e702.1 allowable stress design of rc masonry shear wall 2006-03-29

8/9/2019 E702.1 Allowable Stress Design of RC Masonry Shear Wall 2006-03-29

1/17

American Concrete Institute®

Advancing concrete knowledge

Designing Concrete Structures:

E702.1

Allowable-Stress Design of Reinforced

Concrete Masonry Shear Wall


2/17

March 29, 2006 E702 Example Problems -- Concrete Masonry Shear Wall R. E. Klingner

1

Example #1: Allowable-Stress Design of Reinforced Concrete Masonry Shear Wall

Using the allowable-stress provisions of the 2005 MSJC Code (ACI 530), design the reinforced concrete

masonry shear wall shown below. In the context of an entire building design, design loads would be

obtained using the appropriate provisions of the legally adopted building code (which usually references

ASCE 7). For this facet of the design, unfactored loads are shown below, and are assumed to be due to

earthquake. Corresponding shear and moment diagrams are also shown.

24 ft

10 ft

10 ft

10 ft

10 ft

1 224 ft

10 ft

10 ft

10 ft

10 ft

1 2

30

60

90

120

300

900

1800

3000

Lateral Loads Shear, kips Moment, kip-ft

30 kips

30 kips

30 kips

30 kips

30

60

90

120

300

900

1800

3000


30 kips

30 kips

30 kips

30 kips


3/17


2

Assume an 8-in. nominal concrete masonry wall, grouted solid, with Type S PCL mortar. The total

plan length of the wall is 24 ft (288 in.), and its specified thickness is 7.63 in. Assume an effective

depth d of 285 in.

Concrete MasonryUnit Strength (C90) 1900 psi

Mortar Type S

f ′ m or f g 1,500 psi

E m or E g 1.35 × 106 psi

Reinforcement = Grade 60; E s = 29 × 106 psi

Unfactored axial loads on the wall are given in the table below.

Level DL LL

(Top of Wall) (kips) (kips)

4 90 15

3 180 35

2 270 55

1 360 75

Check shear for the assumed wall thickness.

Use ASCE 7-05 ASD Load Combination 5: 0.6 D + 0.7 E

V = 0.7 x 120,000 lb = 84,000 lb

M = 0.7 x 3000 kip-ft = 2100 kip-ft

P = 0.6 x 360 kips = 216 kips

By Code Section 2.3.5.2.1,

psi6.3828563.7

lb000,84=

×==

bd

V f v

05.1

.in285kips201

/ft.in12ftkip3000=

×

×−=

Vd

M

By Code Section 2.3.5.2.2(b), where shear reinforcement is not provided to resist the calculated

shear:

psi1.381500)05.14(4 3131 =−=′⎟ ⎠

⎞⎜⎝

⎛ −= mv f

Vd

M F


4/17


3

but [ ] GOVERNS 8.32)05.1(4580)(4580 ⇐=−=−≤ psiVd M F v

needed isentreinforcemShear psi6.38 ∴>=vv F f

Now check against the higher allowable shear stresses of Code Section 2.3.5.2.3:

GOVERNS psi13.571500)05.14(4 2121 ⇐=−=′⎟ ⎠

⎞⎜⎝

⎛ −= mv f

Vd

M F

but [ ] OK psi38.6 psi75.72)/(45120 >=−≤ Vd M F v

Design shear reinforcement using Code Section 2.3.2.1 and Code Section 2.3.5.3:

F s = 24,000 psi

2in.29.0.285000,24

.24000,84=

××

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

in psi

inkips

d F

Vs A

s

v

Use No. 5 bars horizontally at 24 inches on center

Now consider the flexural design. The attached spreadsheet for this problem illustrates how to

calculate an allowable-stress moment-axial force interaction diagram for this wall. That spreadsheet

is first used to check the wall with reinforcement consisting of #5 bars @ 4 ft. Because the space

between extreme bars is not quite a multiple of 48 in., the spacing of the interior bars is adjusted

slightly to fit the 16-in. nominal length of the concrete masonry units.

