Axial and Flexure

8/3/2019 Axial and Flexure

1/20

Combined Flexure and Axial Load

Interaction Diagram

Solidly grouted bearing wall

Partially grouted bearing wall

Bearing Walls: Slender Wall Design Procedure Stren th

Serviceability Deflections

Example Pilaster

Prestressed Masonry

Combined Flexural and Axial Loads 1

Interaction Diagram

Assume strain/stress distribution

Compute forces in masonry and steel

Sum forces to get axial force

Sum moment about centerline to get bending moment

Ke oints

Pure axial load

Pure bending

99180.08.02

hhAfAAfP stystnmn

=0.9 rr

9970r80.080.02

hAfAAfP stystnmn

Ast is area of

laterall tied steel



2/20

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.; solid grout

Required: Interaction diagram in terms of capacity per foot

Pure Moment: ys

ysnbfAdfAM

'8.021

ftftkftftkMn /834.0/927.00.9

'8.0

bf

Aa

m

ys


8.0

c

Example 8 in. CMU Bearing WallPure Axial: tr

12

1 4.65

201.2

144

in

in

r

h

ftkPn /7.680.9 99180.08.02

hhAfAAfP stystnmn

ftk/8.61



3/20

Example 8 in. CMU Bearing Wall

Maximum Moment at Pn = 61.8k/ft

a cs:


Example 8 in. CMU Bearing WallChoose strain distribution (alternatively c)

Balanced conditionsT

Strain

Stress

Cm

mC

T

-TCP mn n

nM


nM


4/20

Example - Interaction Diagram

Point c (in) Cm (kip/ft) T (kip/ft) Pn (kip/ft) Mn(kip-ft/ft)

. . . .

a = d 4.76 54.8 0 49.4 7.85

c = d 3.8125 43.9 0 39.5 7.53

2.95 34.0 1.1 29.6 6.71

Balanced 2.09 24.0 3.0 18.9 5.37

1.8 20.7 3.0 16.0 4.81

1.5 17.3 3.0 12.8 4.16

1.2 13.8 3.0 9.7 3.46

. . . . .

0.6 6.9 3.0 3.5 1.85

Pure Moment 0.26 3.0 3.0 0 0.83


Interaction Diagram



5/20

Interaction Diagram Below Balanced

Below the balanced point, the interaction diagram is a straight line:

22

/sp

ys

sp

ysun

tdfA

atfAPM

bf

PfAa

m

uys

80.0

/

These are equations 3-28 and 3-29 in the code except:

modified to account for non-centered steel (ignores any tension

in a possible second layer of steel near the compression face)

corrected Pu to Pu/

For centered bars: / adfAPM sun


Partially Grouted Bearing Wall Small _______ forces

Partially grouted walls act as ______ walls

Compression area is in _____________

Strength design

Hi her axial loads act as _________

Very high axial loads act as ________

Need to calculate rbased on grouted cross-section.



6/20

Interaction Diagram: Solid vs. Partial Grout


Walls: Slenderness EffectsSecond-order procedure (3.3.5): Assumes simple support conditions.

uP 05.0 No height limit mu

m fP

f 20.005.0 h/t 30

nA g

Complementary moment: Design moment

2

p us momen n uce y wa e ec ons

uuu

ufu

u PPM 28

Puf= Factored floor load

Puw = Factored wall load

ufuwu

Assumes maximum moment is at midheight



7/20

Walls: Deflections

Deflection Calculation

Mh52

hMMhM 55 22cr

nmIE

48 crcrm

cr

nm

cr MMIEIE

48

48

Deflection Limit hs 007.0 Calculated under service loads

cr.

3

2 bccd

tPAnI

spuscr

PfA

cuys

'

y m.

For centered bars:32 bccdPAnI u


3fy

Walls: DesignDesign Procedure

1. Solve for Mu. Compare to Mn.. . , .

Solving for M222

cr

crnm

crf

crm

MMIIE

PIE

M

482848

1

22

crf

nm

MMPIE

M

2848

1

.



8/20

Walls - Design Procedure

1. Determine a, depth of compressive stress block

Preliminary estimate of steel; assume steel yields

bf

MtdPdda

m

uu

8.0

2/22

ums

PbafA

/8.0

. s

y

This neglects increase in moment due to second order effects. Can

,

amount of reinforcement.


