4.1 The state of stress at a certain point in a body is given by:[ T ] =

Transcription

1 CHARTER 4 4. The state of stess at a cetain point in a body is given by:[ T ] On each of the coodinate planes (with nomal in e, e, ediections), (a) what is the nomal stess and (b) what is the total sheaing stess Ans. (a) The nomal stess on the e plane (i.e., the plane whose nomal is in the diection e ) is MPa., on the e plane is 4 MPa., and on the e plane is MPa. (b) The total sheaing stess on the e plane is e i +.6 On the e plane is , and on the e plane is MPa. 4. The state of stess at a cetain point in a body is given by: [ T ] 4 (a) Find the stess vecto at a point on the plane whose nomal is in the diection of e + e +e. (b) Detemine the magnitude of the nomal and sheaing stesses on this plane. - Ans. (a) The stess vecto on the plane is ttn, whee n(e+ e + e )/. Thus 5 [] t 4 6, (5 6 5 )/ + + t e e e 5 T n t (/9) e + e + e 5e + 6e + 5e (b) Nomal stess ( ) ( ) n n Magnitude of sheaing stess T t T 86 / O, s ( ) ( )( ) ( ) Ts t-tnn 5e+ 6e + 5 e / / e+ e + e e+ e / Ts 5/.745 _ 4. Do the pevious poblem fo a plane passing though the point and paallel to the plane x x + x 4. - Ans. (a) The nomal to the plane is n ( e e + e )/ 4. [ t ] [ T][ n ] [] t , t ( / 4 )(e 9 e).47e.4e (b) Nomal stess Tn n t [ ] 9 /4. MPa. 4 e i 4- Copyight, Elsevie Inc

2 n Magnitude of sheaing stess ( ) T t T 5 /4..6 O, s Ts ttnn (.47e.4 e) (./ 4)( e e + e).88e.e.77e, Ts The stess distibution in a cetain body is given by x x [ T] x x Find the stess vecto acting on a plane which passes though the point ( /, /, ) and is tangent to the cicula cylindical suface x + x at that point. Ans. Let f x + x, then the unit nomal to the cicle f x e + xe x + x f 4x + 4x f at a point ( x, x ) is given by n e e. At the point ( /, /, ), ( + ) and [ T ] n e e , thus, / 5 [] t 5 / 5 t 5 e+ 5e 5 e Given T, T, and all othe T ij at a point in a continuum. (a) Show that the only plane on which the stess vecto is zeo is the plane with nomal in the e diection. (b) Give thee planes on which thee is no nomal stess acting. n n Ans. (a) [] t n n t nene. t n n te. n n (b) T n t n n. Thus, the plane with n n has no nomal stess. These include n e, n ( e +e)/, n( ee)/ etc Fo the following state of stess [ T ] 5, find T and T whee e 5 is in the diection of e+ e + eand e is in the diection of e+ e e. 4- Copyight, Elsevie Inc

3 e ( e + e + e )/ 4, e ( e + e e )/, thus Ans. ' T [ ] 5 [ ] 5 9 / e' e' e ' ( 5e + 4 e e )/ 4, theefoe, ' T [ ] / MPa αx β 4.7 Conside the following stess distibution [ T ] β whee α and β ae constants. (a) Detemine and sketch the distibution of the stess vecto acting on the squae in the,,,,,,,, and,,. (b) Find the total x plane with vetices located at ( ) ( ) ( ) ( ) esultant foce and moment about the oigin of the stess vectos acting on the squae of pat (a). Ans. (a) The nomal to the plane x is e, thus, te α x e+ βe. On the plane, thee is a constant sheaing stess β in the e diection and a linea distibution of nomal stess α x, (see figue). (b) ( ) F t da α x e + β e dx dx e + 4β e. R ( ) ( α β ) M x tda x e + x e x e + e dx dx o x 4α ( βxe + αxxe αxe) dxdx e e α( ) e e + 4- Copyight, Elsevie Inc

4 4.8 Do the pevious poblem if the stess distibution is given by T T. ij α x and all othe Ans. (a) The nomal to the plane x is e, thus, te α x e. On the plane, thee is a paabolic distibution of nomal stess α x, (see figue). (,-,-) (,-,) α x x (,,-) (,,) x α (b) ( α ) F 4α t da x e dx dx e. R o da ( x x ) ( α x ) dxdx ( αxxe -αxe) dxdx e e M x t e + e e Do poblem 4.7 fo the stess distibution: T α, T T α x and all othe T. Ans. (a) The nomal to the plane x is e, thus, te αe + α xe. On the plane, thee is a constant nomal stess of α and a linea distibution of sheaing stess α xe, (see figue). ij (,-,-) (,-,) α x x (,,-) (,,) x α (b) ( ) F t da α e + αx e dx dx 4α e. R 4-4 Copyight, Elsevie Inc

