Unit I:
Grade of Concrete:
Grade of Concrete is a measure of its strength and is designated by writing the value of the 28days cube strength in KN/Sqm after an alphabet 'M'
Various Grades:
M15, M20, M25, M30, M35,....
As per IS:256-2000, minimum grade of M20 should be used for design
Grades of Steel:
Fe250 - Mild Steel - will have characterestic strength of 250N/Sqm
Fe415 & Fe500 - High Yield Strength Deformed (HYSD) bars
Most commonly used grade is Fe415
Unit II
Flexural Members or Beams:
1. Singly Reinforced Beams - simply supported / fixed/ cantilever beams where depth is not a constraint.
2. Doubly Reinforced Beams - designed when depth of beam is limited. Usually at the supports of beam.
3. T Beams & L Beams- beams can be designed as T-beam or L-beam by taking the advantage of compressible strength of slab. Usually at the mid span of simply supported/fixed beams
Methods of Design:
1. Working Stress Method
2. Limit State Design Method
In Limit State Design Method:
Based on depth of NA, beams are
1. Balanced beams (if Xu = Xumax)
MR = C x Z or T x Z
2. Under Reinforced Beams (if Xu< Xumax)
MR = T x Z
3. Over Reinforced Beams (if Xu > Xumax)
MR = C x Z
Where
C = 0.36 fck b Xu
T = 0.87 fy Ast
Z = d-0.42Xu
Xumax Equation: Xumax / 0.0035 = d-xumax / (0.87fy/Es + 0.002)
Fe250 - 0.53d
Fe415 - 0.48d
Fe500 - 0.46d
d = effective depth = Overall depth of beam - Clear Cover - Half of Dia. of Steel bar
Clear Cover = 25mm for beams
Xu:
Find Xu by Equating C = T
Singly Reinforced Beams:
Type of Problems:
1. Find Ultimate Moment of Resistance of a given Beam
Procedure:
Procedure
Grade of Concrete:
Grade of Concrete is a measure of its strength and is designated by writing the value of the 28days cube strength in KN/Sqm after an alphabet 'M'
Various Grades:
M15, M20, M25, M30, M35,....
As per IS:256-2000, minimum grade of M20 should be used for design
Grades of Steel:
Fe250 - Mild Steel - will have characterestic strength of 250N/Sqm
Fe415 & Fe500 - High Yield Strength Deformed (HYSD) bars
Most commonly used grade is Fe415
Unit II
Flexural Members or Beams:
1. Singly Reinforced Beams - simply supported / fixed/ cantilever beams where depth is not a constraint.
2. Doubly Reinforced Beams - designed when depth of beam is limited. Usually at the supports of beam.
3. T Beams & L Beams- beams can be designed as T-beam or L-beam by taking the advantage of compressible strength of slab. Usually at the mid span of simply supported/fixed beams
Methods of Design:
1. Working Stress Method
2. Limit State Design Method
In Limit State Design Method:
Based on depth of NA, beams are
1. Balanced beams (if Xu = Xumax)
MR = C x Z or T x Z
2. Under Reinforced Beams (if Xu< Xumax)
MR = T x Z
3. Over Reinforced Beams (if Xu > Xumax)
MR = C x Z
Where
C = 0.36 fck b Xu
T = 0.87 fy Ast
Z = d-0.42Xu
Xumax Equation: Xumax / 0.0035 = d-xumax / (0.87fy/Es + 0.002)
Fe250 - 0.53d
Fe415 - 0.48d
Fe500 - 0.46d
d = effective depth = Overall depth of beam - Clear Cover - Half of Dia. of Steel bar
Clear Cover = 25mm for beams
Xu:
Find Xu by Equating C = T
Singly Reinforced Beams:
Type of Problems:
1. Find Ultimate Moment of Resistance of a given Beam
Procedure:
- Write Given Data
- Find Xu by equating C=T
- Find Xumax for given grade of Steel
- Find section of beam (balanced / under reinforced / over reinforced)
- Find MR based on section
- Draw Cross Section of beam and show steel reinforcement details
Procedure
- Write Given Data
- Find Load, Factored Load, and Factored BM
- Find Xumax for a given grade of steel (assume balanced section)
- Find 'd' by equating BM =C x Z
- Find Overall depth of beam
- Find Steel Reinforcement by equating BM = T x Z
- Find No. of bars. (Assume 12mm/16mm/20mm/25mm dia based on Ast required)
- Draw Cross Section of beam and show steel reinforcement details
Check: D is in between 1/10 to 1/15 of the span
Doubly Reinforced Beams:
Type of Problems:
1. Find Ultimate Moment of Resistance of a given Beam
Procedure:
Procedure
1. Find Ultimate Moment of Resistance of a given Beam
Procedure:
- Write Given Data
- Find Mu = Mu limited + Mu2
- Find Mu Limited = C1 x Z1 or T1 x Z1 (assume balanced section like singly reinforced beam)
- Find Mu2 = C2 x Z2 or T2 x Z2
- Find C1 = 0.36fck b Xu
- Find Z1 = d-0.42Xu
- Find T1 = 0.87fy Ast
- Find C2 = fsc Asc - fcc Ac
- Find T2 = d-d'
- Find d' = d - top clear cover - half of dia of top steel bar
- Draw Cross Section of beam and show steel reinforcement details
Procedure
- Write Given Data
- Find Load, Factored Load, and Factored BM
- Find Mu limited = C x Z
- If Mu > Mu limited, then design as Doubly reinforced beam
- Mu = Mu limited + Mu2
- Find Mu2 = Mu - Mu limited
- Find Ast1 by equating Mu limited = T1 x Z1
- Find Ast2 by equating Mu2 = T1 x Z2
- Total Ast = Ast1 + Ast2
- Find Asc by equating Mu2 = T2 x Z2
- Draw Cross Section of beam and show steel reinforcement details
T - Beams & L - Beams:
Type of Problems:
1. Find Ultimate Moment of Resistance of given T-beam or L-Beam
Procedure:
- Write Given Data
- Assume Xu lies in Slab and find Xu by Equating C = T
- Find C = 0.36 fck bf
- Find T = 0.87 fy Ast
- bf for T - beam = L/6 + bw + 6 Df (for multiple spans)
- bf for L - beam = L/12 + bw + 3 Df (for multiple spans)
- Find MR based on Xu lies in 1) xu < Df 2) 3/7 Xu > Df or Df/d < 02 3) 3/7 Xu< Df or Df/d > 0.2
- Draw Cross Section of beam and show steel reinforcement details
2. Design T - Beam or L - Beam:
Procedure:
- Write Given Data
- Find Load, Factored Load, and Factored BM
- Assume Xu lies in Slab
- Assume Depth of Beam (1/2 to 1/15 of span)
- Find Steel Reinforcement by equating BM = T x Z
- Find No. of bars. (Assume 12mm/16mm/20mm/25mm dia based on Ast required)
- Draw Cross Section of beam and show steel reinforcement details
Note: bf for Isolated T-beam and L-beam can be found from IS Code
Tips to get Maximum Marks:
Tips to get Maximum Marks:
- Given Data
- Assumptions
- Design Procedure
- Write references taken from IS Code book
- Formula or Equation with terminology Ex. MR = Momnet of Resistance, d = effective depth
- Draw Cross Sections of structures with Steel Reinforcement Details
- Units
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