textit{unreinforced}

flexural-compression allowable, and the allowable axial stress is

F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]

. For axial force, always apply the

h/r

slenderness factor — a surprising fraction of capacity is lost to it — and use the net area

A_n

, not gross. For combined loading, write the unity equation and the

P \le \tfrac{1}{4}P_e

buckling check as two separate lines and do not stop at the first. When you see

\tfrac{1}{3}f'_m

offered as the

\textit{reinforced}

bending allowable, treat it as a distractor: the reinforced-flexure allowable is

0.45 f'_m

, and

\tfrac{1}{3}f'_m

belongs to unreinforced flexure only.

Key equations

Allowable flexural compressive stress (reinforced)

F_b = 0.45\,f'_m

Reinforced masonry ASD bending limit (TMS 402-16 §8.3.4.2.2, Eq. 8-21). $f'_m$ in psi. The $\tfrac{1}{3}f'_m$ coefficient is the unreinforced-flexure allowable, not this value (the axial coefficient is $\tfrac{1}{4}f'_m$ ).

Allowable steel tensile stress

F_s = 32\,\text{ksi}

Grade 60 reinforcement, ASD (Grade 40/50 → 20 ksi). Used in $M_s = A_s F_s j d$ .

Modular ratio

n = \dfrac{E_s}{E_m} = \dfrac{29{,}000\,\text{ksi}}{900\,f'_m}

$E_m = 900 f'_m$ for CMU. At $f'_m = 2000\,\text{psi}$ , $n \approx 16.1$ (much larger than concrete's ~8).

Radius of gyration, solid wall

r = \dfrac{t}{\sqrt{12}}

Net thickness $t$ (in). 8-in CMU: $t = 7.625\,\text{in}$ , $r \approx 2.20\,\text{in}$ .

Allowable axial force (slender)

P_a = \left(0.25 f'_m A_n + 0.65 A_{st} F_s\right)\!\left[1-\left(\tfrac{h}{140 r}\right)^2\right]

For $h/r \le 99$ . $A_n$ = net area, $A_{st}$ = longitudinal steel. (For $h/r > 99$ , use $(70r/h)^2$ .)

Axial capacity, very slender

P_a = \left(0.25 f'_m A_n + 0.65 A_{st} F_s\right)\!\left(\tfrac{70 r}{h}\right)^2

Use when $h/r > 99$ (Euler-controlled branch).

Allowable axial stress

F_a = \tfrac{1}{4}f'_m\!\left[1-\left(\tfrac{h}{140 r}\right)^2\right]

Allowable axial compressive stress, TMS 402-16 §8.2.4.1 ( $h/r \le 99$ ; above 99 use $\tfrac{1}{4}f'_m(70r/h)^2$ ). The reinforced-masonry check (§8.3.4.2.2) requires $f_a \le F_a$ . It is the $f_a/F_a$ denominator in the unity equation, not a bending allowable.

Unreinforced flexure check

f_b = \dfrac{M}{S} \le F_t

$F_t$ = allowable flexural tension from TMS 402-16 Table 8.2.4.2 (mortar/grout/direction); $S$ = net section modulus. Unreinforced flexural-compression allowable is $\tfrac{1}{3}f'_m$ (§8.2.4.1).

Cracked neutral-axis ratio

k = \sqrt{2\rho n + (\rho n)^2} - \rho n, \quad \rho = \dfrac{A_s}{bd}

Working-stress cracked transformed section for reinforced flexure.

Lever-arm ratio

j = 1 - \dfrac{k}{3}

Internal moment arm $= jd$ . Often $j \approx 0.9$ for typical wall reinforcement.

Allowable reinforced moment

M_{\text{allow}} = \min\!\left(A_s F_s\,j\,d,\ \tfrac{1}{2}F_b\,k\,j\,b\,d^2\right), \quad F_b = 0.45 f'_m

Smaller of steel-controlled and masonry-controlled allowable moment; masonry side uses $F_b = 0.45 f'_m$ .

Unity (interaction) equation

\dfrac{f_a}{F_a} + \dfrac{f_b}{F_b} \le 1.0

$f_a = P/A_n$ , $F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]$ , $f_b$ = flexural compressive stress, $F_b = 0.45 f'_m$ .

Eccentric Euler buckling check

P \le \tfrac{1}{4}P_e, \quad P_e = \dfrac{\pi^2 E_m I_n}{h^2}\!\left(1 - 0.577\tfrac{e}{r}\right)^{3}

Independent stability check on eccentric axial load; $e$ = load eccentricity, $I_n$ = net moment of inertia.

