Understanding of the Bending Theory - Lab Report Example

The bending theory is critical in understanding beam behavior. In the various fields of engineering, moreespecially mechanical and civil engineering, there are vast applications of beams. From concrete structures, to steel structures, and even timber (joists), beams are used to carry loads. It is the behavior of these beams under loads that the theory of bending seeks to demystify hence its importance. If misunderstood in engineering design, the repercussions can be fatal. Storey buildings will collapse, bridges will crumble, even vehicles might breakdown under load because the beam-like structures that are found underneath them . Introduction Bending theory concept must be well understood by engineers. The first step toward understanding this theory is to know the assumptions made. These are: 1. The beam material is homogeneous and isotropic. 2. The value of young’s modulus is the same in tension and in compression. 3. Sections which are transverse are plane before and after bending. 4. The beam bends into an arc with a common center of curvature. 5. The radius is curvature compared to the dimensions of the cross section, is larger. 6. Beam layers are free to expand or contract irrespective of each other (Bansal, 2010). Under load, a beam bends to form an arc. The center of the beam is the neutral axis which does not deform. However, the layer above the neutral axis shortens while the layer below elongates. As a result, beam fibers above the neutral axis are under compression whereas those below are under tension. Objectives The objective of this experiment was to determine young’s modulus of elasticity of steel and brass. Prior to determining young’s modulus: The results from the experiment for the different beams would be used to draw a graph of load (W) against deflection (δ). The cross sectional areas of each of the beams and their second moment of areas would be calculated. Background Information Young’s Modulus Young’s modulus, or modulus of elasticity (E), is a stiffness measure for any material. This is the extent to which the material resists deformation under an applied load. For beams, the kind of deformation that takes place is majorly deflection (degree to which a structural element is displaced under a load) (Bansal, 2010). Consider the beam below. Original beam ………..Deflected beam under load When loaded with a uniformly distributed load or a point load, the beam will be subjected to a bending moment. As a result, it will deflect as shown. Upon removal of the load, the whole strain caused by the load may disappear completely. When this occurs, the material is said to be perfectly elastic.Flexible materials have small values of E whereas stiff materials have large values of E. Graphically; E is equal to the slope of a load against deflection graph. Second Moment of Area The second moment of area (I) is defined as I= Where dA refers to a thin section of area sliced parallel to the neutral axis and the distance separating that slice from the neutral axis is y. Much as the amount that a beam will bend when loaded will be determined by the modulus of elasticity of the material,the second moment of area also determines the degree of deformation in part hence its relevance in the theory of bending. The size and shape of the object determine its second moment of area (Bansal, 2010). Deflection The amount by which a structural element is displaced upon application of a load is called deflection. Depending on how the element is supported, how it will deflect and its degree of deflection will vary. A structural element can be: simply supported, fixed, pinned, supported on rollers or cantilevered. While deflection is least experienced at the supports of the element in question, it is greatest at midspan of the supported element. For this experiment, a simply supported beam was considered. Below is a diagrammatic representation of the simply supported beam together with its relevant equations. W δ = WL3 48EI Where W = Load (N) L = length of beam E = Young’s Modulus GN/m2 I = Second moment of area (m4) δ = Deflection (m) And the second moment of area is I = bd3 12 Procedure 1. Dimensions for each of the two beams were measured. 2. Two knife edge supports were slid on to the rail and set at a distance of 500mm apart. 3. The knife edge supports were firmly secured with their edges pointing upwards. 4. The dial gauge was placed at the center of the supports. 5. The brass beam was placed through the string on the dial gauge and on top of the knife edge supports. 6. The dial was slid down until the plunger got compressed and the smaller of the two dials got fixed at zero. 7. The larger dial was then zeroed before masses of 50g, 100g, 150g, 200g, 250g, and 300g were added to the plunger. 8. Results of the subsequent deflections were recorded upon addition of each mass to the plunger. 9. The masses were removed from the plunger then steps 5-8 were repeated with the steel beam. 10. Results The results were as shown below For brass: Results Tables Variables Units 1 2 3 4 5 6 mass g 50 100 150 200 250 300 Gravity m/s2 9.81 9.81 9.81 9.81 9.81 9.81 Force N δ m 0.00025 0.00077 0.001 0.00123 0.00138 0.00158 Length m I m4 E N/m2 For steel: Results Tables Variables Units 1 2 3 4 5 6 mass g 50 100 150 200 250 300 Gravity m/s2 9.81 9.81 9.81 9.81 9.81 9.81 Force N δ m 0.00034 0.00061 0.00098 0.00128 0.00177 0.00207 Length m I m4 E N/m2 Computations 1. Load Acting on The Beam The load acting on the beam is a force (N). It is arrived at by getting the product of mass in kg and the acceleration due to gravity on the beam (Bansal, 2010). For example, For the 50g mass: load= (50/1000)9.81=0.491N 2. Second Moment of Area The second moment of area for each material was obtained from the formula I = bd3 12 For example for steel I = (0.0129X0.00323)/12 = 3.523X10-1m4 The values are recorded as follows. The recorded values of E are calculated values. For brass: Results Tables Variables Units 1 2 3 4 5 6 mass g 50 100 150 200 250 300 Gravity m/s2 9.81 9.81 9.81 9.81 9.81 9.81 Force N 0.491 0.981 1.472 1.962 2.453 2.943 δ m 0.00025 0.00077 0.001 0.00123 0.00138 0.00158 Length m 0.525 0.525 0.525 0.525 0.525 0.525 I m4 4.645x10-11 4.645x10-11 4.645x10-11 4.645x10-11 4.645x10-11 4.645x10-11 E N/m2 1.275x1011 8.269x1010 9.553x1010 1.035x1011 1.154x1011 1.209x1011 For steel: Results Tables Variables Units 1 2 3 4 5 6 mass g 50 100 150 200 250 300 Gravity m/s2 9.81 9.81 9.81 9.81 9.81 9.81 Force N 0.491 0.981 1.472 1.962 2.453 2.943 δ m 0.00034 0.00061 0.00098 0.00128 0.00177 0.00207 Length m 0.55 0.55 0.55 0.55 0.55 0.55 I m4 3.523x10-11 3.523x10-11 3.523x10-11 3.523x10-11 3.523x10-11 3.523x10-11 E N/m2 Graphs The graph of load against deflection is as attached. A line of best fit has been drawn for each case and the slope computed. Slope for brass: (2-1)/(0.0012-0.0006)=1666.7N/m Slope for steel: (2.8-1.6)(0.0019-0.0011)= 1500N/m N/B: for each material, the slope of each graph is the young’s modulus. Conclusions Steel is the more resilient of the two materials. Its lower value of E means that it is more flexible than brass. For brass, the calculated values of E range from 8.269x1010 to 1.209x1011 whereas the expected values range from 1.02x1011 to 125x1011. For steel, the calculated values of E range from 1.364x1011 to 1.582x1011 compared to the expected values which range from 1.80x1011 to 2.00x1011. The variations between the calculated and the expected values arise from conditions such as inaccuracies in reading and recording values during the experiment, temperature variations, dial gauge malfunctions and even calculation errors. The experiment objectives were succesfully met. Reference Bansal, R. K. (2010). strength of materials. New Delh: Laxmi Publications. Read More