Depending on the building’s Seismic Design Category, prescriptive seismic provisions might require

more reinforcement or a closer spacing of reinforcement than those calculated above.


5/17


4

Al lowable-Stress Interact ion Diagram by Spreadsheet

Concrete Masonry Shear Wall of Example Problem

f'm=1500 psi, 24 ft long, 7.63 in. thick, #5 bars @ 4'

0

100

200

300

400

500

600

700

800

900

0 500 1000 1500 2000 2500 3000

Al lowable M, f t-k ips

A l l o w a b l e P ,

k i p s

The interaction diagram shows that the wall, as designed above, can resist the combination of axial

load and moment (216 kips, 2100 kip-ft).


6/17


5

Example #2: Allowable-Stress Design of Reinforced Concrete Masonry Shear Wall (SI Units)

Using the allowable-stress provisions of the 2005 MSJC Code (ACI 530), design the reinforced concrete





7.32 m

3.05 m

3.05 m

3.05 m

3.05 m

1 27.32 m

3.05 m

3.05 m

3.05 m

3.05 m

1 2

133.44

266.88

400.32

533.76

406.8

1220.4

2440.8

4068

Lateral Loads Shear, kN Moment, kN-m

133.44 kN

133.44 kN

133.44 kN

133.44 kN

133.44

266.88

400.32

533.76

406.8

1220.4

2440.8

4068


133.44 kN

133.44 kN

133.44 kN

133.44 kN


7/17


6

Assume an 203.2-mm nominal concrete masonry wall, grouted solid, with Type S PCL mortar. The

total plan length of the wall is 7.315 m, and its specified thickness is 193.8 mm. Assume an effective

depth d of 7.239 m.

Concrete MasonryUnit Strength (C90) 13.10 MPa

Mortar Type S

f ′ m or f g 10.34 MPa

E m or E g 9308 MPa

Reinforcement = Grade 60; E s = 199,955 MPa


Level DL LL

(Top of Wall) (kN) (kN)

4 400.32 66.72

3 800.64 155.68

2 1200.96 244.64

1 1601.28 333.60

Check shear for the assumed wall thickness.

Use ASCE 7-05 ASD Load Combination 5: 0.6 D + 0.7 E

V = 0.7 x 533.76 kN = 373.63 kN

M = 0.7 x 4068 kN-m = 2847.6 kN-m

P = 0.6 x 1601 kN = 960.77 kN

By Code Section 2.3.5.2.1,

MPa26632.072398.193

kN63.3732 =

×==

mmbd

V f v

05.1

239.763.373

2847=

×

−=

mkN

mkN

Vd

M

By Code Section 2.3.5.2.2(b), where shear reinforcement is not provided to resist the calculated

shear:

MPa26.034.10)05.14(028.04028.0 =−=′⎟ ⎠

⎞⎜⎝

⎛ −= mv f

Vd

M F


8/17


7

but [ ] GOVERNS 22.0)05.1(31.055.0)(31.055.0 ⇐=−=−≤ MPaVd M F v

needed isentreinforcemShearMPai266.0 ∴>=vv F f

Now check against the higher allowable shear stresses of Code Section 2.3.5.2.3:

GOVERNS MPa40.034.10)05.14(42.04042.0 ⇐=−=′⎟ ⎠

⎞⎜⎝

⎛ −= mv f Vd

M F

but [ ] OK MPa266.0MPa49.0)/(31.082.0 =>=−≤ vv f Vd M F

Design shear reinforcement using Code Section 2.3.2.1 and Code Section 2.3.5.3:

F s = 165.48 MPa

2mm190723948.165

6.60963.373=

×

×=⎟⎟

⎠

⎞⎜⎜⎝

⎛ =

mm MPa

mmkN

d F

Vs A

s

v

Use 16-mm bars horizontally at 609.6 mm on center

Now consider the flexural design. The attached spreadsheet for this problem illustrates how to

calculate an allowable-stress moment-axial force interaction diagram for this wall. That spreadsheet

is first used to check the wall with reinforcement consisting of 16-mm bars @ 1.22 m. Because the

space between extreme bars is not quite a multiple of 1.22 m., the spacing of the interior bars is

adjusted slightly to fit the 406.4-mm nominal length of the concrete masonry units.

