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Note for Reinforced Cement Concrete - RCC By ANANDH M

  • Reinforced Cement Concrete - RCC
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CE8501- DESIGN OF REINFORCED CEMENT CONCRETE ELEMENTS results in the prediction of strength in sustained agreement with results of tests .The partial safety γm = 1.5  The tensile strength of concrete is ignored.  The maximum strain in the tension reinforcement in the section at failure shall not be less than fy/1.15Es +0.002  Where fy = characteristic strength of steel Es = modulus of elasticity of steel. 4. Write a short note on doubly reinforced sections. Doubly reinforced sections are generally adopted when the dimensions of the beam have been predetermined from other considerations and the design moments exceed the moments of resistance of a singly reinforced section. 5. What do you understand by limit state of collapse? (Ref IS 456, Page No 67, Clause 35.2) The limit state of collapse of the structure or part of the structure could be assessed from rupture of one or more critical section and from buckling due to elastic or plastic instability or overturning or fatigue etc. The resistance to bending, shear torsion and axial loads at every sections shall not be less than the appropriate value at that section produced by the probable most unfavorable combination of loads on the structure using the structure using the appropriate partial safety factors. 6. Draw the stress-strain curve for mild steel? (Ref IS 456, Page No 70, Fig 23) M.Anandh Asst Prof –SSMIET/Civil Page 2

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CE8501- DESIGN OF REINFORCED CEMENT CONCRETE ELEMENTS 7. Differentiate between WSD and LSD? Sl.No 1 Working stress method Factor of safety for stress is Limit stress method Factor of safety is less more 2 Actual stress strain Actual stress strain relationship is not relationship is considered. considered. 3 Materials strength of not Materials strength is fully fully utilized. utilized. Over reinforcement is Over reinforcement is not possible. possible. 5 Uneconomical. Economical. 6 Compute the application is Compute the application is not done easy. done easy. 4 8. What are the expressions recommended by the IS 456-2000 for modulus of Elasticity and Flexural Strength? (Ref IS 456, Page No 16, Clause 6.2.2, 6.2.3.1) Modulus of elasticity: Ec = 5000√fck Where Ec is the short term static modulus of elasticity in N/mm2. Actual measured values may differ by ±20% from the values obtained from the above equation. Flexural strength: fcr = 0.70√ fck Where fck is the characteristic cube compressive strength of concrete in N/mm2. 9. How to select cross sectional dimensions for beams?  The effective and overall depth of the beam is estimated from span/depth rations to satisfy the limit state of serviceability. Overall depth to width should be in the range of 1.5 to 2 M.Anandh Asst Prof –SSMIET/Civil Page 3

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CE8501- DESIGN OF REINFORCED CEMENT CONCRETE ELEMENTS  The width of the section should accommodate the required number of bars with sufficient spacing between them with minimum side covers of 20mm to the links.  The depth of the beam should be such that the percentage of steel required is around 75% of any one layer  The minimum number of bars on tension face should be not less than two and not more than six in any one layer.  In flanged beams, the depth of the slab is generally taken as 20% of the overall depth.  Common widths of beams are 150, 200, 230; 300mm.Also the width of the beam should be equal to or less the dimensions of the column supporting the beam. 10. Briefly explain about limit state of serviceability. (Ref IS 456, Page No 67, Clause 35.3) The following limit state of serviceability are considered in design 1. Deflection 2. Cracking 11. Briefly explain about partial safety factor. (Ref IS 456, Page No 68, Clause 36.4) When assessing the strength of a structure or structural member for limit state of collapse, the value of partial safety factor γm should be taken as1.5 for concrete and 1.15 for steel. A higher value of partial safety factor for concrete has been adopted because there are greater chances of variation in strength of concrete due to improper compaction, inadequate curing and mixing and variations in the properties of ingredients. 12. Define singly reinforced section In a reinforced concrete, if steel is provided to take up only tension, the section is called as singly reinforced section. 13. Write about limiting neutral axis. (Ref IS 456, Page No 69, Clause 38.1 (f)) Limiting neutral axis, Xu (max) which gets formed when the strain in concrete and strain in steel reaches their maximum permissible values ie 0.0035 and (0.87fy/Es) +0.002 14. Differentiate between design mix and nominal mix. (Ref IS 456, Page No 23, Clause 9.2 & 9.3) M.Anandh Asst Prof –SSMIET/Civil Page 4

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CE8501- DESIGN OF REINFORCED CEMENT CONCRETE ELEMENTS Design mix concrete: For all – important works involving large quantities of concrete, it is preferable to use design mix, which results in considerable economy ensuring the required strength. The design mix uses the following parameters:  Types of cement  Aggregate size and grading.  Water / cement ratio.  Aggregate /cement ratio.  Workability of concrete  Relation between mean and maximum strength and standard deviation.  Grade of concrete. Nominal mix concrete: Nominal mix concrete may be used for concrete of M20 or lower 15. State the assumptions made for design of RC members in working stress method. (Ref IS 456, Page No 80, Clause B-1.3)  At any cross section, plane sections before bending remain plain after bending.  All tensile stresses are taken up by reinforcement and none by concrete, except as otherwise specifically permitted.  The stress-strain relationship of steel and concrete, under working loads, is a straight line.  The modular ratio m has the value 280/(3σcbc)  Where σcbc is permissible compressive stress due to bending in concrete in N/mm 2. 16. What are the advantages in limit state method?  In L.S.D of analysis both elastic and semi plastic theories.  Structure is shape and serviceable does not collapse even under worst loading condition.  Economical one.  Overall size of the flexure member is less by this method.  Actual stress strain relationship considered.  Material strength is fully utilized.  Factor of safety is less. 17. Write the formula for the neutral axis depth factor ‘K’ in working stress design. Kc =(mσcbc / ( mσcbc + σst)) M.Anandh Asst Prof –SSMIET/Civil Page 5

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