Flexural Elements - Engineering Fundamentals Program · Flexural Elements Allowable Stress Design 2 Flexural Members - Allowable Stress Design h Masonry Unit As Grout b d s m Strains

Allowable Stress Design 1

Allowable Stress Design

Flexural Elements


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



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

))(())(( 21 kddbdnkdbkd

bkd

nAs=nbd

Transformed Section

nnnk 2)( 2

31

kj

Steel moment: jdfAM sss

Steel stress:jdA

Mf

ss

Masonry moment: )(2

)( jdf

kdbM mm

Masonry stress:))((

2

jdkdb

Mfm

Allowable masonry stress = 0.45f’m

Allowable steel stress:20 ksi Grade 40 steel32 ksi Grade 60 steel30 ksi Wire joint reinforcement

Allowable stresses (8.3.3.1, 8.3.4.2.2)


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:

Fb = 0.45(2000psi) = 900 psiFs = 32000 psiEm = 900f’m = 1.80 x 106 psiEs = 29 x 106 psin = Es/Em = 16.1ρ = As/bd = 0.44in2/[(7.625in.)(28in.)] = 0.00206nρ = 16.1(0.00206) = 0.0332

227.00332.00332.020332.02)( 22 nnnk

924.03

227.01

31

kj


Example - Masonry Beam, cont

k=0.227 j=0.924

ksiksiininin

inkip

jdA

Mf

ss 328.29

28924.044.0

3402

psipsi

ininin

inkip

jdkdb

Mfm 675543

28924.028227.0625.7

3402

))((

2

Beam is good

OK

OK

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

7

Reinforced Flexure: ASD vs. SD


Example - Masonry BeamGiven: M=125k-in dead load; M=125k-in live load; Grade 60 steel, f’m=1500psi; 8 in CMU; depth of section limited to three courses; Type S masonry cement mortarRequired: Design sectionSolution:

Design Method Reinforcement

Strength Design 1 - #6 (0.34 in2 req’d)

2008 ASD 1 - #8 (0.82 in2 req’d)Masonry controls

2011 ASD 1 - #6 (0.43 in2 req’d)Steel controls


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.


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 = Fb 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)

Fb = 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

Allowable stress design (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 = Fb/Em = 0.0005 for CMU

• Find steel strain

• For k < kbal

• Set steel strain = Fs/Es = 0.00110 for Grade 60

• Find masonry strain

• Compute forces in masonry and steel

• Sum forces to get axial force

• Sum moment about centerline to get bending moment

nFF

F

E

F

E

FE

F

ksb

b

s

s

m

b

m

b

bal /


Example – 8 in. CMU Bearing WallGiven: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; solid grout Required: Construct interaction diagram – ASD; present results in terms of capacity per footSolution:

3118.0

2900032

90045.0

90045.0

ksiksi

ff

ff

EF

EF

EF

k

m

m

m

m

s

s

m

b

m

b

bal

This is kbal for CMU and Grade 60 steel


Example – 8 in. CMU Bearing Wall, ASD

Balanced conditions (allowable stress)

ftkk/ft-TCP m /83.460.143.6

ftkipftininksibkdfC mm /43.6/1219.1900.02

1

2

1

Stress

ftftkftinkin

inftkipM /83.1/0.223

19.181.3/43.6

0.900 ksi32 ksi

ftkipftinksiAfT ss /6.1/05.032 2

Strain0.0005

0.00110

3.82 in

1.19 in

Sum moments about centerline



kd = 1 inch; k = 1/3.81 = 0.262; k<kbal so allowable steel stress controls

Stress

000392.0181.3

100110.0

kdd

kdsm

0.530 ksi32 ksi

Strain0.000392

0.00110

3.81 in

1.00 in

1. Set steel strain = Fs/Es = 32ksi/29000ksi = 0.00110

2. Use similar triangles to find masonry strain

3. Find masonry stress = Emεm = 1800ksi(0.000392) = 0.706 ksi



4. Sum forces and moments

ftkk/ft-TCP m /63.26.123.4

ftkipftininksiCm /23.4/1200.1706.02

1

Stress

ftftkftinkin


00.181.3/23.4

0.706 ksi32 ksi

ftkipftinksiT /6.1/05.032 2

Strain0.000392

0.000110

3.81 in

1.00 in



kd = 2 inch; k = 2/3.81 = 0.526; k>kbal so allowable masonry stress controls

Stress

000453.02

281.30005.0

kd

kddms

0.900 ksi13.1 ksi

Strain0.0005

0.000453

3.81 in

2.0 in

1. Set masonry strain = Fm/Em = 0.45fʹm/900fʹm = 0.0005

2. Use similar triangles to find steel strain

3. Find steel stress = Esεs = 29000ksi(0.000453) = 13.1 ksi



4. Sum forces and moments

ftkk/ft-TCP m /14.1066.080.10

ftkipftininksiCm /8.10/1200.2900.02

1

ftftkftinkin


00.281.3/8.10

ftkipftinksiT /66.0/05.01.13 2

Stress

0.900 ksi13.1 ksi

Strain0.0005

0.000453

3.81 in

2.0 in



Determine maximum axial load:

