section analysis

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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 • The yield criterion for steel reinforcement.

• Different types of stress blocks for ULS and SLS.

• Derivation of design formulae for bending of singly andDerivation of design formulae for bending of singly and doubly reinforced sections using equivalent rectangular stress block.

RC Design and Construction – HKC 2004(2nd_Edition)

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2. Analysis of Section

• The most important principles

are:-– The idealised stress-strain curves for concrete and steel are used

are used.

The distribution of strains is linear and compatible with – 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. resultant forces developed by the section must balancep y the applied loads).

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2.1 Stress-Strain Relations

• Concrete

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

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Fig. 3.8 - Stress-Strain Curve of Concrete

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

by:-cu cu cu m cu f _f _f f 0.45 0.447 = 5 . 1 0.67 = 67 . 0 _≈ γ

where 0.67 allows for the difference between the

m

γ

bending strength and cube crushing strength of concrete and γ_m = 1.5 is the partial safety factor of concrete

concrete.

Th l f 0 45 f ill b f tl d i th • The value of 0.45 f_cu will be frequently used in the

design formulae.

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Fig.2.2 Stress-Strain Curve of Steel Reinforcement

Steel Reinforcement

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

Tension s fy m γ Strain Stre s 200 kN/mm2 ε_y Compression Strain fy ε_y Compression fy m γ Note: f is in N/mm_y 2

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Design Yield Strain for High Tensile and Mild Steel

Design Yield Stress = y _y

m y f f f 87 . 0 15 . 1 = = γ Stress = E_s * ε_s

∴ Design Yield Strain ε_y =

m γ f E y s γ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y

For high-tensile steel (T),

m γ ⎝ ⎠ g ( ), f_y = 460 N/mm2_{, E = 200*10}3 _N/mm2 ε = 460/(1 15 x 200 x 103_{) = 0 00200} ε_y = 460/(1.15 x 200 x 10 ) = 0.00200 F ild t l (R) f 250 N/ 2

For mild steel (R), f_y = 250 N/mm2

ε_y = 250/(1.15 x 200 x 103_{) = 0.00109}

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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 ≥ 0 002 if ε ≥ 0.002,

then the steel yielded and the steel stress = 0.87f_y

if ε < 0.002, then the steel NOT yet yielded, the steel stress < 0.87f_y_y 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

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2.2 Distribution of Strains and Stresses across a Section

Design

Assumptions:-– Concrete cracks in the region of tensile strain.

– After cracking, all tension is carried by the reinforcement

reinforcement.

Plane sections remain plane after straining – Plane sections remain plane after straining.

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• Fig. 2.3 shows the strain diagram and the differentg g types of stress blocks of a member subjected to bending.g d' 9x εcc As' d d' Neutral Axis S= 0. 9 x As (a) (b) (c) T i l Rectangular Equivalent εst εsc Triangular Rectangular Parabolic Equivalent Rectangular

SECTION _STRAINS _{STRESS BLOCK}

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Meanings of the symbols

= Area of compression reinforcement

A_s'

A_s = 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.

= Inset of the compression reinforcement. It is the

d'

depth measured from the top of the section

(subjected to sagging moment) to the centroid of

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compression reinforcement.

Meanings of the symbols

ε_cc = compressive strain of concrete = 0.0035 for ff_cu_cu ≤ 60 MPa

ε_sc = compressive strain of compression reinf. ε_st= tensile strain of tension reinforcement. s = depth of equivalent rectangular stress block

= 0.9x 0.9x

x = depth of neutral axis.

z = lever arm = dist. between the centroids of the conc stress block and the tension reinf

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• 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 (ULS) It the design of ultimate limit state. (ULS) It represents the collapse stage.

• The equivalent rectangular stress block is a

i lifi d l i h l b li

simplified alternative to the rectangular-parabolic block. It can provide more manageable design

f l d i i d d i BS8110 &

formulae and it is adopted in BS8110 & HKC2004.

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Upper Limit of the Neutral Axis

Consider the compatibility of strains of Fig. 2.3, ε_st = ε_cc d x x − ⎛ ⎝⎜ ⎞ ⎠⎟ and ε_sc = ε_cc x ⎝ ⎠ x − d ⎛ ⎝⎜ ⎞ ⎠ ' sc cc x ⎝⎜ ⎠

By rearranging the first equation, x d

st = + 1 εst cc + 1 ε

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At ULS, the maximum compressive strain ε_cc of conc. = 0.0035 for f ≤ 60 MPa.

conc. 0.0035 for f_cu ≤ 60 MPa.

For steel with f_y = 460 N/mm2_{, the yield strain is}

0 002 0.002. d ∴ = 0.636d 0035 0 0020 . 0 1+ = d x

Hence to ensure yielding of the tension steel at ULS

0035 .

