Chapter 2-Doubly Reinforced Beam

 

CE 433 – REINFOR REINFORCED CED CONCRETE DESIGN LECTURE #2

DOUBLY REINFORCED BEAM Melchor M. Famisan, D. Eng. Instructor

 

INTENDED LEARNING OUTCOMES  At the the end of the lesson, lesson, the students shall have have the ability to:  Perform exural analysis of a doubly reinforced reinforced beam.  Describe the procedure of designing doubly reinforced beam. a doubly reinforced beam in accordance with NSCP 2015.

 Design

 

TOPIC TOPIC OUTLINE OUTLINE 1. Flexural Analysis of Doubly Reinforced Beam  

Case 1: Compression steel yields

Case 2: Compression steel does not yield 2. Design of Doubly Reinforced Beam

 

FLEXURAL ANALYSIS OF DOUBLY REINFORCED BEAM Consider the beam section sectio n as shown.

C s = f’  f ’ s A’   A’ s

ϵc = 0.003

d' 

 A’   A’ S 

a

d' 

ϵs’ 

c

c – d’ 

a/2 C c = 0.85f’ c ab

d  d–c

d – d’ 

d – a/2 T = f s As

ϵs

Strain diagram

ϵs = (ϵ y = f  y/E S  ) Stress diagram

 

 ∑F  x = 0 C c + C s = T  0.85’ c ab +  ’s  A  A’ ’ s =   y As

To ensure ductility, the tension steel must yield rst before the concrete reaches a strain of 0.003. Hence,  f s = f  y .

 Assume frst that the the compression steel (CS) yields, i.e., ’ ’ s =   y  CASE 1: Compression steel steel yields a =   y (   As – A  A’ ’ s )/( 0.85’  0.85’ c b) c = a/β 1 Check the strain of CS to verify the assumption ϵ   /(c – d’) = 0.003/c

C s = f’  f ’ s As’ 

 s’

ϵ  s’ = (0.003/c)(c – d’) CS does not yield. Go to case 2.  If ϵ  s’ < (ϵ  y = f  y /E  s ), CS yields (so  ’    ’ s =   y ).  ). Proceed   If ϵ  s’ > (ϵ  y = f  y /E  s ), as follows

d' 

d – d’ 

Take moment tension + C  (d – d’)steel  C (dabout – a/2)the c

d – a/2 T = f  y As

s

 M n =  M n = 0.85 0.85’ ’ c ab(d – a/2) +   y A  A’ ’ s (d – d’)

a/2 C c = 0.85f 0.85f ’ c ab

ϵs = (ϵ y = f  y/E S  )

c ab(d y A s (d – d’)] ≥th φM n = φ[0.8 5f’ ’ o – a/2) + f   yto  A’ ’ determine M   Also solve theφ[0.85f strain tension steel steel the eu reduction actor φ.  φ. 

 

CASE 2: Compression steel does not yield (i.e., f  f’ ’ s ≠ f  y)      ’ ’ s = ϵ  s’  E  s = (0.003/c)(c – d’)(200,000) d’)(200,000)  

= (600/c)(c – d’)

 ∑F  x = 0 C c + C s = T   0.85’ c ab + ’   ’ s A  A’ ’ s =   y As 0.85’ c (  β 1c)b + (600/c)(c – d’) A  A’ ’ s =   y As

equation 1

Solve or c rom equation 1 , then calculate a and ’ s . a = β 1c &  ’    ’ s = (600/c)(c – d’) Take moment about the tension steel   M n = C c(d – a/2 a/2)) + C s(d – d’)  M n = 0.85’ c ab(d – a/2) + ’   ’ s A  A’ ’ s (d – d’)

 Also solve the the strain o tension st steel eel to determine the the reduction actor φ φ..  φM n = φ[0.85f φ[0.85f’ ’ c ab(d – a/2) a/2) + f’  f ’ s A  A’ ’ s (d – d’)] ≥ M u

 

SAMPLE PROBLEM #1 Compute the design moment strength of the beam as shown. Assume    ’ ’ c  = 30 MPa and and      y = 415 MPa.

