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PLEASE USE THE EXAMPLE AS GUIDE AND SHOW ALL THE PROCEDURES, OTHERWISE DON'T ANSWER THIS.   Design a rectangular simply supported flexural member to be used as internal simply supported long beam. Slab has longer span of 8m and shorter span of 5m. Factored slab load is 12KN/m2. Use C-18 concrete and Grade 300 steel. Select SI reinforcement bars and keep maximum width to 300mm. The structure has to be used as commercial building. Architecturally maximum permitted depth should be 550mm. Data L: 8m W: 5m Factored Slab Load: 12 kN/m2 Fc’: 18 Mpa Fy: 300 Mpa b: 300 mm Ha: 375 mm Density of concrete: 2400 kg/m2 Calculations need: Service dead load Factored dead load: Equivalent width of slab supported Factored slab load acting on beam Total factored load Total factored bending moment Effective Depth Depth for deflection Most general depth Dead load Depth of the beam Minimum steel ratio Maximum steel ratio Reinforcement by hit and try please note that l= 8m as is the long beam that is need to be calculated additionally the Equivalent width of slab supported for a long beam needs to be calculated = (1- R²/3)lx the width needs to be increased by 10% thank you

PLEASE USE THE EXAMPLE AS GUIDE AND SHOW ALL THE PROCEDURES, OTHERWISE DON'T ANSWER THIS.   Design a rectangular simply supported flexural member to be used as internal simply supported long beam. Slab has longer span of 8m and shorter span of 5m. Factored slab load is 12KN/m2. Use C-18 concrete and Grade 300 steel. Select SI reinforcement bars and keep maximum width to 300mm. The structure has to be used as commercial building. Architecturally maximum permitted depth should be 550mm. Data L: 8m W: 5m Factored Slab Load: 12 kN/m2 Fc’: 18 Mpa Fy: 300 Mpa b: 300 mm Ha: 375 mm Density of concrete: 2400 kg/m2 Calculations need: Service dead load Factored dead load: Equivalent width of slab supported Factored slab load acting on beam Total factored load Total factored bending moment Effective Depth Depth for deflection Most general depth Dead load Depth of the beam Minimum steel ratio Maximum steel ratio Reinforcement by hit and try please note that l= 8m as is the long beam that is need to be calculated additionally the Equivalent width of slab supported for a long beam needs to be calculated = (1- R²/3)lx the width needs to be increased by 10% thank you

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter9: Composite Construction
Section: Chapter Questions
Problem 9.1.3P
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PLEASE USE THE EXAMPLE AS GUIDE AND SHOW ALL THE PROCEDURES, OTHERWISE DON'T ANSWER THIS.

 

Design a rectangular simply supported flexural member to be used as internal simply supported long beam. Slab has longer span of 8m and shorter span of 5m. Factored slab load is 12KN/m2. Use C-18 concrete and Grade 300 steel. Select SI reinforcement bars and keep maximum width to 300mm. The structure has to be used as commercial building. Architecturally maximum permitted depth should be 550mm.

Data

L: 8m

W: 5m

Factored Slab Load: 12 kN/m2

Fc’: 18 Mpa

Fy: 300 Mpa

b: 300 mm

Ha: 375 mm

Density of concrete: 2400 kg/m2

Calculations need:

  1. Service dead load
  2. Factored dead load:
  3. Equivalent width of slab supported
  4. Factored slab load acting on beam
  5. Total factored load
  6. Total factored bending moment
  7. Effective Depth
  8. Depth for deflection
  9. Most general depth
  10. Dead load
  11. Depth of the beam
  12. Minimum steel ratio
  13. Maximum steel ratio
  14. Reinforcement by hit and try

please note that l= 8m as is the long beam that is need to be calculated additionally the Equivalent width of slab supported for a long beam needs to be calculated = (1- R²/3)lx the width needs to be increased by 10% thank you