5 o da ( x x ) ( α + αx ) dxdx M x t e + e e e ( α α α ) 4α x e + x e - x e dx dx e 4. Conside the following stess distibution fo a cicula cylindical ba: αx αx [ T ] α x α x (a) What is the distibution of the stess vecto on the sufaces defined by (i) the lateal suface x + x 4, (ii) the end face x, and (iii) the end face x l? (b) Find the total esultant foce and moment on the end face x l. Ans. (a) The outwad unit nomal vecto to the lateal suface x + x 4 is given by xe + xe n. The outwad unit nomal vecto to x is n eand that to x l is n e. Thus, αx αx α x x tn t n. α x x t Te αx e + αx e αx e αx e n ( ) ( ) tn Te αxe + αxe αxe + αxe (b) On the end face t Te α xe + α xe n n ( α α ) ( α ) ( α ) FR t da xe + xe da e xda + e xda. [note: the axes ae axes of symmety, the integals ae clealy zeo]. M x tda x e + x e αx e + αx e da ( ) ( ) o ( α x αx ) dae α πde πα de 8παe + 4. An elliptical ba with lateal suface defined by x + x has the following stess x x distibution: [ T ] x. (a) Show that the stess vecto at any point ( x, x, x) on x the lateal suface is zeo. (b) Find the esultant foce, and esultant moment, about the oigin O, of the stess vecto on the left end face x. 4-5 Copyight, Elsevie Inc

6 π π Note: xda and xda. 4 8 Ans. (a) The outwad unit nomal vecto to the lateal suface x + x is given by: x x xe+ xe n, thus, [ t] [ T][ n ] x x x + 4x x 4 x x + x (b) On the left end face x, n e, the stess vecto is t xe xe, FR tda ( xe xe) da ( e) xda ( e ) xda. [note: the axes ae axes of symmety, the integals ae clealy zeo]] Mo x tda ( xe + xe) ( xe xe) da π π π ( x + x ) eda + e e Fo any stess state T, we define the deviatoic stess S to be ST( /) T kk I, whee Tkk is the fist invaiant of the stess tenso T. (a) Show that the fist invaiant of the deviatoic 6 5 stess vanishes. (b) Given the stess tenso [ T ] 5 4 kpa., evaluate S. (c) Show 4 9 that the pincipal diections of the stess tenso coincide with those of the deviatoic stess tenso. / ts t T T / t I T T /. Ans. (a) Fom ST ( ) I, we have, ( ) ( )( ) T kk (b) [ S] ( )[] I kk kk kk / 5 4 kpa (c) Let n be an eigenvecto of T, then Tn λn. Now Sn Tn ( Tkk /) In λn ( Tkk /) n, that is λ λ λ /. Thus, n is also an eigenvecto of S with eigenvalue Sn n whee ( T kk ) ( /) λ T kk. 4. An octahedal stess plane is one whose nomal makes equal angles with each of the pincipal axes of stess. (a) How many independent octahedal planes ae thee at each point? (b) Show that the nomal stess on an octahedal plane is given by one-thid the fist stess invaiant. (c) Show that the sheaing stess on the octahedal plane is given by / Ts ( T T) + ( T T) + ( T T), whee T, T, Tae pincipal values of the stess tenso. Ans. (a) Thee ae fou independent octahedal planes. They ae given by the following unit nomal vectos: 4-6 Copyight, Elsevie Inc

