Properties of Materials - Essay Example

? Properties of Materials Task I Density This is the mass per unit volume of a material or it can be definedalso as the mass of a substance divided by its volume. Examples of density of materials are: 4-6 titanium weighs 0.610 pounds per cubic inch; 6061 aluminum weighs 0.098 pounds per cubic inch; 4130 steel weighs 0.283 pounds per cubic inch. There are two rules of thumb that have been given in comparing the density of materials i.e. titanium is nearly half the density of steel and that of aluminum is about one third the density of steel. Any heavy and weak material is considered bad but for a strong and light material is considered good. Stress and strain Stress has been described as the amount of force applied to a certain material divided by the cross sectional area of the material to the direction of the force. Strain properties, on the other hand, indicate how much a material lengthens under stress. This is measured by dividing the distance the materials over its original length. Below is a typical diagram of a stress-strain curve for different materials. The Figure below denotes the stress-strain curve for steel under different measures of stress: Figure 2: From the above diagram, we can deduce that strain increases faster than stress at any one point beyond point A. Below this point, any steel specimen whether loaded or unloaded would easily return to its original length, which is described as elastic behaviour. Beyond point B, any additional stress results into deformation or inelasticity (George et al, 2002, 26). Tensile and comprehensive strength The strength of a material can be measured by putting a material sample in a powerful materials testing machine, which pulls the materials apart and then records the force required to do so, plus the deformation to the material. Compression strength is determined by subjecting the material under pressure until it breaks i.e. heat, impact, et cetera. Tensile and compressive indicators are then recorded. These are good measure of how much impact the building materials can withstand without breaking if subjected to certain pressures (James, 2011, 35). Elastic and plastic deformation As explained in the stress strain diagrams above, elasticity and plastic deformation occurs prior and after application of a certain amount of stress on a certain material. In figure 2 above, below point A, we find that steel goes back to its original state hence it is elastic. Beyond point B, the steel cannot go back to its original length hence it is deformed or inelastic. Permanent elongation of a material is called plastic deformation. The same case applies to other building materials under different stress conditions. Below is the load deformation curve: Modulus of elasticity This is a coefficient which denotes the ratio of stress per unit area acting on, to cause deformation on a material to the resulting deformation therefrom. The elastic modulus, E, is usually determined after the compression tests are done on buildings. It differs in various types of materials for building. Elasticity modulus for steel is determined during manufacture while that of a concrete wall is calculated depending on the building dimensions. Task II Figure three: Graph of load against extension See the excel attachment Modulus of elasticity, E E = 2G(r+1) where G is the modulus of rigidity and r is the Poisson’s ratio. E= (F) (L1)/ (A) (L2) where F is the force or load, L1 is the original length of material (in this case mild steel), L2 is the amount the length changes on application of the load, A is the cross section area that the force is applied on the material. Area of the rod steel is given by the formula:, then A= 22/7* 12.52 = 491.07 mm2 E = {[50*195] / [491.07*0.09] + [100*195] / [491.07*0.19] + [150*195] / [491.07*0.29] + [160*195] / [491.07*0.34] + [165*195] / [491.07*0.46] + [170*195] / [491.07*0.78] + [180*195] / [491.07*0.84] + [190*195] / [491.07*0.91] + [200*195] / [491.07*0.98] + [210*195] / [491.07*1.07] + [220*195] / [491.07*1.24] +[230*195] / [491.07*1.49] + [240*195] / [491.07*1.88] + [250*195] / [491.07*2.39] + [255*195] / [491.07*3.95] + [260*195] / [491.07*4.94] } / 16 = {220.61 + 209.00 +205.39 + 186.87 + 142.44 + 86.55 + 85.09 + 82.91+ 81.04 + 77.93 + 70.45 + 61.30 +50.70 + 41.54 + 20.90}/16 = 101.42 kN/mm2 b. The stress at the limit of proportionality D is the total elongation L is the original length E = (0.09 + 0.19 + 0.29 + 0.34 + 0.46 + 0.78 + 0.84 + 0.91 + 0.98 + 1.07+1.24+ 1.49 + 1.88 + 2.39 + 3.95 +4.94)/ 195 = 0.112 c. Yield of stress – This is the point at which plastic deformation occurs. The offset rule is used to determine the yield as 0.2% of modulus of elasticity