Moment of inertia for cracked and uncracked section PDF

Preview

Summary

sfs...

Description

560.325 Lecture Notes © T. Igusa

III. FLEXURAL ANALYSIS AND DESIGN OF BEAMS CHAPTER 9, Sections 1–2

1. Introduction The analysis of beams will proceed as follows: Chapter III: Chapter IV: Chapter V: Chapter VI:

Dimensioning of the concrete cross section. Selection and placement of reinforcing steel. Shear reinforcement. Bond and anchorage of rebars. Serviceability, deflections, and cracking.

2. Bending of Homogeneous Beams For elastic analysis, the elementary equations from mechanics of materials is used: f

= My/I

f M I y

= = = =

stress moment applied to the beam moment of inertia of the cross section distance from the neutral axis

Example 3.1 Consider a plain concrete beam that is 10 inches wide and 30 inches deep. The strength of the concrete is given by f'c = 4000 psi and fr = 475 psi. What is the maximum (nominal) moment that can be applied to the beam?

maximum compression strength = .85 f'c = maximum tension strength = fr = 475 psi moment of inertia = I = b h3 / 12 = concrete stress = f = M y / I = maximum (nominal) moment = Mn =

10

560.325 Lecture Notes © T. Igusa

11

3. Reinforced Concrete Beam Behavior Plain concrete beams are not practical because as soon as a crack is formed, the entire beam splits into two parts causing immediate and sudden structural failure. The basic mechanics problem is that concrete is very weak under tension stresses. To increase both the strength and the safety of concrete beams, reinforcing steel is added where the stresses are in tension. The figures below illustrate how the stresses in the beam change as the moment load is increased.

Four-point bending test

Failure configurations

560.325 Lecture Notes © T. Igusa

12

Stages of stress-strain behavior Stresses elastic and section uncracked:

h

d

b no tension cracks if: fct < fr

Stresses elastic and section cracked:

d

no tension stress (fct = 0) because of cracks compression stress is linear elastic if: fc < f'c/2 steel stress is linear elastic if: εs < εy

Stresses inelastic:

d

steel is yielding compression stress is inelastic concrete is still intact

560.325 Lecture Notes © T. Igusa

13

Review of section properties Before beginning the analysis of reinforced concrete beams, some of the properties of the cross section are reviewed.

A 1 , I1

A 2 , I2

The centeroid is given by: y =

1 Area

y =

∑ A i yi ∑ Ai

y dA

The moment of inertia is given by: I =

I =

(y – y)2 dA

∑ Ii

+

∑ A i (y – yi )2

560.325 Lecture Notes © T. Igusa

14

A rectangular stress distribution can be replaced by a single force as follows: f a/2

a C = abf

b

A triangular stress distribution can be replaced by a single force as follows: f a/3

a C = a b f /2

b

a. Stresses elastic and section uncracked To find the stresses in an uncracked beam, it is useful to replace the reinforced concrete cross section with an equivalent, purely concrete, transformed section. Basically, the steel is replaced by an equivalent amount of concrete. Since the modulus of elasticity of steel, Es, is much larger than that of concrete, Ec, the area of the steel must be replaced by a much larger area of concrete. The idea of a transformed section is best explained using diagrams. First, some definitions are introduced:

560.325 Lecture Notes © T. Igusa

As b h d Iut

= = = = =

15

area of reinforcing steel (under tension) width of a rectangular beam total depth of a rectangular beam distance from the top of the beam to the centeroid of the steel moment of inertia of the uncracked, transformed section

fc = compression stress in the concrete fct = tension stress in the concrete fs = tension stress in the steel n

= Es / Ec = modular ratio

In the transformed section, the area of the steel, As, is replaced by a larger area, nAs, of concrete. Some of the concrete "fills" the area, As, originally occupied by the steel. The remaining concrete, which has area (n–1)As, is placed outside of the rectangular dimensions of the beam at the centeroid of the steel.

h

d

b

(n–1)A s

To prove that the transformed concrete section is equivalent necessary only to examine the stresses at the steel: stress at steel rebars

= fs = Es εs =

total tension force from steel rebars

= stress × area =

total tension force from equivalent concrete area

=

One important relation from the preceeding proof is: fs = n fct (at the location of the steel, As)

section, it is

560.325 Lecture Notes © T. Igusa

16

Example 3.2 Consider the same concrete beam examined in Example 3.1 (10" wide, 30" deep, f'c = 4000 psi, and fr = 475 psi). Two inches from the bottom of the beam, two Number 9 bars are inserted. The strength of the steel is fy = 60,000 psi. What is the largest moment that can be applied to the beam before it cracks? Solution First, find the modular ratio, n: Ec = 57,000 √f'c (see page 7) = n

