Structural analysis Essay Example | Topics and Well Written Essays - 1750 words

Structural analysis - Essay Example Assuming the average mass of adult to be 70 kilograms. Therefore, his weight shall be given by Weight= mass* acceleration due to gravity Weight= 70Kg*9.8N/Kg =686N The system is designed to carry three adults but the distance from each one is not specified .Therefore I treat their net weight to acting on the beam as a load equally distributed at 7.5 cm from end R1 indicated by force F1 which is technically equal to F2. Force F1 is given by; [? (the average weight of the three adults)] ? 2 (686?3) N ?2=1029 N Shear forces at the extreme ends of the beam are equal to the opposite reaction forces. The shear forces at any point on the beam is calculatable by using the formula F(x) = Rl –qx= qL/2 –qx= q [L/2 -x] whereby x is the distance from the left end of the beam, L the length of the entire beam and q the load on the beam F (0.075) =1029[1.8?2-0.075] = 848.925 The shearing force (SF) within any given party of the beam illustrates the tendency for the section of the beam on either side of the cross-section to slide or shear laterally in relation to other ends. The diagram shows the sitting beam in which the weights are distributed across it. The three adults will be assumed to be of load W1, W2, and W3 W1 W2 W3 The swing is supported on both ends by the other beams placed vertically to the sitting beam. The two ends in which the beam is supported is of reaction R1 and R2. Assuming the beam is split into two sections at point p. the resultant of the loads, reaction acting on the left of point p is F vertically upwards, and because the entire beam is uniform, the resultant force to the right of P should be F downwards. In this case, F is refers to the shearing force at the reaction P. usually the shearing force at any given part of the beam gives the algebraic sum of the lateral elements of the forces acting on each part of the beam. For bending moments, in the same approach it may that when the bending moments (BM) of the forces acting to the left pa rt of point P are clockwise then the bending moments of the forces on the right side should be in anticlockwise direction as indicated. Bending moment at given point is stated as the algebraic total of moments about a part of the entire forces acting in opposite directions. In a swing the bending moments are taken to be positive because when three adults sits on it, the total weights acts downwards and this is balance by the upward force from the two support chains on either side of the beam. The adults force gives the beam a clockwise push while the side chains give it an anticlockwise force. The results of this system are known as a sagging bending moment because it attempts to make the beam concave at the center. Fixing moments has to be determined in this system and work on the types of loads acting on it. It is also necessary to note that it is not possible to achieve a faultless fixing moment or the joining moment used will be associated to the angular movement of the supporti ng or reaction forces. Assuming that the adults sit adjacent to one another at point x of the beam We will let the shearing force at point X be F and at x = dx be F + dF. Equivalently, the bending moment is at M at x and M + dM at x + dx. Taking w as the mean estimate of loading of the beam length, the sum load is wdx,