Lecture 9 - Flexure

Lecture 9 - Flexure

June 20, 2003CVEN 444

Lecture GoalsLecture Goals

Load EnvelopesResistance Factors and LoadsDesign of Singly Reinforced Rectangular Beam Unknown section dimensions Known section dimensions

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

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)

kip

s

-80

-60

-40

-20

0

20

40

60

80

0 5 10 15 20 25 30 35 40

ft

kip

s

-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

kip

s

-80

-60

-40

-20

0

20

40

0 5 10 15 20 25 30 35 40

ft

k-ft

(Dead Load Only)


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

0102030

4050

0 5 10 15 20 25 30 35 40

ft

kip

s

-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

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 Concrete Beams and Slab SectionsSections

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.)


ACI Code Requirements for Strength Design

Basic Equation: factored resistance factored load effect

Ex.

un M M

ACI Code Requirements for ACI Code Requirements for Strength DesignStrength Design

un M M Mu = Moment due to factored loads (required ultimate moment)

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

material strengths.

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


Required Strength (ACI 318, sec 9.2)

U = Required Strength to resist factored loads

D = Dead Loads

L = Live loads

W = 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.

Factored Load Factored Load CombinationsCombinations

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 +1.6H

U = 0.9 D + 1.0E +1.6H

Resistance Factors, Resistance Factors, ACI ACI Sec 9.3.2 Strength Reduction Sec 9.3.2 Strength Reduction FactorsFactors

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

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


[2] Axial Tension = 0.90

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

*(may increase for very small axial loads)


[4] Shear and Torsion = 0.75

[5] Bearing on Concrete = 0.65

ACI Sec 9.3.4 factors for regions of high seismic risk

Background Information for Background Information for Designing Beam SectionsDesigning Beam Sections

1. Location of Reinforcement locate 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


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


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


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

Minimum Cover DimensionMinimum Cover Dimension

Interior beam.


Reinforcement bar arrangement for two layers.


ACI 3.3.3

Nominal maximum aggregate size.

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

Example - Singly Example - Singly Reinforced BeamReinforced Beam

Design 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.

Example - Singly Example - Singly Reinforced BeamReinforced BeamFrom the calculation of Mn

u

n

c c

2c 1

2c

size

R

2

10.85 0.85 1

2 2

10.85 1 where,

2

10.85 1

2

aM C d

a a af ba d f bd d

d d

a cf bd k k k

d d

f k k bd

Example - Singly Example - Singly Reinforced BeamReinforced BeamSelect c/d =0.275 so that =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 1

2

0.5264 ksi

ck

d

kR f k

Example - Singly Example - Singly Reinforced BeamReinforced BeamCalculate the bd 2

U

2 N

u u

3

12 in225 k-ft

ft0.9

5699 in0.5264 ksi

M

Mbd

R R


Calculate d, if b = 12 in.

32 25699 in

440.67 in 21.79 in.12 in

d d

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

0.275 0.275 22.5 in 6.1875 in.c d

Example - Singly Example - Singly Reinforced BeamReinforced BeamCalculate 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 cA

f

Example - Singly Example - Singly Reinforced BeamReinforced BeamChose one layer of 4 #9 bars

Compute

2 2s 4 1.0 in 4.00 inA

2

s 4.00 in

12.0 in 22.5 in

0.014815

A

bd


Calculate min for the beam

y

min min

c

y

200 2000.005

400000.005

3 3 30000.00411

40000

f

f

f

0.014815 0.005 The beam is OK for the minimum

Example - Singly Example - Singly Reinforced BeamReinforced BeamCheck 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.

Example - Singly Example - Singly Reinforced BeamReinforced BeamCheck 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 - Singly Example - Singly Reinforced BeamReinforced BeamFind a

Find c

2s y

c

4.0 in 40 ksi

0.85 0.85 3 ksi 12.0 in.

5.23 in.

A fa

f b

1

5.23 in.

0.85

6.15 in.

ac

Example - Singly Example - Singly Reinforced BeamReinforced BeamCheck the strain in the steel

Therefore, is 0.9

t cu

22.5 in. 6.15 in.0.003

c 6.15 in.

0.00797 0.005

6.15 in.0.2733

22.5 in.

d c

c

d

Example - Singly Example - Singly Reinforced BeamReinforced BeamCompute the Mn for the beam

Calculate Mu

N s y

2

2

5.23 in.4.0 in 40 ksi 22.5 in.

2

3186.6 k-in

aM A f d

U N

0.9 3186.6 k-in 2863.4 k-in

M M

Example - Singly Example - Singly Reinforced BeamReinforced BeamCheck 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.

c

d Use a smaller c/d

ratio

Documents

Lecture 9 - Flexure