5. Flexural Analysis and Design of Beams 5.1. Reading Assignment - Chapter 5

5. Flexural Analysis and Design of Beams

  5.1. Reading Assignment

  Chapter 3 of text

  5.2. Introduction

  It is of interest in structural practice to calculate those stresses and deformations which occur in a structure in service under design load. For reinforced concrete beams this can be done by the methods just presented, which assume elastic behavior of both materials. It is equally, if not more, important that the structural engineer be able to predict with satisfactory accuracy the ultimate strength of a structural member. By making this strength larger by an appropriate amount than the largest loads which can be expected during the lifetime of the structure, an adequate margin of safety is assured. Until recent times, methods based on elastic analysis like those just presented have been used for this purpose. It is clear, however, that at or near the ultimate load, stresses are no longer

  proportional to strains .

  At high loads, close to ultimate, the distribution of stresses and strains is that of figure 2 rather that the elastic distribution of stresses and strains given in figure 1 below. More realistic methods of analysis, based on actual inelastic rather than an assumed elastic behavior of the materials and results many experimental research, have been developed to predict the ultimate strength.

  Á f f

  c c c

  Á

  c

  Á

  s

  Á

  s

  f f

  s s

  2

  1 As progressively increasing bending moments are applied to the beam, the strains will increase as exemplified by ε , ε , and ε as shown below. Corresponding to these strains and their linear varia-

  1

  2

  3 tion from the neutral axis, the stress distribution will look as shown.

  Stress f

  3

  f

  2

  f

  1

  Á Á Á

  1 Á

  

2

  3

  f Á

  3

  3

  Á f

  2

  2

  Á f

  1

  1 Strain

  Stress

  Stress f

  3

  f

  2

  f

  1

  Á Á Á

  1 Á

  2

  

3

  f f Á

  3

  3

  3

  Á f

  2

  

2

  Á f

  1

  1 Strain

  Stress CI

  VL 4135

  87 Flexur Figure 5.1. Cracks, Strains, and Stresses in test beam (From Nawy’s Book).

  5.1

  e

  Large amount of reinforcement is used. Concrete fails by crushing when strains become so large (0.003 to 0.004). Failure is sudden, an almost explosive nature and occur with no warning ( Brittle Failure).

  Strains Stresses Forces c

  Moderate amount of reinforcement is used. Steel yields suddenly and stretches a large amount, tension cracks become visible and widen and propagate upward (Ductile Fail- ure)

  There are two possible ways that a reinforced beam can fail:

  5.4. Two Different Types of Failure

Figure 5.2. Strain, Stress, and Force Diagrams

  ′bc

  c

  = αf

  c

  βc C

  u

  5.3. Flexure Strength

  Á

  c

  f

  s

  f

  s

  Á

  Actual inspection of many concrete stress-strain curves which have been published, show that the geometrical shape of the stress distribution is quite varied and depends on a number of factors such as cylinder strength, the rate, and duration of loading. Below is a typical stress distribution at the ultimate load.

  As it was mentioned earlier it is important that the structural engineer be able to predict with satisfactory accuracy the ultimate strength of a structural member. It is important to know that at or near the ultimate load, stresses are no longer proportional to strains.

  • Beam will fail by tension of steel
  • Compression failure of concrete
In a rectangular beam the area that is in compression is bc, and the total compression force on this area can be expressed as C = f bc , where f is the average compression stress on the area bc. Evi-

  av av

  dently, the average compression stress that can be developed before failure occurs becomes larger the higher the cylinder strength f ’ of the particular concrete. Let

  c

  f

  av

  (5.7) α = f

  c

  ′ then (5.8)

  C bc

  c av c

  = f = αfbc compression force is applied at βc distance from top fiber, and c is the distance of the N.A. from top fiber. Based on research we have: f

  c

  ′ − 4, 000 and

  0.56 α = 0.72 − × 0.04 < α < 0.72

  1000 f

  c

  ′ − 4, 000 and 0.324

  β = 0.425 − × 0.025 < β < 0.425 1000

  FORCES

  From equilibrium we have C = T or

  c

  (5.9) αf f

  c s s

  ′bc = A (5.10)

  M f (d

  s s

  = TZ = Aβc) or (5.11)

  M Z

  c c

  = C = αfbc (d βc)

