1

CEE 626 MASONRY DESIGN SLIDES

Slide Set 4C Reinforced Masonry

Spring 2015

CODE - COMBINED LOADING – GIVES ASSUMPTIONS AND SAYS USE MECHANICS – THUS INTERACTION DIAGRAMS – CODE SECTION 8.3

(ASD&SD) DESIGN RM WALLS -AXIAL &FLEXURE LOADS

If Load bearing there is a vertical load May be eccentrically Applied

Roof/floor Diaphragm Supports Wall

Flexure and Axial Load with Reinforced Masonry

ASD – Interaction diagrams reinforced masonry walls and pilasters Present design techniques and aids for

design problems involving combinations of flexure and axial force

interaction diagrams by hand and by spreadsheet

Briefly present reinforced masonry columns

Repeat with strength design

Design for Axial Forces and flexure-(OP similar for IP) ASD Interaction Diagrams

COMBINED AXIAL STRESS AND COMPRESSION BENDING STRESS = just add fa+fb (no interaction equation)

Maximum compressive stress in masonry from axial load plus bending must not exceed ( 0.45 ) f’m

Axial compressive stress must not exceed allowable axial stress from Code 8.2.4.1 – Axial forces limited Eq.8-21 or 8-22. Note same result .

Limit tension stress in reinforcing to Fs – usually 32,000 psi – Compression steel similar if tied (not usual – columns)

Code Section 8.3

Limit P allied ≤ P allowable (axial compressive capacity)

Code Equations (8-21) and (8-22) Slenderness reduction coefficients – same as

unreinforced masonry. Compression reinforcing must be tied as per 5.3.1.4 to account for it. Ie FS = 0 (compression) if not tied.

99for 70)65.025.0(

99 for 140

1)65.025.0(

2'

2'

rh

hrFAAfP

rh

rhFAAfP

sstnma

sstnma (8-21)

(8-22)

Axial Compression in Bars Can be accounted for if tied as:

5.3.1.4 Lateral ties ..: (a) Longitudinal reinforcement .. enclosed by lateral ties at least

1/4 in …diameter. (b) Vertical spacing of lateral ≤16 longitudinal bar diameters,

lateral tie bar or wire diameters, or least cross-sectional dimension of the member.

(c) Lateral ties shall be arranged such that every corner and alternate longitudinal bar shall have lateral support provided by the corner of a lateral tie with an included angle of not more than135 degrees. No bar … farther than 6 in. (152 mm) clear… from such a laterally supported bar. Lateral ties … in either a mortar joint or in grout. Where longitudinal bars … circular ties shall be 48 tie diameters.

d) Lateral ties shall be located ….one-half lateral tie spacing above the top of footing or slab in any story, …one-half a lateral tie spacing below the lowest horizontal reinforcement in beam, …

(e) Where beams or brackets frame into a … not more than 3 in…. such beams or brackets.

Allowable – Stress Interaction Diagrams Also See MDG Ch. 9.

stress is proportional to strain ; assume plane sections ; vary strain ( stress ) gradient and position of neutral axis to generate combinations of P and M

Reminder for unreinforced masonry allowable stress in compression is governed by

pure axial capacity , unity equation ( combinations of flexural and axial stress ) or buckling

allowable stress in tension is governed by flexural tensile capacity

Allowable – Stress Interaction Diagrams Walls

Allowable – stress interaction diagram Linear elastic theory – no tension in

masonry it is ignored – Stress proportional to strain

Limit combined compression stress to Fb = 0.45 F’m

P ≤ Pa d usually = t/2 compression Simply assume a stress distribution and

back calculate a P & M combination for Equilibrium

Go through Abrams analysis similar MDG 9.4.6 – pg 9-23

Get equivalent force couple at Centerline of wall thickness

Allowable – Stress Interaction Diagrams Walls

Example 9.4-2 & 9.4-3 & 9.4-4 8” cmu with #5 @ 48 in OC f’m = 1500 psi Based on Effective width = 48 in.

