Bending Beam With Single Reinforcement

BEND-Beam-SingleReinfBEND-Beam-SingleReinf 11

Assembled By : Kawan K. GhaforBuilding Construction Department

Sulaimany University


UNITS

BENDING IN BEAM

ASSUMPTIONS

STRESS-STRAIN RELATIONSHIP

FAILURE MECHANISM

STRESS DIAGRAM

MAXIMUM & MINIMUM REINFORCEMENT

SECTION PROPERTIES

TABLES

EXERCISES


UNITS :

LENGTH Span : m (meter) Element dimension : mm (millimeter)

FORCE, LOAD N (Newton) ; kN (kilo-Newton) 1 kg = 1 x 9.81 = 9.81 N 1 ton = 103 x 9.81 = 9.81 x 103 N = 9.81 kN

STRESSMPa (1 MPa = 1 N/mm2)

DENSITYkN/m3


BENDING IN BEAM

Crack in tension area

Reinforcement in tension area

M+

M+


1

234

5

section strain

1

2

3

4

5

1

2

3

4

5


1. Hypothesis of BernouilliIn bending plane sections remain plane.

2. Law of NavierThe longitudinal strains of fibers are proportional totheir distance to the neutral axis

ASSUMPTIONS

h

b

d

c

’ c

s

NeutralAxis

Where :c = concrete strains = steel strain

’c c

s d - c=

M+


6. The tensile forces are carried entirely by the reinforcement in all sections. The role of concretein tension is neglected.

h

b

d

c’cu

s

NeutralAxis

M+


y

1E

ACTUAL STRESS - STRAIN RELATIONSHIP OF STEEL

Stress

fsu

fy

sh Strain

Not to scale

Elastic

Plastic Strain-hardening

Tensile rupture

Upper yield stress

Est

1


fs (MPa)

s

400

240

fy2 = 400 MPa

fy1 = 240 MPa

Where :fs = stress of steel ; fy = yield stresss = strain of steel ; y = yield strainEs = Modulus of elasticity = 200.000 MPa

For : s < y fs = Es . s

s y fs fy

IDEALIZED STRESS - STRAIN RELATIONSHIP OF STEEL

y1 y2

Es =fs

s


Where:fc’ = compression stress of concrete (based on cylinder specimen)fct = tension stress of concrete 0.5 fc’ (MPa)Ec = Modulus of elasticity of concrete 4700 fc’ (MPa)cu = Ultimate compression strain of concrete

ACTUAL STRESS - STRAIN RELATIONSHIP OF CONCRETE

fc’

Stress(MPa)

Strain ’cu

(0.003)fct


ACTUAL STRESS - STRAIN RELATIONSHIPOF SEVERAL CONCRETE STRENGTH

0.001 0.002 0.003 0.004

700

600

500

400

300

200

100 10

20

30

40

50

60

800

70

80

0

Strain, in/in (mm/mm)

kgf/c

m2

MPa

Co

mp

ress

ion

str

eng

th o

f co

ncr

ete,

fc’

, ks

i12

10

9

8

7

6

5

4

3

2

1

0

11

fc’ = 82.7 MPa

68.9

55.1

41.3

34.4

27.6

20.7


Shaded areaArea of rectangle1 =

1 = 0.5 = 0.333b. Triangle

1 = 0.67= 0.375

c. Parabola

cC

1 = 0.85= 0.425

f’c

a. Hyperbolic

T

c

0 4 8 12

Concrete strength (ksi)

0.2

0.4

0.6

0.8

1.0

0.0

1

•

•••• •

••• • •• •••

•••••• •• •

• • •• •

••

•• •• •• •

•

•

• ••• •

1 = 0.85

1 = 0.65

Values of 1

for = 0.85

COMPRESSION AREA OF BEAM IN BENDING

M+


C

T

½ a

Neutral axis

a = 1cc

0.85 fc’

RECTANGLE SCHEME OF STRESS DIAGRAM

H = 0 C = T M = 0 Mn = C . (d - ½ a)

Mn+

b

d


No (6) about stress - strain relationship (f - ) of concrete(a)

’c

f’c

’cu

(b)’c

f’c

’cu

(c)’c

f’c

’cu

(d)’c

f’c

’cu

No 7(e)

’c

f’c f’c

’cua = 1 cc

Where : = 0.851 = 0.85 for fc’ 30 MPa = 0.85 - 0.008 (fc’ - 30)

for 30 < f’c 55 MPain all cases 1 0.65


y

y

yC

C

C

yC

0.85 fc’

fc’

fc’

fc’

0.67 b.c.fc’ = As.fy 0.37 c Ts (d-0.37c)

