Deflection of a Reinforced Concrete Beam

Deflection of a Reinforced Concrete Beam CE 433, Fall 2006 Overview The deflection of reinforced concrete beams is complicated by several factors. 1...
Author: Bruce Hopkins
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Deflection of a Reinforced Concrete Beam

CE 433, Fall 2006

Overview The deflection of reinforced concrete beams is complicated by several factors. 1) The connections of a cast-in-place reinforced concrete frame usually transfer moment. The midspan deflection of a beam in such a frame is affected by the stiffness of the members framing into the beam ends. The mid-span deflection can be calculated by solving two simultaneous equations for the rotations at each beam end. 2) The flexural stiffness of a reinforced concrete beam changes from the uncracked state to the cracked state. The ACI equation for the effective moment of inertia (Ie) will be used to calculate a moment of inertia somewhere between the uncracked moment of inertia (Ig) and the cracked moment of inertia (Icr) depending on the applied moment (Ma). 3) Concrete continues to deform plastically under long-term loads. The additional deflection due to this deformation, called creep, can be calculated using another ACI equation. The equations necessary to calculate the deflection of a reinforced concrete beam will be developed for each of the effects in the following three sections. 1) Accounting for Joint Rotation An example beam from a multi-story reinforced concrete structure is shown in Figure 1 below. The columns are assumed to be fixed at their far ends, as is the beam to the right of the loaded span. The joint rotations (θleft and θright), and the midspan deflection of the beam (δ) due to the uniform load can be calculated if the flexural stiffness of the connecting members is known (

EI L

).

The connecting members will be assumed to be uncracked (a slightly unconservative assumption) but the effective moment of inertia of the beam will be used. Since the effective moment of inertia (formula shown in the next section) is a function of the applied moment, and the moments in the beam are a function of the joint rotation, an iterative solution is required.

w

θleft δ

θright

Figure 1. Structural model of beam deflection affected by end rotations.

Deflection of a Reinforced Concrete Beam

CE 433, Fall 2006

Equations relating the beam end moments (m) to the joint rotations (θ) are developed by superimposing the following two cases (illustrated in Figure 2): (1) load applied and no joint rotation (2) no load and joint rotation The member end moments for Case (1) are called "fixed-end moments" (FEM) and can be calculated for simple cases using formulas in mechanics of materials and/or structural analysis books. The fixed-end moment due to a uniformly distributed load is: FEM =

w L2 . 12

Equations for the member end moments for Case (2) are developed using available formulas relating end moment to joint rotation at one end only: mA =

4E I θA , L

mB =

2E I θA L

θA

mB

mA

Therefore the end moments for each member framing into the beam are

4E I θ joint L

and the end moments

for the beam are:

mA =

4E I 2E I θA + θB L L

mB =

2E I 4E I θA + θB L L

mA θA

note: θA will be clockwise (negative) and θB will be counter-clockwise (positive).

The total moment at the joint adjacent to the left end of the beam (Joint 2 in Figure 2) is: Σ M jo int 2 = − m21 − m23 − m25 = 0

where m21 =

4 E Ic θ2 Lc

m23 =

4 E Ic θ2 Lc

(

m25 = − FEM 25 + mθ25

)

mB

θB

Deflection of a Reinforced Concrete Beam

CE 433, Fall 2006

and FEM 25 = +

mθ25 =

w L2 12

4 E Ib 2 E Ib θ2 + θ5 Lb Lb

Therefore

Σ M joint 2 = −

4 E Ic 4 E Ic 2 E Ib w L2 4 E I b θ2 − θ2 − − θ2 − θ5 = 0 Lc Lc Lb Lb 12

Similarly

4 E I b _ g w L2 4 E I b 4 E Ic 4 E Ic 2 E Ib θ5 − θ5 − + − θ5 − θ2 = 0 12 Lc Lc Lb Lb Lb is the gross (uncracked) moment of inertia of the beam adjacent to the loaded span.

