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Bending moment calculation (whitney) Flipbook PDF
Bending moment calculation (whitney)
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CHAPTER
Reinforced Concrete Design
Fifth Edition
RECTANGULAR R/C CONCRETE BEAMS: TENSION STEEL ONLY • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
Part I – Concrete Design and Analysis
2b
FALL 2002
By
Dr . Ibrahim. Assakkaf
ENCE 355 - Introduction to Structural Design Department of Civil and Environmental Engineering University of Maryland, College Park
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 1 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Ultimate Moment (Strength)
– The ultimate moment for a reinforced concrete beam can be defined as the moment that exists just prior to the failure of the beam. – In order to evaluate this moment, we have to examine the strains, stresses, and forces that exist in the beam. – The beam of Fig. 1 has a width of b, an effective depth d, and is reinforced with a steel area As.
1
Slide No. 2
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Ultimate Strength
Flexural Strength ACI Approach
Reinforced Concrete Beam
ε c (0.003 as a limit )
b
f c′
NC
N.A.
h
d εs ≥ ε y
Figure 1
Strain
NT
f s = f y as a limit Stress
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Force
Slide No. 3 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Possible Values for Concrete Strains due to Loading (Modes of Failure) 1. Concrete compressive strain is less than 0.003 in./in. when the maximum tensile steel unit equal its yield stress fy as a limit. 2. Maximum compressive concrete strain equals 0.003 in./in. and the tensile steel unit stress is less than its yield stress fy.
2
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 4 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Notes on Concrete Compressive Stresses – The ultimate compressive stress for concrete does not occur at the outer fiber. – The shape of the curve is not the same for different-strength concretes. – The shape of the curve will also depend on the size and dimensions of the beam. – The ultimate compressive stress of concrete develops at some intermediate level near, but not at, the extreme outer fiber.
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 5 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Nominal Moment Strength – The forces NC and NT, and the distance Z separated them constitute an internal resisting couple whose maximum value is termed nominal moment strength of the bending member. – As a limit, this nominal strength must be capable of resisting the actual design bending moment induced by the applied loads.
3
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 6 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Nominal Moment Strength (cont’d) – The determination of the moment strength is complex because of • The shape of the compressive stress diagram above the neutral axis • Not only is NC difficult to evaluate but also its location relative to the tensile steel is difficult to establish
f c′
f s = f y as a limit Stress
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 7 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
How to Determine the Moment Strength of Reinforced Concrete Beam? – To determine the moment capacity, it is necessary only to know 1. The total resultant compressive force NC in the concrete, and 2. Its location from the outer compressive fiber, from which the distance Z may be established.
4
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 8 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams
How to Determine the Moment Strength of Reinforced Concrete Beam? (cont’d)
Q
– These two values may easily be established by replacing the unknown complex compressive stress distribution by a fictitious (equivalent) one of simple geometrical shape (e.g., rectangle). – Provided that the fictitious distribution results in the same total NC applied at the same location as in the actual distribution when it is at the point of failure.
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 9 ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Mathematical Motivation
– Consider the function (1) f (x ) = y = 2 x – Plot of this function is shown in Fig. 2 for x ranges from 0 to 4, and y from 0 to 4. – The area under the curve will be determined analytically. – Note that in real situation this area will be the equivalent,for example, to compressive force NC for concrete per unit length.
