Unit-1

Introduction to Various Design Philosophies

1.1 Introduction to Various Design Philosophies

INTRODUCTION

HISTORIC DEVELOPMENT

 Advances and Future Tends In R.C.C.

INTRODUCTION TO VARIOUS DESIGN PHILOSOPIES OF R.C. STRUTURES

Concrete is strong in compression but week in tension

Steel is reinforced suitable at the place where concrete is weak in tension such combination of steel and concrete

The purpose of R.C structure is increase strength of concrete section and to reduce the size of structural member

Various design philosophies of R.C structure are

Reinforced concrete is used to construct various structures

Fig no 1

Fig no 2   Flyover                 

Following are the various methods for RCC structure

1. Working stress methodology (WSM).

2. Ultimate load methodology (ULM).

3 Limit state methodology (LSM).

 Limitations of WSM strategies over LSM

Advantages of reinforced concrete

1. Strength

It has excellent Tensile likewise as compressive strength a pair of malleability

2. Ductility

Ductility of structures is increase thanks to steel reinforcement

3. Durability

R.C. structures are more durable

 4. Mouldability

The flexibility of reinforcement and therefore the liquidness of concrete makes the R.C. member to be mould in any form thus it's appropriate for architectural demand.

5. Porosity

RC members are impermeable to wet

6. Fireplace resistance

R.C. structures are additional fireplace resistance than alternative materialssuch as steel and wood

7. Economy

R.C.C is cheaper than steel and pre stressed concrete

 Disadvantages of R.C.C

RCC has followed disadvantages

1. Weight of R.C. structures are measure over different materials

2. Significant formwork is needed for R.C. members

3. RC structures take a lot of times to realize its full strength

Reinforced concrete building elements

1. Slab

2.  Beam

3. Column

4. Footing

Design of every parts is depends on load on structure and structure action of member subjected to the current load.

Load transfer mechanism la RC building is load from block is transferred to the supporting beams, and from beam lo column and from column to footing ultimately footing transfer the masses to soil strata.

1. Slabs

Slab is that the 2-dimensional component that is plate component. As a result of the depth of block is thus smaller than its length in x and y direction. It is a flexural member subject to transverse load.

2. Beams

Beam is he one dimensional component that is subjected to transverse beam carries load from slabs, from masonry walls and then self-weights Beams is also supported on another beam old is also on supported on column Failure of beam are native failure, which may lead to failure of that floor

3. Columns

Columns square measure typically the vertical member subjected to axial or eccentric hundreds. The load carries by column is comes from beans and from higher columns if any. The column is incredibly necessary support because failure of column lends to collapse of whole structure.

4. Footing

Footing is the strong base to the structures. It helps to transmit the loads from structure uniformly to the large are beneath

Fig no 4

Working Stress Method (WSM)

Ultimate Load Method

Load factor= Collapse load / working load

 Limit State methods

Permissible Stresses for Materials within the WSM

Permissible stresses for different type of grades of concrete

Permissible stresses in steel in tension

Assumptions created in working stress method

MODULAR RATIO

The ratio between modulus of elasticity of steel to concrete is called as modular ratio.

M = Es/Ec

M= 280/3

BEAMS

Beams are the flexible members are classified into 2 types

Based on geometry

2.     Flanged beam

3.     Based on mode of failure

TYPE OF REACTANGULAR BEAM

Singly reinforcement beams

The beams section in which reinforcements provided to resist only tension is called as singly reinforcement section.

In such beams steel provided only one side of the section

Doubly reinforcement beams

The beam section in which reinforcements in which reinforcements is provided to resist tension and compression is called as doubly reinforcements section.

In such beam steel is provided on both side of the neutral axis

Fig no 5 Singly and doubly reinforced beams

Balanced, under –reinforcement and over reinforcement section for WSM

.

Fig no 6 Types of reinforced section

Balanced section

c b c/ s t/m = X c / d – X c 

Under reinforcement section

(b x) x/2 = m A s t (d – x)

 X < X c

Over reinforced section

                                 X > X c

Comparison between various sections

Need of under reinforced section in LSM design

BASICTERMONOLOGY

EFFECTIVE DEPTH

Effective depth of section is the distance between the center of the area of tension reinforcement and maximum compression fiber

d = overall depth –effective cover

d =D -/2 – clear cover

Fig no 7 Effective depth

CLEAR COVER

The minimum distance between the surface of concrete to the outside of the reinforcements is called clear cover.

