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Information about Singly Reinforce Concrete

Published on November 27, 2007

Author: guest00bb06

Source: slideshare.net

this slide will clear all the topics and problem related to singly reinforced beam by limit state method, things are explained with diagrams , easy to understand .

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DIFFERENT METHODS OF DESIGN OF CONCRETE Working Stress Method Limit State Method Ultimate Load Method Probabilistic Method of Design

Working Stress Method

Limit State Method

Ultimate Load Method

Probabilistic Method of Design

LIMIT STATE METHOD OF DESIGN The object of the design based on the limit state concept is to achieve an acceptable probability, that a structure will not become unsuitable in it’s lifetime for the use for which it is intended, i.e. It will not reach a limit state A structure with appropriate degree of reliability should be able to withstand safely. All loads, that are reliable to act on it throughout it’s life and it should also satisfy the subs ability requirements, such as limitations on deflection and cracking.

The object of the design based on the limit state concept is to achieve an acceptable probability, that a structure will not become unsuitable in it’s lifetime for the use for which it is intended,

i.e. It will not reach a limit state

A structure with appropriate degree of reliability should be able to withstand safely.

All loads, that are reliable to act on it throughout it’s life and it should also satisfy the subs ability requirements, such as limitations on deflection and cracking.

It should also be able to maintain the required structural integrity, during and after accident, such as fires, explosion & local failure. i.e. limit sate must be consider in design to ensure an adequate degree of safety and serviceability The most important of these limit states, which must be examine in design are as follows Limit state of collapse - Flexure - Compression - Shear - Torsion This state corresponds to the maximum load carrying capacity.

It should also be able to maintain the required structural integrity, during and after accident, such as fires, explosion & local failure.

i.e. limit sate must be consider in design to ensure an adequate degree of safety and serviceability

The most important of these limit states, which must be examine in design are as follows

Limit state of collapse

- Flexure

- Compression

- Shear

- Torsion

This state corresponds to the maximum load carrying capacity.

TYPES OF REINFORCED CONCRETE BEAMS Singly reinforced beam Doubly reinforced beam Singly or Doubly reinforced flanged beams

Singly reinforced beam

Doubly reinforced beam

Singly or Doubly reinforced flanged beams

SINGLY REINFORCED BEAM In singly reinforced simply supported beams or slabs reinforcing steel bars are placed near the bottom of the beam or slabs where they are most effective in resisting the tensile stresses .

In singly reinforced simply supported beams or slabs reinforcing steel bars are placed near the bottom of the beam or slabs where they are most effective in resisting the tensile stresses .

Reinforcement in simply supported beam COMPRESSION b STEEL REINFORCEMENT D d TENSION SUPPORT SECTION A - A CLEAR SPAN

Reinforcement in simply supported beam

COMPRESSION b

STEEL REINFORCEMENT D d

TENSION

SUPPORT SECTION A - A

CLEAR SPAN

Reinforcement in a cantilever beam A TENSION D d COMPRESSION SECTION A - A A CLEAR COVER

Reinforcement in a cantilever beam

A

TENSION

D

d

COMPRESSION SECTION A - A

A

CLEAR COVER

STRESS – STRAIN CURVE FOR CONCRETE f ck STRESS .20 % .35% STRAIN

STRESS – STRAIN CURVE FOR CONCRETE

f ck

STRESS

.20 % .35%

STRAIN

STRESS ― STARIN CURVE FOR STEEL

STRESS ― STARIN CURVE FOR STEEL

STRESS BLOCK PARAMETERS ! 0.0035 0.446 f ck X 2 X 2 a X 1 X 1

!

0.0035 0.446 f ck

X 2 X 2 a

X 1 X 1

x = Depth of Neutral axis b = breadth of section d = effective depth of section The depth of neutral axis can be obtained by considering the equilibrium of the normal forces , that is, Resultant force of compression = average stress X area = 0.36 f ck bx Resultant force of tension = 0.87 f y A t Force of compression should be equal to force of tension, 0.36 f ck bx = 0.87 f y A t x = Where A t = area of tension steel

x = Depth of Neutral axis

b = breadth of section

d = effective depth of section

The depth of neutral axis can be obtained by considering the equilibrium of the normal forces , that is,

