Allowable Stress Design - UTK EFP …Allowable Stress Design 3 Allowable Stress Design To find neutral axis, equate moments of areas about neutral axis. (bkd)(21 kd) (n bd)(d kd)b

Allowable Stress Design 1

Allowable Stress Design

Load Combinations

(1) D

(2) D + L

(3) D + (Lr or S or R)

(4) D + 0.75L + 0.75(Lr or S or R)

(5) D + (0.6W or 0.7E)

(6a) D + 0.75L + 0.75(0.6W) + 0.75(Lr or S or R)

(6b) D + 0.75L + 0.75(0.7E) + 0.75S

(7) 0.6D + 0.6W

(8) 0.6D + 0.7E

It shall be permitted to replace 0.6D with 0.9D in combination (8) for the design of Special Reinforced Masonry Shear Walls


Flexural Members - Allowable Stress Design

h

Masonry Unit

As

Grout

b

d

s

m

Strains

kd

fs

fm

Stresses

T=Asfs

jd

C=fm(b)(kd)/2

bd

Asm

s

E

En Assumptions:

1. Plane sections remain plane2. Stress-strain relationship for masonry

is linear in compression3. All masonry in tension is neglected4. Perfect bond between steel and grout5. Member is straight prismatic section

kd

b

nAs=nbd

Transformed Section

Notation:Lower case: calculated stress, fsUpper case: allowable stress, Fs

Fb is allowable compressive stress to resist flexure only. Notes use Fm for allowable compressive stress to resist combinations of flexure and axial load.


Allowable Stress Design

To find neutral axis, equate moments of areas about neutral axis.

))(())(( 21 kddbdnkdbkd

bkd

nAs=nbd

Transformed Section

nnnk 2)( 2

31

kj

Steel: jdFAM sss

Steel stress: jdA

Mf

ss

Masonry: )(2

)( jdF

kdbM mm

Masonry stress:))((

2

jdkdb

Mfm

Allowable stresses (8.3.3.1, 8.3.4.2.2)

Allowable masonry stress = 0.45f′mAllowable steel stress:

20 ksi Grade 40 steel32 ksi Grade 60 steel30 ksi Wire joint reinforcement

Allowable moment:


Example - Masonry BeamGiven: M=340k-in; Grade 60 steel, f’m=2000psi; 8 in CMU; Type S mortar; 4 course high beam (d=28 in.); #6 rebarRequired: Is section adequate?Solution:

Fm = Fs = Em = 900f’m = 1.80 x 106 psiEs = 29 x 106 psin = Es/Em = ρ = As/bd =nρ =

nnnk 2)( 2

3

1k

j


Example - Masonry Beam, cont

k=0.227 j=0.924

jdA

Mf

ss

))((

2

jdkdb

Mfm

What is maximum moment beam could carry?

inkipinksiinjdFAM sss 36428924.03244.0 2

inkipinpsi

ininjdF

kdbM mm 56428924.0

2

90028227.0625.7

2

Mall = 364 kip-in


Masonry Beam - Parametric Study

d=20inch b=7.625inch

9

Reinforced Flexure: ASD vs. SD


ASD Design Procedure

1. Assume value of j (or k). Typically j is between 0.85 and 0.95.

2. Determine a trial value of As. Choose reinforcement.

3. Determine k and j.

4. Determine steel stress and masonry stress.

5. Compare calculated stresses to allowable stresses.

6. If masonry stress controls design, consider other options (such as change of member size, or change of f’m). Reinforcement is not being used efficiently.

jdF

MA

ss

nnnk 2)( 2 3/1 kj

jdA

Mf

ss ))((

2

jdkdb

Mfm

A complete design procedure will be presented later.

11

ASD: Design Method

Calculate

Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n

FF

Fk

sm

mbal

bF

Mddkd

m3

2

223

2

YES

11

2

)(

knF

PbkdF

A

m

m

s

31

kdF

MA

s

s

bF

nFA

s

ss

dkd 222

Iterate. Use (kd)2 as new guess and repeat.

NO

Allowable masonry compression stress

controls

Allowable reinforcement tension stress controls

12

Example: Beam

Given: 10 ft. opening; superimposed dead load of 1.5 kip/ft; live load of 1.5 kip/ft; 24 in. high; Grade 60 steel; Type S masonry cement mortar; 8 in. CMU; f’m = 2000 psiRequired: Design beamSolution:

5.2.1.3: Length of bearing of beams shall be a minimum of 4 in.; typically assumed to be 8 in.