D=d+ /2 +cover

It is provided to protect reinforcements from whether and fire so that if does not corrode or melts

To produce the grip of concrete to steel

To develop the sufficient bond strength along the surface area of the steel bar

Fig no 8 Effective depth

EFFECTIVE COVER

It is distance between extreme compression fiber of centre of tensile reinforcements

EFFECTIVE COVER =OVERALL DEPTH –EFFETIVE DEPTH

Nominal cover is the design depth of concrete cover to all the steel reinforcements including links

It shall not be less than the diameter of bar

Fig no 9 Effective cover

Minimum cover for various structural elements as per clause 26.4

Nominal cover to meet durability requirements as per clause 26.4.2 and table 16 I.S 456:2000

Neutral Axis in the Section

Fig no 10 Neutral axis

Fig no 11 Transformed section

Es =E c

s / Es =e/E e

s =Es/E e *e =me

e =s/m

A equivalent =A+ (m-1) A s t

Fig no 12 Position of N A when stress induced in section

.

E c =maximum strain in concrete

E s = maximum strain in steel

F c d =compressive stress in concrete in bending

F st = stress in steel in tension

E c = c b c /E c

And E s = s t /E s

Stress strain diagram is linear, hence the strain in concrete and steel are proportional to N.A

E c/E s=x/ (d-x)

( c b c/E c)/ ( s t /E s) =x/ (d-x)

( c b c / s t)*(E s /E c) = x/ (d-x)

Where m, E s /E c =m

c b c / s t * m =x/ (d-x)

        X = [1/(1+( s t /(m * c b c))]* d

Fig no 13 Position of N A when section is known

(Moment of area in compression about neutral axis )= (moment of area in tension about neutral axis).

Moment of resistance

The moment of couple formed by total tensile forces in steel acting and the total compressive force in the concrete acting at the counter balance the bending due to external loading is known as moment of resistance 

Fig no 14 Moment of resistance

Moment of resistance = tension force * lever arm

                                        = T*Z

                     MR             = s t A s t (d- x/3)

Moment of resistance = compressive force

                                        =C*Z

                     MR             = c b c/2 b x (d-x/3)

In flexural member

Moment of resistance helps to balance the bending moment due to loading

Key takeaways:

The basically design have different philosophy in which contains

Limit states methods

Working stress methods

Ultimate state methods

1.2 Design of Rectangular Singly and Doubly Reinforced Sections by Working Stress Method

SINGLY REINFORCEMENT SECTION

BEAM

A beam is any structural member which resist load mainly by bending. Therefore, it is also called flexural member. beam may be singly reinforced or doubly reinforced. When steel is provided only in tensile zone

DESIGN CONSTANT OF MATERIAL GRADE

Tensile strength of concrete f ct < modulus of rupture f r

Compression stress of concrete f c << ½ f c

Tensile stress in steel f s< f y yield strength of steel

2.     Stress elastic and section cracked

Tensile strength of concrete f ct > modulus of rupture

Compressive stress of concrete f c < ½ f c

Tensile stress in steel f s < f y yield strength of steel

SINGLY REINFORCED BEAM

Fig no 15 Singly reinforced beam

Compression in concrete C = ½ f c b k d

Tension in steel   T = A s f s

Equilibrium ∑F x = 0

   Compression = tension

½ f c b k d = A s f s

Reinforcement ratio 

= A s / b d

f c / f s  = 2 /k       ……………………(1)

STRAIN COMPABLITY

C /  S = k d / d- k d = k/ 1-k

( fc /E c) /(f s/E s) = k/ 1-k

n f c /f s = k/ 1-k

Know find k

K=)2 –n

Know f c, f sfind k

K = 1/1+ (f s/n f c )

MOMENT OF RESISTANCE

Moment arm distance: j d

J d = d – k d/3

J = 1- k/3

Steel

M = T * j d = A s f s j d

Concrete

M =C * j d = ½ f ck j b d2

R = ½ f c k j

DOUBLY REINFORCED BEAM

Fig no 16 Doubly reinforced section

Top of Form

Such reinforced concrete beams having steel reinforcement both on tensile and compressive faces are legendary as doubly bolstered beams. Doubly reinforced beams, therefore, have moment of resistance quite the singly reinforced beams of a similar depth for explicit grades of steel and concrete.

