RC 09-1354 RC Structural Elements

8/14/2019 RC 09-1354 RC Structural Elements

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C09-1354 RC Structural Elements 1

RC09-1354: RC Structural Elements

Lecture 5 - Flexure


Lecture Goals

Rectangular Beams

Loading and Resistance

Balanced Beams

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Flexural StressThe compressive zone is modeled with a equivalent

.

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Flexural StressThe equivalent rectangular concrete stress distribution

average stress distribution covers.

psi4000for85.0 c1 f

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65.0

1000

4000*05.085.0 c1

f


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Flexural StressRequirements for analysis of reinforced concrete beams

[1] Stress-Strain Compatibility Stress at a point in

member must correspond to strain at a point.

[2] Equilibrium Internal

forces balances with

external forces

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Flexural Stress

Example of rectangular reinforced concrete beam.

.

n

css

x

M2

T0

85.0

CT0

adM

abffA

F

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Flexural StressExample of rectangular reinforced concrete beam.

s y

s

y

y

ac

E

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ycs

1

c

cd

RC09-1354: RC Structural ElementsFlexural Stress Rectangular Example

Example of rectangular reinforced concrete beam.

Given a rectangular beam

fc = 4000 psi

fy = 60 ksi (4 #7 bars)

= = =

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. . . .

Find the neutral axis.

Find the moment capacity of the beam.


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RC09-1354: RC Structural ElementsFlexural Stress

Rectangular Example

Determine the area of steel, #7 bar has 0.6 in2.

The value is = 0.85 because the concrete

2 2s 4 0.6 in 2.4 inA

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has a fc =4000 psi.

1 c 0.85 for 4000 psif

RC09-1354: RC Structural ElementsFlexural Stress Rectangular ExampleFrom equilibrium (assume the steel has yielded)

c s

2

s

c

0.85

60 ksi 2.4 in3.53 in.

0.85 0.85 4 ksi 12 in

y

y

f ba f A

f Aa

f b

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1

3.53 in.4.152 in.

0.85

ac

The neutral axis is


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Rectangular ExampleCheck to see whether or not the steel has yielded.

y

y

s

s0.00207

E 29000 ksi

Check the strain in the steel

s 0.003d c

Steel ielded!

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15.5 in. 4.152 in.

0.003 0.0082 0.0002074.152 in.

RC09-1354: RC Structural ElementsFlexural Stress Rectangular ExampleCompute moment capacity of the beam.

n s y

2

2

3.53 in.2.4 in 60 ksi 15.5 in.

a

M A f d

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1979 k-in. 164.8 k-ft.


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Non-Rectangular ExampleFor a non-rectangular beam

rated at fc = 6 ksi and the steel is

rated at fs = 60 ksi. d = 12.5 in.

(a) Determine the area of the steel for a

balanced system for shown area of concrete.

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(b) Determine the moment capacity of thebeam. Mn

(c) Determine the NA.

RC09-1354: RC Structural ElementsFlexural Stress Non-Rectangular Example

For a non-rectangular beam

e area o e concre e sec on s

c 26 in. 3 in. 10 in. 2 in.38 inA

The force due to concrete forces.

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c c

2

0.85

0.85 6 ksi 38 in

193.8 kips.

C f A


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Non-Rectangular ExampleUsing equilibrium, the area of the steel can be found

c cs s c c s

s

0.850.85

193.8 ki s

T C

f Af A f A A

f

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s

. n60 ksi


Find the center of the area

o concrete area

i i

i

y Ay

A

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n. n. . n. n. n. n.

6 in. 3 in. 10 in. 2 in.

2.8158 in.


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Non-Rectangular Example

The moment ca acit of the beam is

n

193.8 kips 12.5 in. 2.8158 in.

M T d y

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- n. . - .


Com ute the value

c1

4000 psi0.85 0.05* 1000 psi

6000 psi 4000 psi0.85 0.05*

f

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ps

0.75


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Non-Rectangular ExampleFind the neutral axis

1

5.0 in.

ac

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. .

0.75


LoadingThe loading variations are taken into

to determine the ultimate load, U.

r

1.4

1.2 1.6 0.5 or or

U D F

U D F T L H L S R

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r1.2 1.6 0.5 1.0 or or

1.2 1.0 1.0 0.2

etc.

U D W L L S R

U D E L S


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LoadingThe most general equation for the ultimate load,

u

1.2 1.6U D L

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Resistance

The load factors will generate the ultimate load,

structural member.

u nM

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u ma e omen

Mn Nominal Moment

Strength Reduction Factor


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Resistance

The stren th reduction factor varies from member

to member depending whether it is in tension or

compression or the type of member. The code has

been setup to determine the reduction.

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Three possibilities in InelasticBehavior

- -

beam)

Tension Failure - (under-reinforced beam)

Balanced Failure - (balanced reinforcement)

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Inelastic BehaviorCompression Failure

T e concrete w crus

before the steel yields.

