1. CEE 626 MASONRY DESIGN SLIDES Slide Set 4C Reinforced Masonry Spring 2015

2. (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
CODE - COMBINED LOADING – GIVES ASSUMPTIONS AND SAYS USE MECHANICS – THUS INTERACTION DIAGRAMS – CODE SECTION 8.3

3. 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

4. Design for Axial Forces and flexure-(OP similar for IP) ASD Interaction Diagrams
Code Section 8.3
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 – Axial forces limited Eq.8-21 or Note same result .
Limit tension stress in reinforcing to Fs – usually 32,000 psi – Compression steel similar if tied (not usual – columns)

5. 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 to account for it. Ie FS = 0 (compression) if not tied.
(8-21)
(8-22)

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

7. 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

8. Allowable – Stress Interaction Diagrams Walls
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 – pg 9-23

10. Get equivalent force couple at Centerline of wall thickness

15. Allowable – Stress Interaction Diagrams Walls
No. 5bars centered(typ.)
d=3.81”
48”
Example & & 9.4-4
8” cmu with 48 in OC
f’m = 1500 psi Based on Effective width = 48 in.
9.4-2 shows hand calcs. with M = 0 P = kips
For P = 0 , Ms governs and Ms= 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 – and then compared to applied loads in 9.4-4
Complete interaction diagram (see MDG 9.4-4) and over.

16. MDG 9.4-3&4 revised
Fs/n kd fb kd fb fs
Fb= 0.25 f’mR+?

18. Fs/n kd fb

19. MDG 9.4-3&4 revised fs kd fb kd fb fs
Fb= 0.25 f’mR+?

20. MDG 9.4-3&4 revised

21. MDG 9.4-3&4 revised

22. Check .6D + .6W at mid-height Would also have check other load combos
MDG 9.4-3&4
P = 1000 lb/ft
e = 2.48 in
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 cases
and at the top of the wall.
3.33’
Roof/floor Diaphragm Supports Wall
16.67’

23. 0.6D+0.6W

24. MDG Has Three Designs
Ch 16 – TMS Shopping Center

26. MDG Has Three Designs
Ch 18 – RCJ Hotel

28. CHAPTER 17
BIG BOX STORE
Example DPC BOX
- 02 ASD —Reinforced Loadbearing Wall
See Box 00 & 01(Load Dist)
See MDG BOX 02 ASD Ch17

29. Strength Interaction Diagrams MDG 10.4.5
Code Section 9.3
for reinforced masonry
flexural tensile strength of masonry is neglected
equivalent rectangular compressive stress block with maximum stress fm , 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

30. 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 ≤ Pn - Eq 9-19 and Eq 9-20
Eq 9-19 R
Eq 9-20 R

31. Assumptions of Strength Design
MCL
Axial
Load,P
h/2 c
s  y
mu = clay
concrete
Reinforcement a fy
0.80 f’m
1 = 0.80
Locate neutral axis based on extreme - fiber strains
Calculate compressive force C
P = C - Ti (reduced?)
MCL = C(h/2-a/2)± Ti yi Ti C yi
One or more Rebars h

32. 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 )

33. Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5
Maximum reinforcement by
Nominal shear strength by
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
The 2013 code allows three methods for out-of-plane to account for slenderness effects.
Second-order analysis: most computer programs can do this
Moment magnification: this is new in the 2013 Code; it eliminates some of the restrictions/limitations in previous codes, such as simple support, moment approximately at midheight, and an axial load limit.
Complementary moment: this is the “slender wall” method that has been in strength design since the its inception in the MSJC. This is more accurate, but difficult to apply for other than simply supported walls with the moment approximately at mid-height.

34. 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

35. Design of Walls for Out-of–Plane Loads: TMS 402 Section 9.3.5.4.2
Complementary Moment – Iterative?
Eq for Mu < Mcr
Eq for Mu ≥ Mcr

36. 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
Although a lot of engineers iterate to find Mu, Code equations 9-27 and 9-30 are two simultaneous linear equations in two unknowns (Mu and δu), so they can be solved for one of the unknowns.

37. 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.

38. DESIGN of REINF MASONRY WALLS
Do example and in MDG

39. 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.

40. Prescriptive requirements for columns Code 5.3 – Columns . . .
Design Masonry (ASD&SD)
Not used often
Prescriptive requirements for columns Code 5.3 – Columns . . .
See Slide set 3

41. Allowable – Stress Interaction Diagrams ColumnsP
compression
controls
tension
controls b e Fb fb 1 d 5 t d gt 5 4
Fs /n 3 2 1
fs /n
fs / n  Fs / n
fb  Fb
States of Stress - Note the code requires a minimum Eccentricity of 0.1 t

42. Allowable – Stress Interaction Diagrams
allowable – stress interaction diagram by spreadsheet
column example of MDG
CMU
f ‘ m = psi
f ‘ g = psi
16 – in . square column
4 - #7 bars
9.38 in.
in.
9.38 in.

43. . . . Allowable – Stress Interaction Example
150
100
P, kips 50 50
100
150
200
250
300
-50
M, in.-kips
See Column Example in MDG RCJ 03 -

44. . . . 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 – plane
reaction from
diaphragm
gravity load
wind or
earthquake
load
pilaster moment about in – plane axis
wall moment about vertical axis

46. MDG 7.5-18 : Examples of Coursing Layouts for Hollow Unit Pilasters
Alternate Courses
16 x 24 – in. pilaster
16 x 16 – in. pilaster
Fig Coursing Layout for Hollow Unit Pilasters

47. 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.