Click here to Purchase full Solution Manual at http://solutionmanuals.info CHAPTER 3 SOLUTION (3.1) ( a ) We obtain " 4# "x 4 = !12 pxy " 4# "y 4 =0 " 4# "x 2"y 2 = 6 pxy Thus, # 4 " = !12 pxy + 2(6 pxy ) = 0 and the given stress field represents a possible solution. (b) " 2# "x 2 = pxy 3 ! 2 px 3 y Integrating twice px 3 y 3 px 5 y "= 6 ! 10 + f1 ( y ) x + f 2 ( y ) The above is substituted into " 4 ! = 0 to obtain d 4 f1 ( y ) d 4 f2 ( y ) dy 4 x+ dy 4 =0 This is possible only if d 4 f1 ( y ) d 4 f2 ( y ) dy 4 =0 dy 4 =0 We find then f 1 = c 4 y 3 + c5 y 2 + c 6 y + c 7 f 2 = c8 y 3 + c9 y 2 + c10 y + c11 Therefore, px 3 y 3 px 5 y "= 6 ! 10 + ( c4 y 3 + c5 y 2 + c6 y + c7 ) x + c8 y 3 + c9 y 2 + c10 y + c11 ( c ) Edge y=0: a a 4 pa5t Vx = # ! xy tdx = # ( px2 + c3 )tdx = 5 + 2c3 at "a "a a a Py = ! # y tdx = ! (0)tdx = 0 "a "a Edge y=b: a px 4 Vx = " (! 32 px 2b 2 + c1b 2 + 2 + c3 )tdx !a = ! pa 3 (b 2 ! a2 5 )t + 2a ( c1b 2 + c3 )t a Py = " ( pxb3 ! 2 px 3b)tdx = 0 !a ______________________________________________________________________________________ SOLUTION (3.2) Edge x = ±a : " xy = 0 : ! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0 " xy = 0 : ! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0 Adding, ( !3 pa 2 + 2c1 ) y 2 + pa 4 + 2c3 = 0 (CONT.) ______________________________________________________________________________________ ______________________________________________________________________________________ 3.2 (CONT.) or c1 = 23 pa 2 c3 = ! 12 pa 4 Edge x = a : "x = 0: pa 3 y ! 2c1ay + c2 y = 0 or c2 = 2 pa 3 ______________________________________________________________________________________ SOLUTION (3.3) ( a ) Equations (3.6) become !" xy !" xy !# x !x + !y =0 !x =0 Substituting the given stresses, we have c 2 y ! 2 c3 y = 0 Thus c 2 = 2 c3 c1 = arbitrary c2 (b) # x = c1 y + c2 xy " xy = 2 (b 2 ! y 2 ) Assume c1 > 0 and c2 > 0 . y c2 " xy = c2 2 (b ! y ) 2 " xy = 2 (b 2 ! y 2 ) 2 b x b ! x = c1 y ! x = ( c1 + c2 a ) y a ______________________________________________________________________________________ SOLUTION (3.4) Boundary conditions, Eq. (3.6): !# xy !# xy !" y !" x !x + !y =0 !x + !y =0 or ( 2ab ! 2ab) x = 0 ( !2ab + 2ab) y =0 are fulfilled. However, equation of compatibility: 2 !2 ( !!x 2 + !y 2 )(" x + " y ) = 0 or 4ab ! 0 is not satisfied. Thus, the stress field given does not meet requirements for solution. ______________________________________________________________________________________ SOLUTION (3.5) It is readily shown that " 4!1 = 0 is satisfied 4 " !2 = 0 is satisfied (CONT.) ______________________________________________________________________________________ ______________________________________________________________________________________ 3.5 (CONT.) We have " 2 #1 " 2 #1 2 %x = "y 2 = 2 c, %y = "x 2 = 2a , $ xy = ! ""x#"y1 = !b Thus, stresses are uniform over the body. Similarly, for ! 2 : # x = 2cx + 6dy # y = 6ax + 2by " xy = !2bx ! 2cy Thus, stresses vary linearly with respect to x and y over the body. ______________________________________________________________________________________ SOLUTION (3.6) Note: Since ! z = 0 and ! y = 0 , we have plane stress in xy plane and plane strain in xz plane, respectively. Equations of compatibility and equilibrium are satisfied by " x = !" 0 " y = !c "z = 0 (a) ! xy = ! yz = ! xz = 0 We have !y = 0 (b) Stress-strain relations become (" x $!" y ) (" y $!" x ) #x = E , #y = E (c) !% ($ x +$ y ) #z = E , " xy = " yz = " xz = 0 Substituting Eqs. (a,b) into Eqs. (c), and solving ! y = $"! 