of a given material. Sigma = 0.002*E = 0.002 * 101.42 = 0.201 kN/mm2 d. Proof of stress Stress = load / area = 255 / 491.07 = 0.52kN/ mm2 e. % elongation Elongation = P * L / (A * E) P is the force being applied on the material L is the original length A is the area E is the elastic modulus = 255 * 195 / (491.07*101.42) = 0.998 Therefore % elongation = 99.8% f. Failure stress if diameter at failure was 18.5mm – at this point, the area of the rod is calculated at the resultant diameter of the rod after deformation. Stress = Load/ area = 255 / {3.1429 *(18.5 / 2)2} = 255 / 268.91 = 0.95 kN/mm2 g. Permissible stress if a factor of safety of 3 is employed (assume ultimate stress occurs at a load of 255 kN) Permissible stress = tensile strength / factor of safety = 255 / 3 = 85 kN Task III a. Breadth of the bar ‘b’ required, if the member is to be 9mm thick and the allowable stress ‘f’ is not to exceed 15N/ mm2 Stress = force / area 15N/ mm2= 145kN/ Area 15N/ mm2 = 145000N / 9mm *b mm 15N/ mm2 *9b mm2 = 145000 N b = 145000 / 15*9 b = 1074 mm or 1.074m b. Stress in the bar at section y-y Stress of bar = Load / area of bar Area of bar = 2 x (length x width + width x thickness + thickness x length) = 2 *(2000mm * 1074mm + 1074mm *9mm + 9mm * 2000mm) = 4,351,332 mm2 or 4351.332 m2 Stress = 145kN / 4351.332 m2 = 0.033kN/m2 or 33N/m2 1000mm = 1meter 1000N = 1kN Or 145 / 4351332 mm2 = 0.00033kN/ mm2 c. The shear stress in each bolt Shear Stress = Load / area Diameter = 20 mm Radius = half of diameter = 20/2 = 10mm Area of a circle is given by =145kN / 3.1429 *102 = 0.46kN/mm2 d. The amount of extension in the bar if its original length was 4m long (use area at y-y) Elongation = P * L / (A * E) P is the force being applied on the material L is the original length A is the area E is the elastic modulus Elongation = 145kN * 4000mm / (0.00033kN/mm2* 210kN/mm2) Elongation = 8,369,408 mm or 836.94 m e. Strain in the bar (?) E= stress / strain 210 kN/mm2 = 0.00033kN/mm2 / strain Strain = 1.57 *10-6 or 1.57 µ? Strain is a ratio of length to length hence it has no units. It is usually a small figure since most materials are not able to stretch for long before they get damaged. Task IV a. Combining of two materials and the resultant material properties that are superior to its constituent parts. Concrete can be combined with steel to reinforce it. On normal circumstances, concrete is weak in tension but it is brittle and strong in compression. The mixture of cement, sand and gravel is often used to make concrete. An addition of steel to this mixture is meant to strengthen and reinforce the tensile strength of concrete. The mixture hardens on the steel to form reinforced concrete. In this case the steel that is used should have appropriate deformations in order to obtain strong bonds and interlock of the two materials. Various advantages of reinforced have been analysed as described below: It usually has higher compressive strength, resistance to fire is higher as compared to steel, maintenance costs are lower as well as it is long lasting, the reinforced steel has been found to be the only material that can be used in some form of structure e.g. dams, footings etc. i.e. on its own constituents, it would not be utilized effectively; forms shapes easily when cast, and can be used in rigid members as well as in reducing the cross sectional area of some structures e.g. floor columns. The reinforced concrete has disadvantages in that the mixing and casting can reduce the strength of the final product notwithstanding the high costs of casting the steel. An overall comparison to steel alone shows that the reinforced concrete has lower comprehensive strength than steel alone i.e. at a ratio of 1:10. Fiber is usually welded with concrete to avoid shrinkage, to resist temperature, and as reinforcement for columns. The fiber is a mesh of wire welded together either using one way or horizontal, or two way methods i.e. cross sectional. Other fiber materials like glass, plastic et cetera, can be used to reinforce concrete with the aim of increasing resistance and drying shrinkage cracks. These also reduce cracks on widths. The most purpose for this type of reinforcement is to increase durability of concrete. It is however, important to put in the correct amounts of each material for the intended use. For example, steel is combined with concrete at a dosage of 12 is to 71 kilograms per cubic meter. These materials increase the shear strength of the material, impact of resistance and even ductility in comparison to unreinforced steel. A combination of different fibers together with concrete results to a product called polymer fiber reinforced concrete. This material is often very beneficial in