Final Exam 2009 December 2009
December 9, 2009 Final Examination
Mechanics of Deformable Solids MECH 321 Sec. 1 Wednesday, December 9, 2009
Examiner:
Assoc Examiner:
Larry Lessard
Eliot Fried
Student Name:
McGill ID:
INSTRUCTIONS: (Examples) •
This is a CLOSED BOOK examination.
•
You are permitted TRANSLATION dictionaries ONLY.
•
STANDARD CALCULATOR permitted ONLY.
•
This examination consists of 7 questions including 2 short answer questions for a total of 12 pages, including the cover page. Make sure you have 12 pages.
Preview o o o o o o o o
(1) (10%) (2) (20%) (3) (20%) (4) (20%) (5) (20%) (6) (5%) (7) (5%) (8) (0%)
Elasticity Equations Go Habs Go (Boundary Conditions) Three-Dimensional Problem 2D problem, Cartesian Coordinates Energy Methods Rotating Disc Problem Finite Element Analysis Philosophy
•
Attempt all questions
•
There are 6 pages of equations at the end of the exam
•
This examination is PRINTED ON BOTH SIDES of the paper
•
This examination paper M UST BE RETURNED
December 9, 2009
Page 1 of 12
Mech 321 exam
Mechanics of Deformable Solids Mechanical Engineering MECH 321 Examiner: Professor Larry B. Lessard [email protected]
(1) (10%)
Wednesday, December 9, 2009 9:00 - 12:00 398-6305
Assoc Examiner: Professor Eliot Fried [email protected]
398-3739
Elasticity Equations For what body forces will the following stress field describe a state of equilibrium?
" xx = #2x 2 + 3y 2 # 5z " yy = #2y 2 " zz = 3x + y + 3z # 5
(2) (20%)
!
" xy = z + 4 xy # 7 " xz = #3x + y + 1 " yz = 0
Go Habs Go (Boundary Conditions) The figure below shows the loading conditions on a hockey puck due to a slap shot. We are examining the puck when the stick has made contact. The puck is considered an isotropic solid, with Young’s Modulus E, Poisson’s Ratio ν , density ρ , outer radius R and thickness H. The impact force imparted from the left side, is simplified by a constant radial force Q (MPa) over a 40 degree arc of the puck and covers the area from the top to the bottom of the puck in that region. The weight of the puck is countered by an idealized ice load, upward from the bottom, and assumed constant over the entire bottom surface of the puck.
2.1 Referring to the diagram above, set up a complete set of boundary conditions for this 3D problem at all boundaries. The problem has been set up such that there are no displacement boundary conditions. 2.2 Without solving for the stress state, would this problem have a relatively simple solution for stresses or not (state your reasons in one or two sentences).
December 9, 2009
Page 2 of 12
Mech 321 exam
(3) (20%)
Three-Dimensional Problem Consider the following problem, which is a circular rod subjected to two loads. There is a torsion load T and an axial load P. The rod has the following dimensions and load data. Length L = 1m Cross-section area A = 0.1m2 Load P = 5 kN Torque T = 0.8919 kNm J = πR4/2 (polar moment of inertia)
(a) For this 3D problem assume that the axial load P is evenly distributed at each end in tension over area A and that the torque T is causing pure torsion of a cylinder of circular cross section. Find an expression, in cylindrical coordinates for the 3D stress state. (b) The largest stresses will occur at r=R. Evaluate the stress state at r=R. (c) From the results of (b), calculate principal stresses at r=R (4) (20%)
2D problem, Cartesian Coordinates
A stress distribution is given by
" xx = pyx 3 # 2c1 xy + c 2 y " yy = pxy 3 # pyx 3 3 1 " xy = # px 2 y 2 + c1 y 2 + px 4 + c 3 2 2 Where the p’s and c’s are constants.
!
(a) Verify that this field represents a 2D solution for a thin plate of thickness t by verifying the biharmonic equation. (b) Find the resultant normal and shearing boundary forces (Px and Vx) along the edges y=0 and y=b of the plate. December 9, 2009
Page 3 of 12
Mech 321 exam
(5) (20%)
Energy Methods
A vertical load P is applied at B to two bars of equal length L but different cross-sectional areas and moduli of elasticity. (a) Determine the horizontal displacement δB of point B. Use one of the Castigliano energy methods. B. C, and D are all pinned joints Bar BC has modulus EBC, cross-sectional area ABC and cross-sectional moment-of-inertia IBC Bar BD has modulus EBD and cross-sectional area ABD and cross-sectional moment-of-inertia IBD
(b) From the solution that you obtain in part (a), show that it reduces to a predictable result when the two bars are identical (i.e., same modulus and cross-sectional properties)
(6) (5%)
Rotating Disc Problem A thin rotating disc with inner and outer radius undergoes stress due to the rotation.
