Flexure and Shear - Amazon Web Services...Ultimate Limit States: Flexure and Shear 4:30 PM –6:30...
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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)
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© 2018 Canada Masonry Design Centre 2
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)
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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
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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
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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
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Assumptions for Pure Bending
C
T
Mr Mf
Conditions at Ultimate
Moment Capacity1: Strain Compatibility
(Plane sections remain plane)
εmu = 0.003
c
εs = ?
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Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
χf′m
c
T = fy
Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
χ0.85f′m
β1c
T = fy
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Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
β c2
T = ϕsAsfy
C = ϕmχ (0.85f′m)bβ1c
Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
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
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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
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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
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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
C = ϕmχ (0.85f′m)bβ1c
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
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Review: CSA S304 Clauses
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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
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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)
Beam Details• 3-Course beam
• 20 cm units, Type S mortar, 20 MPaBlock strength
• Knock-out depth = 70 mm• Minimum cover = 40 mm• Minimum clearance = 13 mm
• U-Lintel on Bottom course
• 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
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CSA A371
Conditions at Ultimate
Moment Capacity1: Strain Compatibility
(Plane sections remain plane)
εmu = 0.003
c
εs
εs′d′
d
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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′
Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
2
T = ϕsAsfy
C = ϕmχ (0.85f′m)bβ1c
Cs = ϕsAs′fy
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Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
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 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′
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Moment Resistance
• Take moment about any internal point• Moment about tension steel
T = ϕsAsfy
C = ϕmχ (0.85f′m)bβ1c
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
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Review: CSA S304 Clauses
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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
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Intermediate Reinforcement
• Design Considerations• Contributes as tension reinforcement in the
beam• May not yield in tension• Intermediate reinforcement does not
contribute as compression reinforcement
Beam Details• 6-Course beam
• 20 cm units, Type S mortar, 15 MPa block strength
• Knock-out depth = 70 mm• Minimum cover = 40 mm• Minimum clearance = 13 mm
• U-Lintel on Bottom course
• 25M Tension Rebar (As)
• 15M Compression Rebar (As′)
• 10M Compression Stirrups
• 15M Intermediate Reinforcement
d′ = 62 mm
d3
= 462 m
m
d2
= 862 m
m
d1
= 1,108.5 m
m
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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
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Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
2
T1 = ϕsAs1fy
C = ϕmχ (0.85f′m)bβ1c
Cs = ϕsAs′fy
T2 = ϕsAs2fs2
Conditions at Ultimate
Moment Capacity
2: Force Equilibrium
(Summation of Internal Forces is zero)
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 7.5 MPa (Table 4 for a fully-grouted 15 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)
As1 500 mm2(25M bar)
fy 400 MPa
As′ 200 mm2(15M bar)
As2 200 mm2(15M bar)
fs2 ε d2 cc
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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
C = ϕmχ (0.85f′m)bβ1c
Cs = ϕsAs′fy
T2 = ϕsAs2fs2
Mr T1 d1 c T2 d2 c C cβ1c2
Cs c d′
Moment
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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
Review: CSA S304 Clauses
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Shear(Pages 273-279)
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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
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Beam Details• 3-Course beam
• 20 cm units, Type S mortar, 20 MPaBlock strength
• Knock-out depth = 70 mm• Minimum cover = 40 mm• Minimum clearance = 13 mm
• U-Lintel on Bottom course
• 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
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
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Conditions at Ultimate
Shear Capacity
Masonry and Reinforcement Contributions
Variable Value (CSA S304-04 Reference if Applicable)
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
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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
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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
Review: CSA S304 Clauses
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Supplemental eLearning
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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