7 e +e +e e +e e e e + e ee e n, n, n,n 4 We note that e +e +e gives the same plane as n, etc. (b) Using the pincipal diections as the othonomal basis, the matix of T is diagonal, i.e., T [ ] T e± e ± e T. The nomal to an octahedal plane is, thus, T T Tn n Tn [ ± ± ] T ± whee in this equation, the ow matix and column T ± matix of n have the same elements, that is if the ow matix is [ ] then the column matix T is. Thus, Tn [ ± ± ] T ( ) ± T + T + T. T ± (c) ( Ts t n Tn T + T + T ) ( T + T + T + TT + TT + TT) 9 ( T + T + T TT TT TT) ( T T) + ( T T) + ( T T) 9 9 / That is, Ts ( T T) + ( T T) + ( T T) 4.4 (a) Let m and n be two unit vectos that define two planes M and N that pass though a point P. Fo an abitay state of stess defined at the point P, show that the component of the stess vecto t m in the n -diection is equal to the component of the stess vecto tn in the m diection. (b) If meand ne, what does the esults of (a) educe to? Ans. (a) The component of the stess vecto tm in the n -diection is n tm n Tmand the T component of the stess vecto tn in the m diection is m tn m Tnn T m. Since T is T symmetic, theefoe, n T mn Tm, theefoe, n tm m t n. (b) If me and ne, then e te e t e T T. 4.5 Let m be a unit vecto that defines a plane M passing though a point P. Show that the stess vecto on any plane that contains the stess taction t m, lies in the M plane. Ans. Refeing to the figue below, whee m is pependicula to the plane M, and t m is the stess vecto fo the plane. Let N be any plane which contains the vecto t m pependicula to the plane N. Then t n and let n be the unit vecto Tn. We wish to show that tn is pependicula to m. 4-7 Copyight, Elsevie Inc

8 T Now, tn mtn mn T mn Tmn t m, because t m is on the N plane. Thus, tn m, so that tn lies on the M plane. 4.6 Let t m and tn be stess vectos on planes defined by the unit vecto m and n espectively and pass though the point P. Show that if k is a unit vecto that detemines a plane that contains tm and t n, then tk is pependicula to m and n. Ans. Since k is a unit vecto that detemines a plane that contains t m and t n, theefoe, tm tn k. Since tk Tk, tm Tm, and tn Tn, theefoe, t m t n T tm tm tn m tk m Tk k T mk Tmk tm, similaly, tm tn T tn tm tn n tk n Tk k T nk Tnk tn. tm tn 4.7 Given the function f ( xy, ) 4x y, find the maximum value of f subjected to the constaint that x+ y. Ans. Let gxy (, ) 4 x y + λ( x+ y ), then we have the following thee equations to solve fo xy, and λ : g x + λ, g y + λ and x+ y. x y Thus, x + λ λ x, y+ λ λ y, theefoe, x y x + y x x y. That is, fmax occus at x y. That is, f max 4 () () 4.8 Tue o false: (i) Symmety of stess tenso is not valid if the body has an angula acceleation. (ii) On the plane of maximum nomal stess, the sheaing stess is always zeo. Ans. (i) False. (ii) Tue. 4-8 Copyight, Elsevie Inc

9 4.9 Tue o false: (i) On the plane of maximum sheaing stess, the nomal stess is always zeo. (ii) A plane with its nomal in the diection of e + e ehas a stess vecto t 5e + e e It is a pincipal plane. Ans. (i) Not tue in geneal. Maybe tue in some special cases. (ii) Tue. We note that t 5e + e e 5( e + e e ). Theefoe, t is nomal to the plane, so that thee is no sheaing stess on the plane. That is, it is a pincipal plane. 4. Why can the following two matices not epesent the same stess tenso? 4 4 6, Ans. The fist scala invaiant fo the fist matix is 5 MPa. The fist scala invaiant fo the second matix is 6 MPa. They ae not the same, theefoe, they can not epesent the same stess tenso. 4. Given [ T ] (a) Find the magnitude of sheaing stess on the plane whose nomal is in the diection of e +e. (b) Find the maximum and minimum nomal stesses and the planes on which they act. (c) Find the maximum sheaing stess and the plane on which it acts. Ans. (a) Let n ( e+ e ). Then t n i.e., tn n sheaing stess T s. λ (b) The chaacteistic equation is λ λ( λ ) λ λ, λ, λ. The maximum nomal stess is and the minimum nomal stess is MPa. Fo λ, α+ α, so that α α, n ( e+ e )/. Fo λ, α+ α α α, n ( ee )/. ( Tn) ( T ) max n min ( ) (c) ( Ts ) The maximum sheaing stess acts on max the planes n ( n± n )/, i.e., on the planes e and e. 4-9 Copyight, Elsevie Inc

