Reinforced Concrete Beams

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams

Reinforced Concrete Beams

] Mathematical modeling of reinforced concrete is essential to civil engineering

] Mathematical modeling of reinforced concrete is essential to civil engineering

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Concrete as a material Concrete in a structure

Stress distribution in a reinforced concrete beam

Reinforced Concrete Beams

Reinforced Concrete Beams

] Mathematical modeling of reinforced concrete is essential to civil engineering

] Mathematical modeling of reinforced concrete is essential to civil engineering

Blast failure of a reinforced concrete wall Geometric model a reinforced concrete bridge

Reinforced Concrete Beams ] Mathematical modeling of reinforced concrete is essential to civil engineering

Blast failure of a reinforced concrete wall

Reinforced Concrete Beams ] Mathematical model for failure in an unreinforced concrete beam

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams ] In the reinforced concrete beam project, there are three different failure mode we need to investigate

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Reinforced Concrete Beams ] First, lets consider the loading of the beam

P

P

P/2

Reinforced Concrete Beams ] The purpose of RC is the reinforcement of areas in concrete that are weak in tension

P/2

Reinforced Concrete Beams ] Let’s look at the internal forces acting on the beam and locate the tension zones

P

P

∑F

P/2

P/2

Reinforced Concrete Beams ] The shear between the applied load and the support is constant V = P/2

P/2

P/2

P/2

V

=

P 2

−V

⇒ V =

P 2

P/2

Reinforced Concrete Beams ] The shear between the applied load and the support is constant V = P/2

P/2

P/2

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams ] The shear between the applied load and the support is constant V = P/2

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Reinforced Concrete Beams ] Let’s look at the internal moment at section between the supports and applied load

] The shear force V = P/2 is constant between the applied load and the support

P

P

∑M = 2 x

M

X max = 8 inches

P/2

P/2

P/2

Reinforced Concrete Beams ] Let’s look at the internal moment at section between the supports and applied load

x

P/2

M = P/2 4P

(in lbs )

Reinforced Concrete Beams ] The top of the beam is in compression and the bottom of the beam is in tension

] The bending g moment is the internal reaction to forces which cause a beam to bend. Compression force on the upper Bending moment distributed on part of the concrete beam

MCthe cut surface

] Bending moment can also be referred to as torque

T

P

Tension force on the lower part of the concrete beam

2

Reinforced Concrete Beams ] To model the behavior of a reinforced concrete beam we will need to understand three distinct regions in the beam. ] Two are illustrated below; the third is called shear. shear

Reinforced Concrete Beams ] We need models to help us with compression, tension, and shear failures in concrete

P

Bending moment distributed on

cut surface MCthe Compression

P 2

T

Tension

Tension

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams ] We need models to help us with compression, tension, and shear failures in concrete

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Reinforced Concrete Beams ] We need models to help us with compression, tension, and shear failures in concrete

P

P

Compression

Shear

Reinforced Concrete Beams ] We need models to help us with compression, tension, and shear failures in concrete

Shear

Reinforced Concrete Beams ] Compression and tension failures in a reinforced concrete beam

P

Shear

Compression Tension

Shear

Reinforced Concrete Beams ] Compression and tension failures in a reinforced concrete beam

Reinforced Concrete Beams ] Shear failure in a reinforced concrete beam

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams

Whitney Rectangular Stress Distribution

] Shear failure in a reinforced concrete beam

] In the 1930s, Whitney proposed the use of a rectangular compressive stress distribution

Whitney Rectangular Stress Distribution

Whitney Rectangular Stress Distribution

] In the 1930s, Whitney proposed the use of a rectangular compressive stress distribution k3f’c

b

0.85f’c

k2 x

C

c

] Assume that the concrete contributes nothing to the tensile strength of the beam

0.5a

d

h

As

T

k3f’c

P

d

0.85f’c

k2 x

C

c

Shear

Compression Tension T

Compression Tension

Shear T

] The height of the stress box, a, is defined as a percentage of the depth to the neural axis

C

0.85f’c

c

0.5a

C

a

Shear T

C

Whitney Rectangular Stress Distribution

0.5a

a

0.5a

a

T

] Assume that the complex distribution of compressive stress in the concrete can be approximated by a rectangle b

