Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 66

For a tensile test in the x direction σ = σ = 0 and (1) gives

y z

(λ + 2 µ) + λ =− λ

x ,

λ + (λ + 2 µ) =− λ

y z y z x

or

λ

y = z =−

(λ + µ) x 2

Since y = z =− ν x , ν = (λ + µ)λ/ 2 , or λ = 2 νµ/( 1 − 2 ν)

From (1) and the fact that σ x = E x

2 νµ E

E =

− ν (

1 − 2 ν) + 2 µ or G = µ =

1 2 ( 1 + ν)

1.34 As indicated in Eq. (1.37), the two constants G and K can be utilized as the two
independent constants for an isotropic material. Find K as a function of E and

ν.

2
3 G

E

= λ + =

Answer: K

3 ( 1 − 2 ν)

1.35 Show that a rotation

∂v
∂

1
2 x − ∂

u

ω z =

∂ y

would move lines OA and OB of Fig. 1.8 such that the angles they make with the x
and y directions are equal. These angles are of magnitude

∂v
∂

1
2 x + ∂

u

=

xy

∂ y

Furthermore, show that the total derivative du can be expressed as

du =

x dx

+

xy dy

− ω

z dy

For given strain and rotation components, this relationship can be integrated to give
the displacement u .

Hint: For the final request, begin with

= ∂

u

+ ∂

u

du dx

y dy

∂ ∂

x

2 2 2 u i =

1.36 Prove that in the absence of body forces ∇ kk = 0 and further that ∇ ∇ 0

[Eq. (1.67)].

2 2

Hint: From Eq. (1.69) show that ∇ σ ii = 0 . This leads to ∇ kk = 0 .

1.37 Given the expression u 1 ,i u 1 1

(a) Write this in a more compact form using index notation and the summation
convention.
(b) Express this in matrix notation, i.e., with u T

+ u 2 ,i u 2 + u 1 , 1 u 1 ,i 1 + u 1 , 2 u 2 ,i 1 + u 2 , 1 u 1 ,i 2 + u 2 , 2 u 2 ,i 2 ,i =

[ u 1 u 2 ] , find D of u T Du

= .

Answer:

(a) u j,i u j + u j,k u k,i j i = 1 ,j,k = 1or2

i

∂ + ∂∂

i 12

∂∂

1 i 1

(b) D =

∂∂ ∂ + ∂∂

i 2 1 i 2 i 2