Related to the solution of the rotating disc problem, the following three statements are false. Explain why each statement is erroneous (one sentence for each answer). (a) The loading comes from radially applied distributed loads that occur on the outer edge r = b, due to the rotation. (b) It is necessary to impose boundary conditions on this problem that insure there are no infinite stresses at r =0. (c) Boundary conditions will require us to impose that hoop stress
December 9, 2009
Page 4 of 12
"## will be zero at the inner boundary r = a. Mech 321 exam
!
(7) (5%)
Finite Element Analysis (a) In the following welded frame structure (no pinned joints), beam elements will be used in a finite element model.
The global FEA formulation for this problem will be in the form of Where the force vector
[K ]{u} = {F }
{F } = {FAx FAy M A FBx FBy M B etc}
What will be the input value to the global force vector for entry M F ? (if this is impossible to calculate based on the given information, state “impossible”) !
!
(c) Shown below are the element matrices for a bar element and a beam element
!
EA # 1 "1&) ui , ) f i , % (* - = * L $"1 1 '+ u j . + f j .
!
# 12 6L "12 6L &* v i . * Fi . , , , % 2 2 (, EI % 6L 4L "6L 2L (, ) i , , M i , + /=+ / L3 %"12 "6L 12 "6L(,v j , , F j , % 2 2( $ 6L 2L "6L 4L ',-) j ,0 ,- M j ,0
In the bar element, all entries of the stiffness matrix have the same units. This is not the case for the beam element. Explain why in two sentences (maximum).
!
8 (0%)
" Stress is nothing more than a socially acceptable form of mental illness.” ~Richard Carlson, American author
December 9, 2009
Page 5 of 12
Mech 321 exam
Mechanics of Deformable Solids - Useful Equations 1. Elasticity Equations in Cartesian Coordinates I. “Existence of Continuum”
! "F "A ! !B lim = !V" 0 !V : (Bx , By , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector
! B ! B
lim "A # 0
II. Stress Boundary Condition Equation ! Px = l! xx + m! xy + n! xz ! Py = l! xy + m! yy + n! yz ! Pz = l! xz + m! yz + n! zz III. Equation of Motion or Equilibrium Equation for a Deformable Body !" xx !" xy !" xz + + + Bx = 0 !x !y !z !" xy !" yy !" yz + + + By = 0 !x !y !z !" xz !" yz !" zz + + + Bz = 0 !x !y !z IV. Strain-Displacement Relationships (small displacements) "u "v "w ! xx = ! yy = ! zz = "x "y "z "v "u "w "u "w "v ! xy = 12 ( + ) ! xz = 12 ( + ) ! yz = 12 ( + ) "x " y "x " z "y "z V. Symmetry of Stresses and Strains ! xy = ! yx ! xz = ! zx ! yz = ! zy
! xy = ! yx ! xz = ! zx VI. Strain Compatibility Relations ! 2 " yy ! 2" xx ! 2" xy + = 2 !x 2 !y2 !x!y 2 2 ! " zz ! " xx ! 2" xz + = 2 !x 2 !z 2 !x !z 2 2 ! 2 " yz ! " zz ! " yy + =2 !y2 !z 2 !y !z
December 9, 2009
! yz = ! zy ! 2 " yz ! 2 " zx ! 2" zz ! 2 " xy + = + !x!y !z 2 !z!x !y !z ! 2 " yy ! 2 " xz ! 2 " xy ! 2 " yz + = + !x !z !y 2 !y!z !x!y 2 2 ! 2 " xx ! " yz ! 2 " xz ! " xy + = + !y !z !x 2 !x !y !x!z
Page 6 of 12
Mech 321 exam