0

Hence to ensure yielding of the tension steel at ULS, x > 0.636d

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To ensure the tension steel yielding, HKC2004 limits the depth of neutral axis such that

x ≤ (β_b - 0.4)d for f_cu ≤ 45 MPa x ≤ (β(β_b_b - 0.5)d) for 45 < ff_cu_cu ≤ 70 MPa

where β_b= moment at the section after redistribution where β_b

Thus with moment redistribution not greater than 10%, i.e. β_b ≥

moment at the section before redistribution

Thus with moment redistribution not greater than 10%, i.e. β_b ≥ 0.9,

x ≤ 0.5d for ff_cu_cu ≤ 45 MPa

x ≤ 0.4d for 45 < f_cu ≤ 70 MPa

x ≤ 0 33d for 70 < f ≤ 100 MPa itho t moment redistrib tion x ≤ 0.33d for 70 < f_cu ≤ 100 MPa without moment redistribution

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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 f ≤ 45 MPa ) The centroid of this stress 0.9x (for f_cu ≤ 45 MPa ). The centroid of this stress block is s/2 = 0.45x from the top of the section. This location is very close to the centroid of they rectangular-parabolic stress block. Furthermore the the areas of the two types of the stress block are

i t l l Thi i li th t th

approximately equal. This implies that the

compressive forces produced by these two stress blocks are the same

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blocks are the same.

x εcc b =0.0035 0.67 fcu γ_m = 0.45fcu d Neutral_Axis S=0.9 x x Fcc S/2 As Axis εst Fst z = la* d εst

SECTION STRAINS STRESS BLOCK

Fig. 2.4 Singly reinforced section with rectangular stress block

( s = 0 9x for f ≤ 45 MPa ) ( s = 0.9x for f_cu≤ 45 MPa )

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

(Section with tension steel only)

Refer to Fig. 2.4

F = tensile force in the reinforcing steel F_st = tensile force in the reinforcing steel

= 0.87f_y * A_s

F_cc = resultant compressive force in concrete = stress * area = 0.45f_cu_cu* (b*s)

For equilibrium the ultimate design moment M must For equilibrium, the ultimate design moment M must be balanced by the moment of resistance of the section

section.

M = F_cc⋅z = F_st⋅z, z -lever arm

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and z = d - s/2 ⇒ s = 2(d - z)

2.4 Singly Reinforced Rectangular Section in Bending

Take moment about the centroid of tension steel,

∴ M = 0 45f *bs*z

∴ M = 0.45f_cu bs z

= 0.45b⋅2(d - z)z⋅f_cu = 0.9f_cu⋅b(d - z)z Divide the above equation by (bd2f_cu) and let K = M/(bd2_{f )}

and let K = M/(bd f_cu),

2

The equation becomes, (z/d)2 _{- (z/d) + K/0.9 = 0}

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2.4 Singly Reinforced Rectangular Section in Bending

Take moment about the centroid of the stress block of concrete, of concrete, ∴ M = F_st* z F M/ F_st = M/z Also F_st = (f_y / γ_m)⋅A_s With γ_m = 1.15, F_st = 0.87f_y* A_s With γ_m 1.15, F_st 0.87f_y A_s ∴ A

M

∴ A_s =

z

f

_y

⋅

87 .

0

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

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

table as shown below.

F l

(

)

[

]

z d

[

0 5+

(

0 25 K / 0 9

)

]

_Formula z = d 0 5. + 0 25. − K / .0 9 Table: 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

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Limits of Lever Arm z (β_b ≥ 0.9)

• The upper limit of the lever arm curve z = 0 95d • The upper limit of the lever-arm curve, z = 0.95d,

is specified by HKC2004. The lower limit of z = 0 775d is when the depth of neutral axis x = d/2 ( 0.775d is when the depth of neutral axis x = d/2 ( for f_cu ≤ 45 MPa ), which is the maximum value allowed by the code for a singly reinforced section allowed by the code for a singly reinforced section in order to provide a ductile section which will have a gradual tension type failure

have a gradual tension type failure.

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2.4 Singly Reinforced Rectangular Section in Bendingg y g g for f_cu≤ 45 MPa

The max. value of x = 0.5d

When x = 0.5d, z =d - s/2 = d - 0.9x/2 = 0.775d, M =0.45f_cu*bs* z

= 0 45f *b*(0 9*0 5d)(0 775d)0.45f_cu b (0.9 0.5d)(0.775d) ⇒ M = 0.156f_cubd2

(Thi h f i f

(This represents the max. moment of resistance of a singly reinforced section, i.e. without compression

i f )

reinforcement)

The coefficient 0.156 is calculated from the more precise concrete stress block.

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

b 0.0035 0.45fcu As' d'_{x = d/2} _S=0.9x b Fsc F s d Neutral_Axis S x d/2

ε

sc Fcc z As

_ε

st

ε

sc Fst

SECTION STRAINS STRESS BLOCK

Fig. 2.5 Section with compression reinforcement

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2.5 Doubly Reinforced Rectangular Section When M > 0.156f_cubd2_,

cu

the design ultimate moment exceeds the moment of resistance of the singly reinforced section (i.e. resistance of the singly reinforced section (i.e. exceeds the compression resistance of conc.)

