By Maan H Jawad

Evaluation This ebook is written basically for pro engineers drawn to designing plate and shell buildings. It covers easy facets of theories and offers examples for the layout of elements because of inner and exterior so much in addition to different quite a bit resembling wind and lifeless rather a lot. numerous derivations are saved really easy and the ensuing equations are simplified to a degree the place the engineer may well practice them on to layout difficulties. extra complex derivations and extra normal equations should be present in the literature for these attracted to a closer wisdom of the theories of plates and shells. The examples given all through this booklet are meant to teach the engineer the extent of study had to in achieving a secure layout in line with a given required measure of accuracy. This e-book is additionally acceptable for complicated engineering classes.

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**Extra info for Design of Plate and Shell Structures **

**Example text**

2Þ Single Series Solution of Simply Supported Plates 31 The homogeneous solution for the deflection is obtained from Eq. (1- 46). Referring to Fig. 1-17, the deflection in the y-direction due to uniform load is symmetric about the x-axis. Hence, the constants Am and Dm must be set to zero mky since the quantities sinh mky a and y cosh a are odd functions as y varies from positive to negative. Also, m must be set to 1, 3, 5, etc. in order for sin mkx a to be symmetric around x = a/2. Hence, wh ¼ l X mky mky mkx sin Bm cosh þ Cm y sinh a a a m¼1;3;...

The solution is expressed as ð1- 42Þ w ¼ wh þ wp : The homogeneous solution is written as wh ¼ l X fm ðyÞ sin m¼1 mkx a ð1- 43Þ where f ( y) indicates that it is a function of y only. This equation also satisfies a simply supported boundary condition at x = 0 and x = a. Substituting Eq. (1- 43) into the differential equation j4w ¼ 0 Single Series Solution of Simply Supported Plates gives mk 4 a 29 mk 2 d 2 f ðyÞ d 4 f ðyÞ ! mkx m m sin fm ðyÞ À 2 þ ¼0 a dy 2 dy 4 a which is satisfied when the bracketed term is equal to zero.

Hence, the deflection expression becomes w¼ nky l l X sin mkx 16po X a sin b : k 6 D m¼1;3;... n¼1;3;... mn½ðm=aÞ 2 þ ðn=bÞ 2 2 ð1Þ The bending and torsional moment expressions are given by Eq. (1-17) and are expressed as 16po Mx ¼ 4 k " l X l X m¼1;3;... n¼1;3;... Fmn mkx nky sin sin a b # ð2Þ 26 Bending of Simply Supported Rectangular Plates Figure 1-15. Shear and Moment distribution in a uniformly loaded square plate. ) " # l l X 16po X mkx nky My ¼ 4 Gmn sin sin k a b m¼1;3;... n¼1;3;... " # l l X 16po ð1 À AÞ X mkx nky Mxy ¼ Hmn cos cos k4 a b m¼1;3;...