VII. Constitutive Relations (Isotropic Materials) (i) Strain-Stress (Compliance Form) ! xx = 1E (" xx # $" yy # $" zz ) ! yy = 1E (" yy # $" xx # $" zz ) ! zz = 1E (" zz # $" xx # $" yy ) 1 ! xy = 2G " xy = 1+E$ " xy ! xz = 2G1 " xz = 1+E$ " xz ! yz = 2G1 " yz = 1+E$ " yz (ii) Stress-Strain (Modulus Form) E ! xx = "e + 2G# xx = (1+$)(1% 2$ ) [(1 % $ )# xx + $ (# yy + # zz )] E ! yy = "e + 2G# yy = (1+$ )(1%2 $ ) [(1 % $)# yy + $ (# xx + # zz )] E ! zz = "e + 2G# zz = (1+$ )(1% 2$ ) [(1 % $ )# zz + $ (# xx + # yy )] ! xy = 2G# xy = (1+E$) # xy ! xz = 2G# xz = (1+E$ ) # xz ! yz = 2G# yz = (1+E$ ) # yz where e = (! xx + ! yy + ! zz ) G(3" + 2G) "+G " # = 2(" + G) #E " = (1 + # )(1 $ 2# ) E G= 2(1 + # ) E=
2. Elasticity Equations in Cylindrical Coordinates I. “Existence of Continuum”
! "F "A ! !B lim = !V" 0 !V : (Br , B! , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector
! B ! B
lim "A # 0
II. Stress Boundary Condition Equation
! Pr = nr! rr + n"! r" + nz! rz ! P" = nr! r" + n"! "" + nz! "z ! Pz = nr! rz + n"! "z + nz! zz
December 9, 2009
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Mech 321 exam
III. Equation of Motion or Equilibrium Equation for a Deformable Body !" rr 1 !" r# !" zr " rr $ " ## + + + + Br = 0 !r r !# !z r !" r# 1 !" ## !" #z 2" r# + + + + B# = 0 !r r !# !z r !" rz 1 !" #z !" zz " rz + + + + Bz = 0 !r r !# !z r IV. Strain-Displacement Relationships (small displacements) "u 1 "u# ur "u ! rr = r ! ## = + ! zz = z "r r "# r "z 1 "ur "u# u# "u "u "u 1 "u z 1 1 ! r# = 2 ( + $ ) ! rz = 2 ( r + z ) !#z = 12 ( # + ) r "# "r r "z "r "z r "# V. Symmetry of Stresses and Strains ! r" = ! "r ! rz = ! zr ! "z = ! z" ! r" = !"r ! rz = ! zr ! "z = ! z" VI. Strain Compatibility Relations (Not given here) VII. Constitutive Relations (Isotropic Materials) (i) Strain-Stress (Compliance Form)
! rr !%% ! zz ! r% ! rz !%z
= = = = = =
(" rr # $" %% # $" zz ) (" %% # $" rr # $" zz ) (" zz # $" rr # $" %% ) " r% = 1+E$ " r% " rz = 1+E$ " rz " %z = 1+E$ " %z
1 E 1 E 1 E 1 2G 1 2G 1 2G
(ii) Stress-Strain (Modulus Form)
! rr ! && ! zz ! r& ! rz ! &z
E = "e + 2G# rr = (1+$ )(1% 2$ ) [(1 % $ )# rr + $ (#&& + # zz )] E = " e + 2G# && = (1+ $)(1% 2$ ) [(1 % $ )# && + $ (# rr + # zz )] E = "e + 2G# zz = (1+$ )(1% 2$ ) [(1 % $ )# zz + $ (# rr + #&& )] = 2G# r& = (1+E$ ) # r& = 2G# rz = (1+E$) # rz = 2G#&z = (1+E$ ) # &z
where
e = (! rr + ! "" + ! zz )
December 9, 2009
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Mech 321 exam
3. Energy Strain energy "
U = ! U o dV U o = ! # d" V
0
Complementary strain energy "
C = ! C o dV C o = ! # d" V
0
Strain energy Axial L N2 UN = ! dx 2 EA 0
Bending
Shear
L
Torsion L
2
kV dx 2GA 0
M dx 2 EI 0
US = !
UM = !
L
2
T2 dx 2GJ 0
UT = !
Castigliano’s theorem I and II
Fi =
!U !qi
qi =
!C !Fi
where Fi and qi are generalized force and displacement
4. Plane Stress and Plane Strain Plane Stress Constitutive
" xx " yy " xy " xz
1 = (# xx !$# yy ) E 1 = (# yy !$# xx ) E 1 +$ = # xy ! = $ " (# + # ) zz xx yy E E = " yz = 0 " ! zz = $
1 $"
(!