65 mm 2 – 20 mm φ 675 mm 4 – 32 mm φ 75 mm 350 mm

 

 

SOLUTION:

 Assume that bot both h the com compressio pression n steel (CS) (CS) and tension tension st steel eel (TS) yield, i.e., ’   ’ s =  s =   y

Solve the strain of CS and TS to verify the assumption.

< ϵ y , hence CS did not yield  > ϵ y , hence TS yielded 

 

Determine the strain of compression steel and tension steel.

(OK)

(OK and it is in the tension controlled region) Compute the design strength.

 

φM n = 645.06 645.06  kN-m

 

SAMPLE PROBLEM #2 Determine the maximum concentrated live load that the beam can support as shown. Assume  Assume  ’ ’ c = 21 MPa &   &    y  = 415 MPa. Assume that the  weight of the beam is included in the dead load.

65 mm

1.5 m

2 – 25 mm φ

P L

wD = 40 kN/m wL = 50 kN/m

h = 750 mm 7 - 32 mm φ 65 mm 65 mm

b = 400 mm

L=6m

 

SOLUTION: Compute the required strength, M  strength,  M u.

1.5 m

P u = 1.6P  1.6P L wu = 128 kN/m

wu = 1.2(40) + 1.6(50) = 128 kN/m P u = 1.6 P L  kN 6m 384 + 1.2P  1.2P L Find the c.g. of T.S.

384 + 0.4P  0.4P L 384 + 1.2P  1.2P L

 y

c.g.

65 mm

192 + 1.2P  1.2P L 192 – 0.4P  0.4P L

 A  Ayy = ∑(a ∑(a · y) y)

 x

V u

4.5–x

7 y  y =  = 3(65)  y = 27.857 mm

 M u,max

– 384 – 0.4P  0.4P L

Solving for the eective depth.  M u

 

 Assume that both C.S. C.S. & TS yield, such that  415 MPa    ’ ’ s =  s =   y =  = 415

65 mm 2 – 25 mm φ

d = 657.143 mm

Solving for the depth of stress block:

h = 750 mm 7 - 32 mm φ 65 mm

c.g.

65 mm

b = 400 mm

Check the strain of compression steel and tension steel. So both CS & TS yielded > ϵ y > ϵ y

 

Solve the strain of the farthest TS to determine the strength reduction factor, φ. d t = 750 – 65 = 685 mm ϵt = (0.003/ (0.003/cc)( )(d  d t – c) = (0.003/317.835)(685 – 317.835) = 0.00346562 Since ϵ y < ϵt < 0.005, hence TS is within the transition region. φ = 0.65 + 0.25(ϵ 0.25(ϵt – ϵ y)/(0.005 – ϵ y) = 0.65 + 0.25(0.00346562 – 0.002075)/(0.005 – 0.002075) = 0.7689 Compute the design strength.

= 0.7689(1928942.4)(657.14 0.7689(1928942.4)(657.1433 – 270.16/2)(10 270.16/2)(10 -6) + 0.7689(407 0.7689(407425.297 425.297)(657.143 )(657.143 – 65)(10-6) = 959.805 kN-m Solving for PL. 959.805 =

P L = 279.23 kN

 

SAMPLE PROBLEM #3 Given the following data for doubly reinforced beam: Beam width, b = 410 mm Overall depth, h = 650 mm Diameter of tension bars = 28 mm Compression bars: 3 – 20 mm diameter

 A  A’ ’ s 650 mm  As

Compressive strength of concrete:  concrete: ’ ’ c = 20.7 Mpa  Yield strength strength of st steel: eel:      y = 345 Mpa 410 mm Clear cover of 10 mm diameter stirrups = 40 mm a. Compute the required area of tension steel at balanced design condition. b. Compute the design strength of the beam using 3 – 28 mm diameter tension bars. c. Compute the design strength of the beam using 6 – 28 mm diameter tension bars.