Lever arm constant

It is the distance between the compressive force and tensile force developed in the section due to flexural action in a R C member

It is denoted by z

Lever arm (z) = (d- x/3)Bottom of Form

MOMENT OF RESISTANCE

M = M1 + M2

M1 = MC = ½ f c k j b d2

       =A s 1 f s j d

A s 1 = M c / f s j d

M 2 = M – Mc

     = A s 2 f s (d –d’)

    = A s’ f s’ (d – d’)

A s 2 = M –M c/ f c (d –d’)

Area of steel

As = A s 1 + A s 2

STRAIN COMPRASIBILITY

S/ S’ = (d –k d)/(k d- d’)

From hook law,  s = E s f s ,s’= E s f s’

E s f s / E s  fs’ = f s / f s’ = ( d- k d )/( k d – d ‘)

F s’ = f s( (k – d’)/d) /1-k

Design steps

Given data: 

 Size of beam, area of steel and grade of material 

Step 1:  Depth of critical neutral axis (x c). 

            X c / d- X c = c b c/( s t/m )

Step 2:  Actual depth of neutral axis (x), 

                           b x2/2 = m A s t (d – x)

Step 3:  Type of beam section, 

                1) X= X c, balanced section. 

                2) X < X c, under reinforced section. 

                3) X > X c, over reinforced section. 

Step 4:  Moment of resistance, 

                1)  For under reinforced section 

                   MR = T • Z 

                           =   σs t •A s t (d – x/3) 

2) For over reinforced section 

                    MR = C • Z 

= (σc b c /2) •b x (d – x/3) 

3)  For balanced section

MR = T • Z = C • Z  

Problem: -

1)  A reinforced concrete beam section of size 300 mm x 700 mm effective depth is reinforced with 3 bars of 20 mm diameter in tension. The concrete mix and HYSD steel reinforcement used are of grade M20 and Fe415 respectively. 

Find. 

1)  Depth of neutral axis.

2) Type of reinforced section. 

3) Moment of resistance. 

4) Maximum stress induced in material WSM is recommended.

Ans: 

Given, 

b = 300 mm

d = 700 mm

To find: x, MR and σc b c

Sol2 

For M20,      σc b c = 7 N/mm2and m = 280 / 3 * c b c = 230/ 3* 7 = 13.33

For Fe415, σs t = 230 N/mm2

Area of steel = n * /2 * d 2 = 942.5 mm 2

Step 1: Depth of critical neutral axis:

 c b c / (s t /m) = X c / (d –X c)

                  0.406 = X c / (700 – X c)

                   700 = x c + 2.464 X c

X C = 202

Step 2:Actual depth of neutral axis (x) :

                   = m. A s t( d – x)

                 = 13.33x942.5 (700– x)

150 x2 = 8.794 x 106 – 12563.53x

150 x2 + 12563.53x - 8.794 x 106 = 0

 Solving quadratic equation, we get

x = 203.85 mm

Step 3: Type of beam section:

Since, x > X c, then section is over reinforced section and will be fails

in compression.                        

Step 4: Moment of resistance:

          MR = C* Z 

          MR = c b c/ 2 .b .x. (d – x/3) 

                 = 300 X 203.85 (700–203.85 /3) 

= 135.29 x 106 N.mm = 135.29 KN.m

Step 5: maximum stress induced in material:

Since the section fails in compression,

c b c max = 7 N/ mm 2

           Corresponding maximum stress in steel:

 = c b c /s t/m) = x – (d-x)

 = 93.33/ s t = 0.41086

s t max= 227.1 N/mm2

Key take ways

Design of reinforced section by working stress methods find

1.3 Assumptions in Limit State Design Method

F y/1.15Es + 0.002 f y

1.4 Design of Rectangular Singly and Doubly Reinforced beams, T-beams, L-beams by Limit State Design Method

SINGLY REINFORCEMENT BEAM

Fig no 17 Singly supported beamBottom of Form

Top of Form

X = depth of section

b = breadth of section

d = effective depth of section

The depth of neutral axis is obtained by considering the equilibrium of the traditional forces that is,

Resultant force of compression = average stress X area= 0.36 f c k b x

Resultant force of tension = 0.87 f y A t

Force of compression ought to be capable force of tension,

0.36 f c k b x = 0.87 f y A t

The distance between the lines of action of 2 forces C & T is termed the lever arm and is denoted by z.

