Stress and Strain - Structural and Stress Analysis - page 161

τ

τ

τ

A B

A B A B

45 ° 45 °

τ

τ

F IGURE 7.11
Determination of
the relationships
between the elastic
constants

τ

s AC

τ

τ AC

s BD τ BD

D C C

D

τ

(a) (b) (c)

In Fig. 7.11(a), ABCD is a square element of material of unit thickness and is in
equilibrium under a shear and complementary shear stress system τ . Imagine now
that the element is 'cut' along the diagonal AC as shown in Fig. 7.11(b). In order to
maintain the equilibrium of the triangular portion ABC it is possible that a direct
force and a shear force are required on the face AC. These forces, if they exist, will be
distributed over the face of the element in the form of direct and shear stress systems,
respectively. Since the element is small, these stresses may be assumed to be constant
along the face AC. Let the direct stress on AC in the direction BD be σ BD and the
shear stress on AC be τ AC . Then resolving forces on the element in the direction BD
we have

cos 45 ◦ − τ BC cos 45 ◦ = σ BD AC × 1 − τ AB × 1 × × 1 × 0

Dividing through by AC

τ AB τ BC

AC cos 45 ◦ + AC cos 45 ◦

σ BD =

or

τ cos 2 45 ◦ + τ cos 2 45 ◦ σ BD =

from which

σ BD = τ (7.16)

The positive sign indicates that σ BD is a tensile stress. Similarly, resolving forces in the
direction AC

cos 45 ◦ − cos 45 ◦ = τ AC AC × 1 + τ AB × 1 × τ BC × 1 × 0

Again dividing through by AC we obtain

τ AC =− τ cos 2 45 ◦ + τ cos 2 45 ◦ = 0

A similar analysis of the triangular element ABD in Fig. 7.11(c) shows that

σ AC =− τ (7.17)

and

τ BD = 0