Rc doubly-reinforced-beams study material PDF

Preview

Summary

STUDY GUIDE...

Description

Reinforced Concrete Design Module 7 Subject: Reinforced Concrete Design for Architecture Students

1. Title of the Module Beam Reinforced for tension and Compression 2. Introduction These are the reasons for providing compression reinforcement in beams. 1. IT REDUCES, SUSTAINED - LOAD DEFLECTION The addition of compression reinforcement, causes the creep of the concrete in the compression zone to transfer load from the concrete to the compression steel thus reducing the stress in the concrete. Because the lower compression stress in the concrete, it creeps less, leading to a reduction in sustained load deflections. 2. IT INCREASES DUCTILITY. The additional of the compression reinforcement causes a reduction in the depth of the compression stress block. As the depth of the compression block decreases the strain in the tension reinforcement at the failure increases, the resulting in more ductile behavior. The ductility increases significantly when reinforcement is used. 3. IT CHANGES THE MODE OF FAILURE FROM COMPRESSION TO TENSION. When ρmax < ρ, the beam will fail in a brittle manner through crushing of the compression zone before the steel yields. With the additional of compression steel to such a beam, the compression zone is strengthening sufficiently to allow the tension steel to yield before the concrete crushes. *For beams reinforced for tension and compression:

*Effects of

compression reinforcement on the strain distribution in two beams with same area of tension reinforcement.

NOTE: The neutral axis becomes nearer at the top of compression fiber of concrete when it is reinforced for compression.

3. Learning Outcomes At the end of the course, the students should be able to: a. analyze doubly reinforced beams with speed and accuracy and e. be familiar with the NSCP 2015 provisions regarding the same and apply them correctly.

4. Learning Content

ANALYSIS FOR BEAMS REINFORCED FOR TENSION AND COMPRESSION

M1 = As1 fy (d-a/2) As1 = ρbd To obtain a reduction factor of 0.9, let the max. reinforcement ratio corresponding to a net tensile strain of the extreme layer of tensile stress be equal to 0.005.

Assume value of ρ1 =

0.85𝑓`𝑐𝛽(0. 003) (0.008)𝑓𝑦

DERIVATION BY RATIO AND PROPORTION 0.003 0.008 = 𝐶 𝐷 C=T 0.85fc`ab = As fy ρ= As / bd 0.85f `c ab = ρ bd fy a = βc ρ=

0.85𝑓`𝑐𝛽𝑐 𝑏𝑑 𝑓𝑦

𝑑 0.005 = 𝐶 0.008 ρ=

0.85𝑓`𝑐𝛽 0.003 𝑓𝑦

( 0.008)

As1 = ρbd C1=T1 0.85f `c ab = As1 fy M1 =Ø As1 fy (d- a/2) M2 = Ø As1 fy (d- d`)

A. If compression bars will yield: 1. ℇs` > ℇy = fy / Es 2. ρ - ρ` >

0.85𝑓𝑐𝑑`𝛽 (600) 𝑓𝑦 𝑑 (600−𝑓𝑦)

3. fy = fs` 4. ρ = As / bd 0.003+

fy

200000 ρb ρ max > 0.008 ρ > ρmax (compression bars are needed) ρ < ρ max (compression bars are not needed)

5. f `c=28MPa or less use value of β, if it is greater than 28MPa β=0.85-

0.05(𝑓`𝑐−28) 7

B. If compression bars will not yield: 1. ℇs < ℇy = fy / Es 2. ρ - ρ ` <

0.85𝑓𝑐𝑑`𝛽 (600) 𝑓𝑦 𝑑 (600−𝑓𝑦)

600(𝑐−𝑑`)

3. fs`=

𝑐

600(𝑑−𝑐)

; fs=

𝑐

4. ρ = As / bd fy

ρ max =

0.003+ 200000 0.008

ρb

ρ > ρ max (compression bars are needed) ρ < ρ max (compression bars are not needed) NOTE: Adjustment for the concrete area replaced by the compression reinforcement is disregarded as being insignificant for practical design purposes. BALANCED STEEL RATIO OF THE BEAMS REINFORCEMENT FOT BOTH TENSION AND COMPRESSION A. When compression bars will yield ρ 𝑏 = ρb + ῥ B. When compression bars will not yield