 

 SOLUTION: ¿ 650 − 40 − 10 − ¿ 40 + 10 +

1 2

1 2

( 28 )=586 

20 =60

(

)



[  ] 2

   ( 20 ) 2 300 =     of tension steel at balanced design a. Required area 4 ′  = 3

Note: Balanced design is a condition at which the strain of the concrete reaches 0.003 just as the tension steel reaches its yield strain. ∈    =0.003

 ′ 

′ 

∈

 



∈ =

  −  ′ 

 3-20 mm

 −

′ 

∈ =∈ ¿    



( 586 −  ¿) 

  345 200,000

Solve the strain of C.S. ∈=



0.003

 = 372.06 

d = 586 mm  As

Locate the N.A.

 

0.003

/  



0.003 ( 372.06 − 60 ) (   −  ′ ¿)  372.06

¿ 0.00252 yields

410 mm ⸫

use

 

′ 

0.85     

 

 



 3-20 mm d = 586 mm

 ′   

 − 





−′ 

2

 As

  410 mm

∑  

=

    0 :   + =   ❑ 0.85  ′   +  ′    ′ =     









 Where and

  = 0.85 ( 20.7 ) ( 0.85 × 372.06 ) ( 410 )+ 345 ( 300  )   345

2

  7555 . 286  ¿

Note: When the provided area of T.S. is more than 7555.286 mm 2, the concrete will fail rst before the the T.S.provided yields. area These condition must7555.286 be avoided as yield possible. However, when of T.S. is less than mm 2,as themuch T.S. will rst before the concrete fails.

 

b.   When 3 – 28 mm diameter bar is provided for tension steel.  Assume that C.S. DNY. DNY.

∑   =0 :❑  

 

′ 

=

   

600

 (



  +  =   ′ 

T.S. yields yi elds 600

)= −′ 

 (



60 )

−

  

′ 

′ 

0.85     +      =     

0.85 ( 20.7 ) ( 0.85  ) ( 410 ) +

600

(  − 60 ) ( 300  ) =345 ( 1847.26 )



 =0.85 ( 80.472 ) =68.4 

 = 80.472  Check the strain of C.S. ′ 

∈ =

0.003



  (  −  ′  ) ¿   0.003 ( 80.472 − 60 ) 80.472

¿ 0.000763 DNY  Solve for the nominal strength.

(

  =    −

  2

 )

+   (  −  ′  )

′ 

 Where:    =

  600 80.472

( 80.472 − 60 )=152.64  

68.4

[

¿ 0.85 20.7

(

)(

68.4

)(

410

)

(

586 −

2

 )

6

+ 152.64 300   586 − 60

(

)(

)

]

× 10−

¿ 347.947 



 

Determine the design strength. ∈ =

0.003

  (  − ¿)   0.003 ( 586 − 80.472 )



80.472

.005 , eio otrolle ∴ ∅ =0.9 ¿ 0.01885 o trolle ¿ 0 .005 ∅

  = 0.9 ( 347.947 ) ¿ 313 . 153 

∙∙ 

c. When 6 – 28 mm diameter bar is provided for tension steel.

 ( 28 )   =6  

2 2

= 3694.513 

[  ] 4

  T.S. yields

 Assume that C.S. yields.     ′ =    =345 

∑  

=

    0 :   +  =   ❑ 0.85 ′  + ′ 

′ 

=

            0.85 ( 20.7 )  ( 410 ) + 345 ( 300   )= 345 ( 3694.513 ) 





 =131.613 

=

  0.85

=154.84 

Check the strain of C.S. ∈= 0.003 ′ 



  (  −  ′  ) ¿   0.003 ( 154.84 − 60 )= 0.00184 yields 154.84

 

Solve the nominal strength.

