What Causes Moments on Beams - Essay Example

?Question 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. The bending moments are forces in the beam, which lead to its bending. This is calculated by adding up all the external forces at that point, and multiplying that by the perpendicular distance in the direction of the force. Moments can either be positive or negative. 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 – 1.05 Thermal conductivity of plaster - 0.47 Thermal conductivity of air – 0.024 (Yogesh &Jaluria, 2003) U of wall = 1/(brick+wool butt+air+woolbutt+brick+plaster) = 1/(0.024+0.47+0.048+0.048+1.6+1.6) = 0.264 W/m2K U of windows = 1/(glass+air+glass) (Yogesh &Jaluria, 2003) = 1/(1.05+1.05+0.024) = 0.471 W/m2K The temperature difference is (Rathore, 2011) Inside temperature – temperature outside = 22 – 18 = 4K Fabric heat loss = U value * Area * Temp. difference Heat loss of the wall = 0.264 * [(3 * 1.2)-(1.2*1.0)] * 4 = 2.5344 W. Heat loss of the glass = 0.471 * (1.0*1.2) * 4 = 2.2608 W. The discussion Types of heat loss The main types of heat loss from buildings are mainly through the processes of radiation and conduction. The heat loss by radiation takes place through the generation of electromagnetic waves, which are usually in the infra-red region of the light spectrum. A body radiates heat, because the molecules that constitute it are thermally agitated (Rathore, 2011). Due to the agitation, the heat is converted to electromagnetic energy, which is then released by the molecules of that body. In the heat loss through conduction, the heat is distributed through the exchange of kinetic energy between the particles that separate the two environments which bear a temperature difference (Rathore, 2011). As a particle absorbs heat energy, it increases the vibration, as it tries to break free of bonds that hold it in position. In the process, equilibrium is sought, and the adjacent particles also begin to vibrate. The transfer of this kinetic energy results in the heat loss through conduction. Heat gain In the buildings, the most popular type of heat gain is the solar heat gain. Most of the heat gain is gotten through the widows, skylights and glazed doors in buildings. The process is also known as fenestration (Frank, 2010; Ye, 2008). The best way to prevent this, is by installing shades, which prevent the sun’s rays from reaching the glass doors and windows. The other way is to install widows, which are manufactured to have a small solar heat gain coefficient rating. References Bansal, R., 2001. A Textbook of Strength of Materials. New York: OUP. Frank Kreith, R. M., 2010. Principles of Heat Transfer. Stamford: Global Engineering. Kassimali, A., 2010. Structural Analysis. Stamford: Christopher Publishers. Rathore, M. M., 2011. Engineering Heat Transfer. London: James and Bartlett. Ye, J., 2008. Structural and stress analysis: theories, tutorials and examples. New York: Taylor and Francis. Yogesh Jaluria, K. E., 2003. Computational heat transfer . New York: Taylor and Francis. Read More