= Es / Ec =

Next, the basic geometric properties of the cross section are found: No. 9 bar has dia. = 9/8 in. The area of one bar is π (dia.)2 / 4 = As = A1 = b h = A2 = (n–1) As = I1 = b h3 / 12 = I2 ≈ 0 (usual approximation)

(n–1)A s

The neutral axis and uncracked moment of inertia are: ∑ A i yi = A1 y1 + A2 y2 = y = A1 + A2 ∑ Ai Iut =

∑ Ii

+

∑ A i (y – yi )2

=

560.325 Lecture Notes © T. Igusa

17

Next, the stresses are computed using the usual formula, f = M y / I :

(n–1)A s Make sure the concrete tension stress is less than the rupture strength, fr : fct = M y / Iut =

Make sure the steel tension stress is less than the yield strength, fy : fs = n fct (at As) = n M y / Iut = Make sure the concrete compression stress is less than the linear elastic limit, f'c/2: fc = M y / Iut =

560.325 Lecture Notes © T. Igusa

18

b. Stresses elastic and section cracked To find the stresses in a cracked beam, the equivalent, purely concrete, transformed section is still used. The difference between the analysis of cracked and uncracked sections is that no tensile stresses are allowed in the cracked case. The following definitions are useful: kd = distance from the top of the beam to the neutral axis jd = moment arm for the equivalent compression and tension forces Icr = moment of inertia of the cracked, transformed section C = equivalent compression force in the concrete T = equivalent tension force in the steel M = moment applied to beam ρ

= As / bd = ratio between steel and concrete areas

Using basic principles of mechanics, it is possible to find the location of the neutral axis, kd, and the stresses in the concrete and steel. The final results are: k =

(ρ n )2 + 2ρ n – ρ n

(9-3)

j = 1 – k/3 fs =

M As j d

(9-5)

fc =

2M k j bd2

(9-4)

Icr = b k2 j d3 / 2

The derivations are based on the diagram of the concrete stresses on the next page.

560.325 Lecture Notes © T. Igusa

19

kd d

b

nA s

The properties of the two areas, A1 and A2 are: A1 = b k d

y ¯1 = k d / 2

A2 = n As = n ρ b d

y ¯2 = d

¯y = k d

The equations for the forces and moments are: C = area × average stress = T = area × average stress = M = Tjd = = Cjd = Equation (9-3) for the neutral axis is derived from the following: y =

∑ A i yi ∑ Ai

=

A1 y 1 + A2 y2 A 1 + A2

or

y ¯ ( A 1 + A2 ) = A1 y¯1 + A2 y¯2

It is necessary to substitute the terms for A1, A2, y ¯ 1, y ¯ 2, and y¯: kd ( bkd + nρbd ) = bkd ( kd/2 ) + nρbd ( d )

560.325 Lecture Notes © T. Igusa

20

Example 3.3 Consider the same concrete beam examined in Example 3.2 (10" wide, 30" deep, f'c = 4000 psi, fr = 475 psi, fy = 60,000 psi, two No. 9 bars 2" from bottom). Determine the maximum moment that can be carried without stressing the concrete beyond f'c/2 or the steel beyond fy/2. Solution First, the neutral axis is determined: ρ

= As / bd =

ρn = k

=

(ρ n )2 + 2ρ n – ρ n =

kd = j

= 1 – k/3 =

Then, the maximum stresses in the steel and concrete are found: fs =

M = As j d

fc =

2M = k j bd2

Just for checking, the stresses are recalculated using the cracked moment of inertia, Icr, and the usual stress formula, f = My/I: Icr = b k2 j d3 / 2 = fc = M y / Icr = fs = n fct (at As) = n M y / Icr =

560.325 Lecture Notes © T. Igusa

21

4. Design of Tension-Reinforced Rectangular Beams CHAPTER 4, Sections 1–2, 3 (up to "Upper Limit ...", Example 4-2, skip SI units), 4. In this section, the strength of rectangular beams is analyzed and a design method is developed. The strength capacity is the maximum moment a beam can handle before it becomes essentially destroyed. A fundamental assumption is that the steel is yielding and the concrete strains are not high enough to cause crushing. To make sure this type of failure occurs, the steel ratio must be less than the so-called balanced steel ratio, i.e., ρ ≤ ρb. The analysis is more difficult than in Section 3 because the concrete stresses are no longer linear. This means that the linear relation between stress and strain is no longer valid, i.e., fc ≠ Ec εc. However, researchers have found a moderately simple approximation for beam analysis which closely agrees with experiments. This approximation is presented in the following.