xx
+ ! yy )
Plane Strain Constitutive
1 (# xx !$ (# xx + # yy )) 2G 1 (# yy !$ (# xx + # yy )) " yy = 2G 1 " xy = # xy 2G " zz = " xz = " yz = 0
" xx =
5. Airy Stress Functions Stress Function
" 4 # = 0 or
Stresses
$4 # $4 # $4 # + 2 + =0 $x 4 $x 2$y 2 $y 4
$ xx =
! 2# !y 2
$ yy =
! 2# !x 2
$ xy = "
! 2# !x!y
! December 9, 2009
Page 9 of 12
Mech 321 exam
2D Axisymmetric Problems
" rr =
2 2 a 2 pi # b 2 po ( pi # pe ) a b # b2 # a2 r 2 (b 2 # a 2 )
"$$ =
2 2 a 2 pi # b 2 po ( pi # pe ) a b + b2 # a2 r 2 (b 2 # a 2 )
ur =
2 2 1# % a 2 pi # b 2 po 1+ % ( pi # pe ) a b r + E b2 # a2 E r (b 2 # a 2 )
Compound cylinders
!
c
Interference pressure Ee c 2 ! b 2 b 2 ! a 2 p= 2b 3 c 2 ! a 2
(
(
)(
)
b
) a e
Rotating cylinders
Solid
" rr = "33
- ' r *20 3+# $% 2b 2 /1& ) , 2 8 . (b+ 1
- 1+ 3# ' r * 2 0 3+# 2 2 = $% b /1& ) ,2 8 . 3 + # (b+ 1
' r *30 1& # r 2 2 ur = $% b /( 3 + # ) & (1+ # )) , 2 (b+ 1 8E b .
!
Hollow -& a ) 2 & r )2 & a )20 3+# " rr = $% 2b 2 /( + + 1, ( + , ( + 2 'b* ' r * 1 8 .' b *
"33 ur
2 2 - 2 3+# 1+ 3# & r ) & a ) 0 2 2 & a) = $% b /( + + 1, ( + +( + 2 8 3 + # 'b* ' r * 1 .' b *
- & a ) 2 1+ # & a ) 2 1+ # & r ) 2 0 3 + # )(1, # ) ( 2 2 = $% b r 1+ + , 8E
/ .
( + 'b*
( + 1, # ' r *
( +2 3 + # 'b* 1
!
December 9, 2009
Page 10 of 12
Mech 321 exam
Failure theories
Maximum shear failure criteria 1 S # 1 ! # 3 = $ Y = Y " # 1 ! # 3 = SY 2 2
Von Mises failure criteria
1 (" 1 ! " 2 )2 + (" 2 ! " 3 )2 + (" 3 ! " 1 )2 = SY 2
[
]
Torsion Problems (selected results)
Displacements from observations u = "#yz v = #xz w = #$ (x, y) " (x, y) = 0 circular cross # sec tion ! !
!
!
Angle of twist per unit length " = T /GJ
Y
Stresses caused by torsion ! (circular cross-section, cartesian coordinates) " zx = #G$y " zy = +G$x (circular cross-section, cylindrical coordinates) " z# = G#r
?X i , Yi ?
?
Polar moment of inertia (circular cross-section, radius R) J = "R 4 /2
?XX, Y ?
Finite element (Bar element)
!
Stiffness matrix in global coordinate system # l2 lm "l 2 "lm & % ( m 2 "lm "m 2 ( EA % lm [k ] = % 2 L "l "lm l 2 lm ( % ( 2 lm m2 ' $"lm "m l = cos " =
!
m = sin " = L=
(X
X j ! Xi
j
i
L Yj ! Yi L
! Xi ) + (Yj ! Yi ) 2
j
j
Stress in bar
December 9, 2009
2
' ui $ !v ! E ! i! ) = [( l ( m l m]& # L !u j ! !%v j !"
Page 11 of 12
Mech 321 exam
j
FORMULA SHEET Trigonometry sin ! = a c cos! = b c tan ! = a b
c = a 2 + b2 sin 2 " + cos 2 = 1
c
a
θ
sin " 2 = 1 2 (1 ! cos" ) cos" 2 = 1 2 (1 + cos" ) sin 2" = 2 sin " cos"
2
b
2
cos 2" = cos " ! sin "
sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cos b ! sin a sin b a sin A = b sin B c 2 = a 2 + b 2 ! 2ab cos C
B
c A
c 2 = a 2 + b 2 + 2ab cos D
a C
D
b
Derivatives and integrals
dxn = nxn "1 dx
xn +1 ! x dx = n + 1 (u% d & # v du " u dv d (uv) dv du ' v $ = dx dx =u +v dx dx dx dx v2 d sin x d cos x = cos x = " sin x dx dx x sin 2ax 2 ! cos ax dx = 2 + 4a n
Units and constants
1 lb = 4.448 N 1 kip = 1000 lbs 1 ft = 0.3048 m 1 ft = 12 in 1 Pa = 1.45038×10-4 psi Gravitational acceleration g = 9.81 m/s2
December 9, 2009
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Mech 321 exam
Mechanics of Deformable Solids MECH 321 Sec. 1 Wednesday, December 9, 2009
Examiner:
Assoc Examiner:
Larry Lessard
Eliot Fried
Student Name:
McGill ID:
INSTRUCTIONS: (Examples) •
This is a CLOSED BOOK examination.
