تصميم الكمرات بطريقة ultimate.pdf

III. REINFORCED CONCRETE BEAMS

REINFORCED CONCRETE DESIGN ARBT (525-351)

38Yassin S. Sallam

Building Structure

CHAPTER III Reinforced Concrete Beams by Ultimate Strength Method

3.1 ULTIMATE STRENGTH METHOD

(STRENGTH DESIGN METHOD) In the strength design method (formerly called Ultimate Strength method), the service loads are increased sufficiently by factors to obtain the load at which failure is considered to be “imminent”. This load is called the factored load or factored service load. The structure or structural element is then proportioned such that the strength is reached when the factored load is acting. The computation of this strength takes into account the nonlinear stress-strain behavior of concrete. The design strength method may be expressed by the following:

Strength Provided ≥ Strength required to carry factored loads Where the "strength provided " ( such as moment strength ) is computed in accordance with rules and assumptions of behavior prescribed by a building code, and the " strength required " is that obtained performing a structural analysis using factored loads. The strength provide has been commonly referred to by practitioners as necessary" Ultimate strength " However, it is a code-defined value for strength, and is not necessary " Ultimate " in the sense of being a value above which it is impossible to reach. The ACI Code uses a conservative definition of strength; thus the modifier " ultimate " is not appropriate.


39Yassin S. Sallam

Building Structure

3.2 ANALYSIS OF R/C BEAMS

a) INTRODUCTION Moment capacity of beams basic assumptions of design:

1. Plane sections before bending remain plane. 2. Stress sections relations for steel & concrete are nonlinear. 3. Strain in concrete is concrete is to be neglected. 4. Tensile strength of concrete is to be neglected. 5. Tensile stress in reinforcement = fy (neglect steel hardening )

6. The strength of members shall be based on satisfying the applicable conditions of equilibrium and compatibility of strains. 7. Strain in the steel reinforcement and in the concrete shall be assumed directly proportional to the distance from the neutral axis N.A ( except for deep members covered under ACI-10.2)

b) NOMINAL (Mn) AND ULTIMATE MOMENTS (Mu) OF BEAMS: For computation of nominal flexural Mn the following assumptions (ACI -318-10.3) are made as shown in figure (3.1):

1. The strength of members shall be based on satisfying the applicable conditions of equilibrium and compatibility of strains.

2. Strain in the steel reinforcement and in the concrete shall be assumed directly proportional to the distance from the neutral axis N.A ( except for deep members covered under ACI-10.2)

3. The maximum usable strain cuε at the extreme concrete compression fiber shall be assumed equal to 0.003

4. The modulus of elasticity of nonpresressed steel reinforcement may be taken 200 GPa

5. For practical purposes the relationship between the concrete compressive stress distribution and the concrete strain when nominal strength is reached may be taken as an equivalent rectangular stress distribution( ACI-10.2.7), where the average compression stress in concrete = '85.0 cf .

6. The height of the block = a c.1β= Where: c = distance between N.A. and top of section (depth

of the neutral axis measured from extreme compression fibers.


40Yassin S. Sallam

Building Structure

(a) Beam section (b) Strain (c) Actual compression stress distribution (d) Assumed compression stress distribution ( Whitney rectangular stress block)

Figure(3.1 ) : Definition of Whitney rectangular stress distribution.

Table (3.1) values of β1 for different value of 'c

f

For rectangular section: • Force in concrete:

bcfC

bafC

c

c

...85.0...85.0

1

'

β==

• Force in steel : ys fAT = For under-reinforced section The Nominal Moment, from stresses in the beam Since C = T

ysc fAbaf =..85.0 Where bf

fAca

c

ys

..85.0.

'1 == β

)2

(. adfAZTM ysn −==

1β )(' MPafc

0.85 30≤ )30(008.0085.0 ' −− cf 5630 ' ≤≤ cf

0.65 56⟩

ss fAT =

abfC c .85.0 '=

a /2

' 85.0 cf

C

T

N.A

'ckf

003.0=cuε

c

b

h d

As

ca 1β=

Z =d- a/2

(a) (b) (c) (d)


41Yassin S. Sallam

Building Structure

Nominal Moment: To obtain the nominal moment of a beam, the simple steps to follow are used as is illustrated in examples

1. Compute total tensile force ys fAT .= 2. Equal total compression force bafC c ..85.0 '= to ys fAT .= 3. Solve for (a) In this expression (a-b) is the assumed area stressed in compression at '85.0 cf . The compression force C and the tensile force T must be equal to maintain equilibrium at the section 4. Calculate the distance between the centers of gravity of T and C rectangular section it equals (

2adZ −= )

5. Determine Mn, which equals T or C times the distance between their centers of gravity.

Example(3.1) ; Determine the nominal moment of the beam section shown in Figure (3.2) if

MPafc 28' = MPaf y 420=

Solution:

222

6.1472)4

)25((3)

4.(3 mm

xxdAs ===

ππ

T = C baffA cys ..85.0 '=

mmxxbf

fAa

c

ys 8.863002885.0)420)(6.1472(

85.0.

'===

mmNx

xxadCadfAM ysn

.10252

)2

8.86450(4206.1472)2

()2

(

6=

−=−=−=

=252 kN.m The nominal moment in term of ω

)2

( adfAM ysn −= Put bf

fAa

c

ys

..85.0 '=

).85.02

..(

' dbxfxdfA

dfAMc

ysysn −= Since

bdAs=ρ

Figure (3.2)

0.3

mm 253φ

500 mm


42Yassin S. Sallam

Building Structure

)59.01(.'

c

yysn f

fdfAM ρ−=

Let .'.

c

ys

c

y

ff

bdA

ff

== ρω

)59.01(. ω−= dfAM ysn

Table (3.2): Steel bars area (cm2)

3.3 STRENGTH REDUCTION FACTORS φ Ultimate moment, Mu is the section nominal moment multiplied by reduction factor φ . where nu MM .φ= The factors φ for understrength are called strength reduction factor according to ACI-318-9.3. These are also called resistance factors. These factors are used in basic strength equation, which give the nominal strength, assuming:

o Material strengths are as specified. o Member sizes are as shown on the drawings. o Bars are of full weight, and cross-section. o Calculations are accurate.