Σ M joint 5 = − where Ib_g

The two equilibrium equations can be rearranged in matrix form as ⎛ 8 E Ic 4 E Ib ⎜⎜ + Lb ⎝ Lc

⎞ ⎟⎟ θ 2 + ⎠

2 E Ib Lb

2 E Ib Lb ⎛8E I

c θ 2 + ⎜⎜ + L c ⎝

4 E Ib _ g Lb

θ5 +

4 E Ib Lb

⎞ ⎟ θ5 ⎟ ⎠

w L2 = − 12 =

w L2 12

The equations above can be solved for the joint rotations (θ2 and θ5). The midspan deflection can then be calculated from:

δ = δ FEM + δ θ δ FEM =

w L4b 384 E I b

δ θ = −θ left

Lb L + θ right b , 8 8

assuming θ left  is clockwise  (negative)                                         δ =

(

)

w L4b L + − θ left + θ right b 384 E I b 8

Deflection of a Reinforced Concrete Beam

CE 433, Fall 2006

4/9

no joint rotation 3

6

Lc w 5

7

Lc

Lb

m21

+

=

δθ

δFEM

4

1

m23

θ5

θ2

w

2

m25

Lb

m52 m54

actual moments

m56 m57

mθ23

w

+

= FEM25

FEM52

fixed-end moments

mθ25

mθ21

mθ52 m54

mθ56 mθ57

moments due to joint rotation

Figure 2. End moments due to load and no joint rotation, and due to no load with joint rotation.

Deflection of a Reinforced Concrete Beam

CE 433, Fall 2006

5/9

2) Effective Moment of Inertia, Ie

The ACI equation for effective moment of inertia (Ie) accounts for the fact that some of the reinforced concrete beam is cracked, and some of it is uncracked (as shown in Figure 3).

cracked uncracked

cracked

uncracked

cracked

Figure 3. Cracked and uncracked regions of typical reinforced concrete beam.

ACI Equation 9-8 is: 3 ⎡ ⎛ M ⎞3 ⎤ ⎛M ⎞ I e = ⎜⎜ cr ⎟⎟ I g + ⎢1 − ⎜⎜ cr ⎟⎟ ⎥ I cr ⎢ ⎝ Ma ⎠ ⎥ ⎝ Ma ⎠ ⎣ ⎦

where Mcr is the cracking moment =

fr I g yt

and

fr is the modulus of rupture = 7.5 psi f c' , Ig is the gross (uncracked) moment of inertia, and yt is the distance from the neutral axis to the tension face Ma is the moment due to the applied load (MD or MD+L) for example Icr is the moment of inertia of the cracked section. The uncracked response and the cracked response under service loads were studied at the beginning of the semester (see Flexure in RC Beams at 3 Stages). Calculating the moment of inertia is a two-step process: first the neutral axis must be located, then the moment of inertia can be calculated. Two methods of calculating the gross moment of inertia (Ig) are illustrated in Figures 4 and 5: a spreadsheet method and the traditional mechanics of materials method. The spreadsheet method is unchanged to calculate Icr. For the mechanics of materials method, the steel reinforcement is first transformed into and equivalent area of concrete (transformed section method) as shown in Figure 6. The text author suggests using combining the effective moments of inertia at three sections of a continuous beam: left end, mid-span, and right end using the following equation (Eqn. 9-11b on pg 406): I e = 0.70 I e _ mid + 0.15 ( I e _ left _ end + I e _ right _ end )

Deflection of a Reinforced Concrete Beam

CE 537, Spring 2009

y

compression

yt

tension

yb

6/9

C K

NA

T Strain, ε

Stress, σ

ε ( y) = −K y σ ( y) = Ecε ( y) = −Ec K y





C = σ ( y ) dA = − E c K y dA = − E c K



yt

y dA,

0

similarly : T = − E c K



0

− yb

y dA

To find the location of the neutral axis, use the following equilibrium equation: Σ FH = 0, C + T = 0 Spreadsheet Method: 1) select a yt 2) calculate C, T and Σ FH 3) repeat steps 1) and 2) until Σ FH = 0

