Five Storey Building at Porchester - Assignment Example

Five storey building at Porchester Student’s Name College Instructor’s Name Course Name: Introduction The building can be described as circular with a porch like structure entailing columns located under a pediment. It also has a characteristic porch which is linked to the rotunda via a rectangular vestibule which is under a coffered dome with a central opening. Design Foundation Any building stability will be determined by the foundation laid for construction. Building will have Pad Foundations with a square footing that support localised single point loads such as structural columns. The load is then spread to the bearing layer of soil or rock below The foundation should have strong steel to ensure there is a strong foundation for the building. It will have a square footing of 2m by 2m for each column. Whose carrying capacity will be 5624.35kN which is calculated as follows; is Where No is the bearing capacity factor (cohesion), B is the width or diameter of the foundation, D is the effective vertical strength of the soil, Nq is the bearing factor and surcharge friction, is the bearing capacity factor self weight and friction, C is the cohesion strength of the soil. We assume the friction as 350, unit weight as 19kN/m3, and we have depth base as 2m and a square footing of 2m by 2m Where is the ultimate bearing capacity, in this case we will use the formula Where Qu = In this case Nc=45, Nq=35.4, Ny=50 Qu =1.3x2x(45) + 19kN/m3 (2m) x 35.4) + 0.4 (19kN/m3 x 2m x 50 = 117+ 1,345.2+1900 =3,362.2kN/m2 Allowable Bearing Capacity for foot It is Qa = Qu Qa = kN/m2 Shear failure Nc = 16, Nq = 4, N = 1.7 = 229.56 It is Qa = Qu Qa = kN/m2 Allowable total gross load Q= Qa(B)2 = 114.78 X (7)2 = 5624.35kN The load of 2000 kN thus it is acceptable since the allowance load on the clay is 5624.35kN This concrete footing will require steel of 12 – 13 kg /m2 floor area for columns and 30cm x 60cm for all beams, 7.5m distance between columns and a Slab 7.5m x 7.5m with secondary beam at 3.75m. There is a significant impact on the usage of materials in this regard as well; the reason for this is that in order to go green, the materials that should be used in order to change the structure of the property should also be green; materials which are easily renewable as well as recycled. Apart from this there is also the concern of the transportation of the materials. Now for this the suggestion by most environmentalists is that the materials that need to be manufactured should be done in or near the premises Wall construction The walls of the facility will have materials which will include cement mixed with sand but it should be smooth with good decorative finishing. When plastering, hydraulic lime which is available in UK will be used to give velvety texture and a soft look. These materials will make the wall of the facilities more sustainable and attractive to the society. In the facility we will have curved walls which will be made of glass, the glass that is recycled. The recommended glass is 100% recycled in order to ensure sustainability. The walls that are not made of glass will be super insulated by using wraps that are 180mm to ensure heat is kept in the rooms. Glass should be used in curved walls especially for conference sections. The materials should have low carbon levels. The calculations for walls are; a. Effective Ranking Pressure To determine ranking active pressure, the following formulas are employed. Where q is the surcharge load and in this case it is 20kN/m and Ka is while is determined using . To begin with, we are given c’ as 0 therefore, the upper layer of sand its coefficient of ranking active pressure is determined as follows, Ka = The lower layer of sand its ranking active pressure coefficient will be determined as follows Ka = Therefore at Z=0, the surcharge will be = 0.333 x 20kN/m = 6.66kN/m2 At a Z=3, the effective ranking active pressure is = 6.66kN/m2 + 0.333 x 3 x19kN/m3 = 25.641kN/m2 At Z=3 but in the lower layer the force will be 3 x 25.641 x 0.333 = 25.615 At Z=6 metres the effective ranking active pressure is = (25.641 x 3 + 3(21kN/m3- 9.81)) x 0.333 =36.79kN/m2 We have assumed the pressure of water to be 9.81kN/m2 Rankine’s active force per unit length =( ½) H2 K Therefore at Z=0, the Rankine’s active force per unit length will be =( ½) x6.66kN/m2 x 20mx1m =66.66kN/m At a Z=3, Rankine’s active force per unit length is =( ½) x25.641kN/m2 x 19x3m =2,192.31kN/m At Z=3 Rankine’s active force per unit length =( ½) x25.615kN/m2 x 19x3m =2,190.08kN/m At Z=6 metres Rankine’s active force per unit length =( ½) x 36.79kN/m2 x 19x(6mx6m) = 12,582.18kN/m Roofing structure Roofing is essential to this facility; materials that will be used will determine its sustainability. The roof will be of slab and flat. A concrete slab with small resistance will be used in structure where concrete hollow blocks will be arranged between the ribs to ensure a smooth ceiling. This structure shows that reinforced concrete ribs span in the long directions to support topping slab. Some designers opt to run the ribs in a direction that leads to smaller moments and shears in the supporting beams which means much more reinforcement in the ribs. The steel reinforcement roof will have a strength fy = 500N/mm2) and Mild steel reinforcement strength fc = 450N/mm2 while roof slab will have a live load of 250kg/m2 that is supported by continuous beams and slabs. However this live load will not exceed three times dead load of 90kg/m2. The spans will be spaced equally to ensure that load is evenly distributed. The shear strength of rib concrete will be 110% of shear strength of beams and uses 4 by 6 mm stirrups per meter run. The ribs are usually designed in a rectangular manner where there is negative moment at the columns while minimum thickness values are determined as follows; Vu 0.53 Where Φ is the strength reduction factor which is 0.75, b is the width of the strip which is 50cm, d is slab depth which is 26cm that is for 6cm for topping slab and 20cm for shutter. Vu 0.53 = 3,040.29N/mm2 The topping slab thickness will be 1/12 of distance between ribs which is 6.0 cm and continuous beam supported by the ribs. Support edge beam and column is calculated as Mu = (Wu)/24 and (Wu)/16 respectively. Where l = span of the equivalent frame in a bay which is This thickness is usually determined as follows Where Mu is factored bending moment, b is web width; Φ is the strength reduction factor = 36.84cm While the reinforcement ratio for the slab is = 0.431 Bending stress distribution for a beam in pure bending: , where y is the distance from the neutral axis. I= = 0.001378m3 Y= Area A = = 0.023625m2 q = = = 1,111.2N qmax = = = 3,174.6N Lateral stability and fire resistance Fire load- Fire load play a major role in the behaviour of fire. In other words, fire load can be considered as the amount of fire source in the community centre in case of fire. Fire load of each floor of the building should be calculated separately. These are then converted to the fire load density for the enclosure using; (BS 7974 / part 1) qk fire load density for the enclosure (MJ/m2) mctotal mass of each combustible material in the enclosure (kg) Hc effective calorific value of each combustible material (MJ/kg) Aftotal internal floor area of the enclosure (m2) Fire Resistance and Protection of a Steel Beam Applied bending moment MEd = 49,59  52 / 8 = 154,97 kNm Try RHS 25015012, assuming that the design is controlled by strength and not by deflections. Coefficient = (235 / fy)0,5 = 0,92 Compression flange: c / t = (150 – 3  12) / 12 = 9,5  Read More