•
You are permitted TRANSLATION dictionaries ONLY.
•
STANDARD CALCULATOR permitted ONLY.
•
This examination consists of 7 questions including 2 short answer questions for a total of 12 pages, including the cover page. Make sure you have 12 pages.
Preview o o o o o o o o
(1) (10%) (2) (20%) (3) (20%) (4) (20%) (5) (20%) (6) (5%) (7) (5%) (8) (0%)
Elasticity Equations Go Habs Go (Boundary Conditions) Three-Dimensional Problem 2D problem, Cartesian Coordinates Energy Methods Rotating Disc Problem Finite Element Analysis Philosophy
•
Attempt all questions
•
There are 6 pages of equations at the end of the exam
•
This examination is PRINTED ON BOTH SIDES of the paper
•
This examination paper M UST BE RETURNED
December 9, 2009
Page 1 of 12
Mech 321 exam
Mechanics of Deformable Solids Mechanical Engineering MECH 321 Examiner: Professor Larry B. Lessard [email protected]
(1) (10%)
Wednesday, December 9, 2009 9:00 - 12:00 398-6305
Assoc Examiner: Professor Eliot Fried [email protected]
398-3739
Elasticity Equations For what body forces will the following stress field describe a state of equilibrium?
" xx = #2x 2 + 3y 2 # 5z " yy = #2y 2 " zz = 3x + y + 3z # 5
(2) (20%)
!
" xy = z + 4 xy # 7 " xz = #3x + y + 1 " yz = 0
Go Habs Go (Boundary Conditions) The figure below shows the loading conditions on a hockey puck due to a slap shot. We are examining the puck when the stick has made contact. The puck is considered an isotropic solid, with Young’s Modulus E, Poisson’s Ratio ν , density ρ , outer radius R and thickness H. The impact force imparted from the left side, is simplified by a constant radial force Q (MPa) over a 40 degree arc of the puck and covers the area from the top to the bottom of the puck in that region. The weight of the puck is countered by an idealized ice load, upward from the bottom, and assumed constant over the entire bottom surface of the puck.
2.1 Referring to the diagram above, set up a complete set of boundary conditions for this 3D problem at all boundaries. The problem has been set up such that there are no displacement boundary conditions. 2.2 Without solving for the stress state, would this problem have a relatively simple solution for stresses or not (state your reasons in one or two sentences).
December 9, 2009
Page 2 of 12
Mech 321 exam
(3) (20%)
Three-Dimensional Problem Consider the following problem, which is a circular rod subjected to two loads. There is a torsion load T and an axial load P. The rod has the following dimensions and load data. Length L = 1m Cross-section area A = 0.1m2 Load P = 5 kN Torque T = 0.8919 kNm J = πR4/2 (polar moment of inertia)
(a) For this 3D problem assume that the axial load P is evenly distributed at each end in tension over area A and that the torque T is causing pure torsion of a cylinder of circular cross section. Find an expression, in cylindrical coordinates for the 3D stress state. (b) The largest stresses will occur at r=R. Evaluate the stress state at r=R. (c) From the results of (b), calculate principal stresses at r=R (4) (20%)
2D problem, Cartesian Coordinates
A stress distribution is given by
" xx = pyx 3 # 2c1 xy + c 2 y " yy = pxy 3 # pyx 3 3 1 " xy = # px 2 y 2 + c1 y 2 + px 4 + c 3 2 2 Where the p’s and c’s are constants.
!
(a) Verify that this field represents a 2D solution for a thin plate of thickness t by verifying the biharmonic equation. (b) Find the resultant normal and shearing boundary forces (Px and Vx) along the edges y=0 and y=b of the plate. December 9, 2009
Page 3 of 12
Mech 321 exam
(5) (20%)
Energy Methods
A vertical load P is applied at B to two bars of equal length L but different cross-sectional areas and moduli of elasticity. (a) Determine the horizontal displacement δB of point B. Use one of the Castigliano energy methods. B. C, and D are all pinned joints Bar BC has modulus EBC, cross-sectional area ABC and cross-sectional moment-of-inertia IBC Bar BD has modulus EBD and cross-sectional area ABD and cross-sectional moment-of-inertia IBD
(b) From the solution that you obtain in part (a), show that it reduces to a predictable result when the two bars are identical (i.e., same modulus and cross-sectional properties)
(6) (5%)
Rotating Disc Problem A thin rotating disc with inner and outer radius undergoes stress due to the rotation.
