Flexure and Shear - Amazon Web Services...Ultimate Limit States: Flexure and Shear 4:30 PM –6:30 PM Bennett Banting Lecture Outline 1. Overview (5) 2. Design for Flexure a) Tension

Engineered Masonry Design Course Friday April 13, 2018

© 2018 Canada Masonry Design Centre 1

Masonry Beams

Ultimate Limit States: Flexure and Shear

4:30 PM – 6:30 PM

Bennett Banting

Lecture Outline1. Overview (5)

2. Design for Flexurea) Tension Reinforcement (40)b) Compression Reinforcement (20)c) Intermediate Reinforcement (20)

3. Design for Shear (25)

4. Content Review (5)

5. Supplemental eLearning (5)



Overview

• Masonry Beam

• Masonry Lintel

• Movement Joint

• Concrete Block Units for Beams

Fundamental to Masonry

Beams

• Similar to reinforced concrete• Plane sections and compatibility of strains• Equilibrium of internal forces• Perfect bond of reinforcement• After cracking tensile strength of masonry

ignored• Masonry beams may only be solid (fully-

grouted) and reinforced (yielding)• Equivalent stress block is used (different

values from concrete)



Fundamental to Masonry

Beams

• Different from reinforced concrete

• May not have compression reinforcement• May not require skin reinforcement

(intermediate reinforcement)• May not have shear reinforcement• Spacing of reinforcement is restricted by

masonry unit• No flanged beams• Masonry beams are typically simply

supported (frames are rare)• Multi-span beams possible over several

gaps• Relatively short spans in walls such as

windows, doors, garages

Flexure: Beam with Tension Reinforcement(Pages 255-257)Cl. 11 CSA S304



Masonry Beam Loads

• Uniformly Distributed Dead Load of 20.0 kN/m

• Uniformly Distributed Live Load of 15.0 kN/m

• Uniformly Distributed Self-Weight

• 2.4 m centre-to-centre of assumed pin-roller simple supports

Beam Details• 3-Course beam

• 20 cm units, Type S mortar, 20 MPaBlock strength

• Block Strength is NOT design strength!

• U-Lintel on Bottom course

• 20M Tension Rebar

520 mm

45 mm

590

mm

190 mm

15 mm10 mm



Self-Weight Dead Load

Design Moment

Mf = 22.6 kN·m

1.4DL = 1.4 (20.0 kN/m + 2.43 kN/m)

Mf = 36.4 kN·m

1.25DL + 1.5 LL = 1.25 (20.0 kN/m + 2.43 kN/m) + 1.5 (15.0 kN/m)

Load Case #1 Load Case #2



Assumptions for Pure Bending

C

T

Mr Mf

Conditions at Ultimate

Moment Capacity1: Strain Compatibility

(Plane sections remain plane)

εmu = 0.003

c

εs = ?




Moment Capacity

2: Force Equilibrium

(Summation of Internal Forces is zero)

χf′m

c

T = fy


Moment Capacity



χ0.85f′m

β1c

T = fy




Moment Capacity



β c2

T = ϕsAsfy

C = ϕmχ (0.85f′m)bβ1c


Moment Capacity



Variable Value (CSA S304-04 Reference if Applicable)

ϕm 0.6 (Cl. 4.3.2.1)

χ 0.5 (Cl. 11.2.1.6 since stretchers are used grout is NOT continuous horizontally)

f′m 10 MPa (Table 4 for a fully-grouted 20 MPa block)

b 190 mm (For a 20 cm unit and its actual dimensions)

β1 0.8 (Cl. 11.2.1.6 since f′m < 20 MPa)

c

ϕs 0.85 (Cl. 4.3.2.2)

As 300 mm2

fy 400 MPa

cϕ A f

ϕ χ 0.85f′ bβ



Check AssumptionTension reinforcement yields at

εmu = 0.003

c

εs = ?

εc

εd cd

d - cε

Ef

200,000MPa400MPa

0.002

Similar Triangles

Maximum Reinforcement• Masonry beams must be under-

reinforced • Meaning that the extreme

tension reinforcement must yield at ultimate conditions

• Can be checked from similar triangles or using Cl. 11.2.2

cd

600600 f



Special Case

• The balanced condition is defined as when the neutral axis depth, cb, is exactly at the point where the reinforcement will yield at ultimate conditions

• The balance area of steel, Asb, is the required area to satisfy the balanced condition such that

cbd

600600 f

Asbϕ χ 0.85f′ bβ cb

ϕ f

Minimum Reinforcement• Minimum reinforcement required

for masonry beams Cl. 11.2.3.1

ρmin0.8fy



Moment Resistance

• Internal moment couple created by the tension and compression forces acts to resist applied moment

• Take moment about any internal point• Centroid of compression block or

reinforcement saves computation

T = ϕsAsfy


Mr T dβ1c2

ϕsAsfy dβ1c2

β c2

dMoment

Review

• Remember to use masonry-specific values for εmu, β1, χ, ϕm

• Do not flip between reinforced concrete design

• Recall the difference between block strength and masonry strength (always fully-grouted, type S typical)

