MCE 528 APPLIED ELASTICITY

 
 HOMEWORK II — ? — DUE DATE : Next class 

QUESTIONS

Q-1 (TB-Problem 2.2) A two-dimensional displacement field is given by u = k (x2 + y 2 ),

v = k (2x − y), w = 0, where k is a constant. Determine and plot the deformed shape of a
differential rectangular element originally located with its left bottom corner at the origin
as shown. Finally, calculate the rotation component ωz = ωxy .

Solution
x2 + y 2
   
u (x, y, z)
u =  v (x, y, z)  = k  2x − y 
   

w (x, y, z) 0

1
eij = (ui,j + uj,i )
2      
∂υ 1 ∂v ∂u 1 ∂u ∂w
+ +
1    ∂x  2 ∂x ∂y 2  ∂z ∂x  
e = ∇u + (∇u)T =  1 ∂v
+ ∂u ∂v 1 ∂v
+ ∂w
 
2  ∂x ∂y   ∂y 2 ∂z ∂y

2 
1 ∂u ∂w 1 ∂v ∂w

∂w

2 ∂z
+ ∂x 2 ∂z
+ ∂y ∂z

∂υ ∂v ∂v ∂u
ex = = 2kx ey = = −k γxy = 2exy = + = 2k + 2ky = 2k (1 + y)
∂x ∂y ∂x ∂y

!
1 ∂v ∂u 1
ωxy = ωz = − = (2k − 2ky) = k (1 − y)
2 ∂x ∂y 2

Q-2 (TB-Problem 2.3) A two-dimensional problem of a rectangular bar stretched by

uniform end loadings results in the following constant strain field:
 
C1 0 0
eij =  0 −C2 0 
 

0 0 0

1
 where C1 and C2 are constants. Assuming the field depends only on x and y, integrate
the strain-displacement relations to determine the displacement components and identify
any rigid-body motion terms.

Solution: The problem is 2-D problem

∂υ
ex = = C1 u (x, y) = C1 x + f (y)
∂x

∂υ
ey = = −C2 v (x, y) = −C2 y + g (x)
∂y

∂w
ez = =0 w (x, y) = h (x, y)
∂z

!
1 ∂υ ∂u dg (x) df (y) dg df
exy = + + =0 = − = constant = a
2 ∂x ∂y dx dy dx dy

f (y) = ay + C3 g (x) = −ax + C4

!
1 ∂υ ∂w ∂w ∂h
eyz = + =0 = =0 h (x, y) = p (x)
2 ∂z ∂y ∂y ∂y

!
1 ∂u ∂w ∂w ∂h
ezx = + =0 = =0 h (x, y) = q (y)
2 ∂z ∂x ∂x ∂x

h = C5 = constant

Finally writing the all deformation result

u = C1 x + f (y) = C1 x + ay + C3 =⇒ u = C1 x − ωz y + u0

v = C2 y + g (x) = C2 y + ax + C4 =⇒ v = C2 y + ωz x + v0

w = w0

Q-3 (TB-Problem 2.4) A three-dimensional elasticity problem of a uniform bar stretched

2
under its own weight gives the following strain field:
 
Az 0 0
eij =  0 Az 0 
 

0 0 Bz

where A and B are constants. Integrate the strain-displacement relations to determine the
displacement components and identify all rigid-body motion terms.

3
Solution:

∂u
ex = = Az =⇒ u = Azx + f (y, z)
∂x

∂v
ey = = Az =⇒ v = Azy + g (x, z)
∂y

∂w 1
ez = = Bz =⇒ w = Bz 2 + h (x, y)
∂z 2

!
1 ∂υ ∂u dg (x) df (y) dg df
exy = + =0 + =0 = − = p (z)
2 ∂x ∂y dx dy dx dy

!
1 ∂υ ∂w ∂g ∂h ∂g ∂h
eyz = + =0 Ay + =− = −Ay − = q (x)
2 ∂z ∂y ∂z ∂y ∂z ∂y

!
1 ∂u ∂w ∂h ∂f ∂h ∂f
ezx = + =0 = −Ax + = −Ax − = r (y)
2 ∂z ∂x ∂x ∂z ∂x ∂z

From previous relations

p0 (z) = r0 (y) p0 (z) = −q 0 (x) q 0 (x) = −r0 (y)

p0 (z) = q 0 (x) = r0 (y) = 0 p (z) = a q (x) = b r (y) = c

df
= p (z) = a =⇒ f = ay + f1 (x, y)
dy

df
= −Ax − r (y) = −Ax − c =⇒ f = −Axz − cz + f2 (x, y)
dz
To satisfy each form
1
f = −Axz + ay − cz + d1 g = Azx − ax + bz + d2 h = − Ay 2 − by + cx + d3
2
Final form of the displacement functions becomes

u = Azx − ωz y + ωy x + u0 v = Azy − ωx z + ωz x + v0

1 2 4
w= Bz − Ay 2 − ωy x + ωx y + w0
2