5
Slide No. 10
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Mathematical Motivation (cont’d)
Area under the Curve
NC per unit length
y
fc
4 in
4 in
Area
x′
x
x
4 in
x′
x 4 in
A
c
NC
Slide No. 11
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular Beams Q
Mathematical Motivation (cont’d)
4
4
0
0
( )
4
y
1 2
A = ∫ ydx = ∫ 2 x dx = 2 ∫ x dx = 10.7 in 2
x=
~ ∫ x dA
∫ dA
=
~ ∫ x dA
0
4 in
10.7
∫ x dA = ∫ x( ydx ) ~
x′
y
x
4
4 in
0
4
(
)
4
3
= ∫ x 2 x dx =2∫ x 2 dx =25.6 0
dx
A
0
25.6 Therefore, x = = 2.4 in. 10.7
6
Slide No. 12
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular y Beams Actual Curve Q
Mathematical Motivation (cont’d) – Objective
• Our objective is to find a fictitious or equivalent curve results in the same total area A applied at the same location as the actual curve. • Find x’ and y’
4 in
Area = 10.7
x′
x y
x
4 in
A Equivalent Simple Curve Area = 10.7
4 in
x′ 4 in
x′
y′ x
A
Slide No. 13
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular y Beams Actual Curve Q
Mathematical Motivation (cont’d)
– Calculations of x′and y′
4 in
Area = 10.7
x′
x
x′ = 4 − x = 4 − 2.4 = 1.6 in. Area = 2 x′y′
y
Area 10.7 y′ = = = 3.34 in. 2 x′ 2(1.6) 4 in
x
4 in
A Equivalent Simple Curve Area = 10.7
x′ 4 in
x′
y′ x
A
7
Slide No. 14
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Flexural Strength of Rectangular f Beams Actual Stress Distribution c
Q
Mathematical Motivation (cont’d) – If we are dealing with a concrete compressive stress distribution and we let x′ = a / 2 ,then and
Then, 3.2 = 0.80 4
Area = 10.7
x′
x y
y′ = 0.84 f c′
a = 2 x′ = β1c = 2(1.6) = 3.2 in. β1 =
f c′
f c′
c
c
NC per unit length Equivalent Stress Distribution a Area = 10.7 a a 2 2 c
y′ c
NC per unit length
Slide No. 15
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Q
As we saw in our previous mathematical example, any complicated function can be replaced with an equivalent or fictitious one to make the calculations simple and will give the same results. For purposes of simplification and practical application, a fictitious but equivalent rectangular concrete stress distribution was proposed.
8
Slide No. 16
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Q
Q
This rectangular stress distribution was proposed by Whiney (1942) and subsequently adopted by the ACI Code The ACI code also stipulates that other compressive stress distribution shapes may be used provided that they are in agreement with test results. Because of its simplicity, however, the rectangular shape has become the more widely stress distribution (Fig. 2).
Slide No. 17
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Whitney’s Rectangular Stress Distribution Figure 2 0.85 f c′ f c′ a 2
a
N C = 0.85 f c′ab
N.A
d
Z =d−
fy
fy
Actual Compressive Stress Block
Rectangular Equivalent Compressive Stress Block
a 2
N C = As f y Internal Couple
9
Slide No. 18
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Whitney’s Rectangular Stress Distribution – According to Fig. 2, the average stress distribution is taken as
Average Stress = 0.85 f c′ – It is assumed to act over the upper area on the beam cross section defined by the width b and a depth a as shown in Fig. 3.
Slide No. 19
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Whitney’s Rectangular Stress Distribution 0.85 f c′
b
a 2
c a
N .A.
As
Figure 3
N C = 0.85 f c′ab Z
NT = As f y
10
Slide No. 20
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Whitney’s Rectangular Stress Distribution – The magnitude of a may determined by
a = β1c
(2)
Where C = distance from the outer fiber to the neutral axis β1 = a factor dependent on concrete strength, and is given by 0.85 β1 = 1.05 − 5 ×10- 5 f c′ 0.65
for f c′ ≤ 4,000 psi for 4,000 psi < f c′ ≤ 8,000 psi for f c′ > 8,000 psi
(3)
Slide No. 21
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 Determine the nominal moment Mn for a beam of cross section shown, where = 4,000 psi. Assume A615 grade 60 steel that has a yield strength of 60 ksi and a modulus of elasticity = 29 × 106 psi.
10 in.
25 in.
N.A. 23 in.
11
Slide No. 22
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 (cont’d) 10′′
0.85 f c′
εc
c
a 2
a
N C = 0.85 f c′ab
N.A 23′′
Z =d− 3 #8 bars
a 2
N C = As f y
εs
Slide No. 23
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 (cont’d) Area for No. 8 bar = 0.79 in 2 (see Table 1) Therefore, As = 3(0.79 ) = 2.37 in 2
(Also see Table A-2 Text)
Assume that fy for steel exists subject later check. 10 in.
NC = N S 0.85 f c′ab = As f y a=
As f y 0.85 f c′b
=
2.37(60) = 4.18 in. 0.85(4 )(10)
25 in.
N.A. 23 in.