0.64 b.c.fc’ = As.fy 0.35 c Ts (d-0.35c)

0.76 b.c.fc’ = As.fy 0.40 c Ts (d-0.40c)

0.72 b.c.fc’ = As.fy 0.42 c Ts (d-0.42c)

C = T y Mu

Note : d = effective depth of beam section ; c = depth of neutral axisAs = steel reinforcement area ; fy = yield strength of steel

COMPARISON OF SEVERAL SCHEME OF CONCRETE COMPRESSION AREA


Mean 241.2 55.1 22.0 79.3

y

y

yC

C

C

yC

0.85 fc’

fc’

fc’

fc’

241.2 57.6 21.3 79.3

C = T (kN) c (mm) y (mm) Mu (kNm)

241.2 60.3 21.1 79.3

241.2 50.8 20.3 79.5

241.2 53.6 22.5 79.0

The result difference is less than 1%

250

350

As

fc’ = 25 MPa fy = 400 MPaAs = 3D16 = 603 mm2

b = 250 mm h = 350 mm

EXAMPLE


FAILURE MECHANISM DUE TO BENDING

BALANCED FAILURE

UNDER-REINFORCED FAILURE /TENSION STEEL YIELDING

OVER-REINFORCED FAILURE/CONCRETE COMPRESSION FAILURE


s = y s > y s < y

FAILURE MECHANISM

cbalc

c

’cu= 0.003

Balanced Under- Over- Failure Reinforced Reinforced

As = As bal As < As bal As > As bal

MU+

’cu= 0.003 ’cu= 0.003


600600 + fy

cbal = d ……… (1)

1. BALANCED FAILURE ’cu = 0.003 is reached at the same time when s = y

As bal

b

d

’cu = 0.003

cbal

s = y

C

T

d-½a

Strain Stress

abal

0.85 fc’

fs = fy

cbal 0.003 d 0.003 + y

=Multiplied with Es = 2.105 MPa

MU+


C = T

0.85 fc’ . abal . b = As bal . fy

0.85 fc’ . 1 . cbal . b = As bal . fy

0.85 fc’ . 1 . cbal . b fy

0.85 fc’ . 1 . cbal

fy . d

As bal =

bal = … (2)

: bd

0.85 fc’ . 1 600 fy 600 + fy

bal = .

Substitute eq. (1) into eq. (2) found :


2. UNDER-REINFORCED FAILURE ’cu = 0.003 is reached when s >> y ; < bal ; c < cbal

H = 0 C = T 0.85 fc’ . a . b = As . fy

As . fy

0.85 fc’ .a.b

Mn = T (d - ½ a)= As . fy …(2)

a = …. (1)

fs = fys >> y

As

b

d

’cu = 0.003

cC

T

d-½a

Strain Stress

a0.85 fc’

MU+


Mn bd2

Mn = As . fy ( d - )½ . As . fy

0.85 fc’ . b

: bd2

= . fy ( 1 - 0.59 ) fy

fc’

bd2 . fy ( 1 - 0.59 )

fy

fc’Mn =

Substitute eq. (1) into eq.(2) found :


s < y fs < fy

3. OVER-REINFORCED FAILURE ’cu = 0.003 is reached when s < y ; > bal ; c > cbal

As

b

d

’cu = 0.003

cC

T

d-½a

Strain Stress

a

0.85 fc’

This failure mechanism should be avoided. Concrete failure prior to the yielding of steel could lead to a brittle failure condition.

MU+


MAXIMUM REINFORCEMENT (max)

max = 0.75 bal

Where : bal = reinforcement ratio in balanced strain condition

This specification is made in order to ensure that only Under-Reinforced Failure occur in the section.

As bal

b

d

’cu = 0.003

c

y

Strain

’cu + y

c

y

MU+bal =

As bal

b . d


MINIMUM REINFORCEMENT (min)

1.4 fy

min =

This specification is made in order to anticipate a significant raise in steel stress (fs )when the concrete is starting to crack. Significant raise of steel stress make the steel cut off suddenly. NOTE : For slab : min = 0.002 ; fy = 240 MPa

= 0.018 ; fy = 400 MPato anticipate creep and shrinkage in concrete


RECTANGULAR SECTION OF REINFORCED CONCRETE

Compression Reinforcement

Stirrup

Concrete cover

Additional reinforcement(Torsion reinf. : 12mm)