φ/no 1 2 3 4 5 6 7 8 9 10 11 12 14 16 18 20 8 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5.02 5.53 6.03 7.034 8.038 9.04 10

10 0.8 1.6 2.4 3.1 3.9 4.7 5.5 6.3 7.1 7.85 8.64 9.42 10.99 12.56 14.1 15.712 1.1 2.3 3.4 4.5 5.7 6.8 7.9 9 10 11.3 12.4 13.6 15.83 18.09 20.3 22.614 1.5 3.1 4.6 6.2 7.7 9.2 11 12 14 15.4 16.9 18.5 21.54 24.62 27.7 30.816 2 4 6 8 10 12 14 16 18 20.1 22.1 24.1 28.13 32.15 36.2 40.218 2.5 5.1 7.6 10 13 15 18 20 23 25.4 28 30.5 35.61 40.69 45.8 50.920 3.1 6.3 9.4 13 16 19 22 25 28 31.4 34.5 37.7 43.96 50.24 56.5 62.822 3.8 7.6 11 15 19 23 27 30 34 38 41.8 45.6 53.19 60.79 68.4 76 24 4.5 9 14 18 23 27 32 36 41 45.2 49.7 54.3 63.3 72.35 81.4 90.426 5.3 11 16 21 27 32 37 42 48 53.1 58.4 63.7 74.29 84.91 95.5 106 28 6.2 12 18 25 31 37 43 49 55 61.5 67.7 73.9 86.16 98.47 111 123 30 7.1 14 21 28 35 42 49 57 64 70.7 77.7 84.8 98.91 113 127 141 32 8 16 24 32 40 48 56 64 72 80.4 88.4 96.5 112.5 128.6 145 161 34 9.1 18 27 36 45 54 64 73 82 90.7 99.8 109 127 145.2 163 181 36 10 20 31 41 51 61 71 81 92 102 112 122 142.4 162.8 183 203 38 11 23 34 45 57 68 79 91 102 113 125 136 158.7 181.4 204 227 40 13 25 38 50 63 75 88 100 113 126 138 151 175.8 201 226 251


43Yassin S. Sallam

Building Structure

o Strength equation itself is theoretically accurate and applicable to concrete members.

The following table shows Factors of Safety or reduction factor ( φ ) according to different cases. table (3.3).

Table (3.3): value of φ

Ultimate Moment: Ultimate moment of the section: )59.01.(... ωφ −= dfAM ysu

Multiply '

'

..

..

c

c

fdbfdb

)59.01.(......

. 2'' ωφ

ω

−⎟⎟⎠

⎞⎜⎜⎝

⎛= dbf

fdbfA

M cc

ysu

43421

( )ωωφ 59.01... 2' −= dbfM cu The relation between nominal and ultimate moment: 2. bdRMM unu == φ

Or )59.01(.. '2

ωωφ −== cu

u fbdMR

In addition, we can obtain steel ratio ρ from nu RR .φ= as follows:

).211(.

2mf

bdMR y

nn ρρ −==

)..2

11(1

y

n

fRm

m−−=ρ Where: '85.0 c

y

ff

m =

Case φ

Flexure & Axial tension, 0.90 Shear and Torsion 0.85 Compression Members, Spirally Reinforced 0.70 Compression Members (Tied) 0.65 Bearing on Concrete 0.70 Plan Concrete: flexure, compression, shear and bending 0.65


44Yassin S. Sallam

Building Structure

Example (3.2) Find nominal Mn and ultimate moments Mu, for the section, as shown in figure (3.3) Assume f’

c = 28 MPa , fy = 400 MPa Solution:

222

3.3393)3.113(34

)12(3)4

( mmxxxdAs

====ππ

mx 6103.339 −= a) Ultimate moment

024.028

4002.0103.339.

6

' ===−

xxff

bdA

c

ysω

[ ][ ]

mkN

X

dbfM cu

.15.24

)024.0)(59.0(1)024.0()45.0)(2.0)(1028)(9.0(

59.01(...23

2'

=

−=

−= ωωφ

b) The nominal moment:

mkNM

M un .83.26

9.015.24

===φ

Example (3.3) Find As for the shown section in figure (3.4), MPafc 25' = MPaf y 400= Solution: b = 30 cm = 300 mm d = 73 cm = 730 mm Mu= 330 kN.m =330 x106 N.mm

mmNxxMM un .10367

9.010330 6

6

===φ

MPax

xbdMR n

n 29.2)730(300

103672

6

2 ===

8.182585.0

40085.0 '

===xf

fm

c

y

a) Steel ratio: Figure (3.4)

)..2

11(1

y

n

fRm

m−−=ρ

Figure (3.3)

0.2

500 mm123φ=sA

30

80 cm

75 cm

Mu=330 kN m


45Yassin S. Sallam

Building Structure

006.0)400

29.28.18211(8.18

1=−−=

xxρ

= 0.6 % b) Steel area: = 13.14 cm2 Example (3.4) Find the percentage of steel ratio ρ , the ultimate and the nominal moment for the concrete section shown in figure (3.5), MPafc 25' = MPaf y 400= MPaRu 64.4=

)(1000

).(2 mbdxmkNMR n

n =

Solution: b = 32.5 cm = 325 mm d = 61.5 cm = 615 mm As=30 cm2 =30 x102 mm2

a) Steel ratio: 015.0615325

1030 2

===xx

bdAsρ = 1.5 %

156.59.064.4

===φ

un

RR

Since 2bdMR n

n =

Figure (3.5) b)Nominal moment: mmNxxbdRM nn .633791632)615(325156.5. 22 === = 634 kN.m c)Ultimate moment: mkNxMM nu .5706349.0. === φ

3.4 BALANCED R/C RECTANGULAR BEAM The balanced beam in ultimate strength design is not a practical beam, but the concept is fundamental to the philosophy of the code. ACI requires that the concrete strain be taken at 0.003 as also shown in Figure (3.6), for the balance condition to assure ductile failure produced by yielding of steel, which gives sufficient warning.

