Masonry Design · Study · PE Civil: Structural · FE → PE Prep
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
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 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 0.45fm′
with the
31fm′
coefficient used for unreinforced flexure and for 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 fm′. For reinforced masonry in bending, TMS 402-16 §8.3.4.2.2 (Eq. 8-21) sets the allowable flexural compressive stress at Fb=0.45fm′. Do not confuse this with the 31fm′ coefficient, which appears in two other places: it is the allowable flexural compressive stress for unreinforced masonry (§8.2.4.2), and it is the leading coefficient of the reinforced allowable axial stress, Fa=31fm′[1−(h/140r)2] (§8.3.4.2.1). So 31fm′ is real and useful — just not the reinforced bending allowable. (The 0.25fm′ net-area term and the 0.65 steel coefficient in the allowable axial-force equation Pa are still other code-calibrated coefficients — not a flexural allowable.) The allowable tensile stress in Grade 60 reinforcement is Fs=32,000psi (32 ksi). The modular ratio uses Em=900fm′ for concrete masonry and Es=29,000ksi, so n=Es/Em — for fm′=2000psi, n≈16.1, far larger than concrete's n≈8.
A masonry wall or column carries axial load through both the net masonry area An and any longitudinal steel Ast, 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≤99 the factor is [1−(h/140r)2]; for h/r>99 it switches to the Euler-like (70r/h)2. The allowable axial force is then Pa below. For a solid rectangular wall of actual thickness t, r=t/12, so an 8-in CMU wall (t=7.625in) has r≈2.20in and crosses h/r=99 at roughly an 18-ft height.
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 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 163psi normal to bed joints. The allowable moment is simply M=FtS with S the net section modulus. The unreinforced flexural-compression allowable is 31fm′ (§8.2.4.2); at fm′=2000psi that is about 667psi, several times larger than the ∼163psi 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 Fb=0.45fm′.
fb=SM≤Ft,S=6bt2(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 ρ=As/(bd), then the neutral-axis depth ratio is k=2ρn+(ρn)2−ρn 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, Ms=AsFsjd and Mm=21Fbkjbd2 with Fb=0.45fm′. The bar is normally centered in the wall, so d=t/2 for a single curtain.
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, fa/Fa+fb/Fb≤1.0, where fa=P/An is the axial stress, Fb=0.45fm′ is the allowable flexural compressive stress, and fb is the computed flexural compressive stress. The allowable axial stress is Fa=31fm′[1−(h/140r)2] for h/r≤99 per §8.3.4.2.1 — this is the 31fm′ coefficient at work, 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≤41Pe. Both must pass.
Identify the mode first: unreinforced (check fb=M/S≤Ft from the §8.2.4 table) versus reinforced (cracked k–j). Keep fm′ in psi everywhere and remember Fs=32ksi and Em=900fm′. Memorize which coefficient goes where: reinforced flexure uses Fb=0.45fm′ (§8.3.4.2.2); the 31fm′ coefficient is the unreinforced flexural-compression allowable and the leading coefficient of the allowable axial stress Fa=31fm′[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 An, not gross. For combined loading, write the unity equation and the P≤41Pe buckling check as two separate lines and do not stop at the first. When you see 31fm′ offered as the reinforced bending allowable, treat it as a distractor: the reinforced-flexure allowable is 0.45fm′, and 31fm′ belongs to unreinforced flexure and to Fa.
Reinforced masonry ASD bending limit (TMS 402-16 §8.3.4.2.2, Eq. 8-21). fm′ in psi. The 31fm′ coefficient is the unreinforced-flexure allowable and the Fa axial coefficient, not this value.
Allowable steel tensile stressFs=32ksi
Grade 60 reinforcement, ASD (Grade 40/50 → 20 ksi). Used in Ms=AsFsjd
Modular ration=EmEs=900fm′29,000ksi
Radius of gyration, solid wallr=12t
Net thickness
Allowable axial force (slender)Pa=(0.25fm′An+0.65AstFs)[1−(140rh)2]
Axial capacity, very slenderPa=(0.25fm′An+0.65AstFs)(h70r)2
Allowable moment of a reinforced CMU wall (out-of-plane)
Problem. An 8-in nominal, fully grouted CMU wall (t=7.625in, fm′=2000psi) is reinforced with one #5 bar at 24in 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=12in, d=t/2=3.81in. Steel per foot: As=0.31(12/24)=0.155in2/ft
Allowable axial load of a grouted pilaster with slenderness
Problem. A 16-in nominal, fully grouted CMU pilaster (t=15.625in square, fm′=2000psi) is reinforced with four #8
Combined axial and flexure (unity) on the pilaster
Problem. The pilaster of the previous example carries a service axial load P=60kips applied at an eccentricity e=2.0in (so M=Pe=120k-in
Common pitfalls
•Using 31fm′ as the reinforced bending allowable. The TMS 402-16 §8.3.4.2.2 reinforced-flexure allowable is Fb=0.45fm′. The 31fm′ coefficient is the unreinforced flexural-compression allowable (§8.2.4.2) and the leading coefficient of the allowable axial stress Fa (§8.3.4.2.1) — never the reinforced bending limit.
