section analysis

2. Analysis of Section • This chapter will discuss the following topics: • The idealised design stress-strain curves of concrete and steel reinforcement. • The yield criterion for steel reinforcement. reinforcement • Different types of stress blocks for ULS and SLS. • Derivation of design formulae for bending of singly and doubly reinforced sections using equivalent rectangular stress block.

2-1 RC Design and Construction – HKC

2004(2nd

Edition)

2. Analysis of Section • The most important principles are:– The idealised stress-strain curves for concrete and steel are used. used – The distribution of strains is linear and compatible with the distorted shape of the cross-section. – The section shall satisfy the static equilibrium (i.e. p byy the section must balance resultant forces developed the applied loads).

2-2 RC Design and Construction – HKC 2004(2nd Edition)

2.1

Stress-Strain Relations

• Concrete – A parabolic stress-strain curve is used to represent the behaviour of concrete when subjected to loading. loading (see Fig. 3.8). When the strain reaches εo , it is noted that the strain increases while the stress remains constant. The value of εo depends on the concrete grade (i.e. the fcu)

2-3 RC Design and Construction – HKC

2004(2nd

Edition)

Fig. 3.8 - Stress-Strain Curve of Concrete

Reproduce from HK Code 2-4 RC Design and Construction – HKC 2004(2nd Edition)

• From Fig. 3.8, the ultimate strain = 0.0035 is t i l for typical f fcu ≤ 60 MPa. MP Th ultimate The lti t design d i stress is given by:0.67 f cu

γm

=

0.67 f cu = 0.447 f cu ≈ 0.45 f cu 1.5

where 0.67 allows for the difference between the bending strength and cube crushing strength of concrete and γm = 1.5 is the partial safety factor of concrete. concrete • Th The value l off 0.45 0 45 fcu will ill be b frequently f tl usedd in i the th design formulae. 2-5 RC Design and Construction – HKC

2004(2nd

Edition)

Fig.2.2 Stress-Strain Curve of Steel Reinforcement Steel Reinforcement – Fig Fig. 2.2 2 2 shows a typical short short-term term design stress stress-strain strain curve of reinforcement. The behaviour of steel is the same for both tension and compression. fy

Tension

Stress

γm

εy

200 kN/mm2 εy

Compression

Strain

fy

γm Note: fy is in N/mm2

Fig. 2.2 Short term design stress-strain curve for reinforcement 2-6 RC Design and Construction – HKC 2004(2nd Edition)

Design Yield Strain for High Tensile and Mild Steel Design Yield Stress =

fy

γm

=

fy 1.15

= 0.87 f y

Stress = Es * εs ⎛ fy ⎞ ∴ Design Yield Strain εy = ⎜ ⎟ Es ⎝γ m⎠ For high-tensile g steel ((T), ), fy = 460 N/mm2, E = 200*103 N/mm2 εy = 460/(1.15 460/(1 15 x 200 x 103) = 0.00200 0 00200 2 For mild F ild steel t l (R), (R) fy = 250 N/mm N/ εy = 250/(1.15 x 200 x 103) = 0.00109 2-7 RC Design and Construction – HKC

2004(2nd

Edition)

Design Yield Strain for High Tensile and Mild Steel • The value of the design yield strain would be used to determine the design stress of the steel reinforcement. • Take high tensile steel as an example: if ε ≥ 0.002, 0 002 then the steel yielded and the steel stress = 0.87fy if ε < 0.002, then the steel NOT yet yielded, the steel stress < 0.87fy and the steel stress has to be determined from the stress-strain curve of steel reinforcement. Say if ε = 0.0015, then steel stress = ε*E = 0.0015*(200*103) = 300 N/mm2 2-8 RC Design and Construction – HKC 2004(2nd Edition)

2.2 Distribution of Strains and Stresses across a Section

Design Assumptions:– Concrete cracks in the region of tensile strain. – After cracking, reinforcement. reinforcement

all

tension

is

carried

by

the

– Plane sections remain plane after straining. straining

2-9 RC Design and Construction – HKC

2004(2nd

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• Fig. g 2.3 shows the strain diagram g and the different types of stress blocks of a member subjected to bending. g