∴ compression reinforcement is required to provide additional compressive force.dd o co p ess ve o ce.

z = d - s/2 = d - 0.9 * x/2

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2.5 Doubly Reinforced Rectangular Section For equilibrium (see fig. 2.5) ,

y g F_st = F_cc + F_sc ∴ ' 87 0 45 0 87 0 f A f b + f A ∴ and with s = 0.9*d/2 = 0.45d 87 . 0 45 . 0 87 . 0 f_yA_s = f_cubs+ f_yA_s ' 87 . 0 201 . 0 87 . 0 f_yA_s = f_cubd + f_yA_s

Take moment at the centroid of the tension steel A_s,

(

'

)

d d F z F M = +

(

−

)

(

)

(

)

(

)

87 . 0 775 . 0 201 . 0 = f bd d f A' d d' d d F z F M s y cu sc cc − + + =

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(

)

87 . 0 156 . 0 = f_cubd2 + f_yA_s' d −d'

2.5 Doubly Reinforced Rectangular Section

∴ 0.156 ' 2 ' M f bd A_s = − cu d 0 156 ' i h 0 775d 2 bd f ) ( 87 . 0 f_y d d' s − and with z = 0.775d 87 . 0 156 . 0 _' s y cu s A z f bd f A + ⋅ =

The value 0.156 is usually denoted by . K'

If the value of for the section ≤ 0.43, the compression stress shall be taken as f_sc = 0.87f_y

( '/ )d x

sc y

If the value of for the section > 0.43, the compression stress shall be taken as f_sc = E_s*ε_sc.

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2.6 MOMENT REDISTRIBUTION

• For details of moment redistribution consult • For details of moment redistribution, consult

relevant reference book.

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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 f_cu ≤45 MPa? MPa? Choices: (a) 0.95d (a) 0.95d (b) 0.90d (c) 0.775d ( )

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Q3. What is the maximum depth of neutral axis x according to HKC2004 for f_cu ≤ 45 MPa ? Choices: (a) 0.5d (b) 0 636d (b) 0.636d (c) 0.775d

Q4. If K ≤ 0.156 (f_cu ≤ 45 MPa ), state whether the beam section is a singly or doubly reinforced section

section is a singly or doubly reinforced section. Choices:

(a) Singly reinforced section (a) Singly reinforced section (b) Doubly reinforced section

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Q5. Determine the lever arm z by formula or design table. Given that M = 350 kNm, b = 350 mm, d = 480 mm and f_cu = 35 N/mm2_. Choices: (a) 380 mm (b) 400 mm (c) 450 mm

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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

t l steel. (Answer)

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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

d i t l

and compression steel. (Answer)

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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 f_cu = 35 N/mm2_.

(Answer: A_s = 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 d’ = 70 mm and f = 35 N/mm2

480 mm, d = 70 mm and f_cu = 35 N/mm2_. (Answer: A_sc = 363 mm2_{, A}

s = 3320 mm2)

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Assignment No. 2

AQ3 In a doubly reinforced section, it is found that = 0.45. Determine the compressive strain ε_sc of the

(

d x'/

)

compression reinforcement. Hence calculate the design compressive stress of the compression

i f t

reinforcement.

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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

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Maximum size of aggregate: 20 mm

Assignment No. 2 12 10 mm 3T32 500 600 12 mm Link 10 mm Link 3T32 Fig. AQ4a 2T40 +1T20 Fi AQ4b Fig. AQ4b 600 12 mm Link 2T32 4T32

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Assignment No. 2

AQ5. A beam has a section 250 mm wide with an effective depth of 400 mm. Characteristic strengths are: concrete 40 N/mm2_{, steel 460 N/mm}2_{. Given that the} ultimate moment of resistance is to be 225 kNm, show

th t th b b i l i f d d d t i

that the beam can be singly reinforced and determine:

( ) th l

(a) the lever arm

(b) the depth of the neutral axis

( ) h f i f

(c) the area of reinforcement

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Assignment No. 2

AQ6. The design moment for the beam section shown in Fig. AQ6 is 250 kNm. Given that f_cu = 35 N/mm2 _{and f}

y = y 460 N/mm2_{. Determine:}

(a) The lever arm z by using formula. (b) The lever arm z by design table. (c) The area of steel reinforcement.

(d) The size of the steel reinforcement to be provided. (e) The percentage of steel reinforcement.

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Assignment No. 2

300

500

550

Fig. AQ6

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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. 250

450

3T25

Fig AQ7 Fig. AQ7

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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

i l i f d ti i

singly reinforced section assuming no

redistribution,

(b) suitable reinforcement (assuming = 0.10) to provide a moment of resistance of 1300 kNm with

d d'/

provide a moment of resistance of 1300 kNm with no redistribution.