0 # z = " (1+E" )! 0 2 " x = ! (1#!0$E ) "y = 0 Then, Eqs. (2.3) yield, after integrating: 2 u = ! (1!# E)" 0 x v=0 w = " (1+"E )! 0 z ______________________________________________________________________________________ SOLUTION (3.7) Equations of equilibrium, !" xy !# x !x + !y = 0, 2axy + 2axy = 0 "$ xy "# y "y + "x = 0, ay 2 ! ay 2 = 0 are satisfied. Equation (3.12) gives )($ x + $ y ) = "4ay ! 0 2 #2 ( ##x 2 + #y 2 Compatibility is violated; solution is not valid. ______________________________________________________________________________________ ______________________________________________________________________________________ SOLUTION (3.8) We have " 2$ x " 2$ y " 2# xy "y 2 =0 "x 2 = !2ay "x"y = 2ay Equation of compatibility, Eq. (3.8) is satisfied. Stresses are #x = E 1$! 2 (" x + !" y ) = aE 1$! 2 ( x 3 + !x 2 y ) # y = 1$E! 2 (" y + !" x ) = 1$aE! 2 ( x 2 y + !x 3 ) # xy = G" xy = aE 2 (1+! ) xy 2 Equations (3.6) become aE 1!" 2 (3x 2 + 2"xy ) + 1aE +" xy = 0 aE 1!" 2 y 2 + 1!aE" 2 x 2 = 0 These cannot be true for all values of x and y. Thus, Solution is not valid. ______________________________________________________________________________________ SOLUTION (3.9) #x = "u = !2$cx #y = "v = 2ax y "x "y 2Eac # xy = "u "y + "v "x = !2cy + 2cy = 0 Thus 2a x O % x = 1&E# 2 ($ x + #$ y ) = 0 " xy = G! xy 2b $ y = 1!E" 2 (# y + "# x ) = 2 Ecx 2Eac Note that this is a state of pure bending. ______________________________________________________________________________________ SOLUTION (3.10) (a) %x = " 2# "y 2 = 0, % y = 6 pxy $ xy = !3 px 2 Note that " 4 ! = 0 is satisfied. (b) ! y = 6 pbx y ! xy = 3px 2 !x = 0 !x = 0 b ! xy = 0 ! xy = 3pa 2 a x !y =0 ! xy = 3px 2 ( c ) Edge x = 0: V y = Px = 0 Edge x=a: Px = 0 (CONT.) ______________________________________________________________________________________ Click here to Purchase full Solution Manual at http://solutionmanuals.info ______________________________________________________________________________________ 3.10 (CONT.) b V y = " # xy tdy = 3 pa 2 bt ! 0 Edge y = 0: Py = 0 a V x = " # xy tdx = pa 3t ! 0 Edge y = b: V x = pa 3t ! a Py = " # y tdx = 3 pa 2 bt ! 0 ______________________________________________________________________________________ SOLUTION (3.11) ( a ) We have # 4 " ! 0 is not satisfied. 2 p ( x 2 + xy ) p ( 4 xy + y 2 ) % y = !!x"2 = pya 2 , 2 %x = a2 , $ xy = # 2a2 y (b) !y = p ! xy = 2p (1 + 4 ax ) 2p !x = 0 ! x = p(1 + ay ) py 2 a " xy = ! 2a py 2 ! xy = 2a2 ( 4a + y ) a x " y = 0, ! xy = 0 p a py 2 ( c ) Edge x = 0: Vy = ! 2 dy = 16 pat Px = 0 0 2a Edge x=a: a V y = " # xy tdy = 7 6 pat ! 0 a Px = " # x tdy = 3 2 pat ! 0 Edge y = 0: Vx = 0 Py = 0 Edge y = a : a V x = " # xy tdx = 3 2 pat ! 0 a Py = " # y ptdx = pat ! 0 ______________________________________________________________________________________ SOLUTION (3.12) ( a ) We have " 4 ! = 0 is satisfied. The stresses are $ x = ""y#2 = ! bpx3 (6b ! 12 y ) 2 " 2# $y = "x 2 =0 6 py (b ! y ) 2 $ xy = ! ""x"#y = b3 (CONT.) ______________________________________________________________________________________ ______________________________________________________________________________________ 3.12 (CONT.) (b) y pa "x = ! b3 (6b ! 12 y ) b ! xy ! xy x a ______________________________________________________________________________________ SOLUTION (3.13) We have "# "y = ! $P [tan !1 xy + xy x2 + y2 ], "# "x = ! Py $ !y x2 + y2 " 2# ( x 2 + y 2 ) x !2 y 2 x "y 2 = ! $P [ x 2 +x y 2 + ( x 2 + y 2 )2 ] The stresses are thus, ! 