As' d As

SECTION

d'

S=0.99x

εcc x

Neutral Axis

εsc εst

(c) (b) (a) Rectangular Equivalent Ti Triangular l Rectangular Parabolic

STRAINS

STRESS BLOCK

Fig. 2.3 Section with strain diagram and stress blocks 2-10 RC Design and Construction – HKC 2004(2nd Edition)

Meanings of the symbols As' = Area of compression reinforcement

As = Area of tension reinforcement d = effective depth of the section. It is the depth measured from the top of the section (subjected to sagging moment) to the centroid of tension reinforcement.

d ' = Inset of the compression reinforcement. It is the depth measured from the top of the section (subjected to sagging moment) to the centroid of compression reinforcement. 2-11 RC Design and Construction – HKC

2004(2nd

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Meanings of the symbols εcc = compressive strain of concrete = 0.0035 for fcu ≤ 60 MPa εsc = compressive strain of compression reinf. εst = tensile strain of tension reinforcement. s = depth of equivalent rectangular stress block = 0.9x x = depth of neutral axis. z = lever arm = dist. between the centroids of the conc stress block and the tension reinf. conc. reinf 2-12 RC Design and Construction – HKC 2004(2nd Edition)

• The triangular stress block is used in the design of the serviceability limit state. (SLS) • The rectangular parabolic stress block is used in the design of ultimate limit state. state (ULS) It represents the collapse stage. • The equivalent rectangular stress block is a simplified i lifi d alternative l i to the h rectangular-parabolic l b li block. It can provide more manageable design f formulae l andd it i is i adopted d d in i BS8110 & HKC2004. 2-13 RC Design and Construction – HKC

2004(2nd

Edition)

Upper Limit of the Neutral Axis Consider the compatibility of strains of Fig. 2.3, εst = εcc ⎜⎛ d − x ⎞⎟ ⎝

and

x ⎠

εsc = εcc ⎛⎜ x − d ' ⎞ ⎝

x

⎠

By rearranging the first equation, x =

d 1+

ε st ε cc 2-14

RC Design and Construction – HKC 2004(2nd Edition)

Upper Limit of the Neutral Axis At ULS, the maximum compressive strain εcc of conc. = 0.0035 for fcu ≤ 60 MPa. For steel with fy = 460 N/mm2, the yield strain is 0 002 0.002. x=

∴

d 0.0020 1+ 0.0035

= 0.636d

Hence to ensure yielding of the tension steel at ULS, ULS x > 0.636d 2-15 RC Design and Construction – HKC

2004(2nd

Edition)

Upper Limit of the Neutral Axis To ensure the tension steel yielding, HKC2004 limits the depth of neutral axis such that x ≤ (βb - 0.4)d

for fcu ≤ 45 MPa

x ≤ (βb - 0.5)d )

for 45 < fcu ≤ 70 MPa

where βb=

moment at the section after redistribution moment at the section before redistribution

Thus with moment redistribution not greater than 10%, i.e. βb ≥ 0.9, x ≤ 0.5d

for fcu ≤ 45 MPa

x ≤ 0.4d

for 45 < fcu ≤ 70 MPa

x ≤ 0.33d 0 33d for 70 < fcu ≤ 100 MPa without itho t moment redistribution redistrib tion 2-16 RC Design and Construction – HKC 2004(2nd Edition)

2.3

Equivalent Rectangular Stress Block

• The equivalent rectangular stress block shown in Fig. 2.4 is easier to use and provide more manageable design formulae. It is used as an alternative to the more accurate rectangular-parabolic stress block. • It is noted that the stress block extends to a depth s = 0 9x (for fcu ≤ 45 MPa ). The centroid of this stress 0.9x block is s/2 = 0.45x from the top of the section. This location is very y close to the centroid of the rectangular-parabolic stress block. Furthermore the the areas of the two types of the stress block are approximately i t l equal. l Thi implies This i li th t the that th compressive forces produced by these two stress blocks are the same. same 2-17 RC Design and Construction – HKC