Related to the solution of the rotating disc problem, the following three statements are false. Explain why each statement is erroneous (one sentence for each answer). (a) The loading comes from radially applied distributed loads that occur on the outer edge r = b, due to the rotation. (b) It is necessary to impose boundary conditions on this problem that insure there are no infinite stresses at r =0. (c) Boundary conditions will require us to impose that hoop stress
December 9, 2009
Page 4 of 12
"## will be zero at the inner boundary r = a. Mech 321 exam
!
(7) (5%)
Finite Element Analysis (a) In the following welded frame structure (no pinned joints), beam elements will be used in a finite element model.
The global FEA formulation for this problem will be in the form of Where the force vector
[K ]{u} = {F }
{F } = {FAx FAy M A FBx FBy M B etc}
What will be the input value to the global force vector for entry M F ? (if this is impossible to calculate based on the given information, state “impossible”) !
!
(c) Shown below are the element matrices for a bar element and a beam element
!
EA # 1 "1&) ui , ) f i , % (* - = * L $"1 1 '+ u j . + f j .
!
# 12 6L "12 6L &* v i . * Fi . , , , % 2 2 (, EI % 6L 4L "6L 2L (, ) i , , M i , + /=+ / L3 %"12 "6L 12 "6L(,v j , , F j , % 2 2( $ 6L 2L "6L 4L ',-) j ,0 ,- M j ,0
In the bar element, all entries of the stiffness matrix have the same units. This is not the case for the beam element. Explain why in two sentences (maximum).
!
8 (0%)
" Stress is nothing more than a socially acceptable form of mental illness.” ~Richard Carlson, American author
December 9, 2009
Page 5 of 12
Mech 321 exam
Mechanics of Deformable Solids - Useful Equations 1. Elasticity Equations in Cartesian Coordinates I. “Existence of Continuum”
! "F "A ! !B lim = !V" 0 !V : (Bx , By , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector
! B ! B
lim "A # 0
II. Stress Boundary Condition Equation ! Px = l! xx + m! xy + n! xz ! Py = l! xy + m! yy + n! yz ! Pz = l! xz + m! yz + n! zz III. Equation of Motion or Equilibrium Equation for a Deformable Body !" xx !" xy !" xz + + + Bx = 0 !x !y !z !" xy !" yy !" yz + + + By = 0 !x !y !z !" xz !" yz !" zz + + + Bz = 0 !x !y !z IV. Strain-Displacement Relationships (small displacements) "u "v "w ! xx = ! yy = ! zz = "x "y "z "v "u "w "u "w "v ! xy = 12 ( + ) ! xz = 12 ( + ) ! yz = 12 ( + ) "x " y "x " z "y "z V. Symmetry of Stresses and Strains ! xy = ! yx ! xz = ! zx ! yz = ! zy
! xy = ! yx ! xz = ! zx VI. Strain Compatibility Relations ! 2 " yy ! 2" xx ! 2" xy + = 2 !x 2 !y2 !x!y 2 2 ! " zz ! " xx ! 2" xz + = 2 !x 2 !z 2 !x !z 2 2 ! 2 " yz ! " zz ! " yy + =2 !y2 !z 2 !y !z
December 9, 2009
! yz = ! zy ! 2 " yz ! 2 " zx ! 2" zz ! 2 " xy + = + !x!y !z 2 !z!x !y !z ! 2 " yy ! 2 " xz ! 2 " xy ! 2 " yz + = + !x !z !y 2 !y!z !x!y 2 2 ! 2 " xx ! " yz ! 2 " xz ! " xy + = + !y !z !x 2 !x !y !x!z
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Mech 321 exam