This is a sudden

failure.

The beam is known as

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anover-reinforcedbeam.


Inelastic Behavior

Tension Failure

The reinforcement

yields before the

concrete crushes. The

concrete crushes is a

secondary

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compress on a ure.


an under-reinforced

beam.


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Inelastic BehaviorBalanced Failure

The concrete crushes

and the steel yields

simultaneously.


anbalanced-

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reinforced beam.


Inelastic BehaviorWhich type of failure is the most desirable?

-

beam is the most

desirable.

fs = fy

s >>

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You want ductility

system deflects and

still carries load.


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Balanced Reinforcement Ratio, balbal = unique value to get simultaneous c = 0.003

s y

Use similar triangles:

b

y

b cdc

003.0

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RC09-1354: RC Structural ElementsBalanced ReinforcementRatio, bal

The equation can be rewritten to find cb

b y b

b y

bb

y y

0.003d 0.003c c

c 0.003 0.003d

c0.003d 0.003c

d0.003 0.003

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b s

sy y

c E0.003 87000

d E0.003 87000 f


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Nominal Moment EquationThe equation can be rewritten in the form:

c s y

y s

c

. a

Aa

0.85 b

a

f

f

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n s y

2


Nominal Moment Equation

The equation can be rewritten in the form:

y s2sn y

c

A bM d d

bd d 1.7 bdf

f

Use the ratio r = b/d and

2df

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n y

c

r1.7f


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Nominal Moment EquationUse fy/fc and

Use the ratio r = b/d and R

3 3n c cM r d 1 r 1 0.59 d 1.7

f f

2M Rbd

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cR 1 0.59f

RC09-1354: RC Structural ElementsStrain Limits Method forAnalysis

The strength

, ,

will come into the

calculation of thestrength of the beam.

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RC09-1354: RC Structural ElementsLimitations on Reinforcement

Ratio,The selection of the steel will be determined by the

Lower Limit on ACI 10.5.1

ACI Eqn. (10-3)dbf

dbf

fA w

y

w

y

c

s(min) *200

*3

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fc & fy are in psi

RC09-1354: RC Structural ElementsLimitations on ReinforcementRatio,

Lower Limit on ACI 10.5.1

Lower limit used to avoid Piano Wire beams.

Ver small A M < M

yy

c

min

200

3

ff

f

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s is huge (large deflections)

when beam cracks ( Mn > Mcr) beam fails right

away because Mn < Mcr


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Additional Requirements for

Lower Limit onTemperature and Shrinkage reinforcement in structural

s a s an oot ngs . p ace perpen cu ar to

direction of flexural reinforcement.

GR 40 or GR 50 Bars: As (T&S) = 0.0020 Ag

GR 60 or Welded Wire Fabric (WWF):

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As

(T&S) = 0.0018 Ag

Ag - Gross area of the concrete


Example

Given:

fc = 3 ksi & fy = 40 ksi

and As

= 4 in2

Determine:

(1) Determine if the beam will

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(2)

sat s y co e.

If fc = 6 ksi?


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ExampleGiven:fc = 3 ksi & fy = 40 ksi and As = 4 in

2

2

sA 4 in 0.0333bd 8 in. 15 in.

The minimum steel ratio is

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min =0.00411 0.00540000 40000

min 0.005 0.0333 > 0.005 OK!


ExampleGiven:

fc = 3 ksi & fy = 40 ksi and As =4 in2

The neutral axis is

2

y s

c

40 ksi 4 inAa 8.743 in.

0.85 b 0.85 3 ksi 8 in

f

f

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1

. . . .9.23 in. 0.615

0.85 d 15 in.c


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ExampleThe strain in the steel is

t15 in. 7.843 in.

0.003 0.0037.843 in.

0.0027

d c

c

There for the beam is in the com ression zone and

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would be 0.65, however c/d ratio is greater than0.375 so the beam will need to be redesigned.


Examplec/d=0.615

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ExampleGiven:fc = 6 ksi & fy = 40 ksi and As =4 in

2

2

sA 4 in 0.0333bd 8 in. 15 in.

The minimum steel ratio is

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min =0.00581 0.00540000 40000

min 0.00581 0.0333 > 0.00581 OK!


ExampleGiven:

fc = 6 ksi & fy = 40 ksi and As =4 in2

The neutral axis is at

2

y s

c

40 ksi 4 inAa 3.922 in.

0.85 b 0.85 6 ksi 8 in

f

f

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1

. . . .5.22 in. 0.349

0.75 d 15 in.c


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ExampleThe strain in the steel will be

t15 in. 5.22 in.

0.003 0.0035.22 in.

0.0056

d c

c

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There for the beam is in the tension zone and willbe 0.9.


Examplec/d=0.349

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Documents

RC 09-1354 RC Structural Elements