Download solution manual for advanced mechanics of materials and applied elasticity 5th edition by ugural and fenster

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