34 Mechanics of Materials 2 $2.1 Here the particular solution is MM y= n2EI P y =A cos nx+Bsin nx +M/P Now when x =0,dy/dx =0..B=0 and whenx=L,y=0 ）A=-M sec nL 2 M nL M y=-p sec. 2 cos nx+ But when x=L,dy/dx is also zero, 0= nM nL nL 2 别n 2 nM nL The fundamental buckling mode is then given when nL/2 =π 4x2E1 or Pa= 2 (2.4) (d)One end fixed,the other pinned In order to maintain the pin-joint on the horizontal axis of the unloaded strut,it is necessary in this case to introduce a vertical load F at the pin(Fig.2.5).The moment of F about the built-in end then balances the fixing moment. Fig.2.5.Strut with one end pinned,the other fixed. With the origin at the built-in end the B.M.at C is d2y lax=-Py+F(L-x) E F (L-x) F (D2+n2)y= L-)34 Mechanics of Materials 2 $2.1 Here the particular solution is MM y=-- - - n2EI P .. y = Acos nx + Bsinnx + M/P Now when x = 0, dy/dx = 0 :. B = 0 1 -M nL and when x = ,L, y = 0 :.A = - sec - P 2 .. M nL M y= --sec-cosnx+- P2 P But when x = iL, dy/dx is also zero, nM nL nL 0 = ~ sec - sin - P 22 nM nL 0 = - tan - P 2 The fundamental buckling mode is then given when nL/2 = JT '&) 2 =Yt or 4dEI P, = - L= (d) One end Bed, the other pinned In order to maintain the pin-joint on the horizontal axis of the unloaded strut, it is necessary in this case to introduce a vertical load F at the pin (Fig. 2.5). The moment of F about the built-in end then balances the fixing moment. Fig. 2.5. Strut with one end pinned, the other fixed. With the origin at the built-in end the B.M. at C is d2Y dx2 d'v P F -+-y=-(L-x) Et- -Py + F(L - X) dx2 Et' EI F EI (0' + n2)y = -(L - x)