Shear

0.85f’c

k2 x

C

T

Whitney Rectangular Stress Distribution

As

P

c

As

h

k3f’c

b

C

a

d

h

5/10

T

a = β1c

CIVL 1112

Introduction to Reinforced Concrete Beams

Whitney Rectangular Stress Distribution

Whitney Rectangular Stress Distribution

] The height of the stress box, a, is defined as a percentage of the depth to the neural axis

f 'c ≤ 4000 psi

⇒

] The values of the tension and compression forces are: 0.85f’c

β1 = 0.85

0.5a

C

a

f 'c ≥ 4000 psi ⎛ f 'c −4000 ⎞ ⎟ ≥ 0.65 ⎝ 1000 ⎠

T

Whitney Rectangular Stress Distribution

0.85f’c

d

a =

Asfy 0.85f 'c b

a =

Asfy 0.85f 'c b

0.5a

∑ M =T ⎛⎜⎝d

C

a

−

a⎞ ⎟ 2⎠

d

If a > d, then you have too much steel

T

=0

] The internal moment is the value of either the tension or compression force multiplied the distance between them

0.5a

C

∑F

Whitney Rectangular Stress Distribution

] If the tension force capacity of the steel is too high, than the value of a is large

a

C = 0.85f 'c ba

T = Asfy

d

β1 = 0.85 − 0.05⎜

0.85f’c

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M = Asfy ⎛⎜d − ⎝

T

a⎞ ⎟ 2⎠

Whitney Rectangular Stress Distribution

Whitney Rectangular Stress Distribution

] The internal moment is the value of either the tension or compression force multiplied the distance between them

] The internal moment is the value of either the tension or compression force multiplied the distance between them

0.85f’c

0.5a

C

a

a M = Asfy ⎛⎜d − ⎞⎟ 2⎠ ⎝

T

M = 4P

⎛

M = Asfy ⎜⎜d − 0.59 ⎝

⎝

Asfy ⎞ ⎟ f 'c b ⎟⎠

We know that the moment in our reinforced concrete beans is

M = 4P

Substitute the value for a

d

⎛

M = Asfy ⎜⎜d − 0.59

Asfy ⎞ ⎟ f 'c b ⎟⎠

Ptension =

Asfy 4

Asfy ⎞ ⎛ ⎜ d - 0.59 ⎟ f'c b ⎠ ⎝

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams ] There is a “balanced” condition where the stress in the steel reinforcement and the stress in the concrete are both at their yield points ] The amount of steel required to reach the balanced strain condition is defined in terms of the reinforcement ratio:

A ρ= s bd

P

Compression

] The limits of the reinforcement ratio are established as: Reinforcement ratio definition

ρ =

As bd

c > 0.600 d

0.375 <

] Lets consider shear failure in reinforced concrete

P

Vs −Vcr = V =

⎛ Af v yd ⎝ s

P

Af v yd s

2

Shear

⎞ + 2 f 'c bd ⎟ ⎠

Beam failure is controlled by

compression compr ss on

c < 0.600 Transition between tension d and compression control Beam failure is controlled by

tension

Reinforced Concrete Beams ] Lets consider shear failure in unreinforced concrete

Vcr = 2 f 'c bd P V =

Reinforced Concrete Beams

Pshear = 2 ⎜

] The limits of the reinforcement ratio are established as:

ρ as function of c/d

c f 'c ρ = 0.85 β1 d fy

Shear

Reinforced Concrete Beams

c < 0.375 d

Reinforced Concrete Beams

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P

P =Shear 4 f 'c bd

Shear

2

Reinforced Concrete Beams ] Lets consider compression failure in over reinforced concrete ] First, let define an equation that given the stress in the tensile steel when concrete reaches its ultimate strain

⎛d −c ⎞ fsteel = 87, 000 psi ⎜ ⎟ ⎝ c ⎠ ] If fsteel < fy then or

c > 0.600 d

a⎞ ⎛d − c ⎞⎛ ⎟ ⎜ d − 2 ⎟ 87, 000 psi ⎠ ⎝ c ⎠⎝

Mcompression = As ⎜

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beams ] Lets consider compression failure in over reinforced concrete ] First, let define an equation that given the stress in the tensile steel when concrete reaches its ultimate strain

P

M = 4P Pcompression

Reinforced Concrete Beam Analysis ] Let’s use the failure models to predict the ultimate strength-to-weight (SWR) of one of our reinforced concrete beams from lab ] Consider a beam with the following characteristics:

only if fs < fy

Shear

A ⎛d −c = s ⎜ 4 ⎝ c

Concrete strength f’c = 5,000 psi Steel strength fy = 60,000 psi

Shear

The tension reinforcement will be 2 #4 rebars The shear reinforcement will be #3 rebars bent in a U-shape spaced at 4 inches.

a⎞ ⎞⎛ ⎟ ⎜ d − 2 ⎟ 87, 000 psi ⎠ ⎠⎝

Use the minimum width to accommodate the reinforcement

Reinforced Concrete Beam Analysis ] Reinforcing bars are denoted by the bar number. The diameter and area of standard rebars are shown below. Bar #

Reinforced Concrete Beam Analysis ] Based on the choice of reinforcement we can compute an estimate of b and d #4 rebar diameter

Diameter (in) As (in 2 )