Depending on the building’s Seismic Design Category, prescriptive seismic provisions might require

more reinforcement or a closer spacing of reinforcement than those calculated above.


9/17


8

Al lowable-Stress Interact ion Diagram by Spreadsheet (SI)


f'm=10.34 MPa, 7.32 m long, 193.8 mm thick, 16-mm bars @ 1.22 m

0

500

1000

1500

2000

2500

3000

3500

4000

0 500 1000 1500 2000 2500 3000 3500 4000

Al lowable M, kN-m

A l l o w a b l e P ,

k N

The interaction diagram shows that the wall, as designed above, can resist the combination of axial load

and moment (960.77 kN, 2847.6 kN-m).


10/17


9

Example #3: Strength Design of Reinforced Concrete Masonry Shear Wall

Using the strength provisions of the 2005 MSJC Code (ACI 530), design the reinforced concrete





24 ft

10 ft

10 ft

10 ft

10 ft

1 224 ft

10 ft

10 ft

10 ft

10 ft

1 2

30

60

90

120

300

900

1800

3000


30 kips

30 kips

30 kips

30 kips

30

60

90

120

300

900

1800

3000


30 kips

30 kips

30 kips

30 kips


11/17


10

Assume an 8-in. nominal concrete masonry wall, grouted solid, with Type S PCL mortar. The total

plan length of the wall is 24 ft (288 in.), and its specified thickness is 7.63 in. Assume an effective

depth d of 285 in.

Concrete MasonryUnit Strength 1,900

Mortar Type S

f ′ m or f g (psi) 1,500

Reinforcement = Grade 60; E s = 29 × 106 psi


Level DL LL

(Top of Wall) (kips) (kips)

4 90 15

3 180 35

2 270 55

1 360 75

Use ASCE 7-05 Basic Strength Load Combination 7: 0.9 D + 1.0 E

Check shear for assumed wall thickness. By Section 3.3.4.1.2 of the 2005 MSJC Code,

smn

V V V +=

M u = 3,000 x 12 x 1,000 in.-lb = 36.0 x 106 in.-lb

V u = 120,000 lb d v = 285 in.

05.1in.)(285lb120,000

lb-in.1036/

6

=×

=vuu d V M

By Code 3.3.4.1.2.1, (Mu / Vu d v) need not be taken greater than 1.0.

P f Ad V

M V

mnvu

u

m

25.0'75.10.4 +⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −=

( )[ ]

kipskipsV

ininV

m

m

5.2790.905.189

lb)000,360(25.0 psi1500.285.63.70.175.10.4

=+=

+×−=

un V V >φ 80.0=φ kipsV V mn 5.279==


12/17


11

kipsV kipskips u 1206.223)5.279(80.0 =≥=

Shear design is satisfactory so far, even without shear reinforcement. Code Section 3.1.3 will be checked

later.

Now check flexural capacity using a spreadsheet-generated moment-axial force interaction diagram. Try

#5 bars @ 4 ft.

Strength Interaction Diagram by Spreadsheet


f'm=1500 psi, 24 ft long, 7.63 in. thick , #5 bars @ 4'

0

200

400

600

800

1000

1200

1400

1600

1800

2000

0 1000 2000 3000 4000 5000 6000 7000 8000

Mn, ft-kips

P n ,

k i p s

At a factored axial load of 0.9D, or 0.9 x 360 kips = 324 kips, the design flexural capacity of this wall is

about 4000 ft-kips, and the design is satisfactory for flexure.