ρ𝑏 ρb + ρ`(fs`/fy) = Maximum steel area Asmax =ρmax bd

NOTE: Maximum steel ratio 1. For beam reinforced for tension A. ρmax =

0.003+

fy 200000

0.008

ρb

2. For beam reinforced for both tension and compression a. ρ𝑏 = ρb + ρ` b. ρ 𝑏 = ρb + ρ` (fs`/fy)

EXAMPLE PROBLEM 1. (Steel Yield) A rectangular beam has a width of 300mm and an effective depth to the centroid of the tension reinforcement of 600mm. The tension reinforcement consists of 6-32mm diameter bars placed in two rows. Compression reinforcement consisting of 2-25mm diameter bars is placed 62.5mm from the compression face of the beam. 𝒇`𝒄 = 34.6MPa, fy = 414.7MPa. a) Determine the depth of compression block (a). b) Determine the maximum steel ratio (ρmax). c) Determine the design moment capacity of the beam (Mu). SOLUTION: (Method 1) 1. Depth of compression block (a) Check whether compression are really needed As=

𝜋(322 )(6) 4

As=

4825

300(600)

= 4825mm2 = 0.0268 fy

ρmax=

0.003+ 200000 0.008

ρb

bars

0.85fc`β(600) fy(600 + fy)

𝜌𝑏 =

β= 0.85-0.05/7 (34.5-28) = 0.8 0.85(34.6)(0.8)(600) 414.5(600 + 414.5)

𝜌𝑏 =

𝜌𝑏 = 0.0335 0.003+

ρmax=

414.7 200000

0.008

(0.0335)

ρmax= 0.0212 Compression Bars is Needed ρ > ρmax Check if compression bars will yield ρ-ρ`>

0.85𝑓𝑐𝑑`𝛽 (600) 𝑓𝑦 𝑑 (600−𝑓𝑦)

0.0268-0.00545 > ( 0.02135 > 0.01954

,

ρ`=

𝜋252 (2)

= 0.00545

4(300)(600)

0.85(34.6)(0.8 )(62.5)(600) 𝑓𝑦 (600)(600−414.7)

) = 0.01954

fs` = fy (steel in compression yields)

T = C1 + C2 As fy = 0.85 𝑓`𝑐ab + As` fy 4825(414.7) = 0.85(34.6)(a)(300) + (981.75)(414.7) a = 180.64mm (Depth of compression block)

2. Maximum steel ratio (ρmax). fy

ρmax= ρmax=

0.003+ 200000 0.008 414.7 0.003+ 200000 0.008

( ρb + ρ`) (0.0335 + 0.00545)

ρmax= 0.0247 3. Design moment capacity of the beam (Mu). a = βc 180.64 = 0.817c c = 221.1mm ℇs

= C180.64

0.003 221.10

ℇs= 0.0051 > ℇy fs = fy use ∅ = 0.90

NOTE: When ℇs < 0.005, and > 0.002 Value of ∅ = 0.65 + (ℇ𝑠 − 0.002)(

250 3

)

𝑎

Mu = ∅C1(𝑑 − 2 ) + ∅C2 (d-d`) Mu =0.90 [0.85(34.6)(180.64)(300)(600 −

180.64 2

Mu = 927.9 kN.m (Method 2) Assume Steel Yields β = 0.85-0.05/7 (34.6-28) = 0.817 As2 = As’ =

𝜋(25)2 (2) 4

As1 = As – As2 =

= 625 𝜋/2 𝜋(32)2 (6) 4

-

𝜋(25)2 (2) 4

As1 = 2447 𝜋/2 C1 = T1 0.85 𝑓`𝑐 ab = As1 fy 0.85 (34.6) (a) (300) = (2447 𝜋/2) (414.7) a= 180.664mm a = βc

) + 981.75(414.7) (600-62.5)]

147.31 = 0.817 c c = 221.1mm fs =

600(d−c) 𝑐

600(600−221.1) 221.1

=

= 1028.22MPa

fs > fy tension steel yields fs =

600(221.1−62.6) 221.1

= 430.39MPa

fs’ > fy compression bars will yield 𝑎

Mu = ∅C1(𝑑 − 2 ) + ∅C2 (d-d`) Mu = 0.90 [0.85(34.6)(180.64)(300)(600 −

180.64 2

) + 981.75(414.7) (600-62.5)]

Mu = 927.9 kN.m

PROBLEM 2 (DOUBLY REINFORCED COMRESSION BARS ARE NOT NEEDED) A simply supported beam is reinforced with 4896 mm2 at the bottom and 1530 mm2 at the top of the beam. Steel covering to the centroid of reinforced is 62.5 mm at the top and bottom of the beam. The beam has a total depth of 625mm and a width of 350 mm. 𝑓`𝑐 = 34.6 MPa, fy = 414.7 MPa. Balanced steel ratio ρb = 0.03353.