Analysis of the Balanced Beam

32.5 cm

61.5cm 2 30 cm


46Yassin S. Sallam

Building Structure

The analysis of the balanced beam stats from the stain triangles at failure, when the concrete strain is 0.003 . In this case the reinforced steel strain shall be equal to the rate of yield stress to steel elasticity modulus. The neutral axis can be located from the similarity of the strain triangles of Fig. (3.6.b ) (compression strain triangle and the large dotted triangle)as follows: (a) Beam section (b) Strain (c) Actual compression stress distribution (d) Assumed compression stress distribution ( Whitney rectangular stress block)

Figure (3.6) : Definition of Whitney rectangular stress distribution.

a) OVER REINFORCED BEAM:

o ycu εε p o sbs AA f or sbs ρρ f o The failure of this beam could be sudden and violent failure, when the concrete reaches the crushing strain 0030.cu =ε before steel. ( steel does not yield), .figure(3.7.a) o Failure occurs which is not desirable.

b) UNDER REINFORCED BEAM:

o cuy εε p o sbs AA p or sbs ρρ p o The failure of this beam could be failure, when the steel yield first before concrete its the crushing strain 0030.cu =ε . o The yield of steel gives warning of the failure, which give time to evacuate building. figure(3.7.b).

ss fAT =

abfC c .85.0 '=

a /2

' 85.0 cf

C

T

N.A

'ckf

003.0=cuε

c

b

h d

As

ca 1β=

Z =d- a/2

(a) (b) (c) (d)


47Yassin S. Sallam

Building Structure

Figure (3.7) Stress-Strain curve for concrete and steel reinforcement, and definition of εy and εcu

c) BALANCE REINFORCED BEAM: o cuy εε = o sbs AA = or sbs ρρ = o The relative amount of tension steel compared to that in the balanced strain condition will determine whether the failure is ductile. figure(3.7.c).

To find the value of ρ , we analyze the beam assuming:

- εcu = 0.003 and

- Ef y

ys == εε ,

- MPaxGPaE 310200200 == - bcfbafC cx ..85.0.85.0 1

'' β==

- ):(. statebalancedcausesthatareasteelAfAT sbysb=

Since we know εcu = 0.003 and Ef y

ys == εε ,

We can find value of c from strain geometry of d :

s

ycu

cu

scu

cu

Efd

c

+=

+=

ε

εεε

ε

yy f

xfd

c+

=+

=600

600

10200003.0

003.0

3

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

'cf '

cf'cf

yf yfyf

Strain 310 −xε

Stress

2/mmkN

(a) (b) (c)

Concrete

Steel


48Yassin S. Sallam

Building Structure

df

cy

).600

600(+

=

where: d = depth of beam in cm c = the height of stress distributed in curved form MPaxEs

310200= = modulus of elasticity of steel

yf = steel yield stress Calculation of Reinforcement Ratio in Balanced

The ratio bdAs=ρ , where sA is the area of steel reinforcement and b and d

are shown in Fig.( ) where b represents the width of the beam and d the depth of beam to the center of reinforcing steel bars. To find ρb , we equal T = C

{)(

600600.85.0

..600

600..85.0.

...85.0.

'1

'1

'1

bdA

fff

bdA

dbf

ffA

bcffA

sbb

yy

sb

yysb

ysb

c

c

c

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

=

ρβ

β

β

ρ

The Balanced Steel Ratio: ⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

yyb ff

fc

600600.85.0

'

1βρ

The depth of the beam:

10spand = For simply supported beam

12

spand = For continues beams

The wide of the beam: assume that :

beamtheofdepthbeamtheofWidth 21 =

For example the depth of the simply supported beam has a span of 8 m is

Depth mspand 8.0108

10=== = 80 cm

We can increase the depth according to the loads and moments, we can use d = 80 cm with b = 40 cm


49Yassin S. Sallam

Building Structure

Example (3.5 ) For the section given in the figure (3.8) MPafc 25' = MPaf y 300= Find:

a) Balanced steel Ratio bρ , b) Nominal balanced moment nbM , c) Ultimate balanced moment ubM and d) Steel balanced area sbA

Solution We can find bρ , nbM , and ubM in terms of b, d

for MPafc 25' = ⇒ β1 = 0.85

a) Balanced steel Ratio bρ ,

%404.0

300600600

3002585.085.0

)600

600)('

85.0( 1

==+

=

+=

xxx

fff

b

yy

cb

ρ

βρ

b) Nominal Balanced Moment nbM ,

)2

.(. adCZCM nb −==

⎪⎪⎩

⎪⎪⎨

⎧

==

=+

=+

=

=

255.03.085.0

3.045.0300600

600.600

60085.01

xa

xdf

cy

β

CCM nb 3223.0)2255.045.0( =−=

bcfC c ...85.0 '1β=

75.10832.03.0102585.085.0 3 == xxxxxC

0.2 m

0.45 m


50Yassin S. Sallam

Building Structure

mkNM

adCZCM

nb

nb

.5.349)2255.0450.0(75.1083

)2

(.

=−=

−==

c)Ultimate balanced moment ubM nbub MM .φ=

mkNxM ub .5.3145.3499.0 == d) Steel balanced area

dbA bsb .ρ=

2

2

2

3600

36.0

0036.045.02.004.0

mm

cm

mxx

=

=

==

305

5)30(

4

36002

φ

π

Use

AAbarsofNumber

bar

sb ≈==

22

35344

)30(5 mmxAsb ==π

d) MAXIMUM AND MINIMUM AND ECONOMICAL STEEL

RATIOS

To ensure that all beams have the desired characteristics of visible warning if failure is imminent, as well as reasonable ductility at failure, it is recommended by the Code that: - Maximum Steel Ratios. bρρ 75.0max =

)600

600()85.0(75.0'

1maxyy

c

fff

+= βρ

- Minimum Steel Ratios.

yf4.1

min =ρ )( MPainf y


51Yassin S. Sallam

Building Structure

byf

ρρρρ 75.04.1maxmin =≤≤=

- Economical Steel Ratios. If no architecture conditions, for the beam depth and dimensions. Some references give a value of ρ to begin the design cycle. This value is base on the economical design ranges defined from experience or economy studies.