9.4-2 shows hand calcs. with M = 0 P = 137.2 kips

For P = 0 , Ms governs and Ms= 33.07 kip.in (note guessed at j but close)

These were then ÷ 4ft to get capacities per ft P = 34.3 kips/ft and M= 8.27 kip.in/ft Balance point was determined and P and M for this

was determined Interaction diagrams - 3 point and full were

constructed – 9.4-2 and 9.4-3 then compared to applied loads in 9.4-4

Complete interaction diagram (see MDG 9.4-4) and over.

No. 5bars centered(typ.)

d=3.81” 48”

MDG 9.4-3&4 revised

Spreadsheet for calculating allowable-stress M-N diagram for solid masonry wall16.67 Ft Wall w/ No. 5 @ 48in (Centered) NOTE BASED ON 1 ft of Wall and Not EFFECTIVE WIDTHtotal depth, t 7.625 Wall Height, h 16.67 feetf'm, fprimem 1500 Radius of Gyration, r 2.20 inEm 1350000 h/r 90.9Fb 675.00 Reduction Factor, R 0.578Es 29000000 Allowable Axial Stress, Fa 217 psiFs 32000 Net Area, An 91.4 in^2d 3.81 Allowable Axial Compr, Pa 19822 lb (MSJC 2.3.3.2.1)kbalanced 0.311828tensile reinforcement, As/beff 0.0775 #5 @ 48 Centeredwidth, beff 12

because compression reinforcement is not tied, it is not countedk kd fb Cmas fs Axial Force Moment Axial Force

(psi) (lb) (psi) (lb) (lb-in) w/ force limitPoints controlled by steel 0.01 0.04 15 3 -32000 -2477 7 -2477

0.05 0.19 78 90 -32000 -2390 330 -23900.1 0.38 166 378 -32000 -2102 1388 -2102

0.15 0.57 263 901 -32000 -1579 3259 -15790.24 0.91 470 2581 -32000 101 9047 1010.22 0.84 420 2113 -32000 -367 7459 -3670.24 0.91 470 2581 -32000 101 9047 1010.3 1.14 638 4378 -32000 1898 15018 1898

Points controlled by masonry 0.311828 1.19 675 4812 -32000 2332 16433 23320.4 1.52 675 6172 -21750 4487 20392 44870.5 1.91 675 7715 -14500 6592 24512 65920.6 2.29 675 9258 -9667 8509 28241 85090.7 2.67 675 10801 -6214 10320 31577 103200.8 3.05 675 12344 -3625 12063 34520 120630.9 3.43 675 13887 -1611 13763 37072 137631.1 4.19 675 16974 0 16974 41000 169741.3 4.95 675 20060 0 20060 43359 200601.5 5.72 675 23146 0 23146 44151 231461.7 6.48 675 26232 0 26232 43374 26232

2 7.62 675 30861 0 30861 39271 30861 Note Moment Equation not valid after KD >2Pure compression 675 61763 0 61763 0 61763Axial Force Limits 19822 0 19822

19822 44151 19822

kd

fbFs/n

kd

fb

Fb= 0.25 f’mR+?fs

kd

fbFs/n

MDG 9.4-3&4 revised

Spreadsheet for calculating allowable-stress M-N diagram for solid masonry wall16.67 Ft Wall w/ No. 5 @ 48in (Centered) NOTE BASED ON 1 ft of Wall and Not EFFECTIVE WIDTHtotal depth, t 7.625 Wall Height, h 16.67 feetf'm, fprimem 1500 Radius of Gyration, r 2.20 inEm 1350000 h/r 90.9Fb 675.00 Reduction Factor, R 0.578Es 29000000 Allowable Axial Stress, Fa 217 psiFs 32000 Net Area, An 91.4 in^2d 3.81 Allowable Axial Compr, Pa 19822 lb (MSJC 2.3.3.2.1)kbalanced 0.311828tensile reinforcement, As/beff 0.0775 #5 @ 48 Centeredwidth, beff 12

because compression reinforcement is not tied, it is not countedk kd fb Cmas fs Axial Force Moment Axial Force

