Page 220 - Intro to Tensor Calculus

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Figure 2.3-3. One dimensional rod subjected to tension force

The strain associated with the distance ` =∆x = PQ is

∆` P Q − PQ (0Q − 0P ) − (0Q − 0P)
0
0
0
0
e = = =
` PQ PQ
[x +∆x + u(x +∆x, t) − (x + u(x, t))] − [(x +∆x) − x]
e =
∆x
u(x +∆x, t) − u(x, t)
e = .
∆x
Use the Hooke’s law part(i) and write
F u(x +∆x, t) − u(x, t)
= E .
A ∆x
Taking the limit as ∆x approaches zero we find that

F ∂u(x, t)
= E .
A ∂x
Hence, the stress is proportional to the spatial derivative of the displacement function.

Normal and Shearing Stresses

Let us consider a more general situation in which we have some material which can be described as
having a surface area S which encloses a volume V. Assume that the density of the material is % and the
~
material is homogeneous and isotropic. Further assume that the material is subjected to the forces b and ~ t (n)
~
where b is a body force per unit mass [force/mass], and ~ t (n) is a surface traction per unit area [force/area].
The superscript (n) on the vector is to remind you that we will only be interested in the normal component
of the surface forces. We will neglect body couples, surface couples, and concentrated forces or couples that
act at a single point. If the forces described above are everywhere continuous we can calculate the resultant
~
~
force F and resultant moment M acting on the material by constructing various surface and volume integrals
~
which sum the forces acting upon the material. In particular, the resultant force F acting on our material
can be described by the surface and volume integrals:
ZZ ZZZ
~
~
F = ~ t (n) dS + %bdτ (2.3.3)
S V