Now check Code Section 3.1.3. First try to meet the capacity design provisions of that section. At an

axial load of 324 kips, the nominal flexural capacity of this wall is 4000 ft-kips, divided by the strength

reduction factor of 0.9, or 4,444 ft-kips. The ratio of this nominal flexural capacity to the factored designmoment is 4,444 divided by 3,000, or 1.48. Including the additional factor of 1.25, that gives a ratio of

1.85.

kipsV V V V

V V

uuun

un

5.27712031.231.28.0

85.185.1

85.1

=×===≥

≥

φ

φ


13/17


12

The wall requires shear reinforcement. Prescriptive seismic reinforcement (for example, for SDC C) will

probably be satisfied with #5 bars horizontally @ 24 in.

kipskipsV

ininksiin

sd f AkipsV V V

n

yvsmn

9.3894.1105.329

.24.28560.31.0

215.279

215.379 2

=+=

×⎟ ⎠ ⎞⎜

⎝ ⎛ +=⎟

⎠ ⎞⎜

⎝ ⎛ +=+=

Prescriptive seismic reinforcement is sufficient for shear. Use #5 bars at 2 ft.

Check ρmax , assuming that the wall is classified as an intermediate reinforced masonry shear wall.

smu

ymu

mu

ymu

y

y

u

yma

mum

E f

bd

P f

ε ε ε

ε

ε ε

ε ε ε

ε

ρ

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛

+

−⎟⎟

⎠

⎞⎜⎜

⎝

⎛

+=

35.0

3

33

64.0 '

max

In accordance with MSJC Code Section 3.3.3.5.1(d), the governing axial load combination is

D + 0.75 L + 0.525 QE , and the axial load is (360,000 + 0.75 x 75,000 lb), or 416,250 lb.

( )

00127.0

10290025.0)00207.0(30025.0

0025.0

5.0(0.00207)30.0025

3(0.00207)

psi)000,60(

in.285in.)(7.63

lb250,416

(0.00207)30.0025

0.0025 psi)500,1(64.0

max

6

max

=

××⎥⎦

⎤

⎢⎣

⎡

+−⎥⎦

⎤

⎢⎣

⎡

+

−⎥⎦

⎤⎢⎣

⎡

+=

ρ

ρ

psi

Check maximum area of flexural reinforcement per 48 in. of wall length

2

maxmax in.0.46in.48in.)63.7(00127.0.48 ==×= inb As ρ

We have 0.31 in.2 every 48 in., and the design is satisfactory.

Summary:

Use #5 @ 4 ft vertically, #5 @ 2 ft horizontally.

Because the space between extreme bars is not quite a multiple of 48 in., the spacing of the interior

bars is adjusted slightly to fit the nominal 16-in. length of the concrete masonry units.


14/17


13

Example #4: Strength Design of Reinforced Concrete Masonry Shear Wall (SI Units)

Using the strength provisions of the 2005 MSJC Code (ACI 530), design the reinforced concrete





7.32 m

3.05 m

3.05 m

3.05 m

3.05 m

1 27.32 m

3.05 m

3.05 m

3.05 m

3.05 m

1 2

133.44

266.88

400.32

533.76

406.8

1220.4

2440.8

4068


133.44 kN

133.44 kN

133.44 kN

133.44 kN

133.44

266.88

400.32

533.76

406.8

1220.4

2440.8

4068


133.44 kN

133.44 kN

133.44 kN

133.44 kN


15/17


14

Assume an 203.2-mm nominal concrete masonry wall, grouted solid, with Type S PCL mortar. The

total plan length of the wall is 7.315 m, and its specified thickness is 193.8 mm. Assume an effective

depth d of 7.239 m.

Concrete MasonryUnit Strength 13.10 MPa

Mortar Type S

f ′ m or f g 10.34 MPa

Reinforcement = Grade 60; E s = 199,955 MPa


Level DL LL

(Top of Wall) (kN) (kN)

4 400.32 66.72

3 800.64 155.68

2 1200.96 244.64

1 1601.28 333.60

Use ASCE 7-05 Basic Strength Load Combination 7: 0.9 D + 1.0 E

Check shear for assumed wall thickness. By Section 3.3.4.1.2 of the 2005 MSJC Code,

smn

V V V +=

05.1m)(7.239kN533.76

m-kN4068/ ==vuu d V M

By Code 3.3.4.1.2.1, (Mu / Vu d v) need not be taken greater than 1.0.