Example - Slender WallGiven: 18 ft high CMU bearing wall, with 2.5 ft parapet (total height is 20.5

ft); Type S masonry cement mortar; Grade 60 steel in center of wall; Dead

load from roof of 500 lb/ft; Roof live load of 400 lb/ft; Lateral wind load of

ps , ps on parape ; n up o ; oo orces ac on n.

wide bearing plate at edge of wall.

Required: Reinforcing steel.

Determine eccentricit

Solution:

Cross-section

of top of wall

e = 7.625in/2 1.0 in.

= 2.81 in.



9/20

Example - Slender Wall: Estimate Steel

Use 1.2D+1.0W+0.5Lr without second-order effects, parapet, and wall weight

.

b

MtdPdda uu

8.0

2/22

m

um Pbaf /8.0

y

sf


Try #__@__ in., As = ______in2/ft

Example - Slender WallSummary of Strength Design Load Combination Axial Forces

(wall weight is 38 psf for 48 in. grout spacing)

Load CombinationPuf

(kip/ft)

Puw

(kip/ft)

Pu

(kip/ft)

. + . . . .

1.2D+1.6Lr+0.5W 1.240 0.524 1.764

1.2D+1.0W+0.5L 0.440 0.524 0.964

Puf= Factored floor load (just eccentrically applied load)

Puw = Factored wall load (includes wall and parapet weight; found at

midheight of wall between supports (9 ft from bottom)



10/20

Example - Slender Wall: McrFind modulus of rupture; use linear interpolation between no grout and full grout

Ungrouted (Type S masonry cement): 38 psi

Full routed T e S masonr cement : 153 si

psicells

groutedcell

psipsipsifr 2.576

1

)38153(38

Find Mcr, cracking moment:

Commentary allows one to include axial load

,

/ IfAPM nrnu

2/tcr


Example - Slender WallCheck strength; 1.2D+1.0W+0.5Lr

ufe = m - e g momen rom momen a op o wa .

There will be a moment at the top of the wall from eccentric load and

a era orces on parape

Method assumes wind is providing suction on wall

Moments from lateral wind and eccentric load add together Lateral wind load on parapet will cause moment at top to decrease

Decrease in moment from parapet wind is ignored in calculations

(that is, wind load is considered to be zero on parapet)

Earthquake lateral forces on parapet would be included; first mode

has motion in opposite directions.



11/20

Example - Slender Wall: Moment of Inertia

Find In, net moment of inertia

ininftininininin

inftinIn

25.1*2625.7/22/25.181.325.112

25.1/122

32

3

ftinIn /1.331

4

5.2129000 ksiEn s

n cr, crac e moment o nert a.

b

PfAc

uys

'64.0

m

m

3

2 bccd

PAnI uscr

y


Example - Slender WallCheck strength; 1.2D+1.0W+0.5Lr

Solvin for M rucruu MM

hPMeP

hwhPM

1155

1222

crgmcrm

uhP

M5

12

crmIE48

115 22 hPMehw

4828 crgm

ucruf

u

IIEP



12/20

Example - Slender Wall

Check strength; 1.2D+1.0W+0.5Lr

Compare to capacity:

PfAa uys /

/ adfAPM ysun

m.


Example - Slender WallSummary of Strength Design Load Combinations

Load Combination M M M / M Second Order M /

(kip-in/ft) (kip-in/ft) First Order M

0.9D+1.0W 16.6 16.9 0.98 1.06

. + . r+ . . . . .

1.2D+1.0W+0.5Lr 18.2 18.6 0.98 1.12



13/20

Example - Slender Wall

Check deflections:

cr M(in4/ft) (k-in/ft)

D+0.6W 0.466 21.9 10.28 0.767+ +. . . r . . . .

0.6D+0.6W 0.434 20.7 9.71 0.708

22

Deflection Limit ininhs 51.1)216(007.0007.0

cr

crm

cr

nm

cr MMIEIE

48

48


Example - Pilaster DesignGiven: Nominal 16 in. wide x 16 in. deep CMU pilaster; fm=1500 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

n. ns e o ace ; n oa o ps pressure an suc on an up o

8.1 kips (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to

span horizontally between pilasters; No ties.

Solution:

e=5.8 in 2.0 in

ning

Lateral Load

w = 26psf(16ft)de

rticalSpaLoad

Insi


Vd=11.8 in d = 15.625 7.625/2 = 11.8 in


14/20

Example - Pilaster Design

Weight of pilaster:

Weight of fully grouted 8 in wall (lightweight units) is 75 psf. Pilaster is like a

double thick wall. Weight is 2(75psf)(16in)(1ft/12in) = 200 lb/ft

1.2D + 1.6SCritical location is to of ilaster. P = 26.9 ki s M = 156.0 ki -in

bf

MhdPdda

m

uu

8.0

2/2

2

. . .

y

ums

f

aA

.


