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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.