Tension reinforcement

Concrete cover

db or 25 mm

300 mm

MU+


TABLE 5.1.a - CUR IV

Mu/bd2 fy = 240 MPa fy = 400 MPa c/d z/d au au

100 0,0005 0,191 0,0003 0,318 0,012 0,995200 0,0011 0,190 0,0006 0,317 0,023 0,990300 0,0016 0,189 0,0010 0,315 0,035 0,985400 0,0021 0,188 0,0013 0,314 0,047 0,980500 0,0027 0,187 0,0016 0,312 0,059 0,975600 0,0032 0,186 0,0019 0,310 0,071 0,970700 0,0038 0,185 0,0023 0,309 0,084 0,964800 0,0043 0,184 0,0026 0,307 0,096 0,959900 0,0049 0,183 0,0029 0,305 0,109 0,9541000 0,0055 0,182 0,0033 0,303 0,122 0,9481100 0,0061 0,181 0,0036 0,302 0,135 0,9431200 0,0067 0,180 0,0040 0,300 0,148 0,9371300 0,0073 0,179 0,0044 0,298 0,161 0,9321400 0,0079 0,178 0,0047 0,296 0,174 0,926….. ……. ……. …… …… …… …...3000 0,0190 0,158 0,01114 0,263 0,422 0,8213100 0,0199 0,156 0,0119 0,260 0,440 0,8133200 0,0207 0,155 0,0124 0,258 0,458 0,8053300 0,0216 0,153 0,478 0,7973400 0,0225 0,151 0,497 0,7893500 0,0234 0,150 0,518 0,780

CONCRETE STRENGTH f’c 15 = 0,8


MINIMUM CONCRETE COVER (mm)

Structural Un-Exposed to Exposed toComponent ground or weather ground or weather

Floor / wall ØD36 : 20 ØD16 : 40> ØD36 : 40 > ØD16 : 50

Beam all diameter : 40 ØD16 : 40 > ØD16 : 50

Column all diameter : 40 ØD16 : 40 > ØD16 : 50

For all cast in situ concrete which directly contact with ground,minimum concrete cover : 70 mm


AREA OF REINFORCEMENT (mm2)

Number of barØ 1 2 3 4 5 6 7 8 9 10

6 28 57 85 113 141 170 198 226 254 2838 50 101 151 201 251 302 352 402 453 50310 79 157 236 314 393 471 550 628 707 78512 113 226 339 452 565 679 792 905 1018 113114 154 308 462 616 770 924 1078 1232 1385 153916 201 402 603 804 1005 1206 1407 1608 1810 201119 284 567 851 1134 1418 1701 1985 2268 2552 283520 314 628 942 1257 1571 1885 2199 2513 2827 314222 380 760 1140 1521 1901 2281 2661 3041 3421 380125 491 982 1473 1963 2454 2945 3436 3927 4418 490928 616 1232 1847 2463 3079 3695 4310 4926 5542 615832 804 1608 2413 3217 4021 4825 5630 6434 7238 8042


EXERCISE - 1 Analysis of rectangular section due to bending

A rectangular section of reinforced concrete as drawn below :

fc’ = 20 MPafy = 400 MPab = 300 mmh = 500 mm

a. Check the failure mechanism of this section, UNDER or OVER-REINFORCED FAILURE ?

b. Based on (a), find the depth of neutral axis, c, and ultimate moment, Mu that could be resisted by this section.

c. If the reinforcement 4 16 is replaced with reinforcement 6 19,

how much Mu would be increased ?

b

h416MU+


A beam of reinforced concrete resists the uniformed dead load (qD) and live load (qL) as drawn below :

Sectional dimension : 400 x 600 mm, concrete strength fc’ = 25 MPaand steel strength fy = 400 MPa.a. Due to the loads, draw the moment diagram of the beam. b. Design the reinforcement of beam at maximum M+ (in the middle

of span AB) and maximum M- (at support B). c. Check the chosen reinforcement to maximum and minimum

reinforcement requirements as specified in SNI.

qL = 20 kN/mqD = 15 kN/m

A B C7.20 3.60

EXERCISE - 2 Design of rectangular section due to bending


A rectangular section of reinforced concrete as drawn below :

3 22

b

h

a. If fc’ = 20 MPa, fy = 400 MPa, find the ultimate moment Mu that could be resisted by this several value of dimension :b = 300 mmh = 300, 400, 500, 600 mmDrawn your results as graphical model. Mu

(kNm)

h (mm)

b. Compare the result of (a), with another relationship :h=400 mm and b = 300, 400, 500, 600 mm

c. To resist Mu = 400 kNm with beam width b=400 mm,how much hmin ?

EXERCISE - 3 Analysis of rectangular section due to bending


A trapezoid section of reinforced concrete as drawn below :

500

400

600

8 19

fc’ = 25 MPafy = 400 MPa

How much Mu that could be resisted by this section ?

EXERCISE - 4 Analysis of trapezoid section due to bending

End of presentation

Bending Beam With Single Reinforcement

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