a. Equivalent rectangular stress distribution The concrete stresses just before the beam fails has a nonlinear distribution. To make the analysis relatively simple, the nonlinear stresses are replaced with an equivalent rectangular (constant) stress distribution. The magnitude and location of the equivalent stress distribution is defined in the diagram below. ACTUAL STRESS DISTRIBUTION

d

c

EQUIVALENT STRESS DISTRIBUTION a/2

a d

c d – a/2

560.325 Lecture Notes © T. Igusa

22

The relationship between the distance to the neutral axis, c, and the depth of the stress distribution, a, is given by:

Equivalent stress parameters a = β1 c

β1 =

0.85,

if f'c ≤ 4000 psi;

0.85 – 0.05(f'c – 4000)/1000,

if 4000 psi ≤ f'c ≤ 8000 psi;

0.65,

if 8000 psi ≤ f'c

(4-8)

With the preceding information, it is now possible to derive the following:

Location of the neutral axis a =

A s fy 0.85f'c b

=

ρ fy d

(4-11, 4-14a)

0.85 f'c

c = a / β1

Maximum nominal moment Mn = As fy (d – a/2) = R b d2

R = ρ fy 1 – 0.59

(4-12a)

ρ fy f' c

design strength = φ Mn = 0.9 Mn

Minimum and maximum values for the steel ratio, ρ ρmin ≤ ρ ≤ ρmax ρmin = 200 / fy (f'c < 4440psi)

or ρmin = 3 f' c / fy (f'c > 4440psi)

(4-31)

ρmax = 0.75 ρb f' ρb = 0.85 β1 c fy

87,000 87,000 + fy

(4-25)

560.325 Lecture Notes © T. Igusa

23

To determine the location of the neutral axis, it is necessary only to enforce equilibrium. The equivalent forces, C and T, are: C = compression stress × area = T = tensile stress × area = Solving for a: a = The maximum nominal moment is obtained from statics: M n = force × moment arm =

b. Balanced steel ratio The balanced steel ratio, ρb , is the steel ratio where the concrete crushes and the steel yields at the same time. For other steel ratios, we have: ρ < ρb ρ > ρb

under-reinforced, over-reinforced,

εs > εy εs < εy

steel yields before concrete crushes concrete crushes before steel yields

The analytical formula for ρb is derived by looking at the strains instead of the stresses. The strains we are interested are: εu = 0.003 εy = fy / Ey = fy / 29,000,000

compressive strain where concrete crushes tensile strain where steel yields

Using similar triangles, we find the following ratios where the maximum concrete strain is εc = εu and the steel strain is εs = εy:

560.325 Lecture Notes © T. Igusa

24

c. Under-reinforced beams If the steel ratio, ρ, is less than the balanced steel ratio, ρb , then the beam is under-reinforced, and the steel will yield before the concrete crushes. However, if the steel ratio is larger than ρb , then the beam is over-reinforced, and the concrete will crush before the steel yields. As explained in Section 3 (page 11), yielding steel makes the beam fail slowly with large deflections, while crushing concrete makes the beam fail explosively with immediate, catostrophic collapse. To allow for possible variations in the concrete and steel properties, it is safer to make ρ significantly less than ρb . The ACI Code gives the following design limit: ρ ≤ ρmax = 0.75 ρb Thus, the steel ratio must be at least 25% less than the balanced steel ratio.

d. Minimum steel ratio If the moment capacity of an uncracked beam (Section 3a) is greater than the moment capacity of a cracked beam (Section 4), then the beam will fail as soon as a crack is formed. This type of failure is not as catostrophic as concrete crushing, because the steel will still yield. However, it is sudden, because it will occur immediately after the concrete cracks. To eliminate this possibility, the ACI Code gives a lower limit for the steel ratio: ρ ≥ ρmin

where

ρmin = 200 / fy (f'c < 4440psi)

or ρmin = 3 f' c / fy (f'c > 4440psi)

(4-31)

e. Examples of rectangular beam review and design There are two type of problems to consider: Review is where the beam dimensions and reinforcing steel sizes are given, and the beam strength is to be computed. Review is very straightforward since it is necessary only to "plug in" the dimensions into a set of formulas. Design is where the required beam strength (or load) is given, and the dimensions of the beam and reinforcing steel are determined. Design is more challenging than review for two reasons: First, it may be necessary to iterate to find the optimal beam dimensions. Second, there is, in general, no unique solution to the problem. In the following are examples of review and design problems.