10 4. Show the equation fo the nomal stess on the plane of maximum sheaing stess is ( Tn) + ( T ) max n min Tn. Ans. Let { n, n, n } be the pincipal axes of the stess tenso with pincipal values T > T > T, T then [ T ] T. On the plane n ( n ± n ) /, the sheaing stess is a maximum. On T this plane, the nomal stess is: T T ( ) ( ) + T T n + T max n min Tn n Tn Tn [ ± ] T T ± 4. The stess components at a point ae given by: T, T, T 4 T T T. (a) Find the maximum sheaing stess and the planes on which they act. (b) Find the nomal stess on these planes. (c) Ae thee any plane/planes on which the nomal stess is 5 MPa.? Ans. (a) The maximum nomal stess is clealy T 4, acting on the e plane and the minimum nomal stess is clealy T, acting on the e plane. Thus, the maximum 4 sheaing stess is ( Ts ) max 5 MPa., acting on the plane n ( e± e ). (b) Tn [ ± ] [ ] 5 ± 4 ± ± 4 Tmax + Tmin 4 + Note: We can also use the esult of Pob. 4. to obtain Tn 5 (c) No, because Tmax 4MPa 4.4 The pincipal values of a stess tenso T ae T, T and T T If the matix of the stess is given by: [ T ], find the T values of T and T. Ans. I + ( T + + T ) ( T + T ) T T (i) ( )( )( ) ( ) ( ) ( ) ( T )( T ) T T I T T 4T T T 4 T (ii) (i) and (ii) Thus, 4- Copyight, Elsevie Inc

11 T [6 ± 6 ] / ±. Thus, T is eithe 5 o. To detemine which is the coect value, we check ( )( ) ( )( ) ( )( ) ( 4) I + + T + TT + T T + TT + T. (iii) Ty T fist, fom (i), T, so that (iii) is clealy satisfied. Next ty T 5, eq (i) givest 5, then left side of (iii) becomes 5 + ( )(5) 6. Thus, T and T. 4.5 If the state of stess at a point is: [ T ] kpa., find (a) the magnitude of 4 the sheaing stess on the plane whose nomal is in the diection of ( e+ e + e ) and (b) the maximum sheaing stess. Ans. (a) Let n ( e+ e + e ), then 6 [ t ] 4 n ( ) tn e e + e Tn n tn ( 8 + 4) kpa Ts tn Tn 6.76 Ts 6 kpa ( ) (b) ( Ts ) kpa. max Given [ T ] 4 (a) Find the stess vecto on the plane whose nomal is in the diection of e+ e. (b) Find the nomal stess on the same plane. (c) Find the magnitude of the sheaing stess on the same plane. (d) Find the maximum sheaing stess and the planes on which this maximum sheaing stess acts Ans. (a) Let n ( e+ e ), then [ t ] 4 5 n ( ) tn e + e. (b) ( 5 5 ) 5. Tn n t n + MPa (c) Ts t n Tn Copyight, Elsevie Inc

12 λ λ 4 λ 5, λ, λ. Thus, T 5 and T (d) The chaacteistic equation is ( ) ( ) ( n) ( ) max n min Fo ( Tn ) 5 ( 5) α+ 4α α α n ( / )( + ) max e e. Fo ( Tn ) ( +) α+ 4α α α ( / )( ) min 5( ) Thus, ( T ) 4, acting on the plane whose nomal is s max ( )( ) n e e. n / n± n ne and ne. 4.7 The stess state in which the only non-vanishing stess components ae a single pai of sheaing stesses is called simple shea. Take T T τ and all othe T ij. (a) Find the pincipal values and pincipal diections of this stess state. (b) Find the maximum sheaing stess and planes on which it acts. τ Ans. (a) With [ T ] τ, the chaacteistic equation is ( ) λ λ τ λ τ, λ τ, λ. Fo λ τ, ( τ) α+ τα α α n ( / )( e+ e ). Fo λ τ ( + τ) α + τα α α ( / )( ) n e e. Fo λ n e. τ ( τ) (b) ( T s ) max ( )( ) τ, acting on the plane whose nomal is n / n± n ne and ne. 4.8 The stess state in which only the thee nomal stess components do not vanish is called a ti-axial state of stess. Take T σ, T σ, T σ with σ > σ > σ and all othe T ij. Find the maximum sheaing stess and the plane on which it acts. σσ Ans. ( T s ), n ( e± e ). max 4.9 Show that the symmety of the stess tenso is not valid if thee ae body moments pe unit volume, as in the case of a polaized anisotopic dielectic solid. 4- Copyight, Elsevie Inc

