MECH 321 Midterm October 2009
Mechanics of Deformable Solids Mechanical Engineering MECH-321 Prof. Larry Lessard
Wednesday October 21, 2009 3:30 - 4:25 p.m.
Midterm Exam Preview
(1) (10%) Elasticity Equations (2) (40%) Stresses and Boundary Conditions (3) (30%) Boundary Conditions (4) (20%) Short Answer Questions – Energy Methods
(1) (10%) Elasticity Equations The following is the solution for the strains everywhere within a body. Does this solution represent a compatible displacement field?
! 2 2 # 9x y [! ] = # 3xyr 2 # #" 0
3xyr 2 9x 2 y 2 0
$ 0 & 0 & where r2 = x2 + y2 & 0 & %
(assume of course, that constants “3” and “9” are such that strain [ε] has correct units)
(2) (40%)
Stresses and Boundary Conditions. The figure shown in (a) is relatively easy problem involving extension in the x-direction and Poisson contraction in the y- and z-direction due to a uniaxial loading in the x-direction. We would like to know if we can re-create the same strains (and deformations) by loading the same block of material with compressive stresses in the y- and z-directions.
! yy
! zz
y z
x
!xx
! xx
y z
x
! zz (a)
! yy (b)
2.1 Referring to the solution for (a), apply in (b) a load σyy = σzz such that the strains in the y- and zdirection will be identical for both solutions. What are the applied σyy and σzz for (b)? 2.2 For “normal” materials, i.e., Poisson’s Ratio ν=0.3, will the resulting strain in the x-direction be the same for the two solutions? (Show mathematically)
MECH 321 midterm 2009
page 1
(3) (30%)
Boundary Conditions A hockey stick is being modeled to determine the stresses during loading. The blade is isolated and assumptions are made that the heel of the blade is fixed and that the puck causes a total force P, evenly distributed load over one quarter of the front area of the blade as shown in the figure (P is perpendicular to the surface). Set up the stress boundary conditions for this 3-D problem at all surfaces except for the fixed end (The BC at this end is clamped displacements u=v=w=0). Set up BC’s only; do not attempt to solve for the stresses!! Heel of the Blade
Fixed end at x=-L/2 L
L/2
H
H/2
P Area of load application
(4) (20%)
y
x z (origin is at the center of the block)
Short Answer Questions
Short answer Energy Method questions: Answer questions only (DO NOT SOLVE) 4.1 A circular cross-section shaft AB, with diameter 80mm and length 1.0 m, is made of an aluminum allow (E=72 GPa, G=27 GPa). It is attached at point A to a torsional spring with stiffness β= 200 kN-m per rad. A torque T0= 4 kN-m is applied at the free end B.
(a) If we want to determine the rotation of the shaft at B, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s). MECH 321 midterm 2009
t
page 2
4.2 A bar having a circular cross-section is fixed at the origin O and has right angle bends at points A and B. Length OA lies along the z-axis; length AB is parallel to the x-axis; length BC is parallel to the y-axis. MC is a couple lying parallel to the x-axis. G= E/[2(1+ν)]
(a) If we want to determine x, y and z components of the deflection at point C, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
4.3 The structure shown is made up of a cantilever beam AB (E1 , I1 , A1) and two identical rods BC and CD (E2 , A2). Let A1 be large compared to A2 and L1 be large compared to beam depth.
(a) If we want to determine the component of the deflection of point C in the direction of load P, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
MECH 321 midterm 2009
page 3
4.4 The circular curved beam AB has a radius of curvature R, modulus E, and circular cross-section of diameter d. There is a downward vertical load Q.