• Masonry strengths of:

• 5 MPa, 7.5 MPa, 10 MPa, 13.5 MPa, 17 MPa

• For engineering calculations we use actual dimensions

• Beam heights of:

• 190 mm, 390 mm, 590 mm, 790 mm, 990 mm etc.• Beam widths of:

• 140 mm, 190 mm, 240 mm, 290 mm

• Always check assumptions and limits• Reinforcement yields



Review: CSA S304 Clauses





Flexure: Beam with Compression Reinforcement(Page 258)

Compression Reinforcement

• Useful to meet moment demands when other design options exhausted

• Increase block strength or amount of tension reinforcement

• A wider block or deeper beam may be an option in some cases

• Sometimes shear stirrups can act as compression reinforcement stirrups as well



Compression Reinforcement

• Cl. 11.2.6.4• 6.0 mm Diameter Stirrups as a

Minimum (D4.5 Wire)• Spacing of Lesser of 16 Bar

Diameters or 48 Tie Diameters as a Maximum

• Cl. 12.2• Least dimension of member

• 10M Stirrups • Maximum spacing for the following

compression steel sizes• 160 mm – 10M (use 10 mm)• 240 mm – 15M (use 15 mm)• 320 mm – 20M (use 20 mm)• 400 mm – 25M (use 25 mm)



• Knock-out depth = 70 mm• Minimum cover = 40 mm• Minimum clearance = 13 mm


• 25M Tension Rebar (As)

• 15M Compression Rebar (As′)

• 10M Compression Stirrups

508.5 mm

45 mm

190 mm

13 mm11 mm

11 mm

43 mm 62 mm



CSA A371


Moment Capacity1: Strain Compatibility

(Plane sections remain plane)

εmu = 0.003

c

εs

εs′d′

d



Strain Compatibility using Similar Trianglesεmu

c - d′

εs

εs′d′

d - c

c - d′

εs′εmu

c

εs

d - c

εc

εd c

ε ′c d′


Moment Capacity



2

T = ϕsAsfy


Cs = ϕsAs′fy




Moment Capacity




ϕm 0.6 (Cl. 4.3.2.1)


f′m 10 MPa (Table 4 for a fully-grouted 20 MPa block)


β1 0.8 (Cl. 11.2.1.6 since f′m < 20 MPa)

c Unknown

ϕs 0.85 (Cl. 4.3.2.2)

As 500 mm2 (25M bar)

fy 400 MPa

As′ 200 mm2 (15M bar)

cϕ A f ϕ A ′f

ϕ χ 0.85f′ bβ

Strain Compatibility using Similar Triangles

c - d′

εs′εmu

c

εs

d - c

εc

εd c

ε ′c d′



Moment Resistance

• Take moment about any internal point• Moment about tension steel

T = ϕsAsfy


Mr C dβ1c2

Cs d d

β c2

d

Cs = ϕsAs′fy

d′

Mr ϕ χ 0.85f′ bβ c dβ1c2

ϕ A ′f d d

Moment

Review

• Compression reinforcement does not need to yield

• It must be manually checked to verify assumptions

• The location of tension and compression reinforcement is dependent on the unit configuration

• U-Lintel and Knock-out Web units are non-standard and vary by manufacturer

• Specific block manufacturers should be consulted

• Knock-out Web units can facilitate grout continuity• A value of χ = 0.7 may be used if neutral axis lies in

region with horizontal grout continuity

• If specifying knock-out web units for compression reinforcement it is usually best practice to have it throughout beam

• This will also ensure better grout flow in the beam






Flexure: Beam with Intermediate Reinforcement(Page 261)

Intermediate Reinforcement

• Differentiated from “Main Tension” and “Tied Compression” Reinforcement

• Primary purpose is crack control for tall beams (over 3-courses)

• Rationale similar to skin reinforcement for reinforced concrete

• Cl. 11.2.6.3

• 15M bar at 400 mm for 15 cm and 20 cm block

• 2×15M bars at 400 mm for 25 cm and 30 cm block



Intermediate Reinforcement

• Design Considerations• Contributes as tension reinforcement in the

beam• May not yield in tension• Intermediate reinforcement does not

contribute as compression reinforcement


• 20 cm units, Type S mortar, 15 MPa block strength






• 15M Intermediate Reinforcement

d′ = 62 mm

d3

= 462 m

m

d2

= 862 m

m

d1

= 1,108.5 m

m



Assumptions

Main Tension Reinforcement (As1) • Yielded in tension

1First Intermediate Bar (As2)• Under tension but has

NOT yielded

2Second Intermediate Bar (As3)• Under compression but

is NOT tied

3Compression Reinforcement (As′)• Yielded in compression

and is tied

4

1: Strain Compatibility using Similar Trianglesεmu

εs1

εs′

c - d′

εs′εmu

c

d1 - c

εc

εc d3

ε ′c d′

εs3

εs2

c

εs3

c – d3

εs2

d2 - c

εs1

εd2 c

εd1 c




Moment Capacity



2

T1 = ϕsAs1fy


Cs = ϕsAs′fy

T2 = ϕsAs2fs2


Moment Capacity




ϕm 0.6 (Cl. 4.3.2.1)


f′m 7.5 MPa (Table 4 for a fully-grouted 15 MPa block)