12
Slide No. 24
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Table 1. ASTM Standard - English Reinforcing Bars Bar Designation #3 [#10] #4 [#13] #5 [#16] #6 [#19] #7 [#22] #8 [#25] #9 [#29] #10 [#32] #11 [#36] #14 [#43] #18 [#57]
Diameter in 0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257
Area in2 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00
Weight lb/ft 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.650 13.60
Note: Metric designations are in brackets
Slide No. 25
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
10 in.
Example 1 (cont’d) Calculation of Mn a a M n = N C d − = NT d − 2 2 a a M n = 0.85 f c′ab d − = As f y d − 2 2 Based on steel :
25 in.
N.A. 23 in.
4.18 M n = 2.37(60) 23 − = 2,973.4 in. - kips 2 2,973.4 = = 247.8 ft - kips 12
13
Slide No. 26
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 (cont’d) Check if the steel reaches its yield point before the concrete reaches its ultimate strain of 0.003: • Referring to the next figure (Fig. 4), the neutral axis can be located as follows: Using Eqs. 2 and 3 : β1 = 0.85 a = β1c Therefore, a 4.18 = = 4.92 in. c= β1 0.85
Slide No. 27
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 (cont’d) 10′′
Figure 4 0.85 f c′
0.003
a
c
a 2 N C = 0.85 f c′ab
N.A 23′′
d 3 #8 bars
εs
Z =d−
a 2
N C = As f y
14
Slide No. 28
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Equivalent Stress Distribution
ENCE 355 ©Assakkaf
Q
Example 1 (cont’d) By similar triangles in the strain diagram, the strain in steel when the concrete strain is 0.003 can be found as follows: ε 0.003 = s c d −c d −c 23 − 4.92 = 0.003 = 0.011 in./in. ε s = 0.003 c 4.92 The strain at which the steel yields is
εy =
fy Es
=
0.003
c d = 23′′
60,000 = 0.00207 in./in. 29 × 106
Since εs (= 0.011) > εy (= 0.00207) OK
εs
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 29 ENCE 355 ©Assakkaf
Balanced, Overreinforced, and Underreinforced Beams Q
Strain Distribution 0.003
Elastic region
Figure 6
Stress
Figure 5 fy
εy
εy
Strain
Strain
Idealized Stress-Strain Curve
15
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 30 ENCE 355 ©Assakkaf
Balanced, Overreinforced, and Underreinforced Beams Q
Strain Distribution 0.003
εy =
fy E
Underreinforced N.A. Balanced N.A. Overreinforced N.A. εy
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 31 ENCE 355 ©Assakkaf
Balanced, Overreinforced, and Underreinforced Beams Q
Balanced Condition: εs = εy
Q
and
εc = 0.003
Overreinforced Beam εs < εy, and εc = 0.003. The beam will have more steel than required to create the balanced condition. This is not preferable since will cause the concrete to crush suddenly before that steel reaches its yield point.
Q
Underreinforced Beam εs > εy, and εc = 0.003. The beam will have less steel than required to create the balanced condition. This is preferable and is ensured by the ACI Specifications.
16
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 32 ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Q
Q
Although failure due yielding of the steel is gradual with adequate warning of collapse, failure due to crushing of the concrete is sudden and without warning. The first type (Underreinforced beam) is preferred and ensured by the specifications of the ACI. The ACI code stipulates that
As ≤ 0.75 Asb
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
(4)
Slide No. 33 ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Steel Ratio
– The steel ratio (sometimes called reinforcement ratio) is given by b
h
ρ=
N.A.
As bd
(5)
d As
ACI stipulates that ρ max = 0.75 ρ b or Asmax = 0.75 Asb
(6)
17
Slide No. 34
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Example 2
Determine the amount of steel required to create a balanced condition for the beam shown, where = 4,000 psi. Assume A615 grade 60 steel that has a yield strength of 60 ksi and a modulus of elasticity = 29 × 106 psi. Also check the code requirement for ductile-type beam.
10 in. N.A.
25 in.
23 in.
Slide No. 35
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Example 2 (cont’d)
Area for No. 8 bar = 0.79 in 2 (see Table 1) Therefore, As = 3(0.79 ) = 2.37 in 2
10 in.
The strain at which the steel yields is
fy
60,000 = = 0.00207 in./in. εy = Es 29 × 106
25 in.
N.A. 23 in.