Resultant force of compression = average stress X area

= 0.36 f ck bx

Resultant force of tension = 0.87 f y A t

Force of compression should be equal to force of tension,

0.36 f ck bx = 0.87 f y A t

x =

Where A t = area of tension steel

Moment of resistance with respect to concrete = compressive force x lever arm = 0.36 f ck b x z Moment of resistance with respect to steel = tensile force x lever arm = 0.87 f y A t z

Moment of resistance with respect to concrete = compressive force x lever arm

= 0.36 f ck 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 A compression failure is brittle failure. The maximum depth of neutral axis is limited to ensure that tensile steel will reach its yield stress before concrete fails in compression, thus a brittle failure is avoided. The limiting values of the depth of neutral axis x m for different grades of steel from strain diagram.

A compression failure is brittle failure.

The maximum depth of neutral axis is limited to ensure that tensile steel will reach its yield stress before concrete fails in compression, thus a brittle failure is avoided.

The limiting values of the depth of neutral axis x m for different grades of steel from strain diagram.

MAXIMUM DEPTH OF NEUTRAL AXIS f y N/mm 2 x m 250 0.53 d 415 0.48 d 500 0.46 d

LIMITING VALUE OF TENSION STEEL AND MOMENT OF RESISTANCE Since the maximum depth of neutral axis is limited, the maximum value of moment of resistance is also limited . M lim with respect to concrete = 0.36 f ck b x z = 0.36 f ck b x m (d – 0.42 x m ) M lim with respect to steel = 0.87 f ck A t (d – 0.42 x m )

Since the maximum depth of neutral axis is limited, the maximum value of moment of resistance is also limited .

M lim with respect to concrete = 0.36 f ck b x z

= 0.36 f ck b x m (d – 0.42 x m )

M lim with respect to steel = 0.87 f ck A t (d – 0.42 x m )

LIMITING MOMENT OF RESISTANCE VALUES, N MM Grade of concrete Grade of steel Fe 250 steel Fe 450 steel Fe 500 steel General 0.148 f ck bd 0.138 f ck bd 0.133 f ck bd M20 2.96 bd 2.76 bd 2.66 bd M25 3.70 bd 3.45 bd 3.33 bd M30 4.44 bd 4.14 bd 3.99 bd

TYPES OF PROBLEM Analysis of a section Design of a section

Analysis of a section

Design of a section

For under reinforced section, the value of x/d is less than x m /d value. The moment of resistance is calculated by following equation: M u = 0.87 f y A t d – For balanced section, the moment of resistance is calculated by the following equation: M u = 0.87 f y A t ( d – 0.42x m ) For over reinforced section, the value of x/D is limited to xm/d and the moment of resistance is computed based on concrete: Mu = 0.36 f ck b x m ( d – 0.42 x m )

For under reinforced section, the value of x/d is less than x m /d value.

The moment of resistance is calculated by following equation:

M u = 0.87 f y A t d –

For balanced section, the moment of resistance is calculated by the following equation:

M u = 0.87 f y A t ( d – 0.42x m )

For over reinforced section, the value of x/D is limited to xm/d and the moment of resistance is computed based on concrete:

Mu = 0.36 f ck b x m ( d – 0.42 x m )

Analysis of section

Analysis of section

Determine the moment of resistance for the section shown in figure. (i) f ck = 20 N/mm , f y = 415 N/mm Solution: (i) f ck = 20 N/mm , f y = 415 N/mm breadth (b) = 250 mm effective depth (d) = 310 mm effective cover = 40 mm Force of compression = 0.36 f ck b x = 0.36 X 20 X 250x = 1800x N

Determine the moment of resistance for the section shown in figure.

(i) f ck = 20 N/mm , f y = 415 N/mm

Solution:

(i) f ck = 20 N/mm , f y = 415 N/mm

breadth (b) = 250 mm

effective depth (d) = 310 mm

effective cover = 40 mm

Force of compression = 0.36 f ck b x

= 0.36 X 20 X 250x

= 1800x N

Area of tension steel A t = 3 X 113 mm Force of Tension = 0.87 f y A t = 0.87 X 415 X 3 X 113 = 122400 N Force of Tension = Force of compression 122400 = 1800x x = 68 mm x m = 0.48d = 0.48 X 310 = 148.8 mm 148.8 mm > 68 mm Therefore, Depth of neutral axis = 68 mm f y x m 415 0.48d 500 0.46d