560.325 Lecture Notes © T. Igusa

25

Example 3.4. Review of a tension-reinforced beam. Consider the same concrete beam examined in Example 3.2 (10" wide, 30" deep, f'c = 4000 psi, fr = 475 psi, fy = 60,000 psi, two No. 9 bars 2" from bottom). Determine the maximum nominal moment, Mn , and the design strength, φM n . Solution First, the steel ratio must be checked against the limits, ρmin and ρmax : ρ = As/bd = 2.00 / ( 10 × 28 ) = 0.00714 87,000 f' ρb = 0.85 β1 c fy 87,000 + fy

(4-25)

= ρmax = 0.75 ρb = ρmin = 200 / fy (f'c < 4440psi)

or ρmin = 3 f' c / fy (f'c > 4440psi)

(4-31)

Then, the nominal moment and design strength are computed: a =

As fy = 0.85f' c b

(4-11)

Mn = As fy (d – a/ 2) =

(4-12a)

design strength = φ Mn = 0.9 Mn =

Summary of Examples 3.1 to 3.4. Example

neutral axis (inches)

moment (kip-inch)

3.1

No steel

h/2 = 15.0

713

0.3

3.2

Uncracked

y = 15.6

817

0.5

3.3

3.3

Cracked, elastic

kd = 8.0

1,520

1.5

30.0

3.4

Inelastic

a = 3.5

3,150

3.4

60.0

Design

2,830

stresses (ksi) concrete (fc) steel (fs)

560.325 Lecture Notes © T. Igusa

26

Basic design steps There are two sets of well-defined steps which can be used in design. The objective is to satisfy the strength equation, (4-16)

Mu ≤ φ Mn = φ R b d2

The design is based on the strengths of the concrete and steel, as defined by f'c and fy, and the factored load, Mu. DESIGN WITH SPECIFIC CONCRETE DIMENSIONS 1.

Choose the beam width, b, and effective depth, d, using the following rules: a. The total depth, h, and width, b, are in even inches. b. The difference between d and h is usually: h – d = 2.5 in. h – d = 4.0 in.

for a single layer of steel bars; for a double layer of steel bars.

c. The effective depth, d, is usually 2 to 3 times the width, b. Find the required value for R corresponding to b and d: R = Mu / φ b d2 2a. Substitute the preceding value for the equation for R on page 22. The result is a quadratic equation with unknown, ρ. The smallest of the two solutions for the steel ratio, ρ, is the correct solution. 2b. Check the steel ratio with the minimum and maximum values: ρmin ≤ ρ ≤ ρmax Use the equations on the bottom of page 22. 3.

Calculate the required area of steel, As = ρbd. Use Table A-8 (pg. 874) to find suitable rebars. Preference is for fewer and same-sized bars. The rebars should be symmetrically placed (in both size and location). For close cases, redo step 2b.

4.

Use Table A-6 (pg. 870) to check for placement and concrete cover. DESIGN WITH AN INITIAL STEEL RATIO

1'. Find the balanced steel ratio, ρb , using equation (4-25). Then choose an initial value for the steel ratio ρ. An economical and practical choice is ρ = 0.5ρb . 2'. Find the corresponding value for R using the equation on page 22. 3'. Solve the strength equation in terms of bd2: b d2 ≥ Mu / φ R Find b and d which satisfies the above. Then follow the preceding steps for "design with specific concrete dimensions".

560.325 Lecture Notes © T. Igusa

27

Example 3.5. Design of a tension-reinforced beam. A rectangular, tension-reinforced beam is to be designed for dead load of 1400 lb/ft plus self weight and service live load of 800 lb/ft, with a 20 ft simple span. Material strengths will be fy = 60 ksi and f'c = 5 ksi for steel and concrete respectively. The total beam depth (h) must not exceed 20 in. Calculate the required beam width and tensile steel requirement, using a steel ratio of 0.5 ρb . The effective depth (d) may be assumed at 2.5 in. less than the total depth (h). For the first iteration, assume the self weight is 100 lb/ft. Solution Before proceeding with the design steps, it is necessary to find the load. wd = 1400 + 100 = 1500 lb/ft, wl

= 800 lb/ft

w u = 1.4 wd + 1.7 wl = 1.4 (1500) + 1.7 (800) = 3460 lb/ft = 3.46 k/ft M u = wu L2 / 8 = 3.46 (20)2 / 8 = 173 k-ft Also, it is useful to find the parameter using equation (4-8): β1 = 0.85 – 0.05 (f'c – 4000)/1000 = The design then proceeds as follows: 1'. Find the balance...