CE2306 DESIGN OF RC ELEMENTS UNIT II LIMIT STATE DESIGN FOR FLEXTURE Part A (2marks) 1. Explain the check for deflection control in the design of

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1 CE2306 DESIGN OF RC ELEMENTS UNIT II LIMIT STATE DESIGN FOR FLEXTURE Part A (2marks) 1. Explain the check for deflection control in the design of slabs? (NOV-DEC 2012) The deflection of a structure or part thereof shall not adversely affect the appearance or efficiency of the structure or finishes or partitions. The deflection shall generally be limited to the following: a) The final deflection due to all loads including the effects of temperature, creep and shrinkage measured from the as-cast level of the, supports of floors, roofs and all other horizontal members, should not normally exceed span/250. b) The deflection including the effects of temperature, creep and shrinkage occurring after erection of partitions and the application of finishes should not normally exceed span/350 or 20 mm whichever is less. 2. When do you do for doubly reinforced beams? (NOV-DEC 2012) (NOV-DEC 2010) (APRIL MAY 2012) The section reinforced in both tension and compression zone is known as doubly reinforced section. The doubly reinforced beams are adopt when the balanced moment is smaller than the Actual moment. 3. What type of slabs is usually used in practice under reinforced or over reinforced? (NOV-DEC 2009) The depth of slab chosen from deflection requirements will be usually greater than the depth required for balanced design. Hence the area of steel required will be less than the balanced amount. So, the slab is designed as under reinforced section. 4. Why is necessary to provide transverse reinforcement in one way slab? (APRIL MAY 2012) Since the one way slab bends in one direction and also in shorter direction, so it is necessary to provide transvers reinforcement in one way slabs. These slabs adopted when availability of two supports in one direction. 5. Distinguish between under reinforced and over reinforced sections. (MAY JUNE 2009) A beam reaches its permissible stress in steel under the working moment before concrete reaches its stress is called as Under reinforced section. A beam reaches its permissible stress in concrete under the working moment before steel reaches its stress is called as over reinforced section.

2 6. Sketch the edge and middle strip of a two way slab. (MAY JUNE 2009) 7. list the advantages of limit state design. (NOV DEC 12) In this method the structure shall be designed to withstand the safety all loads liable to act on it throughout its life. The aim of design is is to achieve acceptable probabilities that the structure will not become unfit for the use for which it is indented that is, that will be not reaching a limit state. The structure will be designed for safe against durability and serviceability requirements. 8. Mention any two advantages of introducing compression reinforcement in reinforced concrete beam. (APRIL MAY 2012) We introduce compression reinforcement for carry over the additional moments in the beams. The safety against failure of beam due to additional bending moment the compression reinforcement is provided. Part B(16 marks) 1. Design a one way slab with a clear span of 5m, simply supported on 230mm thick masonry walls and subjected to a live load of 4kN/m² and a surface finish of 1kN/mm².Assume Fe 415 steel. Assume that the slab is subjected to moderate exposure conditions. (NOV-DEC 2012) Step 1: Type of Slab. ly/lx = 5/1 = 5>2.it has to be designed as one way slab. Step 2:Effective depth calculation. d = span/(basic value x modification factor) = 5000/(20x0.95) = 270mm D = /2 = 295mm Step 3: Effective Span. Le = clear span + effective depth = = 5.27m (or) Le =c/c distance b/w supports = (230/2) = 5.23m Adopt effective span = 5.23m least value. Step 4: load calculation Live load = 4kN/m² Dead load = 1x1x0.27x25 = 6.75kN/m² Floor Finish = 1kN/m² Total load = 11.75kN/m²

3 Factored load = x 1.5 = kN/m² Step 5: Moment calculation. M = wl²/8 = (17.625x5.232)/8 = 60.26kNm Step 6: Check for effective depth. M = Qbd² d2 = M/Qb = 60.26/2.76x1 = mm say 150mm. For design consideration adopt d = 150mm. Step 7: Area of Steel x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm² Use 10mm dia bars Spacing,S = ast/astx1000 = (78.53/300)1000 = 261mm say 260mmc/c Provide 10mm c/c.