Beam-Theory Method: From Σ FH = 0, ∴



yt

− yb

− Ec K



yt

0

y dA + − E c K



0

− yb

y dA = 0,

or − E c K



yt

− yb

y dA = 0,

y dA = 0, This is true when the origin of the coordinate system coincides with the neutral axis

In general,

∫ y dA = y A,

A

where y = the distance from the reference axis to the centroid of A



So, for y dA = y A to equal 0, y must equal 0.

y

This means that we need to locate the neutral axis at the centroid. We can calculate the centroid for simple shapes using tables. Also, we can calculate y of a shape made up of "n" simple shapes using the formula: yA=



n y A i =1 i i

Figure 4. Two methods for calculating the location of the neutral axis of an uncracked section.

y

Deflection of a Reinforced Concrete Beam

CE 537, Spring 2009

Spreadsheet Method:

Beam-Theory Method:

1) Calculate Mint due to C and T 2) Calculate the moment of inertia (I) from a) the moment vs. curvature relation:

M =

K=

∫∫σ ( y ) y dA =∫∫ ε ( y ) E y dA = ∫∫ K y ) E y dA = KE ∫∫ y where I = ∫∫ y dA

7/9

2

dA =KE I

2

M EI

b) the strain vs. curvature relation: K=

ε M = EI y

ε y

or

I=

M int y Eε

We can calculate the moment of inertia (I) for simple shapes using formulas from tables. For complex shapes, I can be calculating by evaluating the double integral. Also, we can calculate I of a shape made up of "n" simple shapes using the formula: I=

∑ (I n

i =1

i

+ Ai d i2

)

Figure 5. Two methods for calculating the moment of inertia of an uncracked section.

Deflection of a Reinforced Concrete Beam

CE 537, Spring 2009

compression

y

yt

8/9

C K

NA cracked

yb T Strain, ε

Stress, σ

Same as for the uncracked section except: T = As f s

f s = ε s Es < f y

Spreadsheet Method (same as for Ig): To find location of neutral axis: 1) select a yt 2) calculate C, T and Σ FH 3) repeat steps 1) and 2) until Σ FH = 0 To find moment of inertia: 3) Calculate Mint due to C and T 4) Calculate the moment of inertia (I) from I=

M int y Eε

Beam-Theory Method: Since the concrete below the neutral axis is cracked (and has no strength) it is ignored. The steel reinforcement is transformed into an equivalent area of concrete by multiplying As by the modular ratio n (as shown below).

compression

NA cracked tension

A=

Es As = n As Ec

The same procedures as for Ig are then used to find the location of the neutral axis and to calculate the moment of inertia, Icr. Important: Icr is the moment of inertia with the steel transformed into an equivalent area of concrete. When calculating the stress in the steel (e.g. using σ =

My I

), the

stress must be multiplied by the modular ratio, n. Figure 6. Two methods for calculating the location of the neutral axis and the moment of inertia of a cracked section

CE 433, Fall 2006

Deflection of a Reinforced Concrete Beam

9/9

3) Long-term Deflections due to Creep

Concrete continues to deform under long-term loads. This plastic deformation is called creep. The ACI equation to calculate creep deflection (ACI equation 9-11) is: λ=

ξ 1 + 50ρ '

where ξ is the time-dependant factor for long-term load and is equal to Duration of Load

ξ

3 months

1.0

6 months

1.2

12 months

1.4

5 years or more

2.0

Figure 9-16 in the text presents ξ on a semi-log plot, handy for interpolating or deriving an expression for ξ (like for a spreadsheet). ρ' is the compression reinforcement ratio = reinforcement.

As' bd

where As' is the area of compression

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