VII. Constitutive Relations (Isotropic Materials) (i) Strain-Stress (Compliance Form) ! xx = 1E (" xx # $" yy # $" zz ) ! yy = 1E (" yy # $" xx # $" zz ) ! zz = 1E (" zz # $" xx # $" yy ) 1 ! xy = 2G " xy = 1+E$ " xy ! xz = 2G1 " xz = 1+E$ " xz ! yz = 2G1 " yz = 1+E$ " yz (ii) Stress-Strain (Modulus Form) E ! xx = "e + 2G# xx = (1+$)(1% 2$ ) [(1 % $ )# xx + $ (# yy + # zz )] E ! yy = "e + 2G# yy = (1+$ )(1%2 $ ) [(1 % $)# yy + $ (# xx + # zz )] E ! zz = "e + 2G# zz = (1+$ )(1% 2$ ) [(1 % $ )# zz + $ (# xx + # yy )] ! xy = 2G# xy = (1+E$) # xy ! xz = 2G# xz = (1+E$ ) # xz ! yz = 2G# yz = (1+E$ ) # yz where e = (! xx + ! yy + ! zz ) G(3" + 2G) "+G " # = 2(" + G) #E " = (1 + # )(1 $ 2# ) E G= 2(1 + # ) E=
2. Elasticity Equations in Cylindrical Coordinates I. “Existence of Continuum”
! "F "A ! !B lim = !V" 0 !V : (Br , B! , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector
! B ! B
lim "A # 0
II. Stress Boundary Condition Equation
! Pr = nr! rr + n"! r" + nz! rz ! P" = nr! r" + n"! "" + nz! "z ! Pz = nr! rz + n"! "z + nz! zz
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Mech 321 exam
III. Equation of Motion or Equilibrium Equation for a Deformable Body !" rr 1 !" r# !" zr " rr $ " ## + + + + Br = 0 !r r !# !z r !" r# 1 !" ## !" #z 2" r# + + + + B# = 0 !r r !# !z r !" rz 1 !" #z !" zz " rz + + + + Bz = 0 !r r !# !z r IV. Strain-Displacement Relationships (small displacements) "u 1 "u# ur "u ! rr = r ! ## = + ! zz = z "r r "# r "z 1 "ur "u# u# "u "u "u 1 "u z 1 1 ! r# = 2 ( + $ ) ! rz = 2 ( r + z ) !#z = 12 ( # + ) r "# "r r "z "r "z r "# V. Symmetry of Stresses and Strains ! r" = ! "r ! rz = ! zr ! "z = ! z" ! r" = !"r ! rz = ! zr ! "z = ! z" VI. Strain Compatibility Relations (Not given here) VII. Constitutive Relations (Isotropic Materials) (i) Strain-Stress (Compliance Form)
! rr !%% ! zz ! r% ! rz !%z
= = = = = =
(" rr # $" %% # $" zz ) (" %% # $" rr # $" zz ) (" zz # $" rr # $" %% ) " r% = 1+E$ " r% " rz = 1+E$ " rz " %z = 1+E$ " %z
1 E 1 E 1 E 1 2G 1 2G 1 2G
(ii) Stress-Strain (Modulus Form)
! rr ! && ! zz ! r& ! rz ! &z
E = "e + 2G# rr = (1+$ )(1% 2$ ) [(1 % $ )# rr + $ (#&& + # zz )] E = " e + 2G# && = (1+ $)(1% 2$ ) [(1 % $ )# && + $ (# rr + # zz )] E = "e + 2G# zz = (1+$ )(1% 2$ ) [(1 % $ )# zz + $ (# rr + #&& )] = 2G# r& = (1+E$ ) # r& = 2G# rz = (1+E$) # rz = 2G#&z = (1+E$ ) # &z
where
e = (! rr + ! "" + ! zz )
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Mech 321 exam
3. Energy Strain energy "
U = ! U o dV U o = ! # d" V
0
Complementary strain energy "
C = ! C o dV C o = ! # d" V
0
Strain energy Axial L N2 UN = ! dx 2 EA 0
Bending
Shear
L
Torsion L
2
kV dx 2GA 0
M dx 2 EI 0
US = !
UM = !
L
2
T2 dx 2GJ 0
UT = !
Castigliano’s theorem I and II
Fi =
!U !qi
qi =
!C !Fi
where Fi and qi are generalized force and displacement
4. Plane Stress and Plane Strain Plane Stress Constitutive
" xx " yy " xy " xz
1 = (# xx !$# yy ) E 1 = (# yy !$# xx ) E 1 +$ = # xy ! = $ " (# + # ) zz xx yy E E = " yz = 0 " ! zz = $
1 $"
(!
