38117330 Bending Beam With Double Reinforcement

7/29/2019 38117330 Bending Beam With Double Reinforcement

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BEND-DoubleReinf 1

Faculty of Civil Engineering and Planning

Civil Engineering Department

Petra Christian University



BEND-DoubleReinf 2

WHEN IS COMPRESSION REINFORCEMENT NEEDED ?

DESIGN OF DOUBLE REINFORCEMENT

FLOW CHART OF DOUBLE REINFORCEMENT DESIGN

USING TABLE FOR RECTANGULAR SECTION

PROBLEMS

EXERCISES

TABLES



BEND-DoubleReinf 3

WHEN IS COMPRESSION REINFORCEMENT NEEDED ?

1. If a beam section with single reinforcement has areinforcement ratio, , more than its maximum value, maks,

for a certain bending loading, Mu.

2. To accomplish the seismic design provision.(SKSNI T15-1991-03 sec. 3.14.3.2(1)).

3. Practically reasons.







BEND-DoubleReinf 6

IF “YES” : Cs = As’ . f y Compression steel is yield (s’ >y)

Cs

f y

IF “NO” : Compression steel is not yield (s’ y)

f s’ = s’ . Es

Cs = As’ . f s’Cs

f s’ As’ =

As’ =

The value of As needed for the section is :

As = As1 + As2

Where : As1 = As maks

As2 = T2 : f y

d-d’

Cs = As’ f s’

T2 = As2 f y Mn2

2



BEND-DoubleReinf 7

See Condition 2

T2 = As2 f y

d-d’

Cs = As’ f s’

Mn2

2

Cs = T2

* If compression steel is yield :

As’ . f y = As2 . f y As’ . f y

f y

As2 = As’

As2 = then :

* If compression steel is not yield :

As’ . f s = As2 . f y then :

f sf y

As2 = As’ .



BEND-DoubleReinf 8

f s

= f y

DESIGN OF DOUBLE REINFORCEMENT

See Table 5.3.a - 5.3.j (CUR IV) for designing the double

reinforcement in rectangular section with ’ = 0.5 ;

See Table 2.4 for double reinforcement with several ratioof ’/ from Copy Sheet

b

d’

c

’cu = 0.003

s

0.85 f c’

a

Cc

T1

d-½a d-d’

Cs

T2

+

Mn1 Mn2

1 2

As’ ’s

As Mu

d



BEND-DoubleReinf 9

TABLE 5.3.a - CUR IV

= Ratio of tension reinf ; ’= ratio of compression reinf. = 0.5

Mu /bd2 f y = 240 MPa with d’/d = f y = 400 MPa with d’/d =

0.10 0.15 0.20 0.10 0.15 0.20200 0.0011 0.0011 0.0012 0.0007 0.0007 0.0007

400 0.0022 0.0023 0.0023 0.0013 0.0014 0.0014

600 0.0033 0.0034 0.0035 0.0020 0.0020 0.0021

800 0.0044 0.0046 0.0047 0.0027 0.0027 0.0028

1000 0.0056 0.0057 0.0059 0.0033 0.0034 0.0035

1200 0.0067 0.0069 0.0071 0.0040 0.0041 0.0042

1400 0.0078 0.0080 0.0083 0.0047 0.0048 0.00501600 0.0090 0.0092 0.0095 0.0054 0.0055 0.0057

1800 0.0101 0.0104 0.0117 0.0061 0.0062 0.0072

2000 0.0113 0.0116 0.0119 0.0068 0.0070 0.0064

2200 0.0124 0.0128 0.0132 0.0075 0.0077 0.0079

2400 0.0136 0.0140 0.0144 0.0082 0.0084 0.0087

2600 0.0148 0.0152 0.0157 0.0089 0.0091 0.00942800 0.0160 0.0165 0.0170 0.0096 0.0099 0.0102