13 * * * * Ans. Let M Me+ Me + Me be the body moments pe unit volume. Then efeing to the figue shown below, the total moments of all the suface foces and the body foce and body moment about the axis which passes though the cente point A and paallel to the x axis is : M T Δx Δx Δ x / + T +ΔT Δx Δx Δx / ( c ) ( )( ) ( )( )( ) * T ( ΔxΔx )( Δx /) ( T +ΔT )( ΔxΔx )( Δ x /) + M ( ΔxΔx Δx ) ( / )(density)( Δx Δx Δx ) ( Δ x ) + ( Δx ) α x components of the angula acceleation of the element. We now let whee α is the Δx, Δx, Δx and dop all tems of small quantities of highe ode than ( ΔxΔxΔ x), we obtain, ( ) ( ) ( ) * * * * T T M and T T M. T Δx Δx Δx T Δx Δx Δ x + M Δx Δx Δ x T T M, Similaly, one can show that ( ) x+ x T x, x 4. Given the following stess distibution: [ T ] T ( x, x ) x x, find T so x that the stess distibution is in equilibium with zeo body foce and so that the stess vecto on the plane x is given by t ( + x) e+ ( 5x) e. Tij Ans. The equations of equilibium ae + ρbi. Now with B i, we have, x j T T T T T x + f( x ). x x x x T T T T + + T x + g( x), thus, T x x + C. x x x x T T T + +. x x x To detemine C, we have, the stess vecto on the plane x is ( ) ( ) tte T e + T e + T e x + x e + x x + C e. Thus, x ( ) ( ) ( ) ( ) + x e+ x + C e + x e+ 5x e C T x x Copyight, Elsevie Inc

14 x x 4. Conside the following stess tenso: [ T ] α x x. Find an expession x T fot such that the stess tenso satisfies the equations of equilibium in the pesence of body foce vecto B ge, whee g is a constant. Tij Ans. The equations of equilibium ae + ρbi. With B B, B g, we have, x j T T T T T T ρb + + +, ρb + + +, x x x x x x T T T T T ρ g ρb α + ρg + x x x x x α ρ g T + x + f( x, x) α 4. In the absence of body foces, the equilibium stess distibution fo a cetain body is ( ) T Ax, T T x, T Bx + Cx, T T + T /, all othe T ij. Also, the bounday plane x x fo the body is fee of stess. (a) Find the value of C and (b) detemine the value of A and B. Tij Ans. (a) The equations of equilibium ae + ρbi. With B i,, we have, x j T T T ρb + + +, x x x T T T ρb + C+ + C, x x x T T T ρb x x x (b) The unit nomal to the bounday plane x x is n ( e e ) /. Thus, on this plane (note x x), we have, Ax x Ax x [] t x Bx x x Bx x +, thus, ( T + T )/ Ax x A and x Bx+ x B. 4. In the absence of body foces, do the following stess components satisfy the equations of equilibium: 4-4 Copyight, Elsevie Inc

15 ( ) ( ) ( ) T α x + ν x x, T α x + ν x x, T αν x + x, T T αν x x, T T, T T. Tij Ans. The equations of equilibium ae + ρbi. With B i,, we have, x j T T T ρb ανx ανx+ +, x x x T T T ρb ανx + ανx + + x x x T T T ρb Yes, the equations of equilibium ae all satisfied. x x x 4.4 Repeat the pevious poblem fo the stess distibution x+ xxx [ T ] α x x x x x Tij Ans. The equations of equilibium ae + ρbi. With B i,, we have, x T T T ρb α( + ), x x x T T T ρb α( + ) x x x No, the second equation of equilibium is not satisfied. j 4.5 Suppose that the stess distibution has the fom (called a plane stess state) T ( x, x ) T ( x, x ) [ T ] T ( x, x ) T ( x, x ) (a) If the state of stess is in equilibium, can the body foces be dependent on x? (b) If we ϕ ϕ ϕ intoduce a function ϕ ( x, x) such that T, T and T, What should x x x x be the function ϕ ( x, x) fo the equilibium equations to be satisfied in the absence of body foces? T T T T ( x, x ) T ( x, x ) Ans. (a) ρb + + ρb. x x x x x 4-5 Copyight, Elsevie Inc

