The Process of Constructing Shear - Essay Example

? Structural analysis Structural analysis Introduction A beam is any structure that is uniformly and (or) point transversely loaded.Structural analysis utilizes shear and bending moment’s images as tools of analysis. This result can eventually be used to determine deflection using the moment area method. Positive shear forces are usually labeled as those that spin the element clockwise while the bending moment is to warp the element (a) Diagram depicting shear forces (b) Diagram depicting bending moments The process of constructing shear and moment diagrams involves construction of a loading diagram, calculation of the sheer force and bending moment relative to the position of the beam and finally depict the shear and moment diagrammatically. 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 part 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 supporting 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, acting estimably (precisely when in uniform beam) by the center, C. the feature should be in balancing by the action of these forces and couples and where the given equations may be applied. Taking moments about C M + F. + (F + dF) = M + dM Assuming the result dF.dx for the limit: F = , The calculating vertical forces: wdx + F + dF = F Or w = - = - . From this, it is clear that when M is changing persistently zero shearing force matches to both minimum and maximum bending moment. It may be viewed from practice that peaks in bending moment figures usually happen at focused loads or reactions, which these are not stated within the equation F = = 0 though they can ideally illustrate the largest bending moment across the beam. Subsequently, it is not usually adequate to examine the points of zero shearing force when establishing the maximum bending force. At point of action on the beam where the beam will be transforming from sagging to hogging, the bending moment force should be zero and this is referred to as a point of contraflexure or inflection. However, by integrating the above equation when x = a and when x = b then; Mb – Ma =, this indicates that the raise in bending moment across two regions is the area under the shearing force figure. Equivalently, integrating the above equation gives Fa – Fb = gives the area under the load distribution figure. Furthermore, integrating further Ma – Mb = These associations may be highly significant when the rate of loading may not be indicated in an algebraic form as they offer a technique of graphical resolution. The loads across the beam are W1, W2, and W3 with corresponding reactions R1 and R2. Then starting on point x along the action line of R1. The calculation outcome of this system will always come to zero because the loads are at equilibrium with balancing forces. It is also essential to note that maximum bending moments happens at about x = 80 cm. to compute this accurately may require additional points indicated or apply the max and min theory in relation to the given equation. The beam bending moment also need determination of stress from the bending moment (Kharagpur, 2010). Strain is given by the equation: strain = y/R, which shows that strain differs directly as the distance from the neutral axis. Additionally, when the modules of elasticity is constant across the beam, then: E =, stress = E ? strain From these relations bending moment can be computed for any given point along the beam. Graphical representation may be given for bending moments and a graph of M plotted versus x is referred to as bending moment diagram, and it looks like the one given below. M X To develop this bending moment diagram, one can take an arbitrary part of a distant x on the left support beam. This part is used for any value of x only to the right side and an applied force P acting upwards in relation to the adult loads. The developed shear remains constant, and will be positive value of force P. However, the bending moment changes linearly with applied force from the point of support and move up to a maximum of +Pa (Ghali & Neville, 1997). From the equation of bending moment derived earlier, the longitudinal stress ‘f’ within a small section b.dy of part x may be calculated from equation: F = .y, where M1 = the bending moment at that point, I1 second moment at the part on neutral axis, and y = the distance of the area from the neutral axis. b.dy = dA The force acting on dA = f.dA = y.dA. Assume A1 to be the cross-sectional area at x, then the sum force (P1) acting within the area A1 will be given by: P1 = ??y.dA Let P2 be the other force acting within the same area (A2) at the opposite side. Then P2 = ?y?A2, where M2 is the Bending moment across this section, while I2 = second moment of the part on neutral section. Because dx is small then A1 = A2 and I1 = I2. Thus making A to be the area and I to be second moment of the part on neutral part. The shearing force will then be given by; Shearing force at A = P1 – P2 = *y*A = ?, however is the shearing vertical force of the party of the beam. Therefore, P1 – P2 = F ? and letting q be the shearing stress per unit area at A with t being the thickness of the beam, then the shearing force becomes q.t.dx. Therefore, finally we will have q = Stress distributed along the beam may be illustrated more by the model design below. These distribution can further be given by the diagram below depicting the regions under reaction and parallel forces. Y An important equation for determination of this stress distribution is given by t = , this equation is based on given assumptions with related limitations applied to create the flexure stress and strain associations. Deflections Usually, limits should be placed on the sum of deflection beam when subjected to a load. Deflection of this swing is based on the stiffness of the beam as well as its dimensions in relation to the applied force. Deflection due to moment when a strain rode is loaded with action being elastic, the longitudinal centroidal axis changes to a curve stated as elastic curve. In area of constant bending moment, the elastic curve is part of a circle of radius p (Ghali & Neville 1997, p 54). Also, q = However, it is crucial to note the following assumptions while calculating the deflection from curve associations: The beam deflection because of shear stress is negligible (i.e., the plane part remain the same) The square of gradient of the beam is assumed to be minimal in relation to the unity The value of E and I are kept constant for any interval through the beam. Conclusion The loading diagram is established by studying the beam and determining static loads and finding reactions. This varies from service to reaction loads. These loads are factored in a manner that they exert max stress on the beam. Reaction load are easily derived from structural configuration and the service loads by applying various structural analysis methods. On determination of reaction loads then the loading diagram can be drawn (Hibbeler, 1985) References: Ghali, A., and Neville, A. 1997. Structural analysis: a unified classical and matrix approach, New York, NY: Taylor & Francis publishers. Hibbeler, R.C 2005. Structural Analysis, USA: Macmillan. R. 2010. Energy Methods in Structural Analysis, New York, NY Springer publishers. Turner, J., Clough, W., Martin, H., and Topp's T. 2007. "Stiffness and Deflection of Complex Structures." Retrieved on November 21, 2011, from < Read More