Design Analysis Beam ACI

7/31/2019 Design Analysis Beam ACI

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Lecture 10Analysis and Design

September 27, 2001

CVEN 444


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Lecture Goals

Pattern Loading

Analysis and Design

Resistance Factors and Loads

Design of Singly Reinforced Rectangular

Beam

Unknown section dimensions

Known section dimensions


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Member Depth

ACI provides minimum member depth and slabthickness requirements that can be used without adeflection calculation (Sec. 9.5)

Useful for selecting preliminary member sizes

ACI 318 - Table 9.5a:

Min. thickness, h

For beams with one end continuous: L/18.5 For beams with both ends continuous: L/21

L is span length in inches

Table 9.5a usually gives a depth too shallow for design,but should be checked as a minimum.


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Member Depth

ACI 318-99: Table 9.5a


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Member Depth Rule of Thumb:

hb (in.) ~ L (ft.)

Ex.) 30 ft. span -> hb ~ 30 in.

May be a little large, but okay as a start to calc. DL Another Rule of Thumb:

wDL (web below slab) ~ 15% (wSDL+ wLL)

Note: For design, start with maximum moment for

beam to finalize depth. Select b as a function of d

b ~ (0.45 to 0.65)*(d)


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Pattern Loads Using influence lines to determine pattern loads

Largest moments in a continuous beam or frame occur

when some spans are loaded and others are not.

Influence lines are used to determine which spans toload and which spans not to load.

Influence Line: graph of variation of shear,

moment, or other effect at one particular point ina structure due to a unit load moving across the

structure.


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Pattern Loads

Quantitative

Influence Lines

Ordinate are

calculated

(exact)

See Fig. 10-7(a-e)

MacGregor (1997)


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Qualitative Influence Lines

The Mueller-Breslau

principle can be stated as

follows:

If a function at a point on

a structure, such as

reaction, or shear, or

moment is allowed to act

without restraint, the

deflected shape of thestructure, to some scale,

represents the influence

line of the function.


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Pattern Loads

Qualitative Influence Lines

Fig. 10-7 (b,f) from MacGregor (1997)


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Pattern

LoadsFrame Example: Maximize +M at point B.

Draw qualitative influence

lines.

Resulting pattern load:checkerboard pattern


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Pattern Loads

ACI 318-99, Sec. 8.9.1:

It shall be permitted to assume that:

The live load is applied only to the floor or roofunder consideration, and

The far ends of columns built integrally with the

structure are considered to be fixed.

** For the project, we will model the entire frame. **


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Pattern Loads

ACI 318-99, Sec. 8.9.2:

It shall be permitted to assume that the

arrangement of live load is limited tocombinations of:

Factored dead load on all spans with full factored

live load on two adjacent spans.

Factored dead load on all spans with full factored

live load on alternate spans.

** For the project, you may use this provision. **


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Project: Factored Load

Combinations for Beam DesignFactored Load Combinations:

U = 1.4 (DL+SDL) + 1.7 (LLa1)

U = 1.4 (DL+SDL) + 1.7 (LLa2)

U = 1.4 (DL+SDL) + 1.7 (LLb)

U = 1.4 (DL+SDL) + 1.7 (LLc1)

U = 1.4 (DL+SDL) + 1.7 (LLc2)

Envelope Load Combinations:

Take maximum forces from all factored loadcombinations


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Moment

Envelopes

Fig. 10-10; MacGregor (1997)

The moment envelope

curve defines the extreme

boundary values of bendingmoment along the beam

due to critical placements

of design live loading.


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F.Approximate Analysis of Continuous

Beam and One-Way Slab Systems

ACI Moment and Shear Coefficients

Approximate moments and shears permittedfor design of continuous beams and one-

way slabs

Section 8.3.3 of ACI Code


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F. Approximate Analysis of Continuous

Beam and One-Way Slab Systems

ACI Moment and Shear Coefficients - Requirements:

Two or more spans

Approximately Equal Spans

Larger of 2 adjacent spans not greater than shorter by > 20%

Uniform Loads

LL/DL 3 (unfactored)

Prismatic members

Same A, I, E throughout member length

Beams must be in braced frame without significant momentsdue to lateral forces

Not state in Code, but necessary for coefficients to apply

** All these requirements must be met to use the coefficients!**


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F. Approximate Analysis of Continuous

Beam and One-Way Slab SystemsACI Moment and ShearCoefficientsMethodology:

2

)(2

nu

vu

numu

lw

CV

lwCM

wu = Total factored dead and liveload per unit length

Cm

= Moment coefficient

Cv = Shear coefficient

ln = Clear span length for span inquestion forMu at interiorface of exterior support, +Muand Vu

ln = Average of clear span lengthfor adjacent spans forMu atinterior supports

See Fig. 10-11, text


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F. Approximate

Analysis of

Continuous Beam

and One-WaySlab Systems

ACI Moment and

ShearCoefficients

See Section 8.3.3of ACI Code

Fig. 10-11, MacGregor (1997)


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Flexural Design of Reinforced

Concrete Beams and Slab SectionsACI Code Requirements for Strength Design

Basic Equation: factored resistance factored load

effect

Ex.un MM

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

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

= Strength reduction factor (Accounts for variability in dimensions,

material strengths, approximations in strength equations.