ftkip

ftininksi

r

hFAAfP sstnma

/8.35

140

4.651065.0/12625.70.225.0

140

165.025.0

2

2

ininttbt

bt

A

Ir 20.2.625.7289.0289.02

121

3121

994.65.20.2

.144

in

in

r

h


Interaction Diagram: ASD


Partially Grouted Bearing Wall

• Small axial forces

• Partially grouted walls act as solid walls

• Compression area is in face shell

• ASD interaction diagram

• Difficult when neutral axis not in flange

• Trapezoidal stress distribution in flange

• Three point approximation

• Zero axial load; moment capacity

• kd = flange thickness

• Zero moment; axial capacity


Partially Grout vs. Full Grout

Fully Grouted

Partially Grouted


Approximate Interaction Diagram

Three Point Approximation


Allowable Stress - Design Procedure1. Select trial size. For initial guess, consider axial load and moment

independently. Size for axial load alone and moment alone.

2. Assume steel is in tension. Assume compression controls and neglect compression steel. Determine kd.

bF

MtdPddkd

b

sp

3

))2/((2

223

2

If kd>d, check compression or increase size of member.Compare k to kbal.If k ≥ kbal compression controls. n

FF

Fk

sb

bbal

11

2

)(

knF

PbkdF

A

b

b

s

Can result in negative area of steel. Use minimum steel in that case.


Allowable Stress - Design Procedure

If k < kb tension controls. This results in a cubic equation for kd. Use iterative procedure. Start with kd from compression controlling.

32

kdtPM sp

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222

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


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 = Fb, and substitute for Asfs in moment equations using sum of forces.

Mt

dPFbkdkdt

Fbkd spb

spb

22

1

322

1

bF

MtdPddkd

b

sp

3

))2/((2

223

2

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

11

2

)(

knF

PbkdF

A

b

b

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 DesignD + S

Critical location is top of pilaster. P = 19.2 kips M = 19.2kips(5.8in) = 111.4 kip-in

ininksi

inkipininkipinin

bF

MhdPddkd

b

43.2)62.15)(900.0(3

)4.111)2/6.158.11(2.19(2

2

8.11

2

8.113

3

))2/((2

223

2

2

206.08.11

43.2

in

in

d

kdk 312.0

5.2132

675.0

675.0

ksiksi

ksi

nF

F

Fk

sb

bbal

k < kbal Tension controls; iterate


Example - Pilaster DesignD + S

Equation / Value Iteration 1 Iteration 2 Iteration 3

kd (in.) 2.43 3.10 3.12

k 0.206 0.262 0.264

(k-in.) 134.4 130.1 130.0

(in2) -0.066 -0.055 -0.054

(in.) 0.551 0.562 0.562

(in.) 3.10 3.12 3.12

32

kdhPM

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222


Example - Pilaster DesignWhy the negative area of steel?

Sufficient area from just masonry to resist applied forces.

Determine kd from just compression.

inksiin

kip

bF

Pkd

b

74.2)900.0(6.15

)2.19(22

Find the moment

inkipinin

kipkdt

PM

2.132

3

74.2

2

6.152.19

32

Sufficient capacity from just masonry. No steel needed.

Mapp = 111.4 kip-in


Example - Pilaster DesignD + 0.6W Check wind suction

2

22

max 282 wL

MwLMM

wL

MLx

2If x<0 or x>L, Mmax=M

Top of pilaster. P = 9.6k-0.6(8.1)k = 4.74kips M = 4.74k(5.8in) = 27.5kip-in

ininftinftkip

inkipin

wL

MLx 4.139

)12/1)(288(/250.0

5.27

2

288

2

inkinink

inkininkink

wL

MwLMM

6.229

288/0208.02

)5.27(

8

288/0208.0

2

5.27

282 2

22

2

22

max

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

kinftinftkkP 1.712/16.139/20.074.4

Design for P = 7.1 k, M = 230 k-in

kd = 3.45 in. k = 0.293 Tension controls; iterate

As = 0.54 in2


Example - Pilaster Design

D + 0.75(0.6W) + 0.75S At top: P = 13.2 k, M = 76 k-in.

kd = 3.52 in. k = 0.298 Tension controls As = 0.30 in2

x = 127in. M = 202 k-in. P = 15.3 k

0.6D + 0.6W At top: P = 0.9 k, M = 5 k-in.

kd = 2.99 in. k = 0.254 Tension controls As = 0.59 in2

x = 143in. M = 218 k-in. P = 2.3 k

Required steel = 0.59 in2

Use 2-#5 each face, As = 0.62 in2

Total bars, 4-#5, one in each cell


Example - Pilaster Design


Example – Effect of f’m

f’m = 2000 psi

f’m = 1500 psi


Example – ASD vs. SD

ASD

0.6(0.9*SD)