Worked examples

Allowable moment of a reinforced CMU wall (out-of-plane)

Problem. An 8-in nominal, fully grouted CMU wall ( $t = 7.625\,\text{in}$ , $f'_m = 2000\,\text{psi}$ ) is reinforced with one $\#5$ bar at $24\,\text{in}$ on center, centered in the wall (Grade 60). Find the allowable out-of-plane moment per foot of wall by ASD.

Solution. Design a 1-ft strip:

b = 12\,\text{in}

d = t/2 = 3.81\,\text{in}

. Steel per foot:

A_s = 0.31\,(12/24) = 0.155\,\text{in}^2/\text{ft}

. Material constants:

F_b = 0.45 f'_m = 0.45(2000) = 900\,\text{psi}

F_s = 32{,}000\,\text{psi}

E_m = 900(2000) = 1{,}800{,}000\,\text{psi}

n = 29{,}000{,}000/1{,}800{,}000 = 16.1

. Cracked section:

\rho = 0.155/(12 \cdot 3.81) = 0.00339

;

\rho n = 0.0546

k = \sqrt{2(0.0546) + 0.0546^2} - 0.0546 = 0.280

;

j = 1 - 0.280/3 = 0.907

. Steel-controlled:

M_s = A_s F_s j d = 0.155(32{,}000)(0.907)(3.81) = 17{,}140\,\text{lb-in} = 1.43\,\text{k-ft/ft}

. Masonry-controlled:

M_m = \tfrac{1}{2}F_b k j b d^2 = 0.5(900)(0.280)(0.907)(12)(3.81)^2 = 19{,}950\,\text{lb-in} = 1.66\,\text{k-ft/ft}

. Governing

M_{\text{allow}} = \min(1.43, 1.66) = \mathbf{1.43\ k\text{-}ft/ft}

(steel controls). Sanity:

j \approx 0.91

is the expected range, and with the correct

F_b = 0.45 f'_m = 900\,\text{psi}

the masonry side (

1.66

) sits above the steel side, so a lightly reinforced single-curtain wall is steel-controlled here — using the wrong

\tfrac{1}{3}f'_m = 667\,\text{psi}

would have pushed

M_m

down to

1.23

and flipped the answer.

M_{\text{allow}} = \min\!\big(A_s F_s j d,\ \tfrac{1}{2}F_b k j b d^2\big) = \min(1.43,\ 1.66) = 1.43\ \text{k-ft/ft (steel controls)}

Allowable axial load of a grouted pilaster with slenderness

Problem. A 16-in nominal, fully grouted CMU pilaster ( $t = 15.625\,\text{in}$ square, $f'_m = 2000\,\text{psi}$ ) is reinforced with four $\#8$ vertical bars (Grade 60) and is $14\,\text{ft}$ tall, pinned top and bottom. Find the allowable axial load $P_a$ .

Solution. Geometry:

A_n = 15.625^2 = 244\,\text{in}^2

;

A_{st} = 4(0.79) = 3.16\,\text{in}^2

;

r = t/\sqrt{12} = 15.625/3.464 = 4.51\,\text{in}

. Slenderness:

h = 14(12) = 168\,\text{in}

;

h/r = 168/4.51 = 37.2 \le 99

, so use the parabolic branch. Factor

= 1 - (168/(140 \cdot 4.51))^2 = 1 - (0.266)^2 = 0.929

. Capacity:

P_a = [0.25(2000)(244) + 0.65(3.16)(32{,}000)](0.929) = [122{,}000 + 65{,}730](0.929) = 187{,}730(0.929) = 174{,}400\,\text{lb}

P_a = \mathbf{174\ kips}

. Sanity: slenderness trims about 7% off the squash capacity, reasonable for

h/r \approx 37

; net area (not gross) was used throughout.

P_a = [0.25(2000)(244) + 0.65(3.16)(32{,}000)]\left[1-\left(\tfrac{168}{140(4.51)}\right)^2\right] = 174\ \text{kips}

Combined axial and flexure (unity) on the pilaster

Problem. The pilaster of the previous example carries a service axial load $P = 60\,\text{kips}$ applied at an eccentricity $e = 2.0\,\text{in}$ (so $M = P e = 120\,\text{k-in}$ ). Check the unity interaction and the buckling stability limit. Take $f_b$ as the gross flexural compressive stress on the net section.