•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. Pa always carries the [1−(h/140r)2] (or (70r/h)2
•Borrowing concrete's modular ratio or stress block. Masonry uses Em=900fm′ giving n≈16
•Defining Fa as Pa/An in the unity equation. The code allowable axial stress is Fa=31fm′[1−(h/140r)2]
•Keeping fm′ in ksi. Fb=0.45fm′
•Mixing ASD and strength design. The unity equation, Fb, Fs=32ksi, and Pa
•Checking only the unity equation and skipping P≤41Pe. 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
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.
Unlocks with an access pass — one-time payment, no auto-renew. View passes
Pass holders
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.
Unlocks with an access pass — one-time payment, no auto-renew. View passes
Pass holders
.
Em=900fm′ for CMU. At fm′=2000psi, n≈16.1 (much larger than concrete's ~8).
t
(in). 8-in CMU:
t=7.625in
,
r≈2.20in
.
For h/r≤99. An = net area, Ast = longitudinal steel. (For h/r>99, use (70r/h)2.)
Use when h/r>99 (Euler-controlled branch).
Reinforced allowable axial stress (TMS 402-16 §8.3.4.2.1, h/r≤99). This is the 31fm′ coefficient — used in the unity equation as the fa/Fa denominator, not as the bending allowable.
Ft = allowable flexural tension from TMS 402-16 Table 8.2.4.2 (mortar/grout/direction); S = net section modulus. Unreinforced flexural-compression allowable is 31fm′ (§8.2.4.2).
Working-stress cracked transformed section for reinforced flexure.
for typical wall reinforcement.
Smaller of steel-controlled and masonry-controlled allowable moment; masonry side uses Fb=0.45fm′.
) sits above the steel side, so a lightly reinforced single-curtain wall is steel-controlled here — using the wrong
31fm′=667psi
would have pushed
Mm
down to
1.23
and flipped the answer.
vertical bars (Grade 60) and is
14ft
tall, pinned top and bottom. Find the allowable axial load
Pa
.
Solution. Geometry: An=15.6252=244in2; Ast=4(0.79)=3.16in2; r=t/12=15.625/3.464=4.51in.
Slenderness: h=14(12)=168in; h/r=168/4.51=37.2≤99, so use the parabolic branch. Factor =1−(168/(140⋅4.51))2=1−(0.266)2=0.929.
Capacity: Pa=[0.25(2000)(244)+0.65(3.16)(32,000)](0.929)=[122,000+65,730](0.929)=187,730(0.929)=174,400lb.
Pa=174kips. Sanity: slenderness trims about 7% off the squash capacity, reasonable for h/r≈37; net area (not gross) was used throughout.
). Check the unity interaction and the buckling stability limit. Take
fb
as the gross flexural compressive stress on the net section.
Solution. Axial side: fa=P/An=60,000/244=246psi. The allowable axial stress is the code §8.3.4.2.1 value Fa=31fm′[1−(h/140r)2]=31(2000)(0.929)=619psi (not Pa/An). Ratio fa/Fa=246/619=0.397.
Flexure side: In=t4/12=15.6254/12=4,967in4; S=In/(t/2)=4,967/7.81=636in3. fb=M/S=120,000/636=189psi; Fb=0.45fm′=900psi; ratio fb/Fb=189/900=0.210.
Unity: 0.397+0.210=0.607≤1.0 — OK.
Buckling: Pe=h2π2EmIn(1−0.577e/r)3=1682π2(1.8×106)(4967)(1−0.577(2.0)/4.51)3. The bracket =(1−0.256)3=0.412, and π2(1.8×106)(4967)/1682=3,126,000lb, so Pe=3,126,000(0.412)=1,288kips. Limit 41Pe=322kips≥60kips — OK.
Both pass; the pilaster is adequate (0.61≤1.0 and 60≤322kips). Sanity: the interaction sum under 1.0 with ample buckling margin matches a lightly loaded, stocky pilaster.