Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

t
73.7°
τ
y (–16, 53)
R = 55.22
C (–31.5, 0)
70°
n
σ
Begin at point x on Mohr’s circle. Face n of the rotated
stress element is oriented 35° counterclockwise from the x face.
Since angles in Mohr’s circle are doubled, point n is rotated
2(35°) = 70° counterclockwise from point x on the circle. The
coordinates of point n are (σ n , τ nt ). These coordinates will be
determined from the geometry of the circle.
By inspection, the angle between the σ axis and point n is
180° − 73.7° − 70° = 36.3°. Keeping in mind that the coordinates
of Mohr’s circle are σ and τ, we find that the horizontal
component of the line between the center C of the circle and
point n is
∆ σ = Rcos36.3 °= (55.22 MPa)cos36.3°=
44.50 MPa
(– 47, 53) x
τ
and the vertical component is
∆ τ = Rsin36.3 °= (55.22 MPa)sin36.3°=
32.69 MPa
The normal stress on the n face of the rotated stress element can
be computed by using the coordinates of the center C and ∆σ:
R cos 36.3°
(–31.5, 0) C
36.3°
R sin 36.3°
R sin 36.3°
R = 55.22
n
t
R = 55.22
36.3°
C (–31.5, 0)
R cos 36.3°
76.00 MPa
16 MPa
53 MPa
47 MPa
35°
x
The shear stress is computed similarly:
σ =− 31.5 MPa + 44.50 MPa = 13.0 MPa
n
τ = 0 + 32.69 MPa = 32.69 MPa
nt
Since point n is located below the σ axis, the shear stress acting on the n face tends
to rotate the stress element counterclockwise.
A similar procedure is used to determine the stresses at point t. The
stress components relative to the center C of the circle are the same: ∆σ =
44.50 MPa and ∆τ = 32.69 MPa. The normal stress on the t face of the rotated
stress element is
σ =−31.5 MPa − 44.50 MPa = −76.0 MPa
t
Of course, the shear stress acting on the t face must be of the same magnitude
as the shear stress acting on the n face. Since point t is located above the σ
axis, the shear stress acting on the t face tends to rotate the stress element
clockwise.
To determine the normal stress on the t face, we could also use the notion of
stress invariance. Equation (12.8) shows that the sum of the normal stresses acting
on any two orthogonal faces of a plane stress element is a constant
value:
13.00 MPa
32.69 MPa
Therefore,
σn + σt = σx + σ y
σt = σ x + σ y − σn
=− 47 MPa + ( −16 MPa) − 13 MPa
=−76 MPa
The normal and shear stresses acting on the rotated element are
shown in the accompanying sketch.
524