that it reduces cracks on widths and increases resistance of the concrete. It is more superior than concrete alone since it has higher tensile strength and ductility (Joseph, 2006, 45). Synthetic fibers like nylon, polyester et cetera are used to reinforce concrete. These should however, be used with care and correct proportions. They have been found to increase the performance standards of reinforced concrete. Another example of combining materials is zinc and steel in order to reduce corrosion of the metal. This is by galvanizing the zinc material that is used for roofing. a. Physical and chemical factors that affect the strength, hardness and durability of a range of composite materials i. Physical factors Density A strong and light material is said to be of good density whereas a weak and heavy material is said to have bad density. In light of this, materials are usually combined to produce a final product that is of good density. Aluminum is more dense than its composite of aluminum alloys hence more durable and stronger. Electrical conductivity This is the ability of any material to resist electricity. Carbon and cooper are used in electrical wiring since they withstand high electrical voltages without impairing them. A combination of the two materials has been found to increase the durability of the composite material (Buberry, 1988, 316). Magnetic permeability This is the ability of a material to allow magnetic force to pass through it without altering its nature. Impermeable materials usually are affected eventually by magnetic forces while permeable materials withstand the test of time and are stronger. Coefficient of thermal expansion This is the ability of a material to withstand high temperatures. An example is that titanium has a higher thermal conductivity than iron. Therefore titanium coefficient of thermal expansion is higher than iron. Therefore it is considered more durable (McMullan, 1998, 25) ii. Chemical factors Elongation This is the ability of a material to stretch as much as possible or to extend in length when it is subjected to force or a load without breaking. Tests are done for different materials with the application of varying amounts of force. Hardness This is a property that denotes how well a material can withstand force without breaking. Wood is not hard and breaks easily. Concrete alone is subjected to cracks hence reinforcement is usually done to increase its strength, durability and resistance to force. Tests for hardness used lately are Vickers and Rockwell hardness tests (James, 2011, 22). Fatigue limit This is the extent to which a material is considered weakened by force application. Different materials have different fatigue limit. Examples are plastic is not easily fatigued as metal, steel withstands more force than carbon et cetera. This element allows building designers and engineers to combine different materials in construction work to increase durability as well as strength of the building (George et al, 2002, 30). Tensile strength This is the ratio of maximum load to the original cross sectional area. A test of strength with increase of loading rate was done for concrete, mortar, granite, ultra-high strength concrete, which revealed that all the materials increased in tensile strength with increase in loading rate. Granite and the ultra-high strength concreter were found to have higher strength than the rest. Tensile strength can be increased by reinforcing concrete (Tom, 2007, 26) Modulus of elasticity The elastic modulus has been found to different in a range of materials. It is basically the ratio of stress acting on a specific area in order to cause deformation to the resulting deformation. Concrete alone has a low elastic modulus but a composite material of concrete and steel or with fiber increases its elastic modulus. The higher the modulus the stronger and more durable the material becomes. Bibliography Arthur, L., (2010), Materials for Architects and Builders, London, Routledge. Buberry, P., (1988), Environment & Services. Mitchell’s Building Series: London. Environment and services, New York, Longman. McMullan, R., (1998), Environmental Science in Building , Chichester, Wiley & sons. George, E., Maurice, A., & Tatsuo, I., (2002), Handbook of residual stress and deformation of steel, US, ASM International. James, A., (2011), Simplified Mechanics and Strength of Materials Chichester, John Wiley & Sons. Joseph, F., (2006), Significance of tests and properties of concrete and concrete-making materials (5e), California, ASTM International Tom, P., (2011), Dynamic Behavior of Materials, Volume 1: Proceedings of the 2011 Annual Conference on Experimental and Applied Mechanics, London, Springer. Read More