On the deflection of a cantilever beam

Abstract

posed of a material whose modulus of elasticity is E. The problem is to find the deflection of the end-point of the beam due to the vertical load P. The bending moment induced at the point (x, y, s, 9) by the vertical load P is M = P(xl,-x). Therefore dd/ds = ci(xl-x), (1) where a-P/EI. Using the relation dd dd dx dd we obtain-= = cos 0-> ds dx ds dx J' cos 6 dd = j" a(xL-x)dx sin 6 = o(xlx-\x2) + C. (2) The boundary condition at the clamped end of the beam, namely, 6 = 0 when x = 0, reduces Eq. (2) to sin 6 = o,(xlX-%x2). (3) or Thus sin 0£ = \ax\. (4) Combining the latter expression and Eq. (3) we obtain sin 6l-sin 6 = \a{xL-x)2. (5) Thus xl-x = [2o_1 (sin 6l-sin 0)]1/2. Substituting this expression into Eq. (1), we obtain dd dd dy d6 r n-=-= sin 6-= [2a (sin Ol-sin 0)]1/2, ds dy ds dy or sin 6 dd Therefore With the transformation = r Jo [2a(sin Ol-sin 0)]l/2 /'sin 0 dd n [2a(sin dr-sin 0)11/2 ^ cos I / 7T d \ / 7T dL\ I) = cos I) sin d> = k sin -l)d(j> yL = a,-112 I-, (7) J i (1-^sin2^)1'2 W where