Some references give the economic range as yf

fc

/

18.0=ρ

Other references give initial reinforced ratio bρρ 4.0≈ Example :

MPaf y 300= and MPafc 25' =

⎪⎭

⎪⎬

⎫

==

==

016.04.0

015.03002518.0

b

x

ρρ

ρ very close

Example (3.6) For a section of dimension 0.25m by 0.50 m MPafc 28' = , MPaf y 400= Find the ultimate design moment uM this section can carry for the following reinforcement ratios:

1) ρ = ρmin 2) ρ = 0.45 ρb

3) ρ = ρmax a) Find values of ω , Ru and Mu for each steel ratio b) Draw relationships of Mu /ρ and Ru/ ρ


52Yassin S. Sallam

Building Structure

Solution Find : ρmin , ρb , ρmax

1. 0035.0400

4.14.1min ====

yfρρ

2. 0303.0

400600600

4002885.085.0

600600.)85.0(

'

1

=+

=

=+

=

xxx

fffx

yy

cb βρ

0136.00303.045.045.0 === xbρρ 3. 0227.0)0303.0(75.075.0max ==== xbρρρ

a) calculation ω, Ru, and Mu

a.1) for ρ = ρmin= 0.0035

mkNxxxxdbRM

MPaxxxfR

xff

uu

cu

c

y

.3.63)455.0(25.010359.19.0...

359.1)05.059.01(2805.0)59.01(.

05.028

4000035.0

232

'

'

===

=−=−=

===

φ

ωω

ρω

a.2) for ρ = 0.45ρb= 0.0136

mkNxxxxdbRM

MPaxxxfR

xff

uu

cu

c

y

.224)455.0(25.01081.49.0...

810.4)194.059.01(28194.0)59.01(.

194.028

4000136.0

232

'

'

===

=−=−=

===

φ

ωω

ρω

a.3) for ρ = ρmax= 0.0227

mkNxxxxdbRM

MPaxxxfR

xff

uu

cu

c

y

..342)455.0(25.010338.79.0...

338.7)324.059.01(28324.0)59.01(.

324.028

4000227.0

232

'

'

===

=−=−=

===

φ

ωω

ρω

b) For these values we draw the diagram of figure (3.9)


53Yassin S. Sallam

Building Structure

3.5 DESIGN OF R/C BEAMS The nominal moment capacity of R/C Beam with tension reinforcement only: ( )ω59.01 −= dfAM ysn This formula can be expressed in terms of ρ

{

( )ω

ρ

59.01.....

−= dbdfdb

AM ys

n

( )ωρ 59.01... 2 −= dbfM yn This can also be expressed in terms of '

cf and ω

{

( ) ( )ωωωρ

ω

59.01...59.01.'..'

. 2'2 −=−= dbfdbfff

M ccc

yn

Where '

.

c

y

ffρ

ω =

ρ ω Ru (MPa) Mu (kN.m) 0.0035 0.050 1.359 63.3 0.0136 0.194 4.810 224 0.0227 0.324 7.338 342

500

400

300

200

100

0.004 0.008 0.001 0.014 0.018 0.02 0.024 0.026

7.11

5.33

3.56

1.78

0.325

Ru

(Mpa)

P

Mu

(KN

-m)

Ru

Mu

w

Mu (N.m)

ρ

Ru (MPa) ω

0.113

Ru

ω Mu


54Yassin S. Sallam

Building Structure

Any one of these three formulas can be used for design . We may write last one as non-dimension

( )ωω 59.01..'. 2

−=dbf

M

c

n

This formula can be transferred into a table that gives value of for each value of ω. Example (3.7)

Calculate 2.'. dbfM

c

n for

given value of ω 3.6 DESIGN CONSIDERATIONS FOR

MOMENT DESIGN 3.6.a. MINIMUM REINFORCE RATIO :

The minimum reinforcement ratio is used for tension or shrinkage and temperature effects. According to ACI code

for Tension : yf4.1

min =ρ

for Shrinkage or temperature: 002.0min ≥ρ for deferent values of yf as shown in table (3.4):

Table (3.4)

'.

c

y

ff

ρω = 2.'. dbfM

c

n

0.064 0.0616 0.175 0.1569

)(MPaf y y

min f.41

=ρ

320 0.0044 350 0.004 400 0.0035 450 0.0031 500 0.0028 600 0.0023


55Yassin S. Sallam

Building Structure

3.6.b. DEFLECTION REQUIREMENTS FOR BENDING The ACI Code recommends minimum thickness hmin for members designed for bending , see table (3.5). Table minimum thicknesses of non prestressed beams or one way slabs unless deflections are computed

Minimum thickness hmin Member Simply

supports One end

continuous Both ends continuous Cantilever

Solid One-way slabs 20

l 24

l 28

l 10

l

Beams or Ribbed

One - way slabs 16l

5.18l

21l

8l

Table (3.5): Minimum thickness of nonprestressed beams or one-way slabs unldeflections are computed

- If MPaf y 400≠ multiply values by a of:

Minimum thickness; )690

4.0(minminyf

xhfactorxh +=

- For light weight concrete, also modify values - Do not apply when sensitive partitions are present.