(a) If we want to determine the horizontal and vertical components of the deflection of point B, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
(5) (0%) "Pain is inevitable. Suffering is optional." — Haruki Murakami, Japanese author
MECH 321 midterm 2009
page 4
Mechanics of Deformable Solids - Useful Equations 1. Elasticity Equations in Cartesian Coordinates I. “Existence of Continuum”
! " F lim "A # 0 "A ! ! !B lim B = !V" 0 !V ! B : (Bx , By , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector 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
(2.1)
(2.10)
III. Equation of Motion or Equilibrium Equation for a Deformable Body !" xx !" xy !" xz + + + Bx = 0 !x !y !z !" xy !" yy !" yz + + + By = 0 (2.45) !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
(2.81)
V. Symmetry of Stresses and Strains
! xy = ! yx ! xz = ! zx ! yz = ! zy
(2.4)
! xy = ! yx ! xz = ! zx
! yz = ! zy
(2.62)
! 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 2 ! " ! " xy ! " xx ! " xz yz + = + 2 !y !z !x !x !y !x!z
(2.83)
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 2 2 !y !z !y !z MECH 321 midterm 2009
page 5
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 1 ! xz = 2G " xz = 1+E$ " xz ! yz = 2G1 " yz = 1+E$ " yz
(3.30)
(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
(3.28, 3.32)
where e = (! xx + ! yy + ! zz ) G(3" + 2G) "+G " # = 2(" + G) #E " = (1 + # )(1 $ 2# ) E G= 2(1 + # ) E=
(3.31)
2. Elasticity Equations in Cylindrical Coordinates I. “Existence of Continuum” Existence of Stress Vector
Existence of Body Force Vector
! " F lim "A # 0 "A ! ! ! B lim B = !V" 0 !V ! B : (Br , B! , Bz )
! ! =
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 MECH 321 midterm 2009
page 6
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
(2.50)
IV. Strain-Displacement Relationships (small displacements) "u 1 "u# ur "u ! rr = r ! ## = + ! zz = z "r r "# r "z 1 " u " u u "u r "u z "u# 1 "uz (2.85) r # # 1 1 1 ! r# = 2 ( + $ ) ! rz = 2 ( + ) !#z = 2 ( + ) 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 + #&& )] E = 2G# r& = (1+$ ) # r& = 2G# rz = (1+E$) # rz = 2G#&z = (1+E$ ) # &z
where
e = (! rr + ! "" + ! zz )
MECH 321 midterm 2009
page 7
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 L M2 UM = ! dx 2 EI 0 Shear L kV 2 US = ! dx 2 GA 0 Torsion L T2 UT = ! dx 2 GJ 0 Castigliano’s theorem I and II
Fi =
!U !qi
qi =
!C !Fi
where Fi and qi are generalized force and displacement
MECH 321 midterm 2009
page 8
FORMULA SHEET Trigonometry sin θ = a c cosθ = b c tan θ = a b
c
a
c = a2 + b2 sin 2 θ + cos2 = 1
θ b
sin θ 2 = 1 2 (1 − cosθ ) cosθ 2 = 1 2 (1 + cosθ ) sin 2θ = 2 sin θ cosθ cos 2θ = cos2 θ − sin 2 θ sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cosb sin a sin b a sin A = b sin B c 2 = a 2 + b 2 − 2ab cos C c 2 = a 2 + b 2 + 2ab cos D
B c
a
A
C
D
b
Derivatives and integrals dxn xn +1 n "1 n = nx ! x dx = n + 1 dx (u% d & # v du " u dv d (uv) dv du ' v $ = dx dx =u +v 2 dx dx dx dx v d sin x d cos x = cos x = " sin x dx dx x sin 2ax 2 ! cos ax dx = 2 + 4a 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
MECH 321 midterm 2009
page 9
Wednesday October 21, 2009 3:30 - 4:25 p.m.
Midterm Exam Preview
(1) (10%) Elasticity Equations (2) (40%) Stresses and Boundary Conditions (3) (30%) Boundary Conditions (4) (20%) Short Answer Questions – Energy Methods
(1) (10%) Elasticity Equations The following is the solution for the strains everywhere within a body. Does this solution represent a compatible displacement field?
! 2 2 # 9x y [! ] = # 3xyr 2 # #" 0
3xyr 2 9x 2 y 2 0
$ 0 & 0 & where r2 = x2 + y2 & 0 & %
(assume of course, that constants “3” and “9” are such that strain [ε] has correct units)
(2) (40%)
Stresses and Boundary Conditions. The figure shown in (a) is relatively easy problem involving extension in the x-direction and Poisson contraction in the y- and z-direction due to a uniaxial loading in the x-direction. We would like to know if we can re-create the same strains (and deformations) by loading the same block of material with compressive stresses in the y- and z-directions.
! yy
! zz
y z
x
!xx
! xx
y z
x
! zz (a)
! yy (b)
2.1 Referring to the solution for (a), apply in (b) a load σyy = σzz such that the strains in the y- and zdirection will be identical for both solutions. What are the applied σyy and σzz for (b)? 2.2 For “normal” materials, i.e., Poisson’s Ratio ν=0.3, will the resulting strain in the x-direction be the same for the two solutions? (Show mathematically)
MECH 321 midterm 2009
page 1
(3) (30%)
Boundary Conditions A hockey stick is being modeled to determine the stresses during loading. The blade is isolated and assumptions are made that the heel of the blade is fixed and that the puck causes a total force P, evenly distributed load over one quarter of the front area of the blade as shown in the figure (P is perpendicular to the surface). Set up the stress boundary conditions for this 3-D problem at all surfaces except for the fixed end (The BC at this end is clamped displacements u=v=w=0). Set up BC’s only; do not attempt to solve for the stresses!! Heel of the Blade
Fixed end at x=-L/2 L
L/2
H
H/2
P Area of load application
(4) (20%)
y
x z (origin is at the center of the block)
Short Answer Questions
Short answer Energy Method questions: Answer questions only (DO NOT SOLVE) 4.1 A circular cross-section shaft AB, with diameter 80mm and length 1.0 m, is made of an aluminum allow (E=72 GPa, G=27 GPa). It is attached at point A to a torsional spring with stiffness β= 200 kN-m per rad. A torque T0= 4 kN-m is applied at the free end B.