Shear Walls


Shear - Allowable Stress Design

vsvmv FFF

nm

vvm A

Pf

Vd

MF 25.075.10.4

2

1

sA

dFAF

n

svvs 5.0

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

mv fF 2 0.1)/( vVdM

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

mv

v fVd

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


ExampleGiven: 10ft. high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f’m=2000psi; 2-#5 each end; #5 at 48in. (one space will actually be 40 in.); superimposed dead load of 1kip/ft. SDS = 0.4Required: Maximum seismic load based on flexural capacity; shear steel required to achieve this capacity. Solution: Check capacity in flexure using 0.6D+0.7E load combination.

Weight of wall: 38 psf(10ft) = 0.38 k/ft(Wall weight from lightweight units, grout at 48 in. o.c.)

Pu = (0.6-0.7(0.2SDS))D = (0.6-0.7*0.2*0.4)(1k/ft+0.38k/ft)

= 0.544(1.38k/ft) = 0.751 k/ft

Total = 0.751k/ft(16ft) = 12.0 kips


ExampleSolve for stresses, forces and moment in terms of kd, depth to neutral axis.

8 inkd

52 in92 in

140 in188 in

T1 T2 T3 T4 C1C2

kipsinksiT 84.1962.032 21

ininksiC inm 625.7818002

1.81

ininkdksiC in 5.2818002

1.82

22 31.032

188

140inksi

kdin

kdinT

23 31.032

188

92inksi

kdin

kdinT

24 31.032

188

52inksi

kdin

kdinT

ksi

ksi

kd

kdm 29000

32

188

ksi

ksi

kd

kdin 29000

32

188

8.8


Example

• Use trial and error to find kd such that P = 12.0 kips

• P = C1 + C2 – T1 – T2 – T3 – T4

• kd = 38.3 inches

• kdbal = 0.312(284) = 58.6 inches; steel controls

• εm = 0.000282; ε8in. = 0.000224

• C1 = 27.8 kips

• C2 = 15.2 kips

• T1 = 19.8 kips; T2 = 6.7 kips; T3 = 3.6 kips T4 = 0.9 kips


Example

• Sum moments about middle of wall (8 ft from end) to find M

• M = 19.8(188-96) + 6.7(140-96) + 3.6(92-96) + 0.9(52-96) + 27.8[96-(2*0.000224+0.000282)/(0.000224+0.000282)(8/3)] + 15.2[96-8-(38.3-8)/3]

• M = 5816 kip-in. = 485 kip-ft

• Seismic = M/h = (485 kip-ft)/(10ft) = 48.5 kips

• 0.7E = 48.5 kips; E = 69.2 kips

• With strength design, E =102.5 kips

• Interior bars not as effective with ASD

h1h2 b

x

Centroid

3

2

12

12 b

hh

hhx


ExampleCalculate net area, An, including grouted cells. Net area

Net area: 2685)5.262.7)(8(51925.2 ininininininAn

25.0/ VdM mv fF 3Maximum Fv 0.1/ VdM mv fF 2

625.0192/120// ininVdVhVdM vv

kipsinpsiV 4.576858.83 2 Maximum V

psipsifVd

MF m

vv 8.8375.02000625.025

3

225

3

2


Example

in

inpsi

inpsiin

AF

dFAs

sA

dFAF

nvs

sv

n

svvs 6.52

6858.25

1883200031.05.05.0 5.0

2

psi

psiinlbFff vmvvs 8.25

75.0

2.6875.0685/48300 2

Use #5 bars in bond beams. Determine spacing.

Use 3 bond beams, spaced at 48 in. o.c. vertically.

psiin

lbpsi

A

Pf

Vd

MF

nm

vvm

2.68685

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.

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


Example: Special Shear WallIf this were a special shear wall:

• Design for 1.5V, or 1.5(48.5 kips) = 72.8 kips

• fv = (72.8 kips)/(685in2) = 106.2 psi

• But, maximum Fv is 83.8 psi

Cannot build wall. Need to reduce load or fully grout.

If fully grouted:

psiin

lbpsiFvm 9.33

1494

870025.02000625.075.10.4

4

12

21494

.196.625.7

in

ininAn

psi

psiinlbFff vmvvs 3.19

0.1

4.290.11494/72800 2

in

inpsi

inpsiin

AF

dFAs

nvs

sv 3.3214943.19

1883200031.05.05.0

2 Use #5 @ 32 in.


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

No 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)(1494in2) = 149 kips; P = 12.0 kips

00108.0

.188.625.7

31.0*362.0 2

inin

in

bd

As OK

Documents

Flexural Elements - Engineering Fundamentals Program · Flexural Elements Allowable Stress Design 2 Flexural Members - Allowable Stress Design h Masonry Unit As Grout b d s m Strains