Example (3.8): For a simply supported beam 5 m span, and MPaf y 300=

83.06903004.0 =+=Factor

mxfactorxlh 21.083.0205

20min ===

cm21= 3.6.c. CONCRETE COVER (C.C)

Concrete cover is necessary for four reasons:

1. Provide bond between steel & concrete diameterbarcc ≥⇒ .


56Yassin S. Sallam

Building Structure

2. Protect reinforcement against corrosion mmcc 7510. −≥⇒

3. Protect reinforcement from strength loss due to over heating due to fire. 4. In garages and factories cover is needed to account for abrasion due to traffic. 5. For beams: for primary reinforcement mmcc 15).( min =⇒ mmcc 40).( max =⇒ 6. For slabs not in contact with ground exposed (bars < 35 mm) mmcc 15).( min =⇒ Use mmcc 20).( min =

3.6.d. BAR SPACING (S) There are many requirements for bar spacing, in general: For one layer of bars in parallel, spacing should be

diameterbarS ≥min mm25≥ sizeaggregat.max3.1≥ Example( 3.9 ) A slab is reinforced by bars of 10 mm, and using MPafc 25' = with maximum aggregate size in concrete 20mm Solution: The spacing should by: Smin ≥ 10 mm ≥ 25 mm ≥ 1.3 x 20 mm = 26 mm Then Smin =26 mm Also, there is a maximum spacing that we should observe so that reinforcement is effective

• For slabs and walls Smax ≤ 3 x slab thickness ≤ 500 mm

3.7 DESIGN OF REINFORCED CONCRETE

SECTIONS There are two possible methods to design a R/C section:


57Yassin S. Sallam

Building Structure

• When all section dimensions are known : in his case we, find section dimensions based an certain assumption, and find the only unknown: ρ (and hence As ) check that ρ min< ρ < ρ max • When section depth is not known : in this case we need to assume ρ, and find (bd2) assuming b or b/d , we can find value of d→ can find approximate value of hand then recalculate ρ based on the new modified value of d.

I. Design of R/C Sections when its Dimensions are known

In this method, we find estimates of section dimensions that are used for design,

From 2' .. bdf

M

c

u

φ we can find ω sAandρ⇒

This method in detailed step: a. Design loads b. Select section dimensions

Select section width :

- For beams use b = (0.25 ~ 0.6 )d , usually 2db =

- For slabs use b =1m = strip width Find effective depth

Design loads Select section dimensions

c. Design loads d.

2cov

diameterbardiameterbarstirruperconcretdepthoveralldeptheffective −−−=

2

. bsbcchd

φφ −−−=

e. Design of moment:

Mu: type of beam Mn=Mu/φ

For simply supported beam: 8

2UlMu =

For continuous beam /slab : we may use design factors given by ACI f. Steel ratio:

stirrupφcc

S S2barφ

d h


58Yassin S. Sallam

Building Structure

'85.0 c

y

ff

m =

)(1000

).(2 mbdxmkNMR n

n =

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

check that ρ min< ρ < ρ max

g. Reinforcement steels Find steel area dbAs ..ρ= Chose diameter of bars Determine the number of bars for one meter

numberentierAAn

b

s =≥

h. Check spacing S for beam:

maxmin sss ≤≤ snnccb bsb )1(22 −+++= φφ

For Beam 12.2−

−−−=

nnccbs bsb φφ

For Slab 4

1000==

nwidths

cc

barφSbar

b

stirrupφ


59Yassin S. Sallam

Building Structure

Example (3.10) : Design a slab show in figure (3.10), with mh 18.0= , mkNM u .5.29= ,

MPafc 28' = , MPaf y 400= Solution: a. Design Loads b. Find effective depth, d

1482

1425180 =−−=d mm.

c. Steel ratio :

81.162885.0

40085.0 '

===xf

fm

c

y

496.1)148.0(1000

778.321000 22

===xbdx

MR nn

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

400496.181.162

1181.16

1 xxρ

00337.0=ρ Check that ρ

00337.0

0035.0min

=

=

ρ

ρ

ρρ fmin Use ρ = ρmin= 0.0035 d. Reinforcement steels

2min 51814810000035.0.. mmxxdbAs === ρ

Using 222

1544

)14(4

14 mmAmm b ===⇒=πφπφ

Number of bars /m’ '/436.3154518 mbars

AAn

b

s ≈===

b=10 m

mm14=φ

0.18


60Yassin S. Sallam

Building Structure

i. Check spacing

Bar spacing mmn

widthS 2504

1000===

Use 4 mmSbarsmm 250,14 ==φ

Exactly 4 mmSbarsmm 27059.3

1000,14 ===φ

Example (3.11)

Design the beam B1, Mu=125.5 kN.m, dimensions =0.25x0. 45 m, MPafc 28' = , MPaf y 400= , figure (3.11)

a. Design Loads b. Find effective depth

2. b

sbcchd φφ −−−=

mm4052

201025450 =−−−=

md 405.0= mb 25.0= c. Nominal moment

mkNM u .5.125=

mM

M un 44.139

9.05.125

===φ

d. Steel ratio :

81.162885.0

40085.0 '

===xf

fm

c

y

4.3)405.0()25.0(1000

44.1391000 22 ===

xxbdxMR n

n

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

4004.381.162

1181.16

1 xxρ

b

dh


61Yassin S. Sallam

Building Structure


00922.0

0035.0min

=

=

ρ

ρ

02276.000922.00035.0 maxmin === ρρρ pp

e. Reinforcement steels

Using 222

3144

)20(4


25.9314052500092.0.. mmxxdbAs === ρ

Number of bars /m’ '/97.2314

5.931 mbarsAAn

b

s ===

Actual Steel Ratio:

Provided As= 3x314 = 942mm

20093.0405250

942 mmxbd

As ===ρ

Optimum Reinforcement: 1360.00303.045.045.0 === xbopt ρρ ⇒ Used 1360.000930.0 == optρρ p ΟΚ

f. Check Spacing

Bar spacing mmxxxS 602

203102252250=

−−−=

mmsbarsmm 60,20 ==φ

b

ss

StirruMaim steel

b


62Yassin S. Sallam

Building Structure

Example (3.12) Design the same beam using optimum reinforcement ratio, bopt ρρ 45.0= ,

mb 25.0= , MPafc 28' = , MPaf y 400= , mkNM u .5.125= Solution ρopt = 0.45 ρb = 0.45 x(0.0303) = 0.01364 ρb: see example (3.6)

a. Design Loads

b. Determine the depth oh the beam:

1949.028

40001364.0'

=== xff

c

yρω

2

23

2'

676.1086

)1949.059.01(1948.025.010289.05.125

)59.01(....

d

xxxxdxxx

dbfM cu

=

−=

−= ωωφ

md

d

340.011549.0

11549.0676.10865.1252

==

==


63Yassin S. Sallam

Building Structure

c. Section total depth:

2

. bsbccdh φφ +++=

mmh 3852201025340 =+++=

Use h = 400 mm Actual value of reinforcement based on he used depth, h = 400 mm.

d. Steel ratio:

81.162885.0

40085.0 '

===xf

fm

c

y

43.4)340.0()25.0(1000

44.1391000 22 ===

xxbdxMR n

n

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

40043.481.162

1181.16

1 xxρ


01236.0

0035.0min

=

=

ρ

ρ

02276.001236.00035.0 maxmin === ρρρ pp O.K

e. Reinforcement steels

2105134025001236.0. mmxxdbAs === ρ

Using 222

3144

)20(4


Number of bars /m’ '/44.3314

1051 mbarsAAn

b

s ====

f. Actual Steel ratio :

Provided As= 4x314 =1256 mm


64Yassin S. Sallam

Building Structure

20148.0340250

1256 mmxbd

As ===ρ

g. Check spacing

1

2.2−

−−−=

nnccbs bsb φφ

mmS 3.3314

)20(4)10(2)25(2250=

−−−−

=

Figure : (3.12) Minimum spacing:

mms 25min ≥ mmb 20=≥ φ mmSizeAggxMaxb ..3.1=≥ φ mmxb 26193.1 ==≤ φ

2630 min == smms f Bar spacing =30 Spacing is acceptable mmSbarsmm 30,20 ==φ

cc

barφSbar

b

stirrupφ


65Yassin S. Sallam

Building Structure

Example (3.13) Design a rectangular simply supported beam, has 10m of span and :

'/12 mkNWDL = (Including O.W.), '/7 mkNWLL = , Use MPafc 25' = , MPaf y 400= , the maximum aggregate size = 19 mm. The economical reinforcement is possible. Solution:

a. Design Loads U= 1.2 DL+1.6 LL = 1.2 x12+1.6x7 = 25.6 kN/m'

b. Ultimate moment Mu For simply supported beam; mkNxUlM u .320

8)10(6.25

8

22

===

c. Reinforcement ratio

01125.04002518.018.0

'

=== xff

y

cρ (economical reinforcement)

18.025

40001125.0'

=== xff

c

yρω

022.4)18.059.01(2518.0)59.01.(. ' =−=−= xxxfR cu ωω

d. Dimension of beam

326

2

479.14)2(022.49.010320

)2(...

bbxbxxx

bdassumedbRM uu

==

== φ

mmxb

b

280101.22

8.2210097363

3

==

=

Using b = 300 mm d = 2b = 2 x 300 = 600 mm

e. Area of steel 2202560030001125.0.. mmxxdbAs === ρ Using mm25=φ

22

4914

)25( mmAb ==π

Number of bars = barsAAn

b

s 4491

2025=≈==


66Yassin S. Sallam

Building Structure

f. Reinforcement ratio Check

0239.0)400600

600(4002585.075.0)

600600(75.075.0

0035.0400

4.14.101125.0

'

1max

min

=+

=+

==

===

=

xxff

f

f

yy

cb

y

βρρ

ρ

ρ

0239.001125.00035.0 maxmin

maxmin

ρρρ

ρρρ

pp

pp

==

∴ O.K.

g. Section total depth

mm

ccdh sbb

5.6021022520560

2.

=+++=

+++= φφ

Using h = 600 mm h. Design beam

300 mm x 600 mm mm254 φ

i. Check spacing

1

2.2−

−−−=

nnccbs bsb φφ

mm

s

7.4614

)25(4)10(2)20(2300

=

−−−−

=

Minimum spacing: mmS 25min ≥ mmb 25=≥ φ mmSizeAggxMaxb ..3.1=≥ φ

mmxb 26203.1 ==≥ φ mmSmmS 2645 min == f Bar spacing = 45 Spacing is acceptable 4 mmsbarsmm 45,25 ==φ


67Yassin S. Sallam

Building Structure

3.7 DESIGN OF T-BEAMS

Most of concrete floors are cast integrally with the beams in place together (monolithic construction), both beam and adjacent slabs act together to resist applied forces, and hence the stresses resulting from them. For beams in middle, the beam and slabs on both sides form a T-beam, while an edge beam and the adjacent slab form an L-beam. As show in figure (3.13)Since concrete resists mainly compression stresses , the slab shares in resisting forces only when it is subjected to compression stresses. This occurs at mid span (along the beam axis ) for both simple and continuous beams

Figure (3.13)

3.7.a) TYPE OF BEAMS

When the T-beam is subjected to moment, the strain distribution in the beam are similar to those of rectangular beam. However, stress distribution is different, where the compression zone extends across the beams total width (

300

6056

25 45

10

20

bf = effective flange id h

d hf

T-Beam (in the middle) L -Beam (at the edges)

bw


68Yassin S. Sallam

Building Structure

the flange width , fb ). Depth of the compression zone may extend within the flange only (case a), or may extend into the web case (b) Case (1) the beam behaves as a rectangular beam with : a) fta ≤

- fbb = - fta ≤

- fc

y

c

ys tfdf

bffA

a ≤=='' 85.0

..85.0

ρ

- ( )ωωφ 59.01.... 2' −= dbfM fcu

- ⎟⎠⎞

⎜⎝⎛ −=

2. adbfAM fysu φ

This case is the prevalent case in typical construction (multi-story buildings, short –span bridges,…) b) fta =