xx
+ ! yy )
Plane Strain Constitutive
1 (# xx !$ (# xx + # yy )) 2G 1 (# yy !$ (# xx + # yy )) " yy = 2G 1 " xy = # xy 2G " zz = " xz = " yz = 0
" xx =
5. Airy Stress Functions Stress Function
" 4 # = 0 or
Stresses
$4 # $4 # $4 # + 2 + =0 $x 4 $x 2$y 2 $y 4
$ xx =
! 2# !y 2
$ yy =
! 2# !x 2
$ xy = "
! 2# !x!y
! December 9, 2009
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Mech 321 exam
2D Axisymmetric Problems
" rr =
2 2 a 2 pi # b 2 po ( pi # pe ) a b # b2 # a2 r 2 (b 2 # a 2 )
"$$ =
2 2 a 2 pi # b 2 po ( pi # pe ) a b + b2 # a2 r 2 (b 2 # a 2 )
ur =
2 2 1# % a 2 pi # b 2 po 1+ % ( pi # pe ) a b r + E b2 # a2 E r (b 2 # a 2 )
Compound cylinders
!
c
Interference pressure Ee c 2 ! b 2 b 2 ! a 2 p= 2b 3 c 2 ! a 2
(
(
)(
)
b
) a e
Rotating cylinders
Solid
" rr = "33
- ' r *20 3+# $% 2b 2 /1& ) , 2 8 . (b+ 1
- 1+ 3# ' r * 2 0 3+# 2 2 = $% b /1& ) ,2 8 . 3 + # (b+ 1
' r *30 1& # r 2 2 ur = $% b /( 3 + # ) & (1+ # )) , 2 (b+ 1 8E b .
!
Hollow -& a ) 2 & r )2 & a )20 3+# " rr = $% 2b 2 /( + + 1, ( + , ( + 2 'b* ' r * 1 8 .' b *
"33 ur
2 2 - 2 3+# 1+ 3# & r ) & a ) 0 2 2 & a) = $% b /( + + 1, ( + +( + 2 8 3 + # 'b* ' r * 1 .' b *
- & a ) 2 1+ # & a ) 2 1+ # & r ) 2 0 3 + # )(1, # ) ( 2 2 = $% b r 1+ + , 8E
/ .
( + 'b*
( + 1, # ' r *
( +2 3 + # 'b* 1
!
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Mech 321 exam
Failure theories
Maximum shear failure criteria 1 S # 1 ! # 3 = $ Y = Y " # 1 ! # 3 = SY 2 2
Von Mises failure criteria
1 (" 1 ! " 2 )2 + (" 2 ! " 3 )2 + (" 3 ! " 1 )2 = SY 2
[
]
Torsion Problems (selected results)
Displacements from observations u = "#yz v = #xz w = #$ (x, y) " (x, y) = 0 circular cross # sec tion ! !
!
!
Angle of twist per unit length " = T /GJ
Y
Stresses caused by torsion ! (circular cross-section, cartesian coordinates) " zx = #G$y " zy = +G$x (circular cross-section, cylindrical coordinates) " z# = G#r
?X i , Yi ?
?
Polar moment of inertia (circular cross-section, radius R) J = "R 4 /2
?XX, Y ?
Finite element (Bar element)
!
Stiffness matrix in global coordinate system # l2 lm "l 2 "lm & % ( m 2 "lm "m 2 ( EA % lm [k ] = % 2 L "l "lm l 2 lm ( % ( 2 lm m2 ' $"lm "m l = cos " =
!
m = sin " = L=
(X
X j ! Xi
j
i
L Yj ! Yi L
! Xi ) + (Yj ! Yi ) 2
j
j
Stress in bar
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2
' ui $ !v ! E ! i! ) = [( l ( m l m]& # L !u j ! !%v j !"
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Mech 321 exam
j
FORMULA SHEET Trigonometry sin ! = a c cos! = b c tan ! = a b
c = a 2 + b2 sin 2 " + cos 2 = 1
c
a
θ
sin " 2 = 1 2 (1 ! cos" ) cos" 2 = 1 2 (1 + cos" ) sin 2" = 2 sin " cos"
2
b
2
cos 2" = cos " ! sin "
sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cos b ! sin a sin b a sin A = b sin B c 2 = a 2 + b 2 ! 2ab cos C
B
c A
c 2 = a 2 + b 2 + 2ab cos D
a C
D
b
Derivatives and integrals
dxn = nxn "1 dx
xn +1 ! x dx = n + 1 (u% d & # v du " u dv d (uv) dv du ' v $ = dx dx =u +v dx dx dx dx v2 d sin x d cos x = cos x = " sin x dx dx x sin 2ax 2 ! cos ax dx = 2 + 4a n
Units and constants
1 lb = 4.448 N 1 kip = 1000 lbs 1 ft = 0.3048 m 1 ft = 12 in 1 Pa = 1.45038×10-4 psi Gravitational acceleration g = 9.81 m/s2
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