  5.5. Tension Failure

  (5.12) f steel yielding

  s y

  = f From Eq. (5.9) we have

  A f f d f

  s y y y

  A

  s

  d d (5.13) c Ã

  = α d = bd

  αbf αf f

  c c c

  ′ × ′ = ′ Substitute c from Eq. (5.13) in Eq. (5.10) f

  β y (5.14)

  M f d d

  n s y

  = AÃ α f

   c

  ′ with the specific, experimentally obtained values for α and β we always have

  β (5.15) f

  c

  ′ = 4, 000 psi or any other strength α = 0.59 for

  Therefore, Eq. (5.14) simplifies as f y

  (5.16) M f d d

  n s y

  = A − 0.59Ã f

   c

  ′ or f

  y

  2

  (5.17) M f

  1

  n y

  = Ãbd − 0.59Ã

   fc

  ′ where M = nominal moment capacity.

  n

  5.6. Compression Failure

  In this case, the criterion is that the compression strain in the concrete becomes = 0.003, as ε

  u

  previously discussed. The steel stress f , not having reached the yield point, is proportional to the

  s

  steel strain, ; i.e. according to Hooke’s law: ε

  s

  (5.18) Á (ACI 10.2.3), and f

  u s y

  = 0.003 < f (5.19) f Á

  Hooks law, since f < f

  s s s

  = E s y from similar triangles we have Á Á

  u s

  d (5.20)

  − c Á

  s u

  → = Á c = c d

  − c substitute Eq. (5.20) in Eq. (5.19) d

  (5.21) − c f Á Á

  s s s s u y

  = E = E < f c

  From Eq. (5.9) we have d

  (5.22) − c

  αf f E Á

  c s s s s u

  ′bc = A = A c Using Eq. (5.22) solve for c, and then find M the nominal moment capacity.

  n,

5.7. Balance Steel Ratio

  We like to have tension failure, because it gives us warning, versus compression failure which is sudden. Therefore, we want to keep the amount of steel reinforcement in such manner that the failure will be of tension type.

  Balanced steel ratio, represents the amount of reinforcement necessary to make a beam ρ

  b

  fail by crushing of concrete at the same load that causes the steel to yield. This means that neutral axis must be located at the load which the steel starts yielding and concrete starts reaching its compressive strain of ε = 0.003. (ACI 10.2.3)

  u

  Á

  u b

  (5.23) c d

  = Á

  y u

  • Á

  b b

  (5.24) T A f

  s y c

  = C → = α fbc Á

  b u

  (5.25) A f bd f d

  s y y c

  = Ã b = α fb Á

  u y

  • Á

  f

  c Á

  ′ u (5.26)

  Ã = α

  b

  Á f u y

  y + Á

5.8. Strain Limits Method for Analysis and Design (ACI 318).

  In “Strain Limits Method,” sometime referred to as the “Unified Method,” the nominal flex- ural strength of a concrete member is reached when the net compressive strain in the extreme com- pression fiber reaches the ACI code-assumed limit of 0.003 in/in (ACI 10.2.3). It also hypothesized that when the net tensile strain in the extreme tension steel, ε = 0.005 in/in, the behavior is fully duc-

  t

  tile. The concrete beam sections characterized as “Tension-Controlled,” with ample warning of fail- ure as denoted by excessive deflection and cracking.

  If the net tensile strain in the extreme tension fibers, ε , is small, such as in compression mem-

  t

  bers, being equal or less than a “Compression-Controlled” strain limit, a brittle mode of failure is expected with a sudden and explosive type of failure. Flexural members are usually tension-con- trolled. However, some sections such as those subjected to small axial loads, but large bending mo- ments, the net tensile strain, ε , in the extreme tensile fibers, will have an intermediate or transitional

  t

  value between the two strain limit states, namely, between the compression-controlled strain limit of f

  y

  60 ksi (5.27)

  Á 0.002

  t = =

  E 29, 000 ksi =

  s

  and the tension-controlled strain limit ε = 0.005 in/in. Figure 5.3 (ACI Figure R9.3.2 page 118)

  t

  shows these three zones as well as the variation in the strength reduction factors applicable to the total range of behavior.

