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Lecture 9 - Flexure

June 20, 2003CVEN 444

Lecture GoalsLecture Goals

Load EnvelopesResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam

Unknown section dimensionsKnown section dimensions

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MomentMomentEnvelopesEnvelopes

Fig. 10-10; MacGregor (1997)

The moment envelope curve defines the extreme boundary values of bending moment along the beam due to critical placements of design live loading.

MomentMomentEnvelopes ExampleEnvelopes ExampleGiven following beam with a dead load of 1 k/ft and live load 2 k/ft obtain the shear and bending moment envelopes

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MomentMomentEnvelopes ExampleEnvelopes ExampleUse a series of shear and bending moment diagrams

Wu = 1.2wD + 1.6wL

0

1

2

3

4

5

0 5 10 15 20 25 30 35 40

(ft)

kips

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30 35 40

ft

kips

-250

-200

-150

-100

-50

0

50

100

150

0 5 10 15 20 25 30 35 40

ft

k-ft

Shear Diagram Moment Diagram


Wu = 1.2wD + 1.6wL


0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 5 10 15 20 25 30 35 40

ft

k/ft

-20

-15

-10

-5

0

5

10

15

20

0 5 10 15 20 25 30 35 40

ft

kips

-80

-60

-40

-20

0

20

40

0 5 10 15 20 25 30 35 40

ft

k-ft

(Dead Load Only)

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Wu = 1.2wD + 1.6wL


00.5

11.5

22.5

33.5

44.5

5

0 5 10 15 20 25 30 35 40

ft

k/ft

-60-50-40-30-20-10

01020304050

0 5 10 15 20 25 30 35 40

ft

kips

-200

-150

-100

-50

0

50

100

150

200

0 5 10 15 20 25 30 35 40

ft

k-ft

MomentMomentEnvelopes ExampleEnvelopes ExampleThe shear envelope

Shear Envelope

-80-60-40-20

020406080

0 10 20 30 40

ft

kips

Minimum ShearMaximum Shear

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MomentMomentEnvelopes ExampleEnvelopes ExampleThe moment envelope

Moment Envelope

-300

-200

-100

0

100

200

0 5 10 15 20 25 30 35 40

ft

k-ft

Minimum Moment Maximum Moment

Flexural Design of Reinforced Flexural Design of Reinforced Concrete Beams and Slab SectionsConcrete Beams and Slab Sections

Analysis Versus Design:Analysis: Given a cross-section, fc , reinforcement

sizes, location, fy compute resistance or capacity

Design: Given factored load effect (such as Mu) select suitable section(dimensions, fc, fy, reinforcement, etc.)

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ACI Code Requirements for Strength Design

Basic Equation: factored resistance factored load effect

≥Ex.

un M M ≥φ

ACI Code Requirements for Strength ACI Code Requirements for Strength DesignDesign

un M M ≥φMu = Moment due to factored loads (required

ultimate moment)

Mn = Nominal moment capacity of the cross-section using nominal dimensions and specified material strengths.

φ = Strength reduction factor (Accounts for variability in dimensions, material strengths, approximations in strength equations.

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Required Strength (ACI 318, sec 9.2)

U = Required Strength to resist factored loadsD = Dead LoadsL = Live loadsW = Wind Loads E = Earthquake Loads


Required Strength (ACI 318, sec 9.2)

H = Pressure or Weight Loads due to soil,ground water,etc.

F = Pressure or weight Loads due to fluids with well defined densities and controllable maximum heights.

T = Effect of temperature, creep, shrinkage, differential settlement, shrinkage compensating.

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Factored Load CombinationsFactored Load Combinations

U = 1.2 D +1.6 L Always check even if other load types are present.

U = 1.2(D + F + T) + 1.6(L + H) + 0.5 (Lr or S or R)U = 1.2D + 1.6 (Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6 W + 1.0L + 0.5(Lr or S or R) U = 0.9 D + 1.6W +1.6HU = 0.9 D + 1.0E +1.6H

Resistance Factors, Resistance Factors, φ φ −− ACI Sec ACI Sec 9.3.2 Strength Reduction Factors9.3.2 Strength Reduction Factors

[1] Flexure w/ or w/o axial tension

The strength reduction factor, φ, will come into the calculation of the strength of the beam.

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Resistance Factors, Resistance Factors, φ φ −− ACI Sec ACI Sec 9.3.2 Strength Reduction Factors9.3.2 Strength Reduction Factors[2] Axial Tension φ = 0.90

[3] Axial Compression w or w/o flexure(a) Member w/ spiral reinforcement φ = 0.70(b) Other reinforcement members φ = 0.65

*(may increase for very small axial loads)

Resistance Factors, Resistance Factors, φ φ −− ACI Sec ACI Sec 9.3.2 Strength Reduction Factors9.3.2 Strength Reduction Factors

[4] Shear and Torsion φ = 0.75

[5] Bearing on Concrete φ = 0.65

ACI Sec 9.3.4 φ factors for regions of high seismic risk

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Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections

1. Location of Reinforcementlocate reinforcement where cracking occurs (tension region) Tensile stresses may be due to :

a ) Flexureb ) Axial Loadsc ) Shrinkage effects


2. Construction

formwork is expensive - try to reuse at several floors

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3. Beam Depths

• ACI 318 - Table 9.5(a) min. h based on l (span) (slab & beams)

• Rule of thumb: hb (in) l (ft)

• Design for max. moment over a support to set depth of a continuous beam.

≅


4. Concrete Cover

Cover = Dimension between the surface of the slab or beam and the reinforcement

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4. Concrete Cover

Why is cover needed?[a] Bonds reinforcement to concrete[b] Protect reinforcement against corrosion[c] Protect reinforcement from fire (over

heating causes strength loss)[d] Additional cover used in garages, factories,

etc. to account for abrasion and wear.