2004(2nd

Edition)

S=0.9xx

b x Neutral Axis

d

0.67 fcu γm = 0.45fcu

εcc=0.0035

S/2 Fcc z = la* d

As

SECTION

εst

Fst

STRAINS

STRESS BLOCK

Fig. 2.4 Singly reinforced section with rectangular stress block ( s = 00.9x 9x for fcu ≤ 45 MPa )

( s = 0.8x for 45 < fcu ≤ 70 MPa ) ( s = 0.72x for 70 < fcu ≤ 100 MPa ) RC Design and Construction – HKC 2004(2nd Edition)

2-18

2.4 Singly Reinforced Rectangular Section in Bending (Section with tension steel only)

Refer to Fig. 2.4 Fst = tensile force in the reinforcing steel = 0.87fy * As Fcc = resultant compressive force in concrete = stress * area = 0.45fcu* (b*s) For equilibrium, equilibrium the ultimate design moment M must be balanced by the moment of resistance of the section. section M = Fcc⋅z = Fst⋅z, z -lever arm and z = d - s/2 ⇒ s = 2(d - z) 2-19 RC Design and Construction – HKC

2004(2nd

Edition)

2.4 Singly Reinforced Rectangular Section in Bending

Take moment about the centroid of tension steel, ∴ M = 0 45fcu*bs*z 0.45f bs z = 0.45b⋅2(d - z)z⋅fcu = 0.9fcu⋅b(d - z)z Divide the above equation by (bd2fcu) and let K = M/(bd2fcu), ) The equation becomes, ∴ z = d 0.5 +

[

(z/d)2 - (z/d) + K/0.9 = 0

( 0.25 − K / 0.9) ]

2-20 RC Design and Construction – HKC 2004(2nd Edition)

2.4 Singly Reinforced Rectangular Section in Bending

Take moment about the centroid of the stress block of concrete, ∴ M = Fst* z Fst = M/z M/ Also Fst = (fy / γm)⋅As With γm = 1.15, Fst = 0.87fy * As ∴

As =

M 0.87 f y ⋅ z 2-21

RC Design and Construction – HKC

2004(2nd

Edition)

Limits of Lever Arm z (βb ≥ 0.9) for fcu ≤ 45 MPa )

• The lever arm z can be found by using formula or table as shown below.

[

z = d 0.5 +

( 0.25 − K / 0.9) ]

F Formula l

Table: K= M/bd2fcu

≤ 0.043

0.05

0.06

0.07

0.08

0.09

0.10

0.11

0.12

0.13

0.14

0.15

≥ 0.156

la = z/d

0.950

0.941

0.928

0.915

0.901

0.887

0.873

0.857

0.842

0.825

0.807

0.789

0.775

2-22 RC Design and Construction – HKC 2004(2nd Edition)

Limits of Lever Arm z (βb ≥ 0.9) • The upper limit of the lever lever-arm arm curve, curve z = 00.95d, 95d is specified by HKC2004. The lower limit of z = 0 775d is when the depth of neutral axis x = d/2 ( 0.775d for fcu ≤ 45 MPa ), which is the maximum value allowed by the code for a singly reinforced section in order to provide a ductile section which will have a gradual tension type failure. failure

2-23 RC Design and Construction – HKC

2004(2nd

Edition)

2.4 Singly g y Reinforced Rectangular g Section in Bending g for fcu ≤ 45 MPa

The max. value of x = 0.5d When x = 0.5d, z =d - s/2 = d - 0.9x/2 = 0.775d, M =0.45fcu*bs* z 0 45fcu*b*(0 b (0.99*00.5d)(0.775d) 5d)(0 775d) = 0.45f ⇒ M = 0.156fcubd2 (Thi represents the (This h max. moment off resistance i off a singly reinforced section, i.e. without compression reinforcement) i f ) The coefficient 0.156 is calculated from the more precise concrete stress block. 2-24 RC Design and Construction – HKC 2004(2nd Edition)