P f Ad V

M V mn

vu

um 25.0'75.10.4083.0 +

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

( )[ ] ( )kN kN V mmmmV

m

m

5.12424005.842kN)1601(25.0MPa34.1072398.1930.175.10.4083.0

=+=+×−=

un V V >φ 80.0=φ kN V V mn 5.1242==

kN V kN kN u 76.5330.994)5.1242(80.0 =≥=


16/17


15

Shear design is satisfactory so far, even without shear reinforcement. Code Section 3.1.3 will be checked

later.

Now check flexural capacity using a spreadsheet-generated moment-axial force interaction diagram. Try

16-mm bars @ 1.22 m.

Strength Interaction Diagram by Spreadsheet (SI units)


f'm=10.34 MPa, 7.32 m long, 193.8 in. thick, 16-mm bars @ 1.22 m

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 2000 4000 6000 8000 10000 12000

Mn, kN-m

P n ,

k N

At a factored axial load of 0.9D, or 0.9 x 1601 kN = 1441 kN, the design flexural capacity of this wall is

about 5400 kN-m, and the design is satisfactory for flexure.

Now check Code Section 3.1.3. First try to meet the capacity design provisions of that section. At an

axial load of 1441 kN, the nominal flexural capacity of this wall is 5424 kN-m, divided by the strength

reduction factor of 0.9, or 6,027 kN-m. The ratio of this nominal flexural capacity to the factored design

moment is 6,027 divided by 4068, or 1.48. Including the additional factor of 1.25, that gives a ratio of

1.85.

kN V V V V

V V

uuun

un

123376.53331.231.28.0

85.185.1

85.1

=×===≥

≥

φ

φ

The wall requires shear reinforcement. Prescriptive seismic reinforcement (for example, for SDC C) will

probably be satisfied with 16-mm bars horizontally @ 609.6 mm.


17/17


16

kN kipsV

mm

mm MPamm

s

d f AkN V V V

n

yvsmn

17341.4915.1242

6.609

72397.4139.197

2

15.1242

2

15.1242 2

=+=

×⎟ ⎠

⎞⎜⎝

⎛ +=⎟ ⎠

⎞⎜⎝

⎛ +=+=

Prescriptive seismic reinforcement is sufficient for shear. Use 16-mm bars at 609.6 mm.

Check ρmax , assuming that the wall is classified as an intermediate reinforced masonry shear wall.

smu

ymu

mu

ymu

y

y

u

yma

mum

E f

bd

P f

ε

ε ε

ε

ε ε

ε

ε ε

ε

ρ

⎟⎟

⎠

⎞⎜⎜

⎝

⎛

+

−⎟⎟

⎠

⎞⎜⎜

⎝

⎛

+

−⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+=

3

5.0

3

3

364.0

'

max

In accordance with MSJC Code Section 3.3.3.5.1(d), the governing axial load combination is

D + 0.75 L + 0.525 QE , and the axial load is (1601 + 0.75 x 333.6 kN), or 1851.2 kN.

( )

00127.0

995,1990025.0)00207.0(30025.0

0025.05.0

(0.00207)30.0025

3(0.00207)MPa)7.413(

mm7239mm)(193.8

1000kN2.1851

(0.00207)30.0025

0.0025MPa)34.10(64.0

max

max

=

×⎥⎦

⎤⎢⎣

⎡

+−⎥

⎦

⎤⎢⎣

⎡

+

×−⎥

⎦

⎤⎢⎣

⎡

+=

ρ

ρ

MPa

Check maximum area of flexural reinforcement per 1.22 m of wall length

2

maxmax mm27.300mm1220mm)8.193(00127.01220 ==×= mmb As ρ

We have 197.9 mm2 every 1.22 m, and the design is satisfactory.

Summary:

Use 16-mm bars @ 1.22 m vertically, 16-mm bars @ 606.9 mm horizontally.

Because the space between extreme bars is not quite a multiple of 1.22 m, the spacing of the interior

bars is adjusted slightly to fit the nominal 406.4-mm length of the concrete masonry units.

e702.1 allowable stress design of rc masonry shear wall 2006-03-29

Documents