  5.8.1. Variation of Φ as a Function of Strain

  Variation of the φ value for the range of strain between ε = 0.002 in/in and ε = 0.005 in/in can

  t t

  be linearly interpolated:

  0.65

  t

  ≤ φ = 0.65 + (Á − 0.002)(250∕3)) ≤ 0.90 Tied Column

  (5.28)

  0.75 ≤ (φ = 0.75 + (Á t − 0.002)(50)) ≤ 0.90

  Spiral Columnm

  5.8.2. Variation of Φ as a Function of Neutral Axis Depth Ratio c/d

  1

  0.65 ≤ φ = 0.65 + 0.25 − 5 ≤ 0.9

  Tied Column

  3

   c

t

  ∕d

   

  (5.29)

  1

  0.75 ≤ φ = 0.75 + 0.15 − 5 ≤ 0.9

  Spiral Columnm

  3

  

c

  ∕d t

    t

  φ = 0.75 + (Á − 0.002)(50)

0.90 SPIRAL

  0.75

  φ = 0.65 + (Á t − 0.002)(250∕3) φ

0.65 OTHER

  Tension Compression

  Transition Controlled

  Controlled Á

  t

  = 0.002 Á

  t = 0.005

  c c = 0.600

  = 0.375 d

  t d t

Figure 5.3. Example. Calculate Nominal Moment Capacity of a Beam for Fy = 60 ksi

5.9. Example. Calculate Nominal Moment Capacity of a Beam Determine the nominal moment M at which the beam given below will fail.

  n Given

  f

  c

  ′ = 4, 000 psi 23”

  25” f

  y

  = 60, 000 psi

  2 A =2.35 in s

  10”

  Solution

  A

  s

  2.35 Ã 0.0102

  = bd =

  10 × 23 = f

  y

  d c = Ã

  α f

  c

  ′

  23 c 4.89 in

  = 0.0102 × 60 4 × 0.72 = c 4.89 c

  0.213 Tension failure

  < = 0.375 d = 23 = d

  t

  Ã f

  y

2 M bd

  1

  n y

  = Ã f − 0.59 f

  c  ′ 

  2 M

  1

  n

  = (0.0102) × (60 ksi) × (10 in) × (23 in) × − 0.59 × (0.0102) × 60

   

  4 = 2, 950, 000 lb in = 246 k ft

5.10. Prediction of Nominal Strength in Flexure by Equivalent Rectangular Stress Block

  Represents an extension of the empirical method. • Simpler than empirical method - No secondary calculation necessary to locate centroid • (always at stress block center).

  Allows for considerations and analyses of non-rectangular sections. • Must be developed such that it gives the same answer as empirical method - requires • same total compression force and same centroid location.

  • Development of the method:

  c f

  s

  Strains compression side Stresses

  Á

  c

  0.85f

  c

  ′ βc a/2 a c

  = β

  1

  c C

  c

  C cb

  c c

  = αfd h d--a/2

  A

  s

  T T = A f s

  s s y Empirical Approach Equivalent Rectangular

  Á

  s Block. (Whitney Block)

  tension side Figure 5.4. Equivalent Rectangular Block (From Nawy’s Book).

  Require the forces to have the same location: a c (ACI 10.2.7)

  = β

  1

  C from which

  c c c

  = αfcb = γfab γ = α ca

  γ = α β

  1

  β and ACI

  10.2.7 = 2β γ = 0.85

1 ACI 10.2.7.3

  f

  c

  ′ − 4000 β and

  0.65 ACI

  10.2.7.3 = 0.85 − 0.05 ≤ β ≤ 0.85

  1

  1

  1000 C remember c

  c c

  γ = 0.85 and a = β = 0.85fab

  1 For balanced steel ratio we have

  0.85f Á

  u c

  ′ T

  = C

  b

  c

  b b

  Ã f bd b bc

  y c c b = 0.85 fa = 0.85 fβ

  1

  d f

  c Á u

  ′ Ã

  = 0.85 β

  b

  1

  f

  y Á b u y

  • Á

  d-c T

  s

  Á

  y

  substituting = 0.003 and E = 29,000 ksi ε u s f

  c 87, 000

  ′ (5.30)