1. Determine the depth of compression block (a). 2. Determine the design strength using 0.9 as reduction factor. 3. Determine the live load at the mid-span in additional to a DL = 100 KN/m including the weight of the beam if it has a span of 6m. SOLUTION: Depth of compression block (a). Check whether compression bars are really needed

As= 4896 mm2 4896

ρ=350(562.5) = 0.0249 As`=1530 mm2 ρ`=

1530 350(562.5)

= 0.0078

fy

0.003+200000

ρmax=

0.008

ρb

414.7

0.003+ 200000

ρmax=

0.008

(0.03353)

ρmax= 0.0213 SINCE ρ= 0.0249 > ρmax = 0.0213, Therefore, compression bars are needed. Check if compression bars will yield β = 0.85-0.05/7 (34.6-28) = 0.80 Assume steel yields T= 0.85fc’ ab 0.85+C2 As fy = 0.85𝑓`𝑐 ab +As2(fy) 4896 (414.7) =0.85 (34.6) (a) (350) + 1530(414.7) a= 135.608mm c = 169.51mm fs =

600(𝑑−c) 𝑐

=

600(562.5−169.51) 169.51

fs = 1391.033MPa > fy εy =

1391.033 200000

steel yields

= 0.00696

Ø=0.90 𝑎

Mu = Ø 0.85 𝑓`𝑐 ab (d-2 )+ Ø As2fy(d-d’) Mu = 0.90(0.85) (34.6)(135.608)(350) (562.5 Mu = 622.172 kN.m Concentrated live load at mid-span 622.172 =1.2

100(62 ) 8

+1.6

𝑃 (6) 4

P = 34.238 KN

PROBLEM 3. (BALANCE CONDITION)

135.608 )+0.90(1530)(562.5-62.5) 2

A reinforced concrete beam has a width of 250mm and an effective depth 625 mm. It is reinforced for compression having a steel area As’= 1250mm2 with a steel covering of 62.5 mm measured from the center of the steel reinforcement to the top most fibers of the beam. 𝑓`𝑐= 20.68 MPa, fy = 400 MPa, Es=200,000 MPa. 1. Determine the depth of compression block for a balanced condition. 2. Determine the area of balanced steel Asb for the given cross section. 3. Determine the maximum area of flexural steel in tension permitted in the given cross section as section as required by the NSCP Specification.

SOLUTION: 1. Depth of compression block for a balanced condition. fy

ℇ𝒚 = E s = 0.003 𝐶

=

400 200000

0.002 650−𝑐

= 0.002

, c = 375 mm

a= βc (0.85)(375) a = 318.75 mm

2. Area of balanced steel (Asb) ℇ𝑠 ` 0.003 = 312.5 375 ℇs` = 0.0025 > ℇy = 0.002 (steel in compression yields) fs` = fy = 400 MPa C1 + C2 = T 0.85 𝑓`𝑐 ab + As` fy = Asb fy 0.85 (20.68) (318.75) (250) + 1250 (400) = Asb (400) Asb = 4752 mm2 3. Maximum amount of tensile steel permitted: fy

0.003+ 200000

Max. As =

0.008

( ρb + ρ’f )

ρ𝑏 =

0.85𝑓`𝑐β(600) fy(600 + fy)

ρ𝑏 =

0.85(20.68)(0.85)(600) 400(600 + 400)

ρ𝑏 = 0.0224 Max. As =0.625 (0.0224+0.008) Max. As =0.019(625)(250) Max. As = 2968.75 mm2 As =

0.003+

fy 200000

0.008

Asb, (0.625)(4752)

As = 2970 mm2 > Max. As = 2968.75 mm2 Max steel area permitted = 2970 mm2

PROBLEM 4. (STEEL DOES NOT YIELD) A simply supported beam is reinforced with 2463 mm2 at the bottom and 1231.5 mm2 at the top of the beam. Steel covering to the centroid of reinforced is 70 mm at the top and bottom of the beam. The beam has a total depth of 400mm and a width of 300 mm. 𝒇`𝒄 = 30 MPa, fy = 415 MPa, ρb = 0.031. 1. Determine the depth of compression block. 2. Determine the design strength. 3. Design moment capacity of the beam using exact method. SOLUTION: 1. Depth of compression block Check whether compression bars are really needed As= 2463mm2 ρ = 2463/330(300) = 0.024 𝑏 = 0.031 415

ρmax=

0.003+200000 0.008

(0.031)

ρmax= 0.0197 Compression bars are needed

ρ> ρmax Check if compression bars will yield ρ-ρ`> ρ`=

0.85𝑓𝑐𝑑`𝛽 (600) , 𝑓𝑦 𝑑 (600−𝑓𝑦)

1231.5 (330)(300)

= 0.0124

0.0248-0.0124 > (

0.85(30)(0.85)(70)(600) ) 415 (330)(600−415)