(psi) (lb) (psi) (lb) (lb-in) w/ force limitPoints controlled by steel 0.01 0.04 15 3 -32000 -2477 7 -2477

0.05 0.19 78 90 -32000 -2390 330 -23900.1 0.38 166 378 -32000 -2102 1388 -2102

0.15 0.57 263 901 -32000 -1579 3259 -15790.24 0.91 470 2581 -32000 101 9047 1010.22 0.84 420 2113 -32000 -367 7459 -3670.24 0.91 470 2581 -32000 101 9047 1010.3 1.14 638 4378 -32000 1898 15018 1898

Points controlled by masonry 0.311828 1.19 675 4812 -32000 2332 16433 23320.4 1.52 675 6172 -21750 4487 20392 44870.5 1.91 675 7715 -14500 6592 24512 65920.6 2.29 675 9258 -9667 8509 28241 85090.7 2.67 675 10801 -6214 10320 31577 103200.8 3.05 675 12344 -3625 12063 34520 120630.9 3.43 675 13887 -1611 13763 37072 137631.1 4.19 675 16974 0 16974 41000 169741.3 4.95 675 20060 0 20060 43359 200601.5 5.72 675 23146 0 23146 44151 231461.7 6.48 675 26232 0 26232 43374 26232

2 7.62 675 30861 0 30861 39271 30861 Note Moment Equation not valid after KD >2Pure compression 675 61763 0 61763 0 61763Axial Force Limits 19822 0 19822

19822 44151 19822

kd

fb

fs

kdfb

Fb= 0.25 f’mR+?fs

MDG 9.4-3&4 revised

MDG 9.4-3&4 revised

MDG 9.4-3&4P = 1000 lb/ft

e = 2.48 in

Roof/floor Diaphragm Supports Wall

3.33’

16.67’

18 psf If P = all dead load Check .6D + .6W at mid-height Would also have check other load combos 0.6D + .6W at mid-height often governs

Per foot of wall (see Example) P at mid-height including weight of wall P= 756 lb/ft (0.6 D) and M = 7,425 lb.in/ft (0.6D+0.6W)

You would need to look at other load casesand at the top of the wall.

0.6D+0.6W

MDG Has Three Designs Ch 16 – TMS Shopping Center

MDG Has Three Designs Ch 18 – RCJ Hotel

CHAPTER 17BIG BOX STOREExample DPC BOX- 02 ASD —Reinforced Loadbearing Wall See Box 00 & 01(Load Dist)

See MDG BOX 02 ASD Ch17

Strength Interaction Diagrams MDG 10.4.5

for reinforced masonry flexural tensile strength of masonry is

neglected equivalent rectangular compressive stress block

with maximum stress 0.80 fm , 1 = 0.80 stress in tensile reinforcement proportional to

strain , but not greater than fy

stress in compressive reinforcement calculated like stress in tensile reinforcement , but neglected unless compressive reinforcement is laterally tied

Note f = for combined axial and flexure = 0.9

Code Section 9.3

Strength Interaction Diagrams . . . vary strain ( stress ) gradient and position

of neutral axis to generate combinations of P and M – using max stress in masonry = 0.8 f’m over a compression block and max steel stress = fy

Also limit Pu applied to ≤f Pn - Eq 9-19 and Eq 9-20

99for 70))(80(.8.0

99 for 140

1))(80(.8.0

2'

2'

rh

hrfAAAfP

rh

rhfAAAfP

ystsnmn

ystsnmn

R

R

Eq 9-19

Eq 9-20

Locate neutral axis based on extreme - fiber strains

Calculate compressive force C

P = C - Ti (reduced?) MCL = C(h/2-a/2)± Ti yi

mu = 0.0035 clay 0.0025 concrete

s y

fy

Reinforcement

AxialLoad,P

0.80 f’m

1 = 0.80

Assumptions of Strength Design

MCL

Ti C

c

One or more Rebars

a

yi

h

h/2

Strength Interaction Example

vary c / d from zero to a large value when c = cbalanced , steel is at yield strain ,

and masonry is at its maximum useful strain when c < cbalanced , fs = fy when c > cbalanced , fs < fy

