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Unformatted text preview: USE OF DISCONTINUITY FUNCTIONS FOR FINDING BEAM DEFLECTIONS 1 Introduction From the theory of the bending of beams, we know that the bending moment M and the curvature 1 /ρ are related by the equation 1 ρ = dθ dx = M EI , (1) where the convention for a positive bending moment is shown in Figure 1(b) below. In equation (1), θ = du dx (2) is the slope of the beam, u is the vertically upward deflection and EI is the flexural rigidity. F L x F M V (Lx) (a) (b) Figure 1 These equations permit us to find the deflection of the beam if M ( x ) is known as a function of x . For example, Figure 1(a) shows a beam built in at x = 0 and loaded by a concentrated force F at the other end x = L . The freebody diagram Figure 1(b) then shows that M ( x ) = − F ( L − x ) (3) and hence, substituting into (1), we have EI dθ dx = − F ( L − x ) . 1 Integrating with respect to x and using (2) then gives EIθ = EI du dx = − F parenleftBigg Lx − x 2 2 parenrightBigg + A , (4) where A is an arbitrary constant. One more integration gives EIu = − F parenleftBigg L x 2 2 − x 3 6 parenrightBigg + Ax + B , (5) where B is another arbitrary constant. The arbitrary constants in equations (4, 5) are determined from the kine matic conditions describing the way the beam is supported. In the present example, the beam is built in at x = 0, which means that both the slope and the deflection are zero at this point — i.e. u = 0 ; θ = 0 ; at x = 0 . Applying these conditions to equations (4, 5), we obtain A = 0 ; B = 0 (6) and hence the final expression for the deflection is u = − F EI parenleftBigg L x 2 2 − x 3 6 parenrightBigg , from (5, 6). 1.1 Types of Support In beam problems, the only kinds of support allowed are a builtin support, which prevents both deflection and rotation (slope) and a simple support which prevents deflection but permits rotation. Pin joints can also be re garded as simple supports in this context. A summary of the boundary conditions in a beam problem is therefore Builtin support at x = a u ( a ) = 0 ; θ ( a ) = 0 . (7) 2 Simple support at x = b u ( b ) = 0 . (8) A determinate beam must have either a single builtin support or two simple supports, in which case equations (7, 8) will provide exactly two conditions for the two arbitrary constants A,B . If the problem is indetermi nate, there will more than two boundary conditions, but there will also be an equivalent number of additional unknown reaction forces associated with the indeterminacy. 2 Solution starting from the applied loads To use the method presented so far, we need to start by drawing a freebody diagram to determine the moment M as a function of x . An alternative method that avoids this step is to start from the loading on the beam w ( x ) and use the equilibrium relations dV dx = − w (9) dM dx = V , (10) where V is the shear force and a positive value of w corresponds to a down ward load on the beam....
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This note was uploaded on 10/05/2011 for the course MECHENG 211 taught by Professor ? during the Fall '07 term at University of Michigan.
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