Minimum Cover Dimensions (ACI 318 Sec 7.7)

Sample values for cast in-place concrete

• Concrete cast against & exposed to earth - 3 in.

• Concrete (formed) exposed to earth & weather No. 6 to No. 18 bars - 2 in.No. 5 and smaller - 1.5 in

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Minimum Cover Dimensions (ACI 318 Sec 7.7)

•Concrete not exposed to earth or weather- Slab, walls, joists

No. 14 and No. 18 bars - 1.5 inNo. 11 bar and smaller - 0.75 in

- Beams, Columns - 1.5 in


5.Bar Spacing Limits (ACI 318 Sec. 7.6)

- Minimum spacing of bars

- Maximum spacing of flexural reinforcement in walls & slabs

Max. space = smaller of⎩⎨⎧

.in 18 t3

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Minimum Cover DimensionMinimum Cover Dimension

Interior beam.


Reinforcement bar arrangement for two layers.

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ACI 3.3.3

Nominal maximum aggregate size.

- 3/4 clear space - 1/3 slab depth - 1/5 narrowest dim.

Example Example -- Singly Reinforced Singly Reinforced BeamBeamDesign a singly reinforced beam, which has a moment capacity, Mu = 225 k-ft, fc = 3 ksi, fy = 40 ksi and c/d = 0.275

Use a b = 12 in. and determine whether or not it is sufficient space for the chosen tension steel.

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamFrom the calculation of Mn

{

u

n

c c

2c 1

2c

size

R

210.85 0.85 1

2 210.85 1 where, 2

10.85 12

aM C d

a a af ba d f bd dd d

a cf bd k k kd d

f k k bd

β

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞′ ′ ′= − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞′ ′= −⎜ ⎟⎝ ⎠144424443

Example Example -- Singly Reinforced Singly Reinforced BeamBeamSelect c/d =0.275 so that φ =0.9. Compute k’ and determine Ru

( )

( ) ( )

1

u c

0.85 0.275

0.23375

0.85 12

0.233750.85 3 ksi 0.23375 12

0.5264 ksi

ckd

kR f k

β ⎛ ⎞′ = =⎜ ⎟⎝ ⎠

=′⎛ ⎞′= −⎜ ⎟

⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

=

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCalculate the bd 2

U

2 N

u u

3

12 in225 k-ftft

0.9

5699 in0.5264 ksi

MMbdR R

φ⎛ ⎞⎜ ⎟⎝ ⎠= =

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠= =

Example Example -- Singly Reinforced Singly Reinforced BeamBeamCalculate d, if b = 12 in.

32 25699 in 440.67 in 21.79 in.

12 ind d= = ⇒ =

Use d =22.5 in., so that h = 25 in.

( )0.275 0.275 22.5 in 6.1875 in.c d= = =

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCalculate As for the beam

( )( )( )( )

c 1s

y

2

0.85

0.85 3 ksi 12 in. 0.85 6.1875 in.40 ksi

4.02 in

f b cAf

β=

=

=

Example Example -- Singly Reinforced Singly Reinforced BeamBeamChose one layer of 4 #9 bars

Compute ρ

( )2 2s 4 1.0 in 4.00 inA = =

( )( )2

s 4.00 in12.0 in 22.5 in

0.014815

Abd

ρ = =

=

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCalculate ρmin for the beam

y

min minc

y

200 200 0.00540000

0.0053 3 3000 0.00411

40000

f

ff

ρ ρ

⎧ = =⎪⎪= ⇒ =⎨⎪ = =⎪⎩

0.014815 0.005> The beam is OK for the minimum ρ

Example Example -- Singly Reinforced Singly Reinforced BeamBeamCheck whether or not the bars will fit into the beam. The diameter of the #9 = 1.128 in.

( ) ( ) [ ]b stirrup4 3 2 cover

4 1.128 in. 3 1.128 in. 2 1.5 in. 0.375 in.11.65 in

b d s d⎡ ⎤= + + +⎣ ⎦= + + +

=

So b =12 in. works.

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCheck the height of the beam.

Use h = 25 in.

( ) [ ]

bstirrupcover

21.128 in.

22.5 in. 1.5 in. 0.375 in.2

24.94 in

dh d d⎡ ⎤= + + +⎣ ⎦

= + + +

=

Example Example -- Singly Reinforced Singly Reinforced BeamBeamFind a

Find c

( )( )( )( )

2s y

c

4.0 in 40 ksi0.85 0.85 3 ksi 12.0 in.5.23 in.

A fa

f b= =

=

1

5.23 in.0.85

6.15 in.

acβ

= =

=

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCheck the strain in the steel

Therefore, φ is 0.9

( )t cu22.5 in. 6.15 in. 0.003

c 6.15 in.0.00797 0.0056.15 in. 0.273322.5 in.

d c

cd

ε ε− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ≥

= =

Example Example -- Singly Reinforced Singly Reinforced BeamBeamCompute the Mn for the beam

Calculate Mu

( )( )

N s y

2

25.23 in.4.0 in 40 ksi 22.5 in.

23186.6 k-in

aM A f d⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

=

( )U N

0.9 3186.6 k-in 2863.4 k-inM Mφ=

= =

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Example Example -- Singly Reinforced Singly Reinforced BeamBeamCheck the beam Mu = 225 k-ft*12 in/ft =2700 k-in

Over-designed the beam by 6%

2863.4 2700 *100% 6.05%2700

−=

( )6.15 in. 0.273322.5 in.

cd

= = Use a smaller c/d ratio

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