Lever arm z = d – 0.42 x

z = d – 0.42

z = d – (f y A t/f c k b)

Moment of resistance with relevance concrete = compressive force x lever arm= 0.36 f c k b x z

Moment of resistance with respect to steel = tensile force x lever arm

= 0.87 f y A t z

Maximum depth of neutral axis

X m / (d – X m ) = 0.0035/ ((0.87 f y /E s )+ 0.002 )

X m /d = 0.0035 / ((0.87 f y / E s )+ 0.0055)

  Limiting value of tension steel and moment of resistance

Limiting moment of resistance values, N mm

SAMPLE NUMERICAL

Design of a section

Design of rectangular beam to resist a bending moment capable 45 kN m victimization. (1) M15 combine and delicate steel.

The beams are designed in order that below the applied moment each material reach their maximum stresses.

Assume quantitative relation of overall depth to breadth of the beam capable two

Breadth of the beam = b

Overall depth of beam = D

Therefore, D/b =

The balanced design,

Factored BM = moment of resistance with relevance concrete

= moment of resistance with respect to steel

= ratio X B.M

= 1.5 X 45= 67.5 kNm

For balanced section,

Moment of resistance Mu = 0.36 f c k b X m(d - 0.42 X m)

Grade for steel is Fe250For Fe250

For Fe 250 steel,

X m= 0.53d

Mu = 0.36 f c k b (0.53 d) (1 – 0.42 * 0.53) d

= 2.22bd

Since D/b =2 or, d/b = 2 or, b=d/2

Mu = 1.11 d

Mu = 67.5 X ten N mm

d=394 mm and b= 200mm

Adopt D = 450 mm, b = 25 mm, d = 415 mm

Area of tensile steel At = Factored BM /0.87 f y (d -0.42 X m)

                                          = 67.5 * 103 /0.87 * 250 * (1- 0.42 * 0.53 ) 415

In beams the diameter of main strengthened bars is typically hand-picked between twelve mm and 25 mm. Provide 2-20mm and 1-22mm bars giving total area

                                          = 6.28 + 3.80

                                         =10.08 cm > 9.62 cm

DOUBLY REINFORCEMENT BEAM:

Fig no 18 Doubly reinforced section

 If the applied bending moment is bigger than the instant resisting capability or final moment carrying capability of the section (I. e, over strengthened section), then there are often 3 alternatives:

1) If attainable, increase the size of the section, preferred 'depth

 2) Use higher grade of concrete

3) Steel reinforcement is also more in compression zone to extend the instant resisting capability of the section (depth is stay because it is). This is often called a doubly reinforced beam'.

The double reinforced section is often needed underneath the subsequent circumstances:

1) Sectional dimensions area unit restricted thanks to needs ofhead room, look etc. and therefore the strength of given on an individual basis strengthened section is insufficient.

2) The beam that acts as a flanged beam at middle span becomes a rectangular beam at supports of a nonstop beam. At support tension happens at prime creating the projection ineffective and there for the section becomes inadequate to resist giant peak price of support moment,

4) Basement with lower support level and combined with ventilator couples and to style a double strengthened beam.

5) Compression steel is provided typically to scale back the deflection i.e. lo increase the stiffness and additionally to extend the rotation capability.

 6) Compression steel is often utilized in structures in earthquake regions to extend their ductility

7) Compression reinforcement also will and considerably in reducing the future deflection of beams.

The failure theory evolved for on an individual basis strengthened section holds smart for doubly strengthened beam also. The most assumption

1) Plane sections stay plain even when bending

2) R.C. sections in bending fail once the compressive strain within the concrete reaches the worth of zero.0035

3) The strain at any purpose in steel and concrete is taken as adequate the strain resembling the strain at that time of the strain v/s strain graph for the fabric (steel or concrete)

4) X u limited / d' magnitude relation needn't be strictly adhered to in double strengthened beams

5) Even supposing shrinkage, creep and alternative properties of concrete wil have an effect on the particular stat of stress of steel and concrete, these aren't taken into thought in estimating the collapse or final strength

A doubly reinforced section may be looks upon as made up of 2 sections 1 and 2 gives below

Section 1

A singly reinforced section with concrete resisting compressive force c balanced by tensile force t provided by tension steel. this section is assumed so resist part moment

Section2

An imaginary section consisting of compression steel providing additional compression forces which is balanced by tensile force given by tension steel. This section is assumed to resist balancing moment M