16 (, ) (, ) T T T T x x T x x ρb + + ρb. x x x x x Thus, B and Bmust be independent of x. T T T ϕ ϕ ϕ ϕ (b) ρb x + +. x x x x x x x x x x x T T T ϕ ϕ ϕ ϕ ρb + x x x x x x x x x x x x Thus, the equations of equilibium ae satisfied fo any function ϕ ( x, x) which is continuous up to the thid deivatives. 4.6 In cylindical coodinates (, θ, z), conside a diffeential volume of mateial bounded by the thee pais of faces : and + d; θ θ and θ θ + dθ; z z and z z+ dz. Deive the and θ equations of motion in cylindical coodinates and compae the equations with those given in Section 4.8. Ans. Fom the fee body diagam above, we have, F T dθ dz + T + dt + d dθ dz T ddz cos dθ / ( ) ( ) ( ) ( ) ( ) θ + ( Tθ + dtθ ) ddz cos ( dθ / ) Tθθ ddz sin ( dθ / ) ( Tθθ + dtθθ ) ddz sin ( dθ / ) T ( dθ) d + ( T + dt )( dθ) d + ρb ( dθ) ddz ρ( dθ) ddz a. z z z dθ dθ dθ dθ dθ Now, cos...and sin... +! + +! and keeping only tems involving poducts of thee diffeentials (i.e., tems involving poduct of 4 o moe diffeentials dop out in the limit when these diffeentials appoach zeo), we have, T ddθdz + dt dθdz + dt ddz T ddz dθ / ( ) θ θθ ( ) ( dt )( dθ) d ρb ( dθ) ddz ρ( dθ) ddz a + z + Dividing the equation by dθ ddz, we get, T T Tθ Tθθ Tz ρb ρa. This is Eq. (4.8.) dθ z Next, 4-6 Copyight, Elsevie Inc

17 Fθ Tθ( dθ) dz + ( Tθ + dtθ) (( + d) dθ) dz + Tθddzsin ( dθ / ) + ( Tθ + dtθ ) ddz sin ( dθ / ) Tθθ ddz cos ( dθ / ) + ( Tθθ + dtθθ ) ddz cos ( dθ / ) T ( dθ) d + ( T + dt )( dθ) d + ρb ( dθ) ddz ρ( dθ) ddz a. θ z θz θz θ θ dθ dθ dθ dθ dθ Again, cos...and sin... +! + +! and keeping only tems involving poducts of thee diffeentials, we have, T ddθdz + dt dθdz + T ddz dθ / + dt ddz + dt dθd ( θ) ( θ) θ ( ) ( θθ ) ( θz) ρb ( dθ) ddz ρ( dθ) ddz a + θ θ Dividing the equation by dθ ddz, we get, Tθ Tθ Tθ Tθθ Tθz ρbθ ρaθ, this is Eq. (4.8.). θ z 4.7 Veify that the following stess field satisfies the z-equation of equilibium in the absence of body foces: z z Az z z z T A, T, T,, 5 θθ zz A + T 5 z A + T 5 θ Tzθ R R R R R R R R + z Ans. The z equation of equilibium in cylindical coodinate is: Tz Tzθ Tz Tzz R R z ρbz. Now, so that θ z R z R T z z R z 5z R A + 5 A R R R R R R z 5 z z 5 z Tzθ A + A +, R R R R R R R R θ T z z A + 5, R R 4 T zz z R 9z 5z R z 9z 5z A + A z 4 z z R R R R R R R R 4 Tz Tz T zz z z z 9z 5 z 5z Thus, + + A z R R R R R R R R 4 z z z 9z 5 z 5z A R R 5 R 5 R 5 R 5 R 5 R 7 R 7 ( ) ( ) + z 5z 5z + z R 5z 5z R A + A R R R R R R R R 4-7 Copyight, Elsevie Inc

18 4.8 Given the following stess field in cylindical coodinates: Pz Pz Pz T, T,,, 5 zz T 5 z T 5 θθ Tθ Tzθ R + z π R πr πr Veify that the state of stess satisfies the equations of equilibium in the absence of body foces. Ans. T Tθ T Tθθ Tz Pz Pz Pz dθ z πr πr z πr Pz Pz Pz P Pz 5 ( ) + z πr π R πr πr z π z R Pz 5Pz R Pz Pz 5Pz R πr πr πr πr πr z 5Pz 5Pz 5Pz 5Pz 5Pz ( + z ) πr πr πr πr πr Tθ Tθ Tθθ Tθz θ z Tz Tzθ Tz Tzz Pz Pz Pz ρbz θ z πr πr z πr Pz Pz Pz 9Pz Pz πr π R πr πr π z R Pz 5Pz R Pz 9Pz 5Pz R πr πr πr πr πr z 4 Pz 5Pz Pz 9Pz 5Pz 5Pz + z Pz πr πr πr πr πr πr πr 4.9 Fo the stess field given in Example 4.9., detemine the constants A and B if the inne cylindical wall is subjected to a unifom pessue pi and the oute cylindical wall is subjected to a unifom pessue p o. Ans. The given stess field is: B B T A+, T, constant and θθ A T zz Tθ Tz Tθ z. On the oute wall, o, and T po, and on the inne wall, i, and T pi, theefoe, B B we have, po A+ (i) and p i A+ (ii). o i ( ) ( ) ( pii poo ) ( ) B B po pi i o o i, i o o i o i p p B A. 4-8 Copyight, Elsevie Inc