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5 2. Design a simply supported RC beam having an effective span of 5m.the beam has to carry a load of 25 kn/m. sketch the reinforcement details. (NOV-DEC 2010) (NOV-DEC 2012) Step 1: Effective length. Effective span,le = 5m Step 2: Size of the beam. Effective depth = le/10 = 5000/10 = 500mm Assume, b = 2/3d = 2/3x500 = 333.2mm say 340mm Step 3: Load Calculation Live load = 25kN/m Dead load = 1x.340x.500x25 = 4.25kN/m Total load = 29.25kN/m Factored load = 29.25x1.5 = 43.85kN/m Step 4: Moment Calculation. M = wl2/8 = (43.85x52)/8 = kNm Step 5: Check for effective depth. M = Qbd² d2 = M/Qb = /2.76x.340 = 382.2mm say 380mm. d = 380mm > 500mm Hence it is safe. Step 5: Check for effective depth. Mbal = Qbd2 = 2.97x340x5002 = kNm > M Hence it can be designed as singly reinforced beam section. Step 7: Area of Steel x106 = 087x415xAstx500(1-(415 Ast)/(20x340x500)) Ast = mm² Use 20mm dia bars No of bars = Ast/ast = / = 2.45 say 3nos Provide 3#20mm dia as tension reinforcement.

6 3. Design a RC beam 350X700mm effective section, subjected to a bending moment of 300kNm.Adopt M20concrete and Fe415 steel. (NOV-DEC 2009) Step 1: Size of the beam. b = 350mm & D = 700mm d = /2 =665mm Step 2: Moment Calculation. M = 300kNm Step 3: Check for effective depth. Mbal = Qbd² = 2.97x350x6652 = 459kNm > M Hence it can be designed as singly reinforced beam section. Step 7: Area of Steel 459x106 = 087x415xAstx665(1-(415 Ast)/(20x350x665)) Ast = mm² Use 20mm dia bars No of bars = Ast/ast = / = 1.45 say 2nos Provide 2#20mm dia as tension reinforcement.

7 4. Design a one way slab for a clear span 4m simply supported on 230mm thick wall. Subjected to a live load of 4kN/m² and floor finish of 1kN/m².use M20 concrete and F415 steel. (NOV-DEC 2009) Step 1: Type of Slab. ly/lx = 4/1 = 4>2.it has to be designed as one way slab. Step 2:Effective depth calculation. d = span/(basic value x modification factor) = 4000/(20x0.95) = 270mm D = /2 = 295mm Step 3: Effective Span. Le = clear span + effective depth = = 4.27m (or) Le =c/c distance b/w supports = (230/2) = 4.23m Adopt effective span = 4.23m least value. Step 4: load calculation Live load = 4kN/m² Dead load = 1x1x0.27x25 = 6.75kN/m² Floor Finish = 1kN/m2 Total load = 11.75kN/m² Factored load = x 1.5 = kN/m² Step 5: Moment calculation. M = wl2/8 = (17.625x4.232)/8 = 60.26kNm Step 6: Check for effective depth. M = Qbd² d2 = M/Qb = 60.26/2.76x1 = mm say 150mm. For design consideration adopt d = 150mm. Step 7: Area of Steel x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm² Use 10mm dia bars Spacing,S = ast/astx1000 = (78.53/300)1000 = 261mm say 260mmc/c

8 Provide 10mm c/c.

9 5. Deign a rectangular beam of cross section 230 x 600 mm and of effective span 6m.imposed load on the beam is 40 kn/m. Use M20 concrete and Fe415 steel. (APRIL MAY 2012) Step 1: Size of the beam. b = 230mm & D = 600mm d = /2 =565mm Step 4: load calculation Live load = 40kN/m² Dead load = 1x.23x.565x25 = 3.245kN/m² Total load = 43.24kN/m² Factored load = x 1.5 = 64.86kN/m² Step 2: Moment Calculation. M = wl2/8 = (64.86x62)/8 = 291.9kNm Step 3: Check for effective depth. Mbal = Qbd2 = 2.97x230x5652 = 218kNm < M Hence it can be designed as Doubly reinforced beam section. Step 7: Area of Steel Ast = Ast1 + Ast2 218x106 = 087x415xAstx565 (1-(415 Ast)/ (20x230x565)) Ast = 1365mm² Use 20mm dia bars, ast = π/4 (202) = mm² No. of bars = Ast/ast = 1365/ = 4.47 say 5nos. Ast2 = (M-Mbal)/(0.87fy(d-d1)) = (291x x106)/(361x(565-20)) =371.03mm² Use 20mm dia bars, ast = π/4 (202) = mm² No. of bars = Ast/ast = / = 1.8 say 2nos. Step 5: Area of Compression steel: Asc = (M-Mbal) / (fsc.(d-d1)) = (291x x106) / (351.8x(470-30))= mm² Use 20mm dia bars, ast = π/4 (202) = mm² No. of bars = Ast/ast = / = 5.5 say 6nos. Provide 6#20mm dia bars as compression reinforcement.

