Structural Analysis and Design: Design of 5 Storey Hotel Building - Assignment Example

A. Design of 5 Storey Hotel Building 0 General Approach The general approach taken is to obtain the architecture of an office building:- Acquiring design of an office architecture of a multi-storey building basically adopted in the construction design (Alheyad, 2008). . Structural system establishment of the floors from the ground to first, second and third floor Foundation type, column design and system of resisting wind will be affected by taking into consideration the architectural drawings 2.0 Requirements for design Client’s requirements: The ground floor is to have: a reception, conference/meeting rooms, common room and entertainment room, a restaurant, and associated services. The remainder of the floor is to be occupied by the swimming pool. The main building is to include a small scale conference/meeting suite of a maximum 30 persons, a restaurant, a common room, as well as lodging. The minimum clear height for the ground floor is set to be3.0 m. The minimum spacing of vertical structural elements within the ground floor is to be 6.0 m. Access to the 1st, 2nd, 3rd, 4th and 5th floor is by lifts or staircases located at the centre and end of the floors. The 1st and 2nd floor are different size offices with central access corridors. The floor upon completion will have a soffit height at the 1st and the 2nd floor of a minimum of 3.0 m in the floors and 3.6m in corridors, in addition to services allowance. All servicing plant including water tanks, heating boilers and lift mechanisms are to be located on top of the stair/lift towers at the ends of the building. Car parking of 450m^2is to be set near the hotel at ground level over an area of approximately 450 m2. There is need for allowance for future adjustments in demand for different types of services. This is achieved through structural arrangement not compromising the ability to adjust the width and mix of room compartments, though the central corridor arrangement will not be altered as per the client decision. A fire resistance period of 2hrs is to be put in the building. The building is to be exposed to XC-1 environment. 3.0 Structural elements These are framing elements used to transfer the loads from the building to the foundation to ensure the building is stable. These structural elements are: Slabs, beams and columns. Design of a column Columns This is the structural part which is vertical used to support axial compressive loads, by use of moments or without moments (Alheyad, 2008). These vertical loads from the ground and the three floors transmit the compressive loads to the foundation. The column design requires that the spacing between the columns be greater than 6.0 m so will design the spacing used is 7.0m. Reinforcement of Steel in Columns Steel limiting ratio will be from 1% but not exceeding 8% Strength of concrete will range from 25Mpa to 45Mpa (Alheyad, 2008). Steel reinforcement strength will range from 400Mpa to 500Mpa Column self weight =24.5KN/m^3 Floor weight = (3+5) Kpa Procedure of column design 1. The first step is to calculate the factored axial load. 2. The next step is selection of reinforcement ratio (Alheyad, 2008). 4. Taking into mind that the strength of concrete=30 Mpa, Yield strength of steel = 420 Mpa 5. Lastly, will calculate the column reinforcement area, which is As. Q3. Answer. A masorary column of the following dimensions is used in this building Diagram b: Reinforcement of the Columns Q3. Answer for Column and Beam analysis CODE REF. CALCULATIONS OUTPUT Column self weight 2* 24.5 =49 KN 49 KN The ultimate design axial load = 1st floor,2nd floor, and 3rd floor total loads + Column self weight + 4th floor with the other weight (3+5)KN/m 263.25 + 143.1 + 49+3 +5= 463.35 KN 463.35 KN Column design moments: 1st and 2nd column moments - The member stiffness = For Member AB KAB/2 = ½ *bh^3/12LAB ½ *0.3 *0.7^3/12*9 = 0.48*10^-3 0.48* 10^-3 For member BC KBC/ 2 = ½ * bh^3/12LBC ½*0.3 * 0.7^3/12*6 = 0.71 * 10^3 0.71 * 10^3 Column as a member part has stiffness as: K col = ½ * b col* hcol/ 12* H ½ *0.3 * 0.4^3/ 12* 3.0 = 0.53 *10^3 0.53 * 10^3 The Total 1st and 2nd moments = ∑ k = Member stiffness of frame, AB + Member stiffness of BC + 2 * Column stiffness = (0.48 + 0.71 + 2 * 0.53) * 10^3 = 2.25 * 10^-3 2.25 * 10^-3 So the moment distribution factor of the 1st and 2nd floor = K col / ∑ k 0.53 / 2.25 = 0.24 0.24 Fixed End Moment at B = F.E.MBA = Maximum ultimate load * frame AB 47.7 * 9^2 / 12 = 321.975 KN/m 321.975 KN/m Fixed End Moment at B = F.E.MBC = Minimum ultimate load * frame BC 16.2 * 6^2 / 12 = 48.6 KN/m 48.6 KN/m Column moment MED = moment distribution factor of 1st and 2nd floor( F.E.MBA – F.E. MBC) = 0.24 ( 321.975 – 48.6) = 65.61 KN/m 65.61 KN/m The moment of the column in the third floor is = = ∑ k = Member stiffness of frame, AB + Member stiffness of BC + Column stiffness (0.48 + 0.71 + 0.53) * 10^3 = 1.72 * 10^-3 1.72 * 10^-3 So the moment distribution factor of the 3rd floor = K col / ∑ k 0.53 / 1.72 = 0.3031 0.3031 Beams Two kinds of beams are being used in this building design. The first used is the one in all the slab directions. This beam type is simply supported and carries part of the slab load. The support of these beam type is the main beams. The length of these beams is 12m with regards to the construction slab. The other beams are the main structure beams connected to the columns (Muller, 2008) These as stated carry the middle beams load and a part of the slab load. These beams are usually continuous. For the design in the calculation part, the area method will be incorporated. Since the building is exposed to XC-1 environment implying that Cmin = 20mm and the fire resistance of 2hrs meaning minimum breadth (bmin = 200mm) is taken into consideration. CODE REF. CALCULATIONS OUTPUT DESIGN OF A BEAM The number of bars used per each beam = 4 bars as shown on the last diagram on drawings which is diagram 6c. With the support end fixed , the moment at the middle of the end span from the coefficient table is given as; M = 0.086 * Force *Length or span F from above = 76.36KN L / Span = 9.6m The number of bars used per each beam = 4 bars M = 0.086 * 76.36 * 9.6KNm = 63.04 KNm The number of bars used per each beam = 4 bars M = 63.04KNm Bending Reinforcement is given by, K as K = M / bd^2 fck K = (63.04 *10^6) / 1000* 20^2 * 25 = 0.1750 K = 0.1750 Lever arm formula is given by: Z = d (0.5 + √ (0.25 - K / 1.134) Where K from output above is K = 0.1750 Z = d ( 0.5 + √ (0.25 – 0.175 /1.134) = 0.5 + √ ( 0.095655) = 0.5 + 0.30928 = 0.80928 or approximately 0.81 Z = 0.81 Since the Z value is less than 0.95d (0.81d Read More