Circular Plst,es undrr Combined Action of Lateral Load and Tension
orCompression 95. 145 = 0. A Simply Supported Circular Plate under Uniform Load 100. 040 292’0. 5112.
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5. 3). (g) and (j)for any angle (YFIG. The Use of Infinit,e Integrals and
Transforms.
In this manner we find1 a a aw . 118 and 125.
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The maximum bending stress 02 will act at the built-in
edges where the bending moment, is t,hc largest. (d) with q = k6. Flat Slab Having Nine Pnncls and Slab with Two Edges Free 253
56. 2 0.
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(36), and by using
them the bending and twisting moments can be readily calculated for
any value of (Y. 20 surface is therefore the neutral surface. The
diameter of the circle will be equal to Ilr, – AI,, as shown in
Fig. 187 and, from
Fig.
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A Circular Cylindrical Shell Loaded Symmetrically with
Respect to Its Axis 115. Rectangular Plates with Two Opposite Edges
Simply Snpportcd aud theOther Two Edges Free or Supported Elastically 214 49. 0611. , Inc. In considering the curvature of the middle surface in any
direction an (Fig.
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Rectangular Plates with Two Opposite Edges
Simply Pupportcd ant1 theOther Two Edges Clamped 185 43. . 121. If, in addition to lateral loads, there are external forces
acting in the middle plane of the plate, the first assumption does
not hold any more, and it is necessary to take into consideration
the effect on bending of the plate of the stresses acting in the
middle plane of the plate. During bending of the
plate the points a, b, c, and d undergo small displacements.
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Differential Equation for Cylindrical Bending of Plates 2. -l/2 =Substituting expression (a) for w and observing that CZ = Ca =
0, we obtain;+c I sin ,!J sinh p + C4 cos 0 cash p = 0 Cc)from which we find Ci cos fl cash p – Cd sin p sinh p = 0C,=-. edu and the wider internet faster and more securely, please take a few seconds toupgrade your browser. 5320. org/10. , as the plane midway
between the faces oft)he plate.
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The quantity log,,
(lo4 2/U,) then gives the ordinate of the curve in Fig. During bending, the
particles that were in the xy plane undergo small displacements w
perpendicular to the zy plane and form the middle surface of the
plate. dn = cos (Y + – sin (Y a?4 (a)To find the direction (~1 for which the slope is a maximum we
equate to zero the derivative with respect to a! of expression (a). In
such a case the slope of the surface in any direction can be taken
equal to Check This Out angle that the tangent to the surface in that
direction makes with the sy plane, and the square of the slope may
be neglected compared to unity. The unit elongation E= of a fiber at a
distance z from the middle surface (Fig. The
maximum stress occurs at the middle of the strip, where the blog here is a maximum.
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r!) 1 1 Y= r; cos2 (y + – sin2 (y i ry:\nd the ordinateComparing these results with formulas (36)) we conclude that the
coordi- * See S. . For any given plate we begin by
calculating the square root of the lcft- hand side of Eq. (24) for a plate
without initial curvature. Let us now consider the stresses acting on a section of the
lamina abed parallel to the z axis and inclined to the LI: and y
axes.
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We take the zy plane to coincide with the middle
plane of the plnt,e before deflection and the z and y axes along
the edges of the plate as shown. 09. 4BENDING TO you could look here CYLINDRICAL SURFACE 5from the middle surface. (Ic) is equal to the
doubled value of expression (j), we conclude that, if the
directions an and at (Fig. C. For 01 = 7r/2 or 3a/2, weWlFIG.
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