  Ã

  b = 0.85β

  1

  f

  y 87, 000  y

  • f

  ACI 10.3.5. Maximum Net Tensile Strain

  For nonprestressed flexural members and prestressed members with axial load less than 0.10f A

  c g t

  ′ the net tensile strain Á at nominal strength shall not be less than 0.004. f Á

  c u

  ′ Ã

  max = 0.85β

  1

  f

  yÁu

  • 0.004

  f

  y

  = 60ksi

  Á 0.85f u

  c

  ′ Balanced Condition

  b

  c Tension Failure Compression Failure d

  b

  d-c Max Net Tensile Strain

  T

  s

  Á

  y

  0.85f Á

  u c

  ′

  b

  87, 000 c

  = d 87, 000

  t y b

  • f

  c d f

  y

  = 60, 000 psi

  b

  d-c T

  s b

  c

  87 = = 0.60 d

  t

  87 Á + 60

  y

  Á 0.85f u Á

  c u

  ′ c

  b

  = 0.375d c

  b

  c > c = 0.6d d

  T

  s

  Á

  t

  = 0.002 Á

  t

  = 0.005 Á

  y

  Á 0.85f u

  c

  ′

  Max net Tensile Strain b c

  c f

  y

  = 60, 000 psi d c 0.003

  = = 0.429 d-c d 0.003

  t

  • 0.004

  T

  s Á t

  = 0.004

  =

  = 87, 000

  t

  d

  b

  c

  y

  87, 000

  t

  87

  d

  b

  c

  • f

  s

  T

  ′ d

  87

  • 40

  c = 0.428d

  = 0.685d c

  b

  > c

  = 0.0014 c

  t

  Á

  b

  = 0.685 f

  c

  y

  Á

  u

  = 40 ksi Á

  y

  c

  t

  = 0.004 0.85f

  d-c

  ′ d

  c

  0.85f

  b

  c

  b

  y

  s

  Á

  u

  Á

  = 40ksi

  y

  f

  T

  Tension Failure Compression Failure

  Á

  0.85f

  u

  Á

  s

  T

  ′ d

  c

  b

  Balanced Condition Á

  c

  b

  d-c

  y

  Á

  u

  = 0.625 × 0.685 = 0.428

  5.10.1. Example Consider the same example problem given in Section 5.9.

  A

  s

  2.35 Ã 0.0102

  = bd =

  10 × 23 =

  2

  (2.35 in ) × (60, 000 psi)

  0.85f f a

  c s y

  ′ ab = A → = = 4.15 in

  0.85 × (4, 000 psi) × (10 in) c

  = aβ = 4.15∕0.85 = 4.88

  1

  c

  4.88 0.212

  < 0.375 d = 23 =

  Tension failure Therefore the nominal moment capabit will be:

  a

2 M n s f y (d ) ))

  = A − = (2.35 in × (60, 000 psi) × (23 − 2.07) = 246, 000 lb--ft=246 kip--ft

  2

  = 2, 950, 000 lb in = 246 k ft φ = 0.9 M

  u n

  = φM = 0.9 × 246 = 221.4 kip ft

5.10.2. Example. Calculate Nominal Moment Capacity of a Beam Determine if the beam shown below will fail in tension or compression. Given

  f

  c

  ′ = 4, 000 psi 18”

  21” f

  y

  = 60, 000 psi

  2 A =6.00 in s

  10”

  Solution

  A f

  s y

  a =

  0.85f

  c

  ′ b

  6 × 60 ksi a

  = = 10.59 in

  0.85 × 4 ksi × 10 a c

  12.46 = = 10.59

  β 0.85 =

  1

  c

  12.46

  0.69 Compression failure > 0.6 d = 18 = Hence, A does not yield and the strain is smaller than 0.02 in/in. Brittle failure results.

  s This beam does not satisfy ACI Code requirement.

5.10.3. Example. Calculate Nominal Moment Capacity of a Beam Determine if the beam shown below will fail in tension or compression. Given

  f

  c

  ′ = 4, 000 psi 18”

  21” f

  y

  = 40, 000 psi

  2 A =6.00 in s

  10”

  Solution

  A f

  s y

  a =

  0.85f

  c

  ′ b

  6 × 40 ksi a

  = = 7.06 in

  0.85 × 4 ksi × 10 a c 8.31 in

  = = 7.06 β 0.85 =

  1

  c

  8.31

  0.46 > 0.428 < 0.685 Transition Zone d = 18 =

  Hence, the beam is in the transition zone, tension steel yeilds. A reduced value of φ should be used.