(a) If we want to determine the rotation of the shaft at B, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s). MECH 321 midterm 2009
t
page 2
4.2 A bar having a circular cross-section is fixed at the origin O and has right angle bends at points A and B. Length OA lies along the z-axis; length AB is parallel to the x-axis; length BC is parallel to the y-axis. MC is a couple lying parallel to the x-axis. G= E/[2(1+ν)]
(a) If we want to determine x, y and z components of the deflection at point C, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
4.3 The structure shown is made up of a cantilever beam AB (E1 , I1 , A1) and two identical rods BC and CD (E2 , A2). Let A1 be large compared to A2 and L1 be large compared to beam depth.
(a) If we want to determine the component of the deflection of point C in the direction of load P, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
MECH 321 midterm 2009
page 3
4.4 The circular curved beam AB has a radius of curvature R, modulus E, and circular cross-section of diameter d. There is a downward vertical load Q.
(a) If we want to determine the horizontal and vertical components of the deflection of point B, which energy terms should we consider? (b) Do we need to make use of dummy loads? (c) If dummy load(s) are required in the solution, describe their location(s) and direction(s).
(5) (0%) "Pain is inevitable. Suffering is optional." — Haruki Murakami, Japanese author
MECH 321 midterm 2009
page 4
Mechanics of Deformable Solids - Useful Equations 1. Elasticity Equations in Cartesian Coordinates I. “Existence of Continuum”
! " F lim "A # 0 "A ! ! !B lim B = !V" 0 !V ! B : (Bx , By , Bz )
! ! =
Existence of Stress Vector
Existence of Body Force Vector 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
(2.1)
(2.10)
III. Equation of Motion or Equilibrium Equation for a Deformable Body !" xx !" xy !" xz + + + Bx = 0 !x !y !z !" xy !" yy !" yz + + + By = 0 (2.45) !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
(2.81)
V. Symmetry of Stresses and Strains
! xy = ! yx ! xz = ! zx ! yz = ! zy
(2.4)
! xy = ! yx ! xz = ! zx
! yz = ! zy
(2.62)
! 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 2 ! " ! " xy ! " xx ! " xz yz + = + 2 !y !z !x !x !y !x!z
(2.83)
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 2 2 !y !z !y !z MECH 321 midterm 2009
page 5
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 1 ! xz = 2G " xz = 1+E$ " xz ! yz = 2G1 " yz = 1+E$ " yz
(3.30)
(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
(3.28, 3.32)
where e = (! xx + ! yy + ! zz ) G(3" + 2G) "+G " # = 2(" + G) #E " = (1 + # )(1 $ 2# ) E G= 2(1 + # ) E=
(3.31)
2. Elasticity Equations in Cylindrical Coordinates I. “Existence of Continuum” Existence of Stress Vector
Existence of Body Force Vector
! " F lim "A # 0 "A ! ! ! B lim B = !V" 0 !V ! B : (Br , B! , Bz )
! ! =
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 MECH 321 midterm 2009
page 6
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
(2.50)
IV. Strain-Displacement Relationships (small displacements) "u 1 "u# ur "u ! rr = r ! ## = + ! zz = z "r r "# r "z 1 " u " u u "u r "u z "u# 1 "uz (2.85) r # # 1 1 1 ! r# = 2 ( + $ ) ! rz = 2 ( + ) !#z = 2 ( + ) 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 + #&& )] E = 2G# r& = (1+$ ) # r& = 2G# rz = (1+E$) # rz = 2G#&z = (1+E$ ) # &z
where
e = (! rr + ! "" + ! zz )
MECH 321 midterm 2009
page 7
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 L M2 UM = ! dx 2 EI 0 Shear L kV 2 US = ! dx 2 GA 0 Torsion L T2 UT = ! dx 2 GJ 0 Castigliano’s theorem I and II
Fi =
!U !qi
qi =
!C !Fi
where Fi and qi are generalized force and displacement
MECH 321 midterm 2009
page 8
FORMULA SHEET Trigonometry sin θ = a c cosθ = b c tan θ = a b
c
a
c = a2 + b2 sin 2 θ + cos2 = 1
θ b
sin θ 2 = 1 2 (1 − cosθ ) cosθ 2 = 1 2 (1 + cosθ ) sin 2θ = 2 sin θ cosθ cos 2θ = cos2 θ − sin 2 θ sin (a ± b ) = sin a cos b ± cos a sin b cos(a ± b ) = cos a cosb sin a sin b a sin A = b sin B c 2 = a 2 + b 2 − 2ab cos C c 2 = a 2 + b 2 + 2ab cos D
B c
a
A
C
D
b
Derivatives and integrals dxn xn +1 n "1 n = nx ! x dx = n + 1 dx (u% d & # v du " u dv d (uv) dv du ' v $ = dx dx =u +v 2 dx dx dx dx v d sin x d cos x = cos x = " sin x dx dx x sin 2ax 2 ! cos ax dx = 2 + 4a 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
MECH 321 midterm 2009
page 9