DESIGN STEPS

Given

Grade of concrete

Grade of steel

Size of beam

Reinforcement provided

If X u = X u max …………….balanced

If X u< X u max …………….under reinforced

If X u> X u max …………….over reinforced

Equating the compression force and tension force

0.36 * f c k * b * X u * + A s c * f s c = 0.87 * f y * A  s t

From this find X u

M u = 0.36 * f c k * b * X u * (d- 0.42 * X u) + A s c * f s c * (d- d’)

This equation is valid for balanced and under reinforced section

If the section is over reinforced section then take X u = X u max in above equation and use the same equation for finding M u. Do not increase the depth of the beam for over reinforced section.

Given

Grade of concrete and steel

Size of beam

Ultimate moment

M u is given

For balanced section X u = X u max

M u limiting = 0.36 * f c k * b * X u max * (d- 0.42 * X u max)

So M u 2 = M u – M u limiting

A s t limit = M u limit / 0.87 * f y * (d – 0.42 * X u max)

A sc = M u2/ f s c (d- d’)

As t 2 = A s c * f s c/ 0.87 f y

As t =A s t limit + A s t 2

Where

d = effective depth of beam

d’= distance between extreme compression fiber to the center of comp reinforced

b = width of beam

X u = depth of actual neutral axis

X u max = depth of critical neutral axis

A s t = total area of tensile reinforced in mm2

F c k = characteristic compressive strength of concrete

F y = characteristic strength of steel

M u limit = limiting M R of a section without compression reinforced in K N m Top of Form

            M u = ultimate M R

            A s c= area of compression reinforcement

T BEAM

FIG NO 19 T BEAM

A beam is considered as such even if it has a large base to accommodate tension steel

The shape and size of concrete on the tension side assumed to be cracked has no effect on the theoretical moments

R C floors normally consist of slabs and beams that are cast monolithically the two acting together to resist load and because of this interaction the effective section of beam is a T or L beam

The effective flanged width resulting from slab interaction depend on slab type

Effective flanged beam

Exterior beam

Interior beam

Location of neutral axis and rectangular stress block depth

If the stress block falls in the flanged as is very frequent for positive moments the rectangular beam formula apply

In this case tensile concrete is assumed to be cracked and its shapes has no effect

The section is analysis as a rectangular one using the flanged b = bf

Minimum steel for T beam

Minimum steel quantity is imposed by codes for the same reasons as in rectangular beams

A s min =Max [( / 4 f y), (1.4 / f y)] * b w * d

Analysis of T beams

Analysis steps for T beams

As f y/0.85 f y b f

If a <= h f stress block in flanged

If a >= h f stress block in web

Calculate neutral axis depth =a /

and steel strain

Calculate M n ,

and

M n

L BEAM

Fig no 20 L BEAM

In bending the beam take the tension force and slab take compression force since the L beam receive their load from one side only they are subjected to consideration amount of torsion moments so L beam are subjected to bending moments shear force and torsion moments

Moment carried by flange of L beams

The moment M1 causes torsion in the beam and is known as the torsion moments which is restricted by the rectangular portion only and the flanged does not contribute to torsion moment of resistance

Moment carried by flanged of L beam and the slab

The central load W cause bending moments in the beam which is joined restricted by the rectangular portion of the beam as well as the flanged

Changes in floor level may be accommodated by either an L beam or building up one side of an inverted T beam

Code recommends that effective width of the flanged of beam may be taken as follows

2.     for isolated L beams, lesser of

Where

b f = effective width of flanged

b w =breadth of rib

b = actual width of flanged

I o = distance between points of zero moments in beam

The design principles to be followed while designing L beam

Given

Step 1: determine width of flanged

Steps 2: fixed up a suitable depth

Steps 3: calculate MR of the section assuming the failure of concrete to occur .if the moment of resistance is greater than bending moments assumed depth is ok otherwise either increases the depth of the section or provide additional reinforcement in compression side in addition to the tension side

Steps 4: calculate the amount of tension steel for actual bending moments treating it as a balanced section

Steps 5: check for the shear and provide shear reinforced

Steps 6: check of deflection

Steps 7: check for end anchoring and development length

Key takeaways:

In this point we see single reinforced and doubly reinforced beam by LSM method and also the T section beam and L section beam

In this method also same parameter is calculating

References