5.2.1.1.1 Span length of members not built integrally with supports shall be taken as the clear span plus depth of member, but need not exceed distance between center of supports.

• Span = 10 ft + 2(4 in.) = 10.67 ft

5.2.1.2 Compression face of beams shall be laterally supported at a maximum spacing of:

• 32 multiplied by the beam thickness. 32(7.625 in.) = 244 in. = 20.3 ft• 120b2/d. 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft

13

Example: Beam

ftk

ftwLM ft

k

1.458

67.1017.3

8

22

ftk

ftk

ftk

ftk ftLDw 17.35.12083.05.1 2 Load

Weight of fully grouted normal weight: 83 psf

Moment

Determine kd Assume compression controls

.32.9

625.7900.03

1.452

2

20

2

203

3

2

223

1222

ininksi

ftkinin

bF

Mddkd ft

in

b

312.0466.020

32.9

in

in

d

kdkCheck if compression controls Compression

controls

1.160.2900

29000

900

ksi

ksi

f

E

E

En

m

s

m

sCalculate modular ratio, n

14

Example: Beam

Find AsArea of steel

294.1

1466.01

900.01.16

2625.731.9900.0

112

)(

inksi

ininksi

knF

bkdF

A

b

b

s

Bars placed in bottom U-shaped unit, or knockout bond beam unit.

Use 2 - #9 (As = 2.00 in2)

15

Example: Beam

Section 8.3.5.4 allows design for shear at d/2 from face of supports.

Design for DL = 1 k/ft, LL = 1 k/ft

kft

ftftkV 02.9

33.5

17.133.556.11

k

ftftwLV

ftk

ftk

ftk

56.112

67.100.12083.00.1

2

2

ftinin

inft 17.1.4

2

.20.12

1

Shear at reaction

d/2 from face of support

Design shear force

psipsi

fA

Pf

Vd

MF m

nmvm

3.50200025.22

1

25.22

125.075.10.4

2

1

psiinin

k

A

Vf

nvv 2.59

.20.625.7

02.9Shear stress

Allowable masonry shear stress

Suggest that d be used, not dv.

16

Example: Beam

2034.0.20320005.0

.8.20.625.79.8

5.05.0

ininpsi

inininpsiA

dF

sAfA

sA

dFAF

v

s

nvsv

n

svvs

psipsifF mv 4.89200022

psipsipsi 9.83.502.59

Check max shear stress

Req’d steel stress

Determine Av for a spacing of 8 in.

> 59.2psi OK

Use #3 stirrups

Determine d so that no shear reinforcement would be required.

.5.233.50.625.7

02.9in

psiin

k

bF

Vd

vm

Use a 32 in. deep beam if possible;will slightly increase dead load.

17

Summary: ASD vs. SD

Dead Load (k/ft)(superimposed)

Live Load (k/ft)Required As (in2)

ASD SD

0.5 0.50.34

0.34 ( = 1.5 ksi)

0.26

0.26 ( = 1.5 ksi)

1.0 1.00.64

0.65 ( = 1.5 ksi)

0.50

0.52 ( = 1.5 ksi)

1.5 1.51.94

5.09 ( = 1.5 ksi)

0.77

0.80 ( = 1.5 ksi)

ASD: Allowable tension controls for 0.5 k/ft and 1 k/ft.


Partially Grouted Walls - Allowable Stress

djCdjCM wwff f

fmf bt

kd

tkdfC

2

2

)(2 f

fmw tkdb

kd

tkdfC

f

fff tkd

tkd

d

tj

2

23

31

d

kdtj fw 3

21

d

t

b

b

b

bn

d

t

b

b

b

b

d

tnn

b

b

d

t

b

bk ffff

121 2

2

2

sb’

b

d

kd

t f

As

A. Neutral axis in flange; design and analysis for solid sectionB. Neutral axis in web

fm = Fm if masonry controllingfm = Fsk/(n(1-k)) if steel controlling


Partially Grouted Walls - ExampleGiven: 8 in CMU wall; 12 ft high; Grade 60 steel, f’m=2000 psi; Lateral wind load of 42 psf (factored)Required: Reinforcing (place in center of wall)Solution:

ftinlbftftlbftftlbwh

M /5443/4548

12/426.0

8

222

ftininksi

ftinkip

jdF

MA

ss /050.0

)81.3)(9.0(32

/443.5 2

inint

d 81.32

625.7

2

Try #4 @48 in (0.050in2/ft)

