tailieunhanh - Applied Structural Mechanics Fundamentals of Elasticity Part 10

Tham khảo tài liệu 'applied structural mechanics fundamentals of elasticity part 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 256 12 Membrane theory of shells The form of 4 suggests introduction of the sum and the difference of the unknown functions as new functions F1 O F F2 D-Y. 5 If we now divide 4 by sin jp 5 yields by addition and subtraction respectively of the two equations 4 Fi 2 1 2F1 2 Pl 2 0 with X12 2cot P1 2 pBa cosip 1 7 where the index 1 implies 4- and the index 2 implies - . The ordinary inhomogeneous differential equations of the first order with variable coefficients 6 have the following solutions according to The integrals are evaluated by means of 7 Jxj dtp J 2cot p si y d tp 2 In sin p 4 In tan y A1 d p 2 In sino In tinu 2 . 2 _ V 1 e sin2 ỉp tan . In a similar way we determine tp tp IXdv cot2 -2 . p f 2 . tan 2 e _ . eJ sin p cot e J -. 777 - sin2 tp 2 sin3 qp For F we then obtain F -Ẻ Ỉ cos p 1 J sin ip tan cot 2 sin2 p By means of 1 I cos tp 2 cos3 2 2 w 2 v Sin 99 4 sin -F cos the integral can be determined as follows R 3 3 5Ì 0 cos sin 2 2 3 sin p dtp 1 1_ 3 - cos 4- -y- COS tp . 3 If we substitute 2 cos2 I cot ---------- 2 2 Sin COS 1 COS fp sin p Exercise C-12-3 257 we obtain from 9 Fl C1 PQa cos p - y cos3 tp 1 1 cos tp sin3 tp and analogously F2 ca - Po a cos tp - y cos3 tp 1 - cos tp sin3 p Substitution into 5 and solving leads after introduction of two new integration constants Dj Cj C2 and D2 C1 - C2 to y F f2 1 - 2 Dj Dj cos p 4- 2 Po a cos p cos tp - 4- cos3 O __1 _ sin3 qp 10 Y y F1- F2 2 d2 4- DjCostp 4- 2 Po a cos p -ycos3 p 1 sin qp In order to ensure finiteness of the resultant forces at the top p 0 we demand that D3 2p0a 0 . 11 Since sin3 p occurs in the denominator not only the numerator but also its first and second derivative have to vanish at the point p 0 . We obtain from the second equation 10 for the first derivative of the term in square brackets - Dj sin ip 1 2 Po a - sin cp cos2 p sin tp 0 y - 0 and for the second derivative - cos tp 4- 2 p0 a - cos tp - 2 cos tp sin fp 4- cos3 tp 0 . v 0 Whereas the first condition is .