- fc

y

c

ys tfdf

bffA

a ==='' 85.0

..85.0

ρ

- y

cs f

btfA '85.0= or

'85.0 c

ys

ffA

t =

- ( )ωωφ 59.01.... 2' −= dbfM fcu

⎟⎠⎞

⎜⎝⎛ −=

2. adbfAM fysu φ

a

bf = b

hf

bw

c

003.0=cuε

sε

a C

T

N.A


69Yassin S. Sallam

Building Structure

Case (2) the beam behaves as a "true T-beam" where the moment capacity is derived based on the compression zone extending below the flange thickness

ft and fta f btAc f :

case (b): Compression zone extends into web

Figure(3.14 ): Design of T – beam (a)Acts as rectangular beam (b) Direct analysis as T-beam - fta f

- wc

ysfs

bffAA

a'85.0)( −

=

- y

wfcsf f

bbtfA

)(85.0 ' −=

)2()2()1()1( armMomentxForceTensionMomentarmxForceTensionM n +=

)2

().()2

(.

.. 2211

adfAAtdfAM

ZTZTM

ysfsysfn

n

−−+−=

+=

To check a beam to determine its type case (a) or case (b ), we calculate the moment that generates compression zone that extends over the whole flange ( fta = ) Flange moment capacity nfM

)2

.(..85.0 ' fffcnf

tdtbfM −=

)2

.( fysn

tdfAM −=

And compare applied nominal moment capacity nfM If nfn MM ≤ ⇒ T-beam behaves a "rectangular Beam" with fbb =

a

b

hf

bw

a

003.0=cuε

sε

N.A

a C

T

Z

cf φ

1 2 1


70Yassin S. Sallam

Building Structure

If nfn MM f ⇒ T-beam behaves a "True T- Beam" 3.7.b) FLANGE WIDTH FOR T - AND L - BEAMS : It is necessary to determine value of fb for T - and L- beams. ACI code gives fb as the smallest value of the following three equations:

1) Flange width ≤ 44

spanlengthbeams=

4

1Lbf ≤

2) Overhanging part of the slab shall not exceed 8 x flange thickness: ftoverhang 8≤

fwf tbb .16+≤ 3) Overhanging part of the slab shall not exceed half the clear distance between webs :

22loverhang ≤ 2lbb wf +≤

( 2l : clear distance between webs = sl ) Case of Solid Slab on Beams

Exemple (3.14) : Find the flange width for the beam show in Fig. L1 = 6 m, L2 = 4.8 m tf =0.12 m , bw = 0.3 m

1) bf ≤ L1/4 =6/4 =(6/4) = 1.5 m 2) bf ≤ 16 tf + bw

≤ 16(0.12) +0.3 = 2.22 m 3) bf ≤ ℓ2 + bw

≤ (4.8 – 0.3 ) + 0.3 = 4.8 m The flange width for T-beam is the Smallest of (1.5 , 2.22 , 4.8 ) ⇒ bf =1.5 m Figure (3.15)

0.12 m

2l

1l

0.3 m 0.3 m


71Yassin S. Sallam

Building Structure

Case of Ribbed Slab (one - way ) Example (3.15) as shown in figure (3.16) : L1 =4.8 m, L2= 0.5 m

tf =0.1 m , bw = 0.12 m

bf ≤ L1/4 =4.8/4 = 1.2 m bf ≤ 16 tf + bw

≤ 16(0.10) + 0.12 = 1.82 m bf ≤ ℓ2 + bw

≤ (0.5 - 0.1 ) + 0.1 = 0.5 m

Figure (3.16)

The flange width for T-beam is the Smallest of (1.2 , 1.85 , 0.5 ) ⇒ bf =0.5 m The above conditions Table (3.6) become as follows for T-beams and L-beams: Table(3.6 ) fb as the smallest value of the equations, according to ACI code 3.7.C) REINFORCEMENT RATIO LIMITS FOR T-BEAMS : The reinforcement ratio for T-beams and L-beams should be applied carefully, since maximum limit is related to the flange width bf , while minimum reinforcement is related to bw , since minimum reinforcement is provided to prevent tensile rupture of the web(the area subjected to tension).

To simplify the check process, we may use steel areas ; As min , As, and As max dbA fs ..ρ=

dbA ws ..min(min) ρ= dbA fs ..max(max) ρ=

Condition T-Beams L-Beams

1 41lbf ≤

41lbf ≤

2 wff btb +≤16 wff btb +≤ 8

3 wf blb +≤ 2 wf blb +≤

22

Rib RibRib

Beam

Beam

2l1l


72Yassin S. Sallam

Building Structure

and apply the following reinforcement check: Check using reinforcement area:

(max)(min) . sss AAA ≤≤ or (max)(min) ρρρ ≤≤ Simplified Analysis of T Sections: In this simplified analysis, which is good enough for practical applications as it gives reasonable results, we assume that the N.A lies at the top beam web. Therefore

)2

).(..( fysu

tdfAM −= φ

Example (3.16) Mu= 600 kN.m fy= 400 Mc =30 MPa Find Steel area As? Solution:

a) Simplified Analysis

5602201020600

2

=−−−=

−−−= bsbcchd φ

φ Figure (3.17)

)2

).(..( fysu

tdfAM −= φ

)2

100560)(400(9.010600 6 −= sAx

23268 mmAs =

b) From exact solution 23288 mmAs = Example (3.17) Solve the problem (11) the beam B1, shown in Figure (3.18): m.kN.M u 5125= m.bw 250= , mh 45.0= m.t f 180=

MPaf 'c 28= MPaf y 400=

mL 61 = mL 52 = Design this beam

Figure (3.18)

400 mm

600 mm

100 mm

d

2400 mm

bw= 0.25 m

0.45 d

hf =0.18


73Yassin S. Sallam

Building Structure

Solution

2

. bsbcchd φ

φ −−−=

mmd 4052201025450 =−−−=

Find fb ( for L- beam):