3

0.375

0.11

4

0 500 0.500

0 20 0.20

5

0.625

0.31

6

0.750

0.44

7

0.875

0.60

8

1.000

0.79

9

1.128

1.00

10

1.270

1.27

11

1.410

1.56

b

d =6− #4

4.625 in

Space between bars

Reinforced Concrete Beam Analysis ] We now have values for b, d, and As

⎛

M = Asfy ⎜⎜d − 0.59

b

0.5 0 5 −0.75 −0.375 2 #3 rebar diameter

d =

4.0 in

#4

Minimum cover

d

6”

#3 rebar diameter

b ≥ 2(0.5 in ) + 2(0.75 in ) + 2(0.375 in )

d

6”

] If we allow a minimum cover under the rebars were can estimate d

b

Minimum cover

+ 0.75 in =

Reinforced Concrete Beam Analysis Half of #4 bar diameter

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⎝

d

6” #4

Asfy ⎞ ⎟ f 'c b ⎟⎠

] The As for two #4 rebars is:

As = 2(0.20 in 2 ) = 0.40 in 2

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beam Analysis ] Compute the moment capacity

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Reinforced Concrete Beam Analysis ] Let’s check the shear model

⎛ Af v yd

Asfy ⎞ ⎛ ⎟ M = Asfy ⎜⎜d − 0.59 f 'c b ⎟⎠ ⎝

Pshear = 2 ⎜ ⎝

s

⎞ + 2 f 'c bd ⎟ ⎠

Area of two #3 rebars

⎛ 0.4in 2 (60ksi ) ⎞ = 0.4in 2 (60ksi ) ⎜ 4.625in − 0.59 ⎟ 5ksi (4in ) ⎠ ⎝

⎛ 2 ( 0.11in 2 ) ( 60, 000 psi ) 4.625 in ⎞ = 2⎜ + 2 5, 000 psi ( 4 in )( 4.625 in ) ⎟ ⎜ ⎟ 4 in ⎝ ⎠ Shear reinforcement spacing

= 94.0 k ⋅ in

⇒P =

M 4

= 23.5 kips

Reinforced Concrete Beam Analysis ] Let’s check the reinforcement ratio

ρ =

As bd

= 35,757 lbs = 35.76 kips Since Ptension < Pshear therefore Ptension controls

Reinforced Concrete Beam Analysis ] An β1 estimate is given as:

Reinforcement ratio definition

ρ = 0.85 β1

f 'c ≤ 4000 psi ρ as function of c/d

c f 'c d fy

f 'c ≥ 4000 psi ⎛ f 'c −4000 ⎞ ⎟ ≥ 0.65 ⎝ 1000 ⎠

β1 = 0.85 − 0.05⎜

⎛ 5, 000 − 4, 000 ⎞ ⎟ = 0.80 1, 000 ⎝ ⎠

β1 = 0.85 − 0.05 ⎜

To compute ρ, first we need to estimate β1

Reinforced Concrete Beam Analysis ] Check the reinforcement ratio for the maximum steel allowed for tension controlled behavior or c/d = 0.375

ρ = 0.85 β1

5 ksi c f 'c = 0.85(0.80)0.375 d fy 60 ksi

= 0.021 ρ=

0.4 in 2 As = = 0.022 bd 4in (4.625in )

] The amount of steel in this beam is just a bit over the allowable for tension controlled behavior.

β1 = 0.85

⇒

Reinforced Concrete Beam Analysis ] However, the maximum about of steel where compression is in control is c/d = 0.600

ρ = 0.85 β1

5 ksi c f 'c = 0.85(0.80)0.600 d fy 60 ksi

= 0.034 ρ=

0.4 in 2 As = = 0.022 bd 4in (4.625in )

] Therefore, the beam is in the lower part of the transition zone and for our purposes is OK.

CIVL 1112

Introduction to Reinforced Concrete Beams

Reinforced Concrete Beam Analysis ] An estimate of the weight of the beam can be made as: Size of concrete beam

W =

Unit weight of concrete

Reinforced Concrete Beam Analysis ] In summary, this reinforced concrete beam will fail in tension

(4in )(6in )(30in ) ⎛ 145lb ⎞ 1728in 3 /ft 3 ⎜⎝ ft 3 ⎟⎠

Additional weight of rebars

4”

⇒ P = 23.5 23 5 kips

Unit weight of steel

(0.4in 2 )(30in ) ⎛ 490lb − 145lb ⎞ + ⎜ ⎟ 1728in 3 /ft 3 ⎝ ft 3 ⎠

= 60.42lb + 2.39lb = 62.81lb

Reinforced Concrete Beam Analysis

Questions?

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4.625”

6”

SWR = As = 0.4 in2

23,500lb = 62.81lb

374.2