Lecture 9 - Flexure June 20, 2003 CVEN 444

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Lecture 9 - Flexure

June 20, 2003

CVEN 444

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Lecture Goals Lecture Goals

Load Envelopes

Resistance Factors and Loads

Design of Singly Reinforced Rectangular Beam

 Unknown section dimensions

 Known section dimensions

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Moment Moment

Envelopes Envelopes

The moment envelope

curve defines the extreme boundary values of

bending moment along the beam due to critical

placements of design live loading.

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Moment Moment

Envelopes Example Envelopes Example

Given following beam with a dead load of 1 k/ft and live load 2 k/ft obtain the shear and bending moment envelopes

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Moment Moment

Use a series of shear and bending moment diagrams W_u = 1.2w_D + 1.6w_L

0 1 2 3 4 5

0 5 10 15 20 25 30 35 40

(ft)

kips

-80 -60 -40 -20 0 20 40 60 80

0 5 10 15 20 25 30 35 40

kips

-250 -200 -150 -100 -50 0 50 100 150

0 5 10 15 20 25 30 35 40

k-ft

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Moment Moment

Shear Diagram Moment Diagram

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

kips

-80 -60 -40 -20 0 20 40

0 5 10 15 20 25 30 35 40

ft

k-ft

(Dead Load Only)

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Moment Moment

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0 5 10 15 20 25 30 35 40

ft

k/ft

-60 -50 -40 -30 -20 -10 0 10 20 30 40 50

0 5 10 15 20 25 30 35 40

kips

-200 -150 -100 -50 0 50 100 150 200

0 5 10 15 20 25 30 35 40

k-ft

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Moment Moment

The shear envelope

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Moment Moment

The 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

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Flexural Design of Reinforced Flexural Design of Reinforced

Concrete Beams and Slab Sections Concrete Beams and Slab Sections

Analysis Versus Design:

Analysis: Given a cross-section, f_c , reinforcement sizes, location, f_y compute

resistance or capacity

Design: Given factored load effect (such as M_u) select suitable section(dimensions, f_c, f_y, reinforcement, etc.)

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Flexural Design of Reinforced Flexural Design of Reinforced

ACI Code Requirements for Strength Design

Basic Equation: factored resistance factored load effect



Ex.

u

n M

M 



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ACI Code Requirements for Strength ACI Code Requirements for Strength

Design Design

u

n

M

M 



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

M_n = 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.

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Flexural Design of Reinforced Flexural Design of Reinforced

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

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Flexural Design of Reinforced Flexural Design of Reinforced

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.

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Factored Load Combinations Factored Load Combinations

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 (L_r or S or R) U = 1.2D + 1.6 (L_r or S or R) + (L or 0.8W)

U = 1.2D + 1.6 W + 1.0L + 0.5(L_r or S or R) U = 0.9 D + 1.6W +1.6H

U = 0.9 D + 1.0E +1.6H

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Resistance Factors,

Resistance Factors,   ACI Sec ACI Sec 9.3.2 Strength Reduction Factors 9.3.2 Strength Reduction Factors

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

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

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Resistance Factors,

[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)

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Resistance Factors,

[4] Shear and Torsion = 0.75

[5] Bearing on Concrete = 0.65

ACI Sec 9.3.4  factors for regions of high seismic risk

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Background Information for Background Information for

Designing Beam Sections Designing Beam Sections

1. Location of Reinforcement locate reinforcement where cracking occurs

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

b ) Axial Loads

c ) Shrinkage effects

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Background Information for Background Information for

2. Construction

formwork is expensive - try to reuse at several floors

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3. Beam Depths

• ACI 318 - Table 9.5(a) min. h based on l (span) (slab & beams)

• Rule of thumb: h_b (in) l (ft)

• Design for max. moment over a support to set depth of a continuous beam.



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Background Information for Background Information for

4. Concrete Cover

Cover = Dimension between the surface of the slab or beam and the reinforcement

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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.

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Background Information for Background Information for

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

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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 in No. 11 bar and smaller - 0.75 in - Beams, Columns - 1.5 in

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Background Information for Background Information for

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

t

3

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Minimum Cover Dimension Minimum Cover Dimension

Interior beam.

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Minimum Cover Dimension Minimum Cover Dimension

Reinforcement bar arrangement for two layers.

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Minimum Cover Dimension Minimum Cover Dimension

ACI 3.3.3

Nominal maximum aggregate size.

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

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Example - Singly Reinforced Example - Singly Reinforced Beam Beam

Design a singly reinforced beam, which has a moment capacity, M

_u

= 225 k-ft, f

_c

= 3 ksi, f

_y

= 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.

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From the calculation of M_n

n

c c

2

c 1

2

0.85 0.85 1 1

2 2

0.85 1 1 where, 2

0.85 1 1 M C d a

a a a

f ba d f bd d

d d

a c

f bd k k k

d d

f k k bd



 

   

 

 

     

            

   

  

       

   

 

 

   

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Select c/d =0.275 so that  =0.9. Compute k’ and determine R_u

 

   

1

u c

0.85 0.275 0.23375

0.85 1

2

0.23375 0.85 3 ksi 0.23375 1

2 0.5264 ksi

k c

d

R f k k

 ^{ }

     



 

   

 

 

   



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Calculate the bd ²

U

2 N

u u

3

12 in 225 k-ft

ft 0.9

5699 in M

bd M

R R



 

 

 

 

   

 

   

 

 

 

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Calculate d, if b = 12 in.

2 5699 in3 2

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  

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Example - Singly Reinforced Example - Singly Reinforced Beam Beam

Calculate A_s for the beam

       

c 1

s

y

2

0.85

0.85 3 ksi 12 in. 0.85 6.1875 in.

40 ksi 4.02 in

f b c

A f

 



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Chose one layer of 4 #9 bars

Compute 



²



²

s 4 1.0 in 4.00 in

A  

   

2

s 4.00 in

12.0 in 22.5 in 0.014815

A



 bd 



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Calculate _min for the beam

y

min min

c y

200 200

0.005 40000

0.005

3 3 3000

0.00411 40000

f f f

 

  



   

  



0.014815 0.005 The beam is OK for

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Check whether or not the bars will fit into the beam.

The diameter of the #9 = 1.128 in.

     

b stirrup

4 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.

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Example - Singly Reinforced Example - Singly Reinforced Beam Beam

Check the height of the beam.

Use h = 25 in.

   

b cover stirrup

2

1.128 in.

22.5 in. 1.5 in. 0.375 in.

2 24.94 in

h d  d    d 

   



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Find a

Find c

   

2 s y

c

4.0 in 40 ksi 0.85 0.85 3 ksi 12.0 in.

5.23 in.

a A f

 f b 



1

5.23 in.

0.85 6.15 in.

c a

  



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Check 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

  ^ ^ ^  ^ ^ ^

 

 

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Compute the M_n for the beam

Calculate M_u

   

N s y

2

5.23 in.

4.0 in 40 ksi 22.5 in.

2 3186.6 k-in

M  A f d  a 

 

   



 

U N

0.9 3186.6 k-in 2863.4 k-in M  M

 

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Example - Singly Reinforced Example - Singly Reinforced Beam Beam

Check the beam M_u = 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.}^{22.5 in.}



^0.2733

c

d   Use a smaller c/d

ratio