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Flexural Design of Reinforced

Concrete Beams and Slab SectionsRequired Strength (ACI 318, sec 9.2)

U = Required Strength to resist factored loads

D = Dead LoadsL = Live loads

W = Wind Loads

E = Earthquake Loads

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 Combinations

Similar combination for earthquake, lateral pressure,fluid pressure, settlement, etc.

U = 1.05 D + 1.28 L + 1.4 E

U = 0.9 D + 1.43 E

U = 1.4 D + 1.7 L + 1.7 H

U = 0.9 D + 1.7 H

U

=1.4 D + 1.7 L + 1.4 F

U = 0.9 D + 1.4 F

U = 0.75(1.4 D + 1.4 T +1.7 L)

U = 1.4 (D + L)


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Resistance Factors, ACI Sec 9.3.2

Strength Reduction Factors

[1] Flexure w/ or w/o axial tension = 0.90

[2] Axial Tension = 0.90

[3] Axial Compression w or w/o flexure

(a) Member w/ spiral reinforcement = 0.75(b) Other reinforcement members = 0.70

*(may increase for very small axial loads)

[4] Shear and Torsion = 0.85

[5] Bearing on Concrete = 0.70

ACI Sec 9.3.4 factors for regions of high seismic risk


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Background Information for Designing

Beam Sections

1.

2.

Location of Reinforcement

locate reinforcement where cracking occurs

(tension region) Tensile stresses may be due to :a) Flexure

b) Axial Loads

c ) Shrinkage effects

Construction

formwork is expensive -try to reuse at several

floors


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Beam Sections

4. Concrete Cover

Cover = Dimension between the surface of the slab or

beam and the reinforcement

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.


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Beam Sections

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

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

No. 14 and No. 18 bars - 1.5 in

No. 11 bar and smaller - 0.75 in

- Beams, Columns - 1.5 in


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Beam Sections

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

.in18

t3


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Minimum Cover Dimension

Interior beam.


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Reinforcement bar arrangement for two layers.


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ACI 3.3.3

Nominal maximumaggregate size.

3/4 clear space.,

1/3 slab depth,

1/5 narrowestdim.


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Design Procedure for section dimensions

are unknown (singly Reinforced Beams)

1) For design moment

Substitute:

-

-

-

b0.852

bddbd

b0.852

AdA2

adTMM

c

y

y

c

ys

ysnu

f

ff

f

ff

bd

Aand s

c

y

f

f


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Let

59.01bd

d59.0dbdM

d59.0dbdMM

2c

cu

ynu

-

-

-

f

f

f


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Let

R

M

bd

R

59.01bd

M

u

2

c2

u

-

f


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Assume that the material properties, loads, and span length are all known.

Estimate the dimensions of self-weight using the following rules of

thumb:

a. The depth, h, may be taken as approximate 8 to 10 % of the

span (1in deep per foot of span) and estimate the width, b,

as about one-half of h.

b. The weight of a rectangular beam will be about 15 % of thesuperimposed loads (dead, live, etc.). Assume b is about

one-half of h.

Immediate values of h and b from these two procedures should be selected.

Calculate self-weight and Mu.


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Design Procedure for section dimensions are

unknown (singly Reinforced Beams)

1 Select a reasonable value for based on

experience or start with a value of about 45% to

55 % ofbal.2 Calculate the reinforcement index,

3 Calculate the coefficient

c

y

f

f

59.01R c - f


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4 Calculate the required value of

5 Select b as a function of d. b ~ (0.45d to 0.65d)

6 Solve for d. Typically round d to nearest 0.5 inch

value to get a whole inch value for h, which is

approximately d = 2.5 in.

R

M

bd

u

2


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7 Solve for the width, b, using selected d value.

Round b to nearest whole inch value.

8 Re-calculate the beam self-weight and Mu basedon the selected b and h dimensions. Go back to

step 1 only if the new self weight results in

significant change in Mu.

9 Calculate required As = bd. Use the selectedvalue of d from Step 6. And the calculated (not

rounded) value of b from step 7 to avoid errors

from rounding.


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Select steel reinforcing bars to provide

As (As required from step 9). Confirm that

the bars will fit within the cross-section. It may benecessary to change bar sizes to fit the steel in one

layer. If you need to use two layers of steel, the

value of h should be adjusted accordingly.

Calculate the actual Mn for the section dimensions

and reinforcement selected. Check strength,

(keep over-design within 10%)

un

MM

10

11


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are known (singly Reinforced Beams)


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known (singly Reinforced Beams)

1 Calculate controlling value for the design moment,

Mu.2 Calculate d, since h is known.

d h - 2.5in. for one layer of reinforcement.

d h - 3.5in. for two layers of reinforcement.


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3 Solve for required area of tension reinforcement,

As , based on the following equation.

-

2

adAMM ysnu f


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Rewrite the equation:

-

2

ad

M

dreq'A

y

u

s

f

Assume (d-a/2) 0.9d to 0.95d and solve for As(reqd)

Note = 0.9 for flexure without axial load

(ACI 318-95, Sec. 9.3)


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4 Select reinforcing bars so As(provided) As(reqd)

Confirm bars will fit within the cross-section. It

may be necessary to change bar sizes to fit the steel

in one layer or even to go to two layers of steel.

5 Calculate the actual Mn for the section dimensions

and reinforcement selected. Verify .Check strength (keep over-design

with 10%)

un MM ys


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