Example - Pilaster DesignWhy the negative area of steel?

Sufficient area from just masonry to resist applied forces.

Determine a from ust com ression.

in

inksi

kip

bf

Pa

m

u 44.16.155.18.0

9.26

8.0

Find the moment

ininat 44.16.15inipipPM u

190

229.26

22

Sufficient ca acit from ust masonr . No steel needed.

u -

. .



15/20


0.9D + 1.0W Check wind suction

At top of pilaster. Pu = 0.9(9.6) 1.0(8.1) = 0.54 kips

Mu = 0.54kips(5.8in) = 3.1 kip-in

2

22

max 282 wL

MwLMM

wL

MLx

2

If xL, Mmax=M

. .


Example - Pilaster Design0.9D + 1.0W At top: Pu=0.5 k Mu=3 k-in

a = 2.04 in A = 0.59 in2

x=144 in Pu=2.7 k Mu=361 k-in

1.2D + 1.0W + 0.5S At top: Pu=8.2k Mu=48k-in

a = 2.39 in A = 0.54 in2

x = 139in Pu=11.0k Mu=384k-in

1.2D + 1.6S + 0.5W At top: Pu=22.8k Mu=132k-inx=117in P

u=25.2k M

u=252k-in

. .

a = 1.92 in As = 0.14 in2

Re uired steel = 0.59 in2

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

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



16/20



Columns Isolated vertical member; width 3 x thickness; height 4 x thickness

Minimum side dimension is 8 in. (1.14.1.1)

Effective height / least dimension 25 (2.1.6.1)

Distance between lateral supports 30 width (3.3.4.4.2)

Minimum eccentricity is 0.1t (2.1.6.2)

Minimum reinforcement is 0.0025An (1.14.1.2)

Maximum reinforcement is 0.04An (1.14.1.2) Additional maximum reinforcement requirements in strength design

Minimum of 4 bars, one in each corner(1.14.1.2)

. . . .

Allowable compressive stress is smaller of 0.4fy or 24 ksi (2.3.2.2.2)

Solid grouted (3.3.4.4.1)

: . . .

1/4 in. diameter; located in mortar joint or grout

spacing 16 longitudinal bar diameter, 48 tie diameter, or least cross-



17/20

Slenderness Effects

Long column reduction factor

Reduce ordinate (axial force) of interaction diagram

Piers, columns, and in-plane walls

3.3.4.1.1 Nominal axial and flexural strength The nominalaxial strength, Pn , and the nominal flexural strength, Mn , of

a cross section shall be determined in accordance with the

design assumptions of Section 3.3.2. Using the

slenderness-dependent modification factors of Eq. (3-17) [1-

r an q. - r , as appropr a e, e

nominal axial strength shall be modified for the effects of

slenderness.


Interaction Diagram: Slenderness Effects

80

.,

h = 0 ft

60

70

h = 12 ft

40

50

P(kip-ft/ft)

20

30h = 20 ft

0

10

0 1 2 3 4 5 6 7 8


M (kip-ft/ft)


18/20

Bearing Walls

Location of Reaction:Wall section

Bearing area (1.9.5):

2A

w/3

1

1

1A

o A

Members that rotate will cause

reaction to shift towards edge

A2

A2 ends at edge

o mem er or

head joint in

stack bond

w/2

Members that experience

A1A2 Strength Design:

= 0.6 (3.1.4.5)

Bn = 0.6fmAbr(3.1.7)


little rotation (deep truss) Plan view

Bearing WallsDistribution of Concentrated Loads Along Wall: (1.9.7)

Load is dispersed along a 2 vertical: 1 horizontal line.

Load

Load is dispersed

at 2:1 slope

Check bearing

on hollow wall

(a) Distribution of conenctrated load through bond beam



19/20

Bearing1

Load

1

Load Load

Walls 2

h

h/2

Effective Length Effective

Length

Effective

Length

1

2

Load

1

2

Load

Effective

Length

Effective

Length


(b) Distribution of conenctrated load in wall

Prestressed Masonry


www.masoncontractors.org/newsandevents/masonryheadlines/892004950.php

www.durowal.com/prod/pdf/catalog/07-sure-stress_post_tensioning.pdf


20/20

Prestressed Masonry

Load indicating washer (LIW)


Documents

Axial and Flexure