10 6. A hall has clear dimensions 3 m x 9m with wall thickness 230 mm the live load on the slab is 3kN/m² and a finishing load of 1kN/m² may be assumed. Using M20 concrete and Fe415 steel, design the slab. (APRIL MAY 2012) Step 1: Type of Slab. ly/lx = 9/3 = 3>2.it has to be designed as one way slab. Step 2:Effective depth calculation. d = span/(basic value x modification factor) = 3000/(20x0.95) = 270mm D = /2 = 295mm Step 3: Effective Span. Le = clear span + effective depth = = 3.27m (or) Le =c/c distance b/w supports = (230/2) = 3.23m Adopt effective span = 3.23m least value. Step 4: load calculation Live load = 4kN/m² Dead load = 1x1x0.27x25 = 6.75kN/m² Floor Finish = 1kN/m²

11 Total load = 11.75kN/m² Factored load = x 1.5 = kN/m² Step 5: Moment calculation. M = wl²/8 = (17.625x3.232)/8 = 60.26kNm Step 6: Check for effective depth. M = Qbd² d2 = M/Qb = 60.26/2.76x1 = mm say 150mm. For design consideration adopt d = 150mm. Step 7: Area of Steel x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm² Use 10mm dia bars Spacing,S = ast/astx1000 = (78.53/300)1000 = 261mm say 260mmc/c Provide 10mm c/c.

12 7. Design a two way slab panel for the following data. Size = 7mx5m Width of Supports = 230 mm Edge condition = interior Live load = 4kN/m Floor finish = 1kN/m² Consider M 20 grade concrete and Fe 415 grade steel. (MAY JUNE 2009) Step 1: Type of Slab. ly/lx = 7/5 = 1.4>2.it has to be designed as two way slab. Step 2:Effective depth calculation. For Economic consideration adopt shorter span to design the slab. d = span/(basic value x modification factor) = 5000/(20x0.95) = 270mm D = /2 = 295mm Step 3: Effective Span. For shorter span: Le = clear span + effective depth = = 5.27m (or) Le =c/c distance b/w supports = (230/2) = 5.23m Adopt effective span = 5.23m least value. For longer span: Le = clear span + effective depth = = 7.27m (or) Le =c/c distance b/w supports = (230/2) = 7.23m Adopt effective span = 7.23m least value. Step 4: load calculation Live load = 4kN/m² Dead load = 1x1x0.27x25 = 6.75kN/m²

13 Floor Finish = 1kN/m² Total load = 11.75kN/m² Factored load = x 1.5 = kN/m² Step 5: Moment calculation. Mx = αx. w. lx = 0.103x17.625x5.23 = 9.49kNm My = αy. w. lx = x17.625x5.23 = 4.425kNm Step 6: Check for effective depth. M = Qbd² d2 = M/Qb = 9.49/2.76x1 = mm say 150mm. For design consideration adopt d = 150mm. Step 7: Area of Steel. For longer span: 4.425x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 180mm² Use 10mm dia bars Spacing,S = ast/astx1000 = (78.53/300)1000 = 261mm say 260mmc/c Provide 10mm c/c. For shorter span: 9.49x106 = 0.87x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 200mm² Use 10mm dia bars Spacing, S = ast/astx1000 = (78.53/300)1000 = 281mm say 300mmc/c Provide 10mm c/c

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15 8. Describe the procedure for design of shear reinforcement. (APRL MAY 11) The nominal shear stress value is determined using trhe formula=shear force / (bxd) in KN/mm². The design shear strength of concrete is determined by using the table in IS ,page no 73,table no 19. Copare the nominal shear strength and design shear strength,the nominal shear stress value is less then design shear strength the minimum shear reinforcement is provided in the form of vertical stirrups,in the case IS is referred. The nominal shear stress value is exceeds the design shear strength,shear reinforcement shall be provided in any of the following forms. a) vertical stirrups b) Bent up bars along with stirrups, c) Inclined stirrups. Where bent up bars are provided their contribution towards shear resistance shall not be more than half that of the total shear reinforcement. Shear reinforcement shall be provided to carry a shear equal to the strength of shear reinforcement.