DERIVATIONS FOR
COMPLEMENTARY
STRESS
Fig. 4.10
„
Figure 4.10 shows a general three-dimensional
stress element, showing three normal stresses σx,
σy and σz, all positive and six shear stresses τxy,
τyx, τyz, τzy, τzx and τxz
„
„
Consider the element of unit thickness subjected
to shearing stresses along its edges, τxy and τyx
Shear forces along the sides AB and CD are (τxy
x 1 x dy) and along AD and BC are (τyx x 1 x dx)
To maintain rotational equilibrium of the
element, the above forces must balance out
„ Taking moments about z-axis through the centre
of the element :
2 (τxy x 1 x dy) dx/2 - 2 (τyx x 1 x dx) dy/2 = 0
τxy = τyx
„
„
These are termed complementary shear stresses
„
Thus a shear stress on one plane is always
accompanied by a complementary shear stress of
the same sign and magnitude on a perpendicular
plane
„
Similarly,
τyz = τzy, τzx = τxz
„
Outwardly directed normal stresses are called
tension or tensile stresses and are considered
positive
„
Shear stresses on a positive face of an element
are positive if they act in the positive direction
of a reference axis
„
„
„
„
First suffix of a shear stress component is the
coordinate normal to the element face
The shear stress component is parallel to the
axis of the second subscript
Shear stresses will be taken as positive when
they are in clockwise direction (cw)
In fig., τyx is cw (clockwise) and positive ; τxy is
ccw (counterclockwise) and negative
MOHR’S CIRCLE
„
„
„
Suppose the element of the last fig., is cut by an
oblique plane at an angle φ to the y-axis
For equilibrium of ABC, the forces on AB, BC
and CA must be in balance
As the element is of constant unit thickness, the
areas of the faces are proportional to the lengths
of the sides of the triangle
„
Resolving forces normal to the plane AB,
σAB-σxBCcosφ-σyACsinφ-τxyBCsinφ-τyxACcosφ = 0
Dividing by AB,
σ-σxcos2φ-σysin2φ-τxycosφsinφ-τxysinφcosφ = 0
σ-σxcos2φ-σysin2φ-2τxycosφsinφ = 0
σ − σ x cos 2φ − σ y sin 2φ − τ xy cos φ sin φ − τ xy sin φ cos φ = 0
σ −σx
AC
AC
BC
BC
cos φ − σ y
sin φ − τ xy
sin φ − τ xy
cos φ = 0
AB
AB
AB
AB
1 + cos 2 φ ⎞
⎛ 1 − cos 2 φ ⎞
−
σ
⎟
⎟ − τ xy sin 2 φ = 0
y⎜
2
2
⎠
⎠
⎝
⎝
σ − σ x ⎛⎜
1 + cos 2φ ⎞
⎛ 1 − cos 2φ ⎞
⎟ +σ y⎜
⎟ + τ xy sin 2φ = 0
2
2
⎝
⎠
⎝
⎠
σ = σ x ⎜⎛
1
2
1
2
1
2
1
2
σ = σ x + σ x cos 2φ + σ y − σ y cos 2φ + τ xy sin 2φ
σ=
σ x +σ y σ x −σ y
2
+
2
cos 2φ + τ xy sin 2φ