Example 3.6: Design a rectangular singly reinforced flexural member to be used as an interior simply supported short beam. The slab system has a longer span of 5 m and a shorter span of 3.5 m. Factored stab load is 10.00 kN/m, 17.25 MPa and = 300 MPa. Select SI reinforcing bars and keep the width "of member equal to 228 mm. Architecturally, maximum permitted depth is 375 mm. Solutton: Data (- 3,5 m factored slab load 10.00 kN/m" S'- 15 MPa I,- 300 MPa b- 228 mm hu= 375 mm Approximate Self Weight - 2400 x 2283.5 9.81 1000" 12 "1000 Service dead load = 1.57 kN/m Factored dead load- 1.2 x 1.57 - 1.88 kN/m Selected Depth - 300 mm = h-60 - 240 mm The figure of 60 mm is corresponding to expected one layer of steel. E Factored Slab Load Acting On Beam Minimum Steel Ratio Factored slab load on beam - width of slab x slab load per unit arca - 234 x 10.00 = 23.4 kN/m Anie - 141, - 14/300 - 0.00467 Total Factored Load Maximum Steel Ratio - 23.4 + 1.88 - 25.3 kN/m An - 0375 x 0.85 A 17.25 Total Factored Bending Moment - 0.375 x 0.85 x 0.85 x 300 - 0.0156 - M. Calculation Of Steel 25.3x3.5 38.74 AN-m Trial 1: Assume a d/3- 240/3 - 80 mm 8 - 717 mm? M 38.74 x10 Minimum Effective Depth For Singly Reinforced Section 0.9x 300 240- hat - dain + 60 (assuming one layer of steel) 4,1, 0.85 Ca Trial 2: 39.28.74 x10 Vo.205 x17.25x 228 M. + 60- + 60- 279 mm V0.205, 77x 300 64 mm Depth For Deflection Control 0.85 x17.25x 228 38.74 x 10" - 690 mm Minimum depth of beam for deflection control (A) e/ 20 - 3500/ 20 0.9× 300 240- - 175 mm Maximum Architectural Depth 690x 300 0.85x17.25 x 228 Trial 3: - 62 mm hamar - 375 mm 38.74 x 10** - 687 mm Most General Depth 0.9x300 240- -// 12 - 3500/12 - 292 mm The depth may be selected in multiples of the brick height, if possible. (sufficiently close to the previous answer) A reinforcement of 2 20 +1 IS provides the required arca of steel (Fig. 3.27).
Transcribed Image Text:Example 3.6: Design a rectangular singly reinforced flexural member to be used as an interior simply supported short beam. The slab system has a longer span of 5 m and a shorter span of 3.5 m. Factored stab load is 10.00 kN/m, 17.25 MPa and = 300 MPa. Select SI reinforcing bars and keep the width "of member equal to 228 mm. Architecturally, maximum permitted depth is 375 mm. Solutton: Data (- 3,5 m factored slab load 10.00 kN/m" S'- 15 MPa I,- 300 MPa b- 228 mm hu= 375 mm Approximate Self Weight - 2400 x 2283.5 9.81 1000" 12 "1000 Service dead load = 1.57 kN/m Factored dead load- 1.2 x 1.57 - 1.88 kN/m Selected Depth - 300 mm = h-60 - 240 mm The figure of 60 mm is corresponding to expected one layer of steel. E Factored Slab Load Acting On Beam Minimum Steel Ratio Factored slab load on beam - width of slab x slab load per unit arca - 234 x 10.00 = 23.4 kN/m Anie - 141, - 14/300 - 0.00467 Total Factored Load Maximum Steel Ratio - 23.4 + 1.88 - 25.3 kN/m An - 0375 x 0.85 A 17.25 Total Factored Bending Moment - 0.375 x 0.85 x 0.85 x 300 - 0.0156 - M. Calculation Of Steel 25.3x3.5 38.74 AN-m Trial 1: Assume a d/3- 240/3 - 80 mm 8 - 717 mm? M 38.74 x10 Minimum Effective Depth For Singly Reinforced Section 0.9x 300 240- hat - dain + 60 (assuming one layer of steel) 4,1, 0.85 Ca Trial 2: 39.28.74 x10 Vo.205 x17.25x 228 M. + 60- + 60- 279 mm V0.205, 77x 300 64 mm Depth For Deflection Control 0.85 x17.25x 228 38.74 x 10" - 690 mm Minimum depth of beam for deflection control (A) e/ 20 - 3500/ 20 0.9× 300 240- - 175 mm Maximum Architectural Depth 690x 300 0.85x17.25 x 228 Trial 3: - 62 mm hamar - 375 mm 38.74 x 10** - 687 mm Most General Depth 0.9x300 240- -// 12 - 3500/12 - 292 mm The depth may be selected in multiples of the brick height, if possible. (sufficiently close to the previous answer) A reinforcement of 2 20 +1 IS provides the required arca of steel (Fig. 3.27).
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