Fm = 0.45(2000psi) = 900 psiFs = 32000 psiEm = 1.80 x 106 psiEs = 29 x 106 psin = Es/Em = 16.1

Solve as solid section = 0.00109k = 0.171kd = 0.651 in < tf = 1.25 in OKj = 0.943fm = 387 psi OKfs = 30.3 ksi OK

Use #4 @ 48 inches


Flexural and Axially Loaded Elements

Allowable Compressive Force (8.3.4.2.1)

99 140

165.025.02

r

h

r

hFAAfP sstnma

99 70

65.025.02

r

h

h

rFAAfP sstnma

Ast is area of laterally tied steel


Interaction Diagram

• Assume strain/stress distribution

• For k > kbal Set masonry strain, find steel strain

• Masonry strain = Fm/Em = 0.0005 for CMU

• For k < kbal Set steel strain, find masonry strain

• Steel strain = Fs/Es = 0.00110 for Grade 60

• Compute forces in masonry and steel

• Sum forces; sum moments about centerline

312.0

2900032

900

45.0900

45.0

/

ksiksi

f

ff

f

E

F

E

FE

F

nFF

Fk

m

m

m

m

s

s

m

m

m

m

sm

mbal

Grade 60 steel


Example – 8 in. CMU Bearing Wall

Given: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; partial grout; f’m = 2000 psi Required: Interaction diagram in terms of capacity per foot

Pure Moment: n = 16.1 ρ = 0.00109 nρ = 0.0176

ftftk

ftk-inM 479.0.755

ftink

ftin

sss inksijdFAM 75.5.81.3943.03205.02

ftink

ftinm

m inksi

injdF

kdbM

65.12

81.3943.02

90.081.3171.012

2

171.00176.00176.020176.02)( 22 nnnk

943.03

171.01

31

kj

Find k

Find j

Find Ms

Find Mm

Allowable M


Example – 8 in. CMU Bearing Wall

Pure Axial:

ftkP 2.17

991.5466.2

144

in

in

r

h

NCMA TEK 14-1B Section Properties of Concrete Masonry Wallsr = 2.66 in. An = 40.7in2/ft In = 332.0 in4/ft

ftk

ftin

sstnma

ksi

r

hFAAfP

2.17140

1.54107.400.225.0

140

165.025.0

2

2

2

Find h/r

Find Pa

4.3.2 Radius of gyration Radius of gyration shall be computed using average net cross-sectional area of the member considered.


Example – 8 in. CMU Bearing Wall, ASD

Find Cm

Find T

Find P

Find M

Balanced:

ftftk

ftk

M

P

83.1

.834

ftk

ftk

m-TCP 83.460.143.6

ftk

ftin

mm inksibkdfC 43.61219.1900.02

1

2

1

Stress

ftftk

ftink

ftk in

inM

83.10.22

3

19.181.343.6

0.900 ksi32 ksi

Strain0.0005

0.00110

3.82 in

1.19 inkd < 1.25 in.N.A. in face shell

ftk

ftin

ss ksiAfT 6.105.0322



Find Cm

Find T

Find P

Find M

BelowBalanced:kd = 1.00 in.

ftftk

ftk

M

P

22.1

.622

ftk

ftk

m-TCP 62.260.122.4

ftk

ftin

mm inksibkdfC 22.41200.1704.02

1

2

1

Stress

ftftk

ftink

ftk in

inM

22.17.14

3

00.181.322.4

1800ksi(0.000391) =0.704 ksi

32 ksi

ftk

ftin

ss ksiAfT 6.105.0322

Strain0.000391

0.00110

3.82 in

1.00 in



AboveBalanced:kd = 2.00 in. Stress

0.900 ksi

13.1 ksiStrain0.0005

0.00045

3.82 in

2.00 in

Neutral axis is in web

ftk

ftinksiT 66.005.01.13

2

2.00 in

0.338 ksi

1.25 in

ftk

ftin

m inksi

C 28.91225.12

338.0900.01

ftk

ftin

m inksiC 25.0275.0338.02

12

Face Shell

Web

Steel



ftk

ftk

mm -TCCP 87.866.025.028.9 21

ft

ftkftink

ftk

ftk ininininM 58.20.315.181.325.053.081.328.9

9.28 k/ft

0.66k/ft

1.25+0.75/3=1.5in.