In reference to the strain diagram of Fig. 7, and from similar triangles,
cb d − cb = 0.003 0.00207
18
Slide No. 36
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Reinforcement Ratio Limitations and Guidelines
ENCE 355 ©Assakkaf
Table 1. ASTM Standard - English Reinforcing Bars Bar Designation #3 [#10] #4 [#13] #5 [#16] #6 [#19] #7 [#22] #8 [#25] #9 [#29] #10 [#32] #11 [#36] #14 [#43] #18 [#57]
Diameter in 0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257
Area in2 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00
Weight lb/ft 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.650 13.60
Note: Metric designations are in brackets
Slide No. 37
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Reinforcement Ratio Limitations and Guidelines
ENCE 355 ©Assakkaf
Q
Figure 7
Example 2 (cont’d) 10′′
0.85 f c′
0.003
N.A
a
cb
23′′
d − cb 3 #8 bars
0.00207
a 2 N C = 0.85 f c′ab
Z =d−
a 2
N C = As f y
Strain
19
Slide No. 38
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Reinforcement Ratio Limitations and Guidelines
ENCE 355 ©Assakkaf
Q
Example 2 (cont’d) cb 23 − cb = 0.003 0.00207 From which, cb = 13.6 in. Using Eqs. 2 and 3 : β1 = 0.85 bbcause f c′ = 4,000 psi
10 in.
25 in.
N.A. 23 in.
a = β1c = 0.85(13.6) = 11.6 in.
Slide No. 39
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Example 2 (cont’d) N Cb = 0.85 f c′ab b = 0.85(4 )(11.6 )(10 ) = 394.4 kips N Cb = NTb = Asb f y Therefore,
N 394.4 Asb = Cb = = 6.57 in 2 fy 60 Hence, required steel for balanced condition = 6.57 in2
10 in.
25 in.
N.A. 23 in.
From Eq. 6,
Asmax = 0.75 Asb = 0.74(6.57 ) = 4.93 in 2 > As = 2.37 in 2 OK
20
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 40 ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Steel Ratio Formula for Balanced Beam Instead of using laborious techniques for determining the balanced steel of beam, the following formula can be used to determine the steel ratio ρb at the balance condition: ρb =
0.85 f c′β1 87,000 f + 87,000 fy y
(7)
where f c′ = compressive strength of concrete (psi) fy = yield strength of steel (psi) β1 = factor that depends on f c′ as given by Eq. 3
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
Slide No. 41 ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines Q
Lower Limit for Steel Reinforcement – The ACI Code establishes a lower limit on the amount of tension reinforcement. The code states that where tensile reinforcement is required , the steel area As shall not be less than that given by As , min =
3 f c′ 200 bw d ≥ bw d fy fy
(8)
Note that for rectangular beam bw = b
21
Slide No. 42
CHAPTER 2b. RECTANGULAR R/C BEAMS: TENSION STEEL ONLY
ENCE 355 ©Assakkaf
Reinforcement Ratio Limitations and Guidelines (Table A-5 Text) f c′ (psi )
Table 1. Design Constants
3 f c′ 200 ≥ f y f y
3,000 4,000 5,000 6,000
0.0050 0.0050 0.0053 0.0058
3,000 4,000 5,000 6,000
0.0040 0.0040 0.0042 0.0046
3,000 4,000 5,000 6,000
0.0033 0.0033 0.0035 0.0039
3,000 4,000 5,000 6,000
0.0027 0.0027 0.0028 0.0031
Recommended Design Values ρmax = 0.75 ρb Fy = 40,000 psi 0.0278 0.0372 0.0436 0.0490 Fy = 50,000 psi 0.0206 0.0275 0.0324 0.0364 Fy = 60,000 psi 0.0161 0.0214 0.0252 0.0283 Fy = 75,000 psi 0.0116 0.0155 0.0182 0.0206
ρb
k (ksi)
0.0135 0.0180 0.0225 0.0270
0.4828 0.6438 0.8047 0.9657
0.0108 0.0144 0.0180 0.0216
0.4828 0.6438 0.8047 0.9657
0.0090 0.0120 0.0150 0.0180
0.4828 0.6438 0.8047 0.9657
0.0072 0.0096 0.0120 0.0144
0.4828 0.6438 0.8047 0.9657
22