1) mlbf 5.146

41 ==≤

2) mbtb wff 69.125.0)18.0(8.8 =+=+≤

3) mblb wf 65.225.02

25.0522 =+

−=+≤

The flange width for T-beam is the Smallest of (1.5, 1.69, 2.65) Use mbf 5.1=

Check Mn against =Mnf :

mkNMM un .4.139

9.05.125

===φ

)2

(...85.0 ' fffcfn

tdtbfM −=

mkNxM fn .2.2024)218.0(405.0)18.0)(5.1)(1028(85.0 3 =⎥⎦

⎤⎢⎣⎡ −=

mkNMmkNM nfn .2.2024.4.139 == p Then the Section is Rectangular with: mbb f 5.1==

81.162885.0

40085.0 '

===xf

fm

c

y

5666.0)405.0()5.1(1000

4.1391000 22

===xxbdx

MR nn

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

4005666.081.162

1181.16

1 xxρ


74Yassin S. Sallam

Building Structure

001437.0=ρ Design the beam: 26 87310)405.0)(5.1(001437.0.. mmxdbA fs === ρ

using mm20=φ

( ) 22

314420 mmAb ==

π

No of bares : 378.2314873

≈=== barsAA

nb

s

Then mm203 φ

( ) 22

9423420)( mmxprovidedAs ==

π

873)(942)( == requiredAprovidedA ss f O.K Example (3.18) The floor of figure (3.19) consists of 100 mm slab supported by beams of 6.70 m span, cast monolithically with the slab at spacing of 2.50 m center to center. These beams have 300 mm wide webs and 500 mm depth to the centre of steel plus the necessary cover to steel. Dead load moment 86 kN.m, and fy = 400 MPa and fc=28 MPa. Calculate the required area of steel sA Figure (3.19) Solution: Use mbf 675.1=

a) Ultimate moment LLDLu MMM 6.1.2.1 +=

mkNxxM u .3691666.1862.1 =+= b) Nominal moment

Effective depth b= 1.675

0.3 0.3 0.3 2.5 m 2.5 m

d=0.5


75Yassin S. Sallam

Building Structure

mkNMM un .410

9.0369

===φ

Nominal moment for flange:

)2

(85.0 fffcnf

tdtxbxfxM −=

mkNxxxxM nf .1794)21.05.0(1.0675.1102885.0 3 =−=

nnf MM f The section is rectangular

c) Reinforcement ratio for web:

81.162885.0

40085.0 '

===xf

fm

c

y

491.1)5.0()1.1(1000

4101000 22

===xxbdx

MR nn

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

y

n

fmR

m2111ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

400491.181.162

1181.16

1 xxρ

d) Design the beam:

26 217310)5.0)(1.1(00385.0.. mmxdbA fs === ρ Use mm25=φ

( ) 22

314425 mmAb ==

π

No of bares n = 54.4491

2173≈== bars

AA

b

s

Then 2

2

2454)(54

)25(255 mmprovidedxAmm s ===⇒ πφ

e) Check bar spacing:

1

2.2−

−−−=

nnccbs bsb φφ

00385.0=ρ


76Yassin S. Sallam

Building Structure

mms 8.284

)25(5)10(2)20(2300=

−−−=

Minimum spacing: mmS b 20min =≤ φ mmS 25min ≤ SizeAggxMaxS ..3.1min ≤ mmxS 26203.1min =≤

mmSmmS 2826min == p

Example (3.19) Calculate the design moment of rectangular section with the following details: b=250 mm, d= 440 mm d' 60 mm, tension steel is six bars 25 mm in diameter ( in two rows) , compression steel is three bars 25 mm in diameter, MPafc 20' = , MPaf y 350= , Solution: 1) Check if the compression steel yields:

22

294564

)25( mmxAs ==π 2

2' 9423

4)20( mmxAs ==

π

0268.0440250

2945===

xbdAsρ 0086.0

440250942;

' ===xbd

Asρ

20039422945' =−=− ss AA 0182.00086.00268.0' =−=− ρρ

For compression steel to yield:

0135.0)350600

600)(44060)(

35020(85.085.0

)600

600(85.0.

'

'''

=−

=−

−=−

xx

fdd

ffx

yy

c

ρρ

φρρ

0135.00182.0' f=− ρρ Therefore, compression steel yields. 2) Calculate nM


77Yassin S. Sallam

Building Structure

mmxx

xff

bAAa

c

yss 1652502085.0

350200385.0

)('

'

==−

=

mkNxx

xxxM

ddfxAadfAAM

n

ysyssn

.4331043312510250

10)60440(350942)2

165440(3502003

)()2

()(

66

6

'''

==+=

−+−=

−+−−=

−

Example (3.20): Design for the bending the shown cantilever. MPafc 25' = , MPaf y 400= Solution For the section a-a Mu(a) = - (67 x 3) x 1.5 = - 301.5 kN.m

mkNMM un .335

9.05.301

===φ

73.2)1064.0(103.0

10335233

6

2===

xxx

bdM

R nn

8.182585.0

40085.0 '

===xf

fm

c

y

)211(1

y

n

fmR

m−−=ρ

%73.0

0073.0)400

73.28.18211(8.18

1

=

=−−=xxρ

0035.0400

4.14.1min ===

yfρ

= 0.35%

)600

600()85.085.0(75.0'

maxyy

c

fff

x+

=ρ

02.0)400600

600(4002585.085.075.0max =

+= xxxρ

=2% maxmin ρρρ pp 21464300073.0.. cmxxdbAs === ρ

Figure (4.13 )

a Wu=67 kN/m

3 m 301.5 kN.m

d=64 cm

h =70 cm

30 cm


78Yassin S. Sallam

Building Structure

As =1400 mm2 2

2

3804

)22( mmAb ==π

Use mm224 φ

Figure (3.14) Types of stirrups

Open stirrups for beams with significant torsion

l3.1≥

Close stirrups for beams with significant torsion

Concrete confinement one side

Concrete confinement

(a) (b) (c) (d) (e)

Documents

تصميم الكمرات بطريقة ultimate.pdf