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Question 1 A joist is a beam in a building, which is made parallel to its adjacent member from one wall to the next, in order to support a ceiling or floor. As a result, the distribution of the weight on the beam is assumed to be uniform (Bansal, 2001). Due to this fact, there will be occurrence of bending moments on the beams or joists…
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Download file to see previous pages This is determined by the direction of the moment (Bansal, 2001). When the moment on the left of the force is clockwise and the moment on the right of the force is anticlockwise, the moment generated is considered positive. However, this causes the beam to bend and is also called the sagging moment. There are two types of loads, which cause moments on beams. These are either concentrated loads or distributed loads. Distributed loads have the weight spread over the significant length of the beam. However, the concentrated loads have their weight placed on one point of the entire beam. The shearing force in the beam is the chance that at that point, the beam is likely to slide laterally against the other portion of the beam. The diagrams below best explain the relationship between beams which have uniform weight distribution, their bending moments, and their shearing force distribution along the length. For a beam whose length is denoted as l with a distributed weight of w supported on both ends: The total weight acting on the beam is W*L= WL The reactive forces at both supports of the beam are obtained to be WL/2 (for each end of the beam) In order for the moments to be calculated, the force on the beam is assumed to be acting on the middle of the beam (L/2) (Bansal, 2001; Kassimali, 2010). The moments calculated about a point X length from the left of the beam, will be denoted as: (WL/2)*X – (WX)*X/2 = WL/2 * (L - X) The maximum shearing force will be WL/2, and the minimum shearing force will be –WL/2. There is no shear at the centre L /2. The moment is greatest here, according to the analysis. This can be found by replacing X with L/2 to give: M= (WL/4)*(L - L/2) = WLL/8 The ultimate limit state allows that the load allowed on the beam be 1.33*W, where W is the weight of the first plastic deformation of the beam. Thus, the initial load allowed on the beam, considering the ultimate limit state design, should consider that the beam is subjected to elasticity up to a certain extent (Kassimali, 2010). The solution W = 5KN + 1 KN = 6 KN L = 4000mm = 4M Shearing force: Wl/2 =6000N * 4m* 0.5 = 12000NM 12 KNM The maximum moment: WLL/8 = ( 6000*4*4) / 8 = 12000NM The ultimate weight allowed is, 1.33 * 6000 = 7980 N Question 2 The U-values are sum of all the thermal resistances of the materials used in the construction of the walls of buildings. This is also described as the sum of the inverse of all the thermal resistances in the materials used in constructing buildings. Thermal resistivity is a measure of a material’s property to fight the transfer of heat across a material, with a temperature difference across it. These values are obtained from already set British standards, published by the British Standard Institute. The units for this property are (m2k)/W. Fabric heat loss in materials occurs when there is a temperature difference between two different sides of that material. Due to the difference, the material experiences a process of heat transfer from the hot side to the cold side. In the construction industry, it is important to obtain the heat loss values in order to know how to heat up buildings in winter, to a desirable temperature. The units for the U value are W/m2K, which is the reciprocal of the thermal resistivity (Yogesh & Jaluria, 2003). The thermal resistivity of the dense brickwork – 1.6 Thermal resistivity of wool batts - 0.048 Thermal conductivity of glass – ...Download file to see next pagesRead More
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