Strength Engineering Materials - Assignment Example

Introduction This paper consists of 5 tasks where various aspects of strength engineering materials have been looked into. In the first task the deflection and slope in a simply supported beam were calculated. In the second task there was calculation involving reinforced concrete where the moment capacity and area of reinforcement were calculated. The report also looks at eccentric loading of a circular beam in task 3, theoretical buckling load in task 4 and then experiment on buckling load is in task 5. Task 1 A beam A,B,C is simply supported at A and C. It carries a vertical downward load of 1KN at B. The beam also carries a uniformly distributed (UDL) load of 500N/m across it's entire length. The left hand support is at position A. Distances: AB = 0.3m and BC = 0.7m. The rectangular aluminium section for the beam is 40mm wide and 60mm deep. The beam is 1m long. i. Determine the slope at support A. Slope at A due to concentrated load Where P=1000N a=0.7m b=0.3 l=1m substituting Slope at C due to UDL Total slope at A =0.00118+0.00041=0.00159rad 0.090 ii. Determine the slope at support C. Slope at C due to concentrated load Where P=1000N b=0.3 l=1m substituting Slope at C due to UDL Total slope at C =0.000903+0.00041=0.0001313rad 0.00790 iii. Determine the deflection in the middle of the beam. E = Youngs Modulus = 70GN/M2 This provides evidence for P2.1 Deflection due to concentrated load Substituting we have Deflection due to UDL Making relevant substitution Total deflection  Task 2 Information has been supplied on a reinforced concrete beam, rectangular in section as follows: D = depth of section = 460mm B = breadth of section = 250mm n = modular ratio = 16 Maximum allowable stress in the steel = 150MN/m² Maximum allowable stress in the concrete = 8MN/m² A = total area of steel reinforcement = 0.0013m² a) Determine the moment of resistance of the beam In the beam compression force =Tension force Substituting required values Moment of resistance  Substituting relevant values we have This provides evidence for P2.2 b) If the service loads were to be increased by 25% and the breadth of the beam is to remain at 250mm, determine the new depth of the beam and the area of steel for tension reinforcement required This provides evidence for M1 Increasing service load by 25% result to an increase in moment of resistance by the same margin Thus new value of  Minimum reinforcement ration  if  whichever is greater Applying  Also   gives a bigger value thus we use 0.01 in finding the solution Rule of thumb is d=1.8b =1.8x0.25 =0.45m From  Task 3 A column of diameter Ø0.6m carries a load of 400KN which is offset from the columns centroid by 0.15m. a) Determine both the Tensile and Compressive stresses due to the offset load. Total for at any side of the column is given by  Where is direct compression stress is bending stress due to moment y varies from -3 to 3 through zero Compressive stress is at y=3 Tensile stress is at y=-3 Compressive b) Using graphical methods, determine the neutral axis of the column This provides evidence for P2.3 The values of stress from the centre of the beam to outer surface of the beam are as shown in table 3.1. From the table it can be seen that the stress is maximum 4245.189N which is a compressive side and it is on the surface towards which the eccentric force is applied. The minimum force is on the opposite side of the maximum compressive force and it is a tensile force of -1415.19N. Table 3.1 -0.3 1415 -2830.19 -1415.19 -0.2 1415 -1886.79 -471.792 -0.1 1415 -943.396 471.6038 0 1415 0 1415 0.1 1415 943.3962 2358.396 0.2 1415 1886.792 3301.792 0.3 1415 2830.189 4245.189 The data in the table is used to draw figure 1 . From the figure it can be seen that at the centre of the beam where x=0 we have a compressive stress of 1415N and the neutral axis is where we have 0 stress and from the diagram this point -0.04m meaning that it is 0.04m from the centre in the opposite side (left side) of the side where the force is applied. Figure 1 Task 4 A strut is 0.8m long and has a rectangular cross section of breadth = 12mm and depth = 12mm. The bottom of the strut is fixed rigid into a ground socket and the top is unrestrained. E = 200 GN/m². (Strut end conditions: One end fixed, the other end unrestrained, n = 0.25; Both ends pinned, n =1) a) Calculate the buckling load of the strut. This provides evidence for P2.4\ Buckling load  For n=0.25 F=1331N b) If the end conditions of the same strut were changed to pinned at both the top and bottom, calculate the new buckling load and compare the result to that calculated in task 4a above. For the case pinned at both the top and bottom, n=1 F=1221N Task 5 a) Using a strut with cross section and length the same as the one used in task 4, setup the apparatus for loading struts and determine the critical load: Setup the apparatus as follows using the correct method and procedure: Load the strut into the apparatus with the end conditions pinned at both ends of the strut Gradually apply load to the strut until it fails and record the critical load value This provides evidence for P2.5 The WP 120 buckling device is use in performing the experiment. The force gauge is turned is attached to the system. The thrust piece was secured in the machine by inserting it in socket and fastening the clamp screw. The long thrust piece together with V-notch were inserted into the guide bush then held firmly after which the specimen rod was inserted in the v-notch. In setting the specimen it was aligned such that the buckling direction was pointing in the direction of the direction of the lateral column. The rod specimen was pretightened with low not measurable load before being subjected slow loading by use of load nut. The deflection of the rod was read from the measuring gauge with recording being made after every 0.25mm up to 1mm. At the point when the was no more change in the loading force the test was concluded. Test results Load Deflection (mm) 0 0 100 0.1 200 0.2 300 0.3 400 0.4 500 0.5 600 0.6 700 0.7 800 0.8 900 0.9 1100 1.0 From the table it can be seen that the specimen buckled at 1100N The buckling load is lower than what was recorded in the theoretical calculation. This can be attributed to errors in the gouges used in measurements and also the fact that the specimen could be having slightly different buckling load due to variation brought about in manufacture process. Discussion In this paper we first looked at deflection and slope due to a beam being subject to both UDL and point loads. In this task it was important to handle each load separately and then adding the effect. In the task 2, reinforced concrete was dealt with and in reinforcing concrete the steel is assumed to take all the tensile stress. The design of a reinforced concrete beam involve iteration before arriving at the desired design. When a beam is eccentrically load it was found that the eccentricity result to a bending moment stress one side of the column being subjected to tensile stress and the opposite side being under compressive stress. Due to the introduction of the bending moment stress the eccentric loading increases the stress in the column considerably compared to non- eccentrically loaded column. In conclusion research paper involved successful application of engineering formulas with reasonable answers being obtained. References Courtney, T.H., (1990). Mechanical Behavior of Materials, McGraw-Hill, New York, Hayden, H.W., Gere, James M.; Goodno, Barry J. Mechanics of Materials (Eighth ed.). pp. 1083–1087. ISBN 978-1-111-57773-5 William D. Callister, Jr, Materials Science and Engineering – An introduction, sixth edition, John Wiley & Sons, Inc. 2004. Read More