19 4.4 Veify that Eq. (4.8.4) to (4.8.6) ae satisfied by the stess field given in Example 4.9. in the absence of body foces. B B Ans. The given stess field is: T A, Tθθ Tφφ A+, Tθ Tφ Tθφ. ( T ) ( Tθ sinθ ) T φ - T θθ + T φφ + + sinθ θ sinθ φ B B B B A A A A ( ) ( T sinθ ) Tθ θθ Tθφ Tθ Tθ Tφφ cotθ sinθ θ sinθ φ T cot Tφφ cotθ θθ θ ( Tφ ) ( Tφθ sinθ ) Tφφ Tφ Tφ + Tθφ cotθ sinθ θ sinθ φ 4.4 In Example 4.9., if the spheical shell is subjected to an inne pessue pi and an oute pessue p o, detemine the constant A and B. B B B Ans. Fom the example, we have, T A, thus, p o A and p i A ( ) ( ) oop i i po pi oi and B o i o i p A ( ). 4.4 The equilibium configuation of a body is descibed by: x 6 X, x X, x X. If the Cauchy stess tenso is given by: 4 4 T,and all othe T ij, (a) calculate the fist Piola Kichhoff stess tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is e plane and (b) calculate the second Piola-Kichhoff tenso and the coesponding pseudo stess vecto fo the same plane. Ans. Fom x 6 X, x X, x X, we obtain the defomation gadient F and its 4 4 invese as: o i 4-9 Copyight, Elsevie Inc

20 6 /6 F / 4, 4 F and detf. /4 4 [ ] (a) The fist Piola-Kichhoff stess tenso is, fom ( ) ( ) T T det F T F : /6 /6 T [ To ] ( det F)[ T] ( ) () 4 F 4 Fo a unit aea in the defomed state in the e diection, its undefomed aea is 6 6 T ( daono) F n () /4 dao o 6 det n e. F /4 That is, its undefomed plane is also e plane. The coesponding pseudo stess vecto is given by /6 to Ton o, whee no e. Thus t [ ] o Tono to to ( /6) e We note that the pseudo stess vecto is in the same diection as the Cauchy stess vecto and the intensity of the pseudo stess vecto is /6 of the Cauchy stess vecto simply because the undefomed aea is 6 times the defomed aea and both aeas have the same nomal diection. (b)the second Piola-Kichhoff stess tenso is, fom T% ( F) F T( F ) T /6 /6 T ( det ) [ ] ( ) () 4 4 T% F F T F 4 4 /6 /6 / The coesponding pseudo stess vecto is given t % Tn % o, whee no e. Thus, t% Tn % o t% ( / 56) e. 4.4 Can the following equations epesent a physically acceptable defomation of a body? Give eason. x X, x X, x 4X. / Ans. [ F] / detf. The given equations ae not acceptable as a 4 physically acceptable defomation because it gives a negative atio of defomed volume to the undefomed volume. o det 4- Copyight, Elsevie Inc

21 4.44 The defomation of a body is descibed by: x 4 X, x ( /4 ) X, x ( /4) X. (a) Fo a unit cube with sides along the coodinate axes what is its defomed volume? What is the defomed aea of the e face of the cube? (b) If the Cauchy stess tenso is given by: T,and all othe T ij, calculate the fist Piola Kichhoff stess tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is e plane. (c) Calculate the second Piola-Kichhoff tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is e plane. Also, calculate the pseudo diffeential foce fo the same plane. x 4 X, x /4 X, x /4 X, we have (a) Ans. Fom ( ) ( ) 4 [ F] /4 detf /4, thus dv ( detf ) dvo dv ( /4 )() /4. /4 /4 /4 T da dao( det F)( F ) no ()( / 4 ) 4 ( / 4 ) da ( /6) e 4 That is, the defomed volume is /4 of its oiginal volume and the e face of unit aea defomed into an aea /6 of it oiginal aea and emain in the same diection. These esults ae quite obvious fom the geomety of the defomation. (b) The fist PK stess tenso is: / 4 /6 [ ] ( )[ ] ( ) T To detf T 4 F 4 4 The coesponding pseudo stess vecto fo e -plane in the defomed state, whose undefomed plane is also e -plane, is given by to Ton o, whee n o e, that is to ( /6) e The Cauchy stess vecto on the e face in the defomed state is t e Clealy the Cauchy stess vecto has a lage magnitude because the aea in the defomed state is /6 of the undefomed aea. (c) The second PK stess tenso is: / 4 /6 / 64 [ o ] 4 T% F T 4 The coesponding pseudo stess vecto fo the e -plane in the defomed state, whose undefomed plane is also e -plane, is given by t % Tn % o, whee n o e. Thus, t% ( / 64 ) e The 4- Copyight, Elsevie Inc