β1 0.8 (Cl. 11.2.1.6 since f′m < 20 MPa)

c

ϕs 0.85 (Cl. 4.3.2.2)

As1 500 mm2(25M bar)

fy 400 MPa

As′ 200 mm2(15M bar)

As2 200 mm2(15M bar)

fs2 ε d2 cc



Verify Assumptionsεmu

εs1

εs′

c - d′

εs′εmu

c

d1 - c

εc

εc d3

ε ′c d′

εs3

εs2

c

εs3

c – d3

εs2

d2 - c

εs1

εd2 c

εd1 c

Moment ResistanceTake moment about neutral axis

T1 = ϕsAs1fy


Cs = ϕsAs′fy

T2 = ϕsAs2fs2

Mr T1 d1 c T2 d2 c C cβ1c2

Cs c d′

Moment



Review

• Reinforcement depths are typically based on the knock-out web unit configuration

• It is difficult to place reinforcement elsewhere in masonry

• Most efficient to place it on the webs of knock-out web units

• Problems may be iterative• Solution is dependent on the location of neutral axis

• Intermediate reinforcement should concentrate close to the tension side of the beam

• Explicitly specified in the 2014 CSA S304

• We took moment about the neutral axis to avoid changing signs in our calculation

• If moment was calculated about the lowest tension steel, C, Cs acts against As2 moments




Shear(Pages 273-279)



Changes to the 2004

Design Standard

• Masonry beam shear design has adapted to the modified compression field theorem used in reinforced concrete design

• Course work is based on 2004 standard

• See Supplemental eLearning for a more detailed description of designing with the 2014 standard

Shear Failure










508.5 mm

45 mm

190 mm

13 mm11 mm

11 mm

43 mm 62 mm

Shear Resistance

• Vm• Aggregate interlock

• Vs• Shear Stirrups

• d• Shear Crack Length

• s• Shear Stirrup Spacing

Vm

Vs

s

d

d

T

C

Vs

Vf




Shear Capacity

Masonry and Reinforcement Contributions


ϕm 0.6 (Cl. 4.3.2.1)

λ 1.0 (Concrete density over 2,000 kg/m3, 0.85 or 0.75 for Lightweight, Cl. 11.3.3)

f′m 10.0 MPa (Table 4 for a fully-grouted 20 MPa block)

d 508.5 mm

bw 190 mm (For a 20 cm unit and its actual dimensions)

ϕs 0.85 (Cl. 4.3.2.2)

Av 100 mm2 (10M bar)

fy 400 MPa

s 200 mm

Vr Vm Vs 0.16ϕmλ f m 1.0d 4002000

bwdϕsAvfyd

s

Shear Reinforcement Detailing

• Minimum shear reinforcement • Must be provided when Vf > 0.5 Vm

• Av > 0.35bws/fy• Maximum Spacing of Shear Reinforcement

• s ≤ d/2 or 600 mm



Determining Vf

Mf

VmaxVf

ℓ2 d

Vmaxℓ2

Shear Detailing• Stirrups are not required where

Vf < 0.5Vm

• Limited options with masonry compared with concrete design

• If counting on stirrups as compression ties then limited applicability



Review

• Shear may be resisted entirely by masonry if Vf ≤ 0.5 Vm

• Shear reinforcement is very different from reinforced concrete

• Spacing generally limited to increments of 200 mm to fit in cells

• Vsmax limits most beams to a single 10M stirrup

• Rarely is it ever practical to change shear reinforcement over beam span

• Substantial changes were made to shear design in 2014 standard

• Focus in eLearning module






Supplemental eLearning



General Overview and Materials

Changes from 2004 to 2014 CSA

Masonry Material, Construction and Design Standards

Specialty Mortars, Clay Brick,

Connectors and Stone Products

Case Studies and Diagnostics of

Masonry Veneers

Masonry Beams: Ultimate Limit States

Modified Compression Field Theory and Shear

Design of Masonry Beams using the 2014

Standard

Support of Masonry and Bearing Design, Using Movement Joints for

Structural Applications and Arching of Masonry

over Openings

Design of Brick Beams, Deep Beams and

Prestressed Beams

Documents

Flexure and Shear - Amazon Web Services...Ultimate Limit States: Flexure and Shear 4:30 PM –6:30 PM Bennett Banting Lecture Outline 1. Overview (5) 2. Design for Flexure a) Tension