0.0124 < 0.036 Therefore, steel does not yield β = 0.85-0.05/7 (30-28) = 0.84 T = C1 + C2 As fy = 0.85 f’c ab + As’ fs’ 2463(415) = 0.85(30)(0.84c)(300) + (1231.5)(

600(c−70) 𝑐

)

c = 114.423 mm, a = βc a = 96.115 mm (Depth of compression block) 2. Mu fs =

600(𝑐−d`)

𝜀𝑠 =

𝑐 fy 𝐸𝑠

=

=

600(114.423−70)

232.94 200000

114.423

= 232.94

= 0.00116

εs = 0.00116 < 0.002 ; use Ø = 0.65 𝑎

Mu = ∅C1(𝑑 − ) + ∅C2 (d-d`) 2 Mu =0.65 [0.85(30)( 96.115)(300)(330 − Mu = 183.229 kN-m 3. Moment capacity (using exact method)

96.115 2

) + 1231.5 (232.94) (330-70)]

2463(415) = 0.85(30)(0.84c)(300) + 1231.5[

600(𝑐−70) 𝑐

– 0.85(30)]

c = 117.479 mm a = βc a = 98.862 mm fs =

600(117.479−70) 117.479

fs = 242.489 ; ℇ = 0.00121 Ø = 0.65 Mu = 186.199 KN.m

5. Teaching and Learning Activities Solve the following problems for mastery of the procedures and specifications: PROBLEM 1. A rectangular beam has a width of 400mm and Effective depth is 600mm. The tension reinforcement consists of 3-32mm and 3-32mm diameter bars placed in two rows. Compression reinforcement consisting of 3-25mm diameter bars is placed 62.5mm from the compression face of the beam. 𝒇`𝒄= 20.7MPa, fy = 415MPa. a) Determine the depth of compression block (a). b) Determine the maximum steel ratio (ρmax). c) Determine the design moment capacity of the beam (Mu). PROBLEM 2. A simply supported beam is reinforced with 4896 mm2 at the bottom and 2 – 32 mm at the top of the beam. Steel covering to the centroid of reinforced is 62.5 mm at the top and bottom

of the beam. The beam has a total depth of 500 and a width of 300 mm. 𝒇`𝒄= 27 MPa, fy = 276 MPa. 1. Determine the depth of compression block (a). 2. Determine the design strength.

PROBLEM 3. A reinforced concrete beam has a width of 300 mm and an effective depth 600 mm. It is reinforced for compression having a steel area As`= 1200mm2 with a steel covering of 62.5 mm measured from the center of the steel reinforcement to the top most fibers of the beam. 𝒇`𝒄= 30 MPa, fy = 415 MPa, Es=200,000 MPa. 1. Determine the depth of compression block for a balanced condition. 2. Determine the area of balanced steel Asb for the given cross section. 3. Determine the maximum area of flexural steel in tension permitted in the given cross section as section as required by the NSCP Specification. PROBLEM 4. A rectangular beam reinforced for both tension and compression bars has an area of 1250 mm2 for compression bars and 4032 mm2 for tension bars. The tension bars are placed at a distance of 75 mm from the bottom of the beam while the compression bars are placed 62.5 mm from the top the beam 𝑓`𝑐 = 20 MPa, fy = 416.6MPa. Width of the beam is 350 mm with a total depth of 675mm. 1. Determine the depth of compression block. 2. Determine the ultimate moment capacity of the beam. 3. Determine the safe live concentrated load that the beam could support at its mid-span if it has a span of 6 m. Assume weight of concrete to be 23.5kN/m3.

PROBLEM 5. A doubly reinforced concrete beam has a max. effective depth of 630mm and is subjected to a factored moment of 1062 kn-m (including its own weight). 𝑓`𝑐= 27.58 MPa, fy = 413.4 MPa. Use 62.5 mm steel covering. 1. Determine the width of the beam. 2. Determine the reinforcement for compression. 3. Determine the reinforcement for tension.

6. Recommended learning materials and resources for supplementary reading.

Reinforced Concrete Design by Gillesania, Design of Concrete Structures by Nilson et. al. 7. Flexible Teaching Learning Modality (FTLM) adopted Remote Asynchronous (modules, exercises, problem sets, etc…) 8. Assessment Task In this part, students are given Self-assessment Questions (SAQs) and asked to consider broader aspects of the different topics taken up. Quizzes have been prepared for this part and can be found in the original module of the author.

9. References Besavilla, V., Reinforced Concrete Design, 2016 Esplana, Dindo Civil Engineering Review Manual, 2015 Gillesania, DIT Reinforced Concrete Design, 3rd Edition, 2015 National Structural Code of the Philippines 2015 Nilson, W. Reinforced Concrete Design, 2010...