22 pseudo foce df % is elated to the foce df( t odao / 6 e fo da o ) by the fomula / 4 /6 - dff % df, Thus, - [ ] d d 4 f % F f df % 64 e The defomation of a body is descibed by: x X+ kx, x X, x X. (a) Fo a unit cube with sides along the coodinate axes what is its defomed volume? What is the defomed aea of the e face of the cube? (b) If the Cauchy stess tenso is given by: T T,and all othe T ij, calculate the fist Piola Kichhoff stess tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is e plane and compae it with the Cauchy stess vecto in the defomed state. (c) Calculate the second Piola-Kichhoff tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is e plane. Also, calculate the pseudo diffeential foce fo the same plane. Ans. Fom x X+ kx, x X, x X, we have (a) k [ F] detf, thus dv ( det F ) dvo dv dvo. T da dao( det F)( F ) no ()() k k daeke That is, the defomed volume is the same as its oiginal volume and the e face of unit aea defomed into an aea + k of it oiginal aea and whose nomal is in the diection of e ke. These esults ae quite obvious fom the geomety of the defomation. (b) The fist PK stess tenso is: k [ ] ( )[ ] ( ) T To det F T k F The coesponding pseudo stess vecto fo the ( e ke) plane, whose undefomed plane is the e plane, is given by to Ton o, whee n o e. Thus, to ( ke+ e ) The Cauchy stess vecto on the ( e ke) face in the defomed configuation is [] t [ T][ n] k ( k ) t e + e + k + k The Cauchy stess vecto has a smalle magnitude because the defomed aea is undefomed aea. + k times the 4- Copyight, Elsevie Inc

23 (c) The second PK stess tenso is: k k k [ o ] T% F T The coesponding pseudo stess vecto fo the e keplane, whose undefomed plane is the e plane, is given by t % Tn % o, whee no e. Thus, t% ( ke+ e) The pseudo foce df % - is elated to the foce df todao ( ke + e ) fo dao by the fomula dff % df, ( ) k k - Thus, d [ d ] f % F f df % ( ke +e) The defomation of a body is descibed by: x X, x X, x X. (a) Fo a unit cube with sides along the coodinate axes, what is its defomed volume? What is the defomed aea of the e face of the cube? (b) If the Cauchy stess tenso is given by: Mpa., calculate the fist Piola Kichhoff stess tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is the e plane and compae it with the Cauchy stess vecto on its defomed plane, (c) calculate the second Piola-Kichhoff tenso and the coesponding pseudo stess vecto fo the plane whose undefomed plane is the e plane. Also, calculate the pseudo diffeential foce fo the same plane. Ans. Fom, x X, x X, x Xwe have (a) [ F] detf 8, thus dv ( det F ) dvo dv 8dVo 8. / / T da dao( det F)( F ) no ()( 8 ) / ( 8 ) da 4e / (b) The fist PK stess tenso is: / 4 T [ To ] ( det F)[ T] ( ) ( 8 ) / 4 F / 4 The coesponding pseudo stess vecto fo the e plane in the defomed state, whose undefomed plane is also e plane, is to 4 e The Cauchy stess vecto on the e plane is t e The Cauchy stess vecto has a smalle magnitude because the aea is fou times lage. (c) The second PK stess tenso is: 4- Copyight, Elsevie Inc

24 / 4 [ o ] / 4 T% F T / 4 The coesponding pseudo stess vecto fo the e plane in the defomed state, whose undefomed plane is also e plane, is t% e The pseudo foce df % is elated to the foce - df ( todao 4 e fo dao ) by the fomula dff % df. Thus, / - d [ d ] 4 / f % F f df % e. / 4-4 Copyight, Elsevie Inc