Compute nominal axial force and moment Pn = fi Mn = fi x yi ( measured from centerline )

Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5

Maximum reinforcement by 9.3.3.5 Nominal shear strength by 9.3.4.1.2 Three procedures for computing out – of –

plane moments and deflections - Load Side Second – order analysis - Computer Moment magnification method (new) Complementary moment method; additional

moment from P – δ effects – Slender wall analysis

Slide 33

Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5.4.3

Moment Magnification

Mu < Mcr: Ieff = 0.75In

Mu ≥ Mcr: Ieff = Icr

Slide 34

)339(

)329(1

1

)319(

2

2

0,

hIE

P

PP

MM

effme

e

u

uu

Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5.4.2

Complementary Moment – Iterative?

Eq. 9-29 for Mu < Mcr

Eq. 9-30 for Mu ≥ McrSlide 35

)309(48

5485

)299(485

)289(

)279(28

22

2

2

crm

cru

nm

cru

nm

uu

ufuwu

uuu

ufu

u

IEhMM

IEhM

IEhM

PPP

PePhwM

Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5.4.2

Or Solve Equations for Mu - 2 eq. 2 unknows

Mu < Mcr

Mu ≥ Mcr

Slide 36

4851

282

2

nm

u

uuf

u

u

IEhP

ePhw

M

crm

u

crnm

ucruuf

u

u

IEhP

IIEhPMePhw

M

4851

1148

528

2

22

Design of Walls for Out-of–Plane Loads: Design Procedure

Estimate amount of reinforcement from the following equations.

This neglects P – δ effects. Can estimate increase in moment, such as 10%, for a preliminary estimate of amount of reinforcement.

Slide 37

bf

MtdPddam

uu

8.0

2/22

f

y

ums f

PbafA f/8.0

DESIGN of REINF MASONRY WALLS

Do example 10.4-3 and 10.4-5 in MDG

DESIGN of REINF MASONRY WALLS - Out of Plane loads – Axial and flexure

Should check shear but never a problem out of plane for LB and non LB walls.

Prescriptive requirements for columnsCode 5.3 – Columns . . . See Slide set 3

Design Masonry (ASD&SD) Not used often

41

Allowable – Stress Interaction Diagrams Columns

States of Stress - Note the code requires a minimum Eccentricity of 0.1 t

t

b

345

2 1

compressioncontrols

fb

P

e

gt

d

fs /n

tension controls

Fs /n

Fb

d15

fs / n Fs / n fb Fb

42

Allowable – Stress Interaction Diagrams

• allowable – stress interaction diagram by spreadsheet

• column example of MDG » CMU

» f ‘ m = 1500 psi

» f ‘ g = 2700 psi

» 16 – in . square column» 4 - #7 bars

9.38 in. 15.63 in.

15.63 in.

9.38 in.

43

. . . Allowable – Stress Interaction Example

See Column Example in MDG RCJ 03 -

-50

0

50

100

150

0 50 100 150 200 250 300

M, in.-kips

P, k

ips

RFAAfP sstnma )65.025.0( '

. . . Distinction between columns and pilasters -

Pilasters are thickened sections of masonry walls , are not isolated members , and are not columns

Pilasters need not be reinforced nor have 4 bars

If pilaster capacity includes the effect of longitudinal reinforcement , however , that reinforcement must be tied laterally

MASONRY COLUMNS MUST BE REINFORCED

45

Forces on Pilaster – Wall System

See MDG – Chapter 7

mid - height of pilaster

out – of – planereaction fromdiaphragm

gravity load

wind orearthquakeload

pilaster moment about in – plane axis

wall moment about vertical axis

46

MDG 7.5-18 : Examples of Coursing Layouts for Hollow Unit Pilasters

Fig. 7.5-18 Coursing Layout for Hollow Unit Pilasters

Alternate Courses

16 x 24 – in. pilaster16 x 16 – in. pilaster

Pilasters Flexural/axial design of pilasters

by allowable – stress provisions is as discussed previously.

Flexural/axial design of pilasters by strength provisions is as discussed previously.