0.25 k/ft

1.5 in

0.53 inh2

h1b

x

Centroid

3

2

21

21 b

hh

hhx

ftftk

ftk

M

P

58.2

.878


Interaction Diagram: ASD


Approximate Interaction Diagram

Three Point Approximation

Three point approximation

• Zero axial load; moment capacity

• kd = flange thickness

• Zero moment; axial capacity


Allowable Stress - Design Procedure

Calculate

Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n

FF

Fk

sm

mbal

bF

MtdPddkd

m3

))2/((2

223

2

YES

11

2)(

knF

PbkdF

A

m

m

s

32

kdtPM

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222

Iterate. Use (kd)2 as new guess and repeat.

NO

Compression controls

Tension controls


Derivation of Design Equations

Sum forces: PfAfbkd ssm 2

1

Sum moments: Mt

dfAkdt

fbkd spss

spm

2322

1

If the masonry stress controls, set fm = Fm, and substitute for Asfs in moment equations using sum of forces.

Mt

dPFbkdkdt

Fbkd spb

spb

22

1

322

1

bF

MtdPddkd

m

sp

3

))2/((2

223

2

This is quadratic equation in kd, which can be solved to obtain:

11

2

)(

knF

PbkdF

A

m

m

s



kdd

kd

s

m

Find a ratio of strains:

If the steel stress controls, set fs = Fs, and find fm in terms of Fs.

kdd

kd

n

f

E

f

kdd

kdE

kdd

kdEEf s

s

smsmmmm

Substitute into sum of forces, and solve for Asfs.

Pkdd

kd

n

fbkdPfbkdfA s

mss

2

1

2

1

Now substitute into sum of moments and set fs = Fs.

Mt

dPkdd

kd

n

Fbkd

kdt

kdd

kd

n

Fbkd spssps

22

1

322

1



This is cubic equation in kd:

02226

23

dM

tdPkdM

tdPkd

n

bdFkd

n

bF spspss

Although there are analytical ways to solve a cubic equation, an iterative solution might be the easiest. Solve for As from:

s

s F

P

kdd

kd

nbkdA

1

2

1


Example - Pilaster DesignGiven: Nominal 16 in. wide x 16 in. deep CMU pilaster; f’m=2000 psi; Grade 60 bar in each corner, center of cell; Effective height = 24 ft; Dead load of 9.6 kips and snow load of 9.6 kips act at an eccentricity of 5.8 in. (2 in. inside of face); Factored wind load of 26 psf (pressure and suction) and uplift of 8.1 kips (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between pilasters; No ties.Required: Determine required reinforcing using allowable stress design.Solution:

Ver

tical

Spa

nnin

g

d=11.8 in

xLoad

e=5.8 in 2.0 in

Em = 1800ksin = 16.1

Lateral Loadw = 0.6(26psf)(16ft)=250 lb/ft

Insi

de


Example - Pilaster Design

0.6D + 0.6W

Top of pilaster. P = 0.6(9.6)k-0.6(8.1)k = 0.9kips M = 0.9k(5.8in) = 5.2kip-in

inin

inkipin

wh

Mhx

inft

ftk

1.143288250.0

2.5

2

288

2121

inkin

inkinink

wh

MwhMM

ink

ink

3.2182880208.02

)2.5(

8

2880208.0

2

2.5

282 2

22

2

22

max

Find axial force at this point. Include weight of pilaster (200 lb/ft).

kinkPinft

ftk 3.21.14320.06.09.0

.121

Design for P = 2.3 k, M = 218 k-in

Find location of maximum moment

Usually load combination with smallest axial load and largest lateral load controls. Try 0.6D + 0.6W to determine required reinforcement and then check other load combinations.