3000 0.0172 0.0177 0.0182 0.0103 0.0106 0.0109

3200 0.0184 0.0189 0.0195 0.0110 0.0114 0.0117

3400 0.0196 0.0202 0.0208 0.0118 0.0121 0.0125

3600 0.0208 0.0214 0.0221 0.0125 0.0129 0.0133

3800 0.0220 0.0227 0.0234 0.0132 0.0136 0.0141

4000 0.0233 0.0240 0.0247 0.0140 0.0144 0.0148

CONCRETE STRENGTH f’c 15 with compression reinforcement

= 0,8



BEND-DoubleReinf 10

TABLE 2.4 - COPY SHEET

RECTANGULAR BEAM WITH DOUBLE REINFORCEMENT Steel yield stress, fy = 240 MPa

RATIO OF REINFORCEMENT : (%) Concrete strength, fc’ = 20 MPa

Rn = Mn / (b.d2) d’/d = 0.15

Units : Mn (kNm), b and d (m) min = 0.58%

’/

Rn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2400 0.85 0.79 0.74 0.68 0.62 0.57 0.52 0.48 0.44 0.40

2600 1.02 0.98 0.93 0.89 0.84 0.79 0.74 0.69 0.65 0.61

2800 1.15 1.12 1.08 1.04 1.00 0.95 0.91 0.86 0.82 0.77

3000 1.27 1.24 1.21 1.17 .13 1.09 1.05 1.00 0.96 0.913200 1.38 1.35 1.32 1.29 1.25 1.21 1.17 1.13 1.09 1.04

3400 1.48 1.46 1.43 1.40 1.37 1.33 1.29 1.25 1.21 1.16

3600 1.82 1.56 1.54 1.51 1.48 1.44 1.40 1.36 1.32 1.28

3800 2.80 1.66 1.64 1.62 1.59 1.55 1.51 1.47 1.43 1.39

4000 1.89 1.74 1.72 1.69 1.66 1.62 1.58 1.54 1.49

4200 2.38 1.84 1.82 1.79 1.76 1.72 1.68 1.64 1.60

4400 2.87 1.94 1.92 1.89 1.86 1.83 1.79 1.74 1.70

4600 3.36 2.24 2.02 1.99 1.96 1.93 1.89 1.85 1.80

4800 3.85 2.57 2.11 2.09 2.06 2.03 1.99 1.95 1.90

5000 2.89 2.21 2.19 2.16 2.13 2.09 2.04 2.00

5200 3.22 2.42 2.29 2.26 2.22 2.19 2.14 2.10

...



BEND-DoubleReinf 11

FLOW CHART OF DOUBLE REINFORCEMENT DESIGN

USING TABLE FOR RECTANGULAR SECTION

1

As = . b . d

As’ = (’/) . As

Finish

START

Define : b, d, d’ (trial & errors)

Calculate loads and Mu

Calculate : Mn = Mu /

Mn

bd2

Select appropriate ’/

See Table to find

1

Rn =

Not found

found



BEND-DoubleReinf 12

SOLVE THESE PROBLEMS

Time frame : 15 Minutes

As’

As

d’

h

b

d

c s’ cu’ = 0.003 c = 120 mm

d = 720 mm

d’ = 80 mm

f c’ = 25 MPa

f y = 400 MPab = 400 mm

h = 800 mm

a. Check the compression steel whether it is yield or not.b. Check the section, is it in “UNDER-REINFORCED” -

condition ?

1.

The solutions are on the last

pages of this presentation







BEND-DoubleReinf 15

s’ 40

0.003 120

s’ = 0.001f y 400

Es 200.000

s’ < y Compression steel is not yield

1. a. Is compression steel yield yet ?

120

600

As’

As

d’

h

b

d

s’

cu’ = 0.003

40

=

= =y

SOLUTION

= 0.002

s



BEND-DoubleReinf 16

b. Is section in UNDER-REINFORCED Condition ?

s 0.003

600 120

s 0.015 > y ( = 0.002)

Tension steel is yield.

The section is in UNDER-REINFORCED Condition.

=

=

2. h = minimum, if As = maximum.

max = 0.75 bal

= …….

As max = max . b . d

= . 0.8 . f y (1 - 0.588 )

..... See CUR 1 p. 54

f y

f c’

Mu

bd2



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38117330 Bending Beam With Double Reinforcement