Area of tension steel A t = 3 X 113 mm

Force of Tension = 0.87 f y A t

= 0.87 X 415 X 3 X 113

= 122400 N

Force of Tension = Force of compression

122400 = 1800x

x = 68 mm

x m = 0.48d

= 0.48 X 310

= 148.8 mm

148.8 mm > 68 mm

Therefore,

Depth of neutral axis = 68 mm

Lever arm z = d – 0.42x = 310 – 0.42 X 68 = 281 mm As x < x m ( It is under reinforced ) Since this is an under reinforced section, moment of resistance is governed by steel. Moment of resistance w.r.t steel = tensile force X z M u = 0.87 f y A t z = 0.87 X 415 X 3 X 113 X 281 Mu = 34.40kNm

Lever arm z = d – 0.42x

= 310 – 0.42 X 68

= 281 mm

As x < x m ( It is under reinforced )

Since this is an under reinforced section, moment of resistance is governed by steel.

Moment of resistance w.r.t steel = tensile force X z

M u = 0.87 f y A t z

= 0.87 X 415 X 3 X 113 X 281

Mu = 34.40kNm

Design of a section

Design of a section

Question : Design a rectangular beam to resist a bending moment equal to 45 kNm using (i) M15 mix and mild steel Solution : The beam will be designed so that under the applied moment both materials reach their maximum stresses. Assume ratio of overall depth to breadth of the beam equal to 2. Breadth of the beam = b Overall depth of beam = D therefore , D/b = 2 For a balanced design, Factored BM = moment of resistance with respect to concrete = moment of resistance with respect to steel = load factor X B.M = 1.5 X 45 = 67.5 kNm

Question : Design a rectangular beam to resist a bending moment equal to 45 kNm using (i) M15 mix and mild steel

Solution :

The beam will be designed so that under the applied moment both materials reach their maximum stresses.

Assume ratio of overall depth to breadth of the beam equal to 2.

Breadth of the beam = b

Overall depth of beam = D

therefore , D/b = 2

For a balanced design,

Factored BM = moment of resistance with respect to concrete

= moment of resistance with respect to steel

= load factor X B.M

= 1.5 X 45

= 67.5 kNm

For balanced section, Moment of resistance M u = 0.36 f ck b x m (d - 0.42 x m ) Grade for mild steel is Fe250 For Fe250 steel, x m = 0.53d M u = 0.36 fck b (0.53 d) (1 – 0.42 X 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 10 Nmm d=394 mm and b= 200mm f y x m 250 0.53d 415 0.48d

For balanced section,

Moment of resistance M u = 0.36 f ck b x m (d - 0.42 x m )

Grade for mild steel is Fe250

For Fe250 steel,

x m = 0.53d

M u = 0.36 fck b (0.53 d) (1 – 0.42 X 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 10 Nmm

d=394 mm and b= 200mm

Adopt D = 450 mm b = 250 mm d = 415mm Area of tensile steel At = = = 962 mm = 9.62 cm Minimum area of steel A o = 0.85

Adopt D = 450 mm

b = 250 mm

d = 415mm

Area of tensile steel At =

=

= 962 mm

= 9.62 cm

Minimum area of steel A o = 0.85

= = 353 mm 353 mm < 962 mm In beams the diameter of main reinforced bars is usually selected between 12 mm and 25 mm. Provide 2-20mm and 1-22mm bars giving total area = 6.28 + 3.80 = 10.08 cm > 9.62 cm

=

= 353 mm

353 mm < 962 mm

In beams the diameter of main reinforced bars is usually selected between 12 mm and 25 mm.

Provide 2-20mm and 1-22mm bars giving total area

= 6.28 + 3.80

= 10.08 cm > 9.62 cm

!

!

SLIDES BY : HARSIMRAN SINGH TIWANA ROLL NO :5059 UNIVERSITY : 514010017 GROUP MEMBERS : DILRAJ SINGH D3/CIVIL/5051 HARSIMRAN SINGH D3/CIVIL/5059 AMNINDER SINGH D3/CIVIL/5060

SLIDES BY :

HARSIMRAN SINGH TIWANA

ROLL NO :5059

UNIVERSITY : 514010017

GROUP MEMBERS :

DILRAJ SINGH D3/CIVIL/5051

HARSIMRAN SINGH D3/CIVIL/5059

AMNINDER SINGH D3/CIVIL/5060

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