in

inksi

inkipininkipinin

bF

MhdPddkd

m

00.36.15900.03

)2182/6.158.113.2(2

2

8.11

2

8.113

3

))2/((2

223

2

2

254.08.11

00.3

in

in

d

kdk k < kbal Tension controls; iterate



Equation / Value Iteration 1 Iteration 2 Iteration 3

kd (in.) 3.00 3.38 3.40

k 0.254 0.286 0.288

(k-in.) 15.6 15.3 15.3

(in2)

0.585 0.593 0.593

(in.)0.678 0.686 0.686

(in.)3.38 3.40 3.40

32

kdhPM

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222

Try 2-#5, 4 total, one in each cell

inkinin

kM

6.15

3

00.3

2

62.153.2

2585.0

3254.0

18.1132

6.15218in

inksi

inkAs

in

inksi

ksiink678.0

62.1532

1.1632585.03.2 2

inininininkd 38.3678.08.11678.02678.0 22



0.6D+0.6WP = 2.3kM = 18.2k-ft

D+0.75(0.6W)+0.75SP = 15.3kM = 16.8k-ft

D+SP = 19.2kM = 9.25k-ft

D+0.6WP = 7.1kM = 19.2k-ft


Example – Effect of f’m

f’m = 2000 psi

f’m = 1500 psi


Example – ASD vs. SD

ASD

0.6(0.9*SD)


Example: Wind Loads ASD

16 ft

32 p

sf

D = 500 lb/ftLr = 400 lb/ftW = -360 lb/ft

2.67

ft

Given: 8 in. CMU bearing wall; Grade 60 steel; Type S masonry cement mortar; f’m=2000psi; roof forces act on 3 in. wide bearing plate at edge of wall.

Required: Reinforcement

Solution:

Estimate reinforcement

~8

0.6 0.032 168

0.62

As,req’d = 0.065 in2/ft

Try #4 @ 40 in. (0.06 in2/ft) Cross-sectionof top of wall

Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.



Load Comb. Pf (kip/ft) P (kip/ft) Mtop (k-ft/ft) M (k-ft/ft) As (in2)

0.6D+0.6W 0.084 0.340 -0.049 0.590 0.051

D+0.6W 0.284 0.711 -0.002 0.613 0.040

ft

ftkftftk

top ftksfMwhM

590.02

049.0

8

16032.06.0

28

122

Check 0.6D+0.6W

084.036.06.05.06.0 ftk

ftk

ftk

fP

/340.0867.2040.06.0084.0 ftkftftksfP ftk

049.0

2

67.2032.06.0.81.2084.0

2

121

ftftk

inft

ftk

top

ftksfinM

Use #4@ 40 in. (0.06 in2/ft)

Find force at top of wall

Find force at midheight

Find moment at top of wall

Find moment at midheight

17% more steel than required; with SD #4@40 in. is 14% more steel than required.



Sample Calculations: 0.6D+0.6W

1. kbal = 0.312; kdbal = 1.19in.2. Assume masonry controls.

Determine kd.Since 0.355 in. < 1.18 in.

tension controls. 3. Iterate to find As.

.355.0

12900.03

590.02

2

.81.3

2

.81.33

3

2

223

.122

2

inksi

inin

bF

Mddkd

ftinftin

ftftk

m

3/2/ kdtPM sp

3/1 kdF

MMA

ss

bF

nFAP

s

ss

dkd 222

Equation / Value Iteration 1 Iteration 2

kd (in.) 0.355 0.707

(k-ft/ft) 0.1047 0.1013

(in2/ft) 0.049 0.051

(in.) 0.0804

(in.) 0.707


Shear - Allowable Stress Design

gvsvmv FFF

nm

vvm A

Pf

Vd

MF 25.075.10.4

2

1

sA

dFAF

nv

svvs 5.0

Maximum Fv is: gmv fF 3 25.0)/( vVdM

gmv fF 2 0.1)/( vVdM

Interpolate for values of M/(Vdv) between 0.25 and 1.0 gm

vv f

Vd

MF

25

3

2


Shear – Special Shear Walls

nm

vvm A

Pf

Vd

MF 25.075.10.4

4

1

Masonry allowable shear stress decreased by a factor of 2, from ½ to ¼.

Accounts for degradation of masonry shear strength that occurs in plastic hinging regions.