2.5 Doubly Reinforced Rectangular Section for fcu ≤ 45 MPa (Section with both tension and compression reinforcement)

As' d

d' x = d/2 Neutral Axis

As

SECTION

0.45fcu S S=0.9x

0.0035

b

Fsc F Fcc z

εsc

εst STRAINS

Fst STRESS BLOCK

Fig. 2.5 Section with compression reinforcement 2-25 RC Design and Construction – HKC

2004(2nd

Edition)

2.5 Doubly Reinforced Rectangular Section When M > 0.156fcubd2, the design ultimate moment exceeds the moment of resistance of the singly reinforced section (i.e. exceeds the compression resistance of conc.) ∴ compression reinforcement is required to provide dd o co compressive p ess ve force. o ce. additional z = d - s/2

= d - 0.9 * x/2 = d - 0.9⋅* 0.5d/2 = 0.775d 2-26

RC Design and Construction – HKC 2004(2nd Edition)

2.5 Doublyy Reinforced Rectangular g Section For equilibrium (see fig. 2.5) , Fst = Fcc + Fsc ∴

0.87 f y As = 0.45 f cu bs b + 0.87 f y As'

and with

s = 0.9*d/2 = 0.45d 0.87 f y As = 0.201 f cu bd + 0.87 f y As'

Take moment at the centroid of the tension steel As,

(

M = Fcc z + Fsc d − d '

)

( )

= 0.201 f cu bd (0.775d ) + 0.87 f y As' d − d '

(

= 0.156 f cu bd 2 + 0.87 f y As' d − d '

) 2-27

RC Design and Construction – HKC

2004(2nd

Edition)

2.5 Doubly Reinforced Rectangular Section ∴

and d

As'

=

M − 0.156 f cu bd 2 0.87 f y ( d − d ' )

0 .156 f cu bd 2 As = + As' 0 .87 f y ⋅ z

with i h z = 00.775d 775d

The value 0.156 is usually denoted by K'. If the value of ( d '/ x ) for the section ≤ 0.43, the compression stress shall be taken as fsc = 0.87fy If the value of ( d '/ x ) for the section > 0.43, the compression stress shall be taken as fsc = Es*εsc. 2-28 RC Design and Construction – HKC 2004(2nd Edition)

2.6

MOMENT REDISTRIBUTION

• For details of moment redistribution, redistribution consult relevant reference book.

2-29 RC Design and Construction – HKC

2004(2nd

Edition)

Self-Assessment Questions Q1.

What is the upper limit of the lever arm z? Choices: (a) 0.95d (b) 0.90d (c) 0.775d

Q2.

What is the lower limit of the lever arm z for fcu ≤45 MPa? Choices: (a) 0.95d (b) 0.90d ((c)) 0.775d 2-30

RC Design and Construction – HKC 2004(2nd Edition)

Self-Assessment Questions Q3.

What is the maximum depth of neutral axis x according to HKC2004 for fcu ≤ 45 MPa ? Choices: (a) 0.5d (b) 0 636d 0.636d (c) 0.775d

Q4.

If K ≤ 0.156 (fcu ≤ 45 MPa ), state whether the beam section is a singly or doubly reinforced section. section Choices: (a) Singly reinforced section (b) Doubly reinforced section

2-31 RC Design and Construction – HKC

2004(2nd

Edition)

Self-Assessment Questions Q5.

Determine the lever arm z by formula or design table. Given that M = 350 kNm, b = 350 mm, d = 480 mm and fcu = 35 N/mm2. Choices: (a) 380 mm (b) 400 mm (c) 450 mm

2-32 RC Design and Construction – HKC 2004(2nd Edition)

Self-Assessment Questions Q6.

Draw an annotated diagram of a singly reinforced section showing the strain & stress distribution at ultimate limit state. Based on the diagram drawn, derive the formula for calculating the area of tension steel. t l (Answer)

2-33 RC Design and Construction – HKC

2004(2nd

Edition)

Self-Assessment Questions Q7.