2" x3 %x = !y 2 = # 2$P ( x 2 + y 2 )2 ! 2" xy 2 %y = !x 2 = # 2$P ( x 2 + y 2 )2 2 x2 y % xy = # !!x!"y = # 2$P ( x 2 + y 2 )2 P !x L 2 P !L ! xy ______________________________________________________________________________________ SOLUTION (3.14) Various derivatives of ! are: 2 3 "# "x = $0 4 ( y ! yh ! hy2 ), " 2# "x 2 =0 2 2y " 4# "x 2"y 2 = 0, " 2# "x"y = $40 (1 ! h ! 3hy2 ) (a) 6 xy 6 Ly " 2# "y 2 = $0 4h ( !2 x ! h + 2L + h2 ) 4 4 ! " !x 4 = 0, ! " !y 4 =0 It is clear that Eqs. (a) satisfy Eq. (3.17). On the basis of Eq. (a) and (3.16), we obtain (CONT.) ______________________________________________________________________________________ ______________________________________________________________________________________ 3.14 (CONT.) "x = #0 4h ( !2 x ! 6hxy + 2 L + 6 Ly h ), "y =0 2 (b) " xy = ! "40 (1 ! 2hy ! 3hy2 ) From Eqs. (b), we determine Edge y = h : "y = 0 ! xy = ! 0 Edge y = !h : #y =0 " xy = 0 2 Edge x = L: # x = 0, " xy = ! "40 (1 ! 2hy ! 3hy2 ) It is observed from the above that boundary conditions are satisfied at y = ±h , but not at x = L . ______________________________________________________________________________________ SOLUTION (3.15) (a) For # 4 " = 0, e = !5d and a, b, c are arbitrary. Thus " = ax 2 + bx 2 y + cy 3 + d ( y 5 ! 5 x 2 y 3 ) (1) ( b ) The stresses: $x = " 2# "y 2 = 6cy + 10d ( 2 y 3 ! 3x 2 y ) (2) = 2a + 2by ! 10dy 3 2 $y = " # "x 2 (3) $ xy = ! ""x"#y = !2bx ! 30dxy 2 2 (4) Boundary conditions: # y = !p " xy = 0 (at y=h) (5) Equations (3), (4), and (5) give b = !15dh 2 2a ! 40dh 3 = ! p (6) h h h ! $ x dy = 0 ! y$ x dy = 0 ! # xy dy = 0 (at x=0) (7) "h "h "h Equations (2), (4), and (7) yield c = !2dh 2 (8) Similarly "y =0 ! xy = 0 (at y=-h) give a = 20dh 3 (9) Solution of Eqs. (6), (8), and (9) results in a = ! 4p b = ! 163ph c = ! 40ph d= p 80 h 3 e = ! 16ph3 (10) The stresses are therefore " x = ! 320pyh + 8hp3 ( 2 y 3 ! 3x 2 y ) 3 " y = ! 2p ! 38pyh ! 8pyh3 3 px y2 " xy = 8h (1 ! h2 ) ______________________________________________________________________________________ ______________________________________________________________________________________ SOLUTION (3.16) We obtain " 2# p ( x 2 !2 y 2 ) $x = "y 2 = a2 ! 2" py 2 #y = !x 2 = a2 (a) 2 2 pxy $ xy = ! ""x"#y = ! a2 Taking higher derivatives of ! , it is seen that Eq. (3.17) is not satisfied. Stress field along the edges of the plate, as determined from Eqs. (a), is sketched bellow. y !y = p ! xy = 2 p ax 2p p ! xy = 0 ! xy = 2 p ay 2 a ! x = 2 p ay2 a 2 " x = p(1 ! 2 ay2 ) 2 a x p " y = 0, ! xy = 0 ______________________________________________________________________________________ SOLUTION (3.17) The first of Eqs. (3.6) with Fx = 0 !" xy pxy !y = I Integrating, pxy 2 ! xy = 2I + f1 ( x ) (a) The boundary condition, pxh 2 (! xy ) y =h = 0 = 2I + f1 ( x ) gives f1 ( x ) = ! pxh 2 2 I . Equation (a) becomes " xy = ! 2pxI ( h 2 ! y 2 ) (b) Clearly, (" xy ) y = ! h = 0 is satisfied by Eq. (b). Then, the second of Eqs. (3.6) with Fy = 0 results in "# y 2 2 p(h ! y ) "y = 2I Integrating, p y2 "y = 2I y (h 2 ! 3 ) + f 2 ( x) (c) 2 Boundary condition, with t = 3I 2h , (CONT.) ______________________________________________________________________________________ Click here to Purchase full Solution Manual at http://solutionmanuals.info
Report "Solution Manual for Advanced Mechanics of Materials and Applied Elasticity, 5th Edition Ansel C. Ugural Saul K. Fenster Sample"