Seismic design load required to be increased by 1.5 for shear

Maximum reinforcement: Shear walls having• M/(Vd) ≥ 1 and• P > 0.05f′mAn

m

yy

m

f

fnf

fn

2max

For distributed reinforcement, ρ is the total area of tension reinforcement divided by bd.


Example

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 100 kips. SDS = 0.4

Required: Design the shear wall; ordinary reinforced shear wall

Solution: Check using 0.6D+0.7E load combination.

• Try #6 in each of last 3 cells; #6 @40in.

P = (0.6-0.7(0.2)(SDS))D = (0.6-0.7(0.2)(0.4))(1k/ft(16ft)+7.8k) = 12.9 kips

Weight of wall: [40 psf(12ft)+2(2ft)75psf]10ft = 7800lb

Lightweight units, grout at 40 in. o.c. 40 psf; full grout 75 psf

From interaction diagram OK; stressed to 90% of capacity

Strength Design:2-#5 at end; #5 @ 40 in.


Example


Example


Example

Calculate net area, Anv, including grouted cells. 2849)5.262.7)(8(91925.2 ininininininAnv

Maximum Fv

625.0192/120// ininVdVhVdM vv

psipsifVd

MF gm

vv 8.8375.02000625.025

3

225

3

2

OK

psiin

lb

A

Vf

nvv 4.82

849

1000007.02

Shear ratio

Shear stress


Example

Strength Design: Req’d s = 28.0 in.

in

inpsi

inpsiin

Af

dFAs

sA

dFAF

nvs

sv

n

svvs

9.258494.42

1883200031.05.05.0

5.0

2

psipsipsiFfF vmgvvs 4.425.6775.0/4.82/

Use #5 bars in bond beams.Determine spacing.

Use #5 at 24 in. o.c.

psiin

lbpsi

A

Pf

Vd

MF

nm

vvm

5.67849

870025.02000625.075.10.4

2

1

25.075.10.42

1

2

s ≤ min{d/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code 8.3.5.2.1

Top of wall P = (0.6-0.7(0.2SDS))D = 0.544(1k/ft)(16ft) = 8.70 kips

(critical location for shear):

Determine Fvm

Required steel stress


Example: Special Shear Wall

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 100 kips. SDS = 0.4

Required: Design the shear wall; special reinforced shear wall

Solution: Check using 0.6D+0.7E load combination.

• Design for 1.5V, or 1.5(70 kips) = 105 kips (Section 7.3.2.6.1.2)

• fv = (105 kips)/(849in2) = 123.7 psi

• But, maximum Fv is 83.8 psi

• Fully grout wall


Example: Special Shear Wall

Strength Design:Req’d s = 54.6 in. using 1.25Mn

Req’d s = 31.6 in. using 2.5Vu

psiin

lbpsiFvm 0.34

1464

870025.02000625.075.10.4

4

12

21464.192.625.7 inininAnv

psipsiFff vmvvs 7.370.347.71

in

inpsi

inpsiin

AF

dFAs

nvs

sv 6.1614947.37

1883200031.05.05.0

2

Use #5 @ 16 in.

psi

in

lb

A

Vf

nvv 7.71

1464

1000007.05.12

Calculate net area, Anv, including grouted cells.

Shear stress

Determine Fvm

(special wall)

Use #5 bars in bond beams.Determine spacing.

Required steel stress


Example: Maximum Reinforcing

If we needed to check maximum reinforcing, do as follows.

00582.0

200060000

1.16600002

20001.16

2max

psipsi

psi

psi

f

fnf

fn

m

yy

m

Section 8.3.3.4 Maximum ReinforcementNo need to check maximum reinforcement since only need to check if:

• M/(Vdv) ≥ 1 and M/(Vdv) = 0.625

• P > 0.05f′mAn 0.05(2000psi)(1464in2) = 146 kips; P = 12.2 kips

00184.0

.188.625.7

44.06 2

inin

in

bd

As OK

For distributed reinforcement, the reinforcement ratio is obtained as thetotal area of tension reinforcement divided by bd. Assume 6 bars in tension.

Documents

Allowable Stress Design - UTK EFP …Allowable Stress Design 3 Allowable Stress Design To find neutral axis, equate moments of areas about neutral axis. (bkd)(21 kd) (n bd)(d kd)b