Solution. Axial side:

f_a = P/A_n = 60{,}000/244 = 246\,\text{psi}

. The allowable axial stress is the §8.2.4.1 value

F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2] = \tfrac{1}{4}(2000)(0.929) = 465\,\text{psi}

(not

P_a/A_n

). Ratio

f_a/F_a = 246/465 = 0.529

. Flexure side:

I_n = t^4/12 = 15.625^4/12 = 4{,}967\,\text{in}^4

;

S = I_n/(t/2) = 4{,}967/7.81 = 636\,\text{in}^3

f_b = M/S = 120{,}000/636 = 189\,\text{psi}

;

F_b = 0.45 f'_m = 900\,\text{psi}

; ratio

f_b/F_b = 189/900 = 0.210

. Unity:

0.529 + 0.210 = 0.739 \le 1.0

, OK. Buckling:

P_e = \dfrac{\pi^2 E_m I_n}{h^2}(1 - 0.577 e/r)^3 = \dfrac{\pi^2 (1.8\times10^6)(4967)}{168^2}(1 - 0.577(2.0)/4.51)^3

. The bracket

= (1 - 0.256)^3 = 0.412

, and

\pi^2(1.8\times10^6)(4967)/168^2 = 3{,}126{,}000\,\text{lb}

, so

P_e = 3{,}126{,}000(0.412) = 1{,}288\,\text{kips}

. Limit

\tfrac{1}{4}P_e = 322\,\text{kips} \ge 60\,\text{kips}

— OK. Both pass; the pilaster is adequate (

\mathbf{0.74 \le 1.0}

and

60 \le 322\,\text{kips}

). Sanity: the interaction sum under 1.0 with ample buckling margin matches a lightly loaded, stocky pilaster.

\dfrac{f_a}{F_a} + \dfrac{f_b}{F_b} = \dfrac{246}{465} + \dfrac{189}{900} = 0.529 + 0.210 = 0.74 \le 1.0

Common pitfalls

•Using $\tfrac{1}{3}f'_m$ as the reinforced bending allowable. The TMS 402-16 §8.3.4.2.2 reinforced-flexure allowable is $F_b = 0.45 f'_m$ . The $\tfrac{1}{3}f'_m$ coefficient is the unreinforced flexural-compression allowable (§8.2.4.1), never the reinforced bending limit; the allowable axial stress uses $\tfrac{1}{4}f'_m$ .
•Citing a handbook page for masonry. The NCEES PE Civil Reference Handbook contains no masonry design — cite TMS 402-16 (the supplied edition), and never TMS 402-22.
•Dropping the slenderness factor on axial capacity. $P_a$ always carries the $[1-(h/140r)^2]$ (or $(70r/h)^2$ ) reduction; omitting it overstates capacity by 5–40%.
•Borrowing concrete's modular ratio or stress block. Masonry uses $E_m = 900 f'_m$ giving $n \approx 16$ (not 8), and ASD masonry flexure uses the elastic cracked $k$ – $j$ method, not the $0.85 f'_m$ stress block.
•Defining $F_a$ as $P_a/A_n$ in the unity equation. The code allowable axial stress is $F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]$ (§8.2.4.1); using $P_a/A_n$ overstates $F_a$ (it folds in the steel term) and is not the §8.3.4.2.2 definition.
•Keeping $f'_m$ in ksi. $F_b = 0.45 f'_m$ , $E_m = 900 f'_m$ , and the §8.2.4 tension table are all in psi; mixing ksi and psi corrupts $n$ and every allowable.
•Mixing ASD and strength design. The unity equation, $F_b$ , $F_s = 32\,\text{ksi}$ , and $P_a$ are service-load ASD; do not combine them with factored loads or $\phi$ from §9 strength design.
•Checking only the unity equation and skipping $P \le \tfrac{1}{4}P_e$ . Eccentric slender members can satisfy the stress interaction yet fail the independent buckling-stability limit.

References

TMS 402-16 — Building Code Requirements for Masonry Structures, §8.2 (Unreinforced) and §8.3 (Reinforced), Allowable Stress Design
TMS 402-16 §8.3.4.2.2 — Reinforced flexural compressive stress $F_b = 0.45 f'_m$ (Eq. 8-21) and combined unity interaction
TMS 402-16 §8.2.4.1: allowable axial stress $F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]$ ; §8.3.4.2.1: allowable axial force $P_a$
TMS 402-16 §8.2.4: unreinforced allowables (flexural compression $\tfrac{1}{3}f'_m$ in §8.2.4.1, flexural tension table in §8.2.4.2)
ASCE 7-16 — Minimum Design Loads (service-load combinations applied to ASD masonry checks) — Loads come from ASCE 7-16; the resistance side is TMS 402-16.

Masonry Shear & Reinforcement

ASD shear in concrete masonry per TMS 402-16: the shear stress, the masonry and steel allowable-shear contributions, shear and flexural reinforcement sizing, and minimum/maximum steel and spacing.

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Slender Walls & Brick Veneer

Out-of-plane slender-wall design per TMS 402-16 §9.3.5: P-delta moment magnification, the service-deflection limit and slenderness gate, and anchored brick-veneer ties.

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