Draw an annotated diagram of a doubly reinforced section showing the strain & stress distribution at ultimate limit state. Based on the diagram drawn, derive the formulae for calculating the area of tension andd compression i steel. t l (Answer)

2-34 RC Design and Construction – HKC 2004(2nd Edition)

Assignment No. 2 AQ1 Determine the area of tension steel required of the singly reinforced section. Given that M = 350 kNm, b = 350 mm, d = 480 mm and fcu = 35 N/mm2. (Answer: As = 2186 mm2) AQ2 Determine the areas of tension and compression reinforcement required of the doubly reinforced section. Given that M = 500 kNm, b = 350 mm, d = 480 mm, mm dd’ = 70 mm and fcu = 35 N/mm2. (Answer: Asc = 363 mm2, As = 3320 mm2)

2-35 RC Design and Construction – HKC

2004(2nd

Edition)

Assignment No. 2 AQ3 In a doubly reinforced section, it is found that ( d '/ x ) = 0.45. Determine the compressive strain εsc of the compression reinforcement. Hence calculate the design compressive stress of the compression reinforcement. i f t (Answer: εsc = 0.001925, fsc = 385 N/mm2)

2-36 RC Design and Construction – HKC 2004(2nd Edition)

Assignment No. 2 AQ4 Determine the effective depths of the beam sections as shown in Fig. AQ4a, AQ4b & AQ4c. (a)

Nominal cover: 40 mm Link size: 10 mm

(b)

Cover to main reinforcement: 50 mm Link size: 12 mm.

((c))

Nominal cover: 30 mm Link size: 12 mm Maximum size of aggregate: 20 mm 2-37

RC Design and Construction – HKC

2004(2nd

Edition)

Assignment No. 2

12 mm Link 600

500

10 mm Link

3T32 2T40 +1T20

Fig. AQ4a

12 mm Link

600

Fi AQ4b Fig.

2T32 4T32

Fig. AQ4c

2-38 RC Design and Construction – HKC 2004(2nd Edition)

Assignment No. 2 AQ5. A beam has a section 250 mm wide with an effective depth of 400 mm. Characteristic strengths are: 2 concrete 40 N/mm , steel 460 N/mm2. Given that the ultimate moment of resistance is to be 225 kNm, show th t the that th beam b can be b singly i l reinforced i f d andd determine: d t i ((a)) (b) ( ) (c)

the lever th l arm the depth of the neutral axis the h area off reinforcement i f

2-39 RC Design and Construction – HKC

2004(2nd

Edition)

Assignment No. 2 AQ6. The design moment for the beam section shown in Fig. AQ6 is 250 kNm. Given that fcu = 35 N/mm2 and fy = 460 N/mm2. Determine: (a) (b) (c) (d) (e)

The lever arm z by using formula. The lever arm z by design table. The area of steel reinforcement. The size of the steel reinforcement to be provided. The percentage of steel reinforcement. Does it meet the code’s requirements?

2-40 RC Design and Construction – HKC 2004(2nd Edition)

Assignment No. 2

550

500

300

Fig. AQ6

2-41 RC Design and Construction – HKC

2004(2nd

Edition)

Assignment No. 2 AQ7 A singly reinforced beam has the section in Fig. AQ7. Given that the concrete grade is C35, nominal cover is 30 mm and the size of link is 10 mm. Determine: (a) the effective depth (b) the depth of the neutral axis (c) the ultimate moment of resistance.

450

250

3T25

Fig AQ7 Fig. 2-42 RC Design and Construction – HKC 2004(2nd Edition)

Assignment No. 2 AQ8 A beam has width 400 mm, effective depth 700 mm , is made from C35 concrete, and uses high-yield reinforcement. Determine: (a) the maximum moment of resistance and the corresponding area of tension reinforcement for the singly i l reinforced i f d section ti assuming i no redistribution, (b) suitable reinforcement (assuming d '/d = 0.10) to provide a moment of resistance of 1300 kNm with no redistribution.

2-43 RC Design and Construction – HKC

2004(2nd

Edition)