(

  =    −

  2

)

′ 

+   (  −  ′  )

 Where:     = 345 

¿ 0.85 ( 20.7 ) ( 131.613 ) ( 410 ) 586 −  131.613 + 345 ( 300   ) ( 586 − 60 )

[

¿ 664.929 

∙∙ 

(

2

 )

Determine the design strength. 0.003 ∈ =



  0.003

  (  − ¿) 154.84 ( 586 − 154.84 ) ¿ 0.00835 ¿ 0 .005 005 , eio otrolle o trolle ∴ ∅ =0.9

∅

  = 0.9 ( 664.929 ) ¿ 598 . 436  ∙ 

]

× 10 − 6

 

DESIGN OF DOUBLY REINFORCED BEAM  When it is restricted or not allowed allowed to have a bigger sec section tion of the b beam, eam, such that when it is design as singly reinforced cannot be possible (i.e., the design strength, φM nn11 , of a singly reinforced is less than the required strength, M  strength,  M u), it is necessary then to provide compression steel in order to increase its strength. Subsequently, additional tension steel is also required to compensate the provided compression steel. ϵc = 0.003

a/2 d' 

d 

C s = f’  f ’ s A  A’ ’ s

C c = 0.85f c’ ab

a

ϵs’  c

c – d’ 

 A’   A’ s =

d–c

+

T  1= f  y As1

 As= As1 + As2

   M n

d – d’ 

d – a/2

M n1 

T  2= f  y As2 ϵs = ( Stress diagrams

f  yn/E   ) n1   M n2ϵ y= =M   - M   - M  S 

ϵs Strain diagram

 

Design Data: b, d, d, ’ ’ c ,   y and M u Design Procedure:  1.   Calculate frst the nominal strength ( M n1n1) and the corresponding area o  1. tension steel ( As1) considering the section as si singly ngly reinorced. T To o ensure ductility,, a ductility assume ssume that ϵ  s = 0.005. ϵ  s = ( 0.003/c)(d-c) 0.003/c)(d-c) = 0.005

c = ( 3/8  3/8))d  a = β 1c   ∑F  x = 0: T  1 = C c

Note: If M n1 > M u/φ , use SRB.  If M  If M n1  0.005 , tension controlled (φ = 0.9) Check the design strength. ∅

  = 0.9 ( 648.026 )= 583.22  ∙ ¿ [  = 539  ∙ ] (therefore safe)

Check the horizontal spacing of tension bars. 350 − 2 ( 40 ) − 2 (10 ) − 3 ( 28 )   h = ¿ 83  >  = 28  ∴  

[

2

Check the steel ratio.      1176   = 0.02111 =   = ( ) 

′ 

  =

  ′  

=

]

  39 392 2  

(

350 35 0 500

)

=0.00704



350 ( 500 )

 

0.02 0211 11 1 − 0.00704 ¿ 0.

(

 )

400.793 415

= 0.01431< ρmax (OK)

2 – 28 mm φ

500 mm 600 mm 6 – 28 mm φ

350 mm

Final Beam Details

 

REFERENCES 

  Association of Structural Engineers of the Philippines,  National Structural Code o the Philippines (NSCP C101,Vol. I Buildings, Towers and Other Vertical Vertical Structures), 7 th ed ., ., 2016



Darwin, D., Dolan, C. & Nilson, A. (2016). Design o Concrete

Structures, 15th ed.  ed. New  New York: McGraw-Hill Education  Hassoun, N. & Al-Manaseer, A. (2015). Structural Concrete: Theory and Design, 6th ed . New Jersey: John Wiley & Sons 

McCormac, J. & Brown, R. (2014). Design o Reinorced Concrete, 9th ed .

USA: John Wiley & Sons  Nawy, E. (2010). Prestressed Concrete, 5th ed . New Jersey: Prentice hall  Wight,

J. & MacGregor MacGregor,, J. (2012). Reinorced Concrete: Mechanics & Design, 6th ed . New Jersey: Pearson Education