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

element counterclockwise; therefore, point x is
located to the left of the τ axis and below the σ axis.
On the y face, the normal stress is σ y = 12 MPa (T),
and the shear stress τ acting on the y face rotates the
element clockwise; therefore, point y is located to the
right of the τ axis and above the σ axis.
Note: In this introductory example, the subscript
“ccw” has been added to the shear stress at point x
simply to give further emphasis to the fact that the
shear stress on the x face rotates the element counterclockwise.
Similarly, the subscript “cw” added to the
shear stress at point y is meant to call attention to the
fact that the shear stress on the y face rotates the element
clockwise.
The center of Mohr’s circle can be found by averaging
the normal stresses acting on the x and y faces:
27
(–60, 27ccw) x
36
36.86°
C
R = 45
(–24, 0)
τ
τ
y (12, 27 cw)
σ
σx
+ σ
C =
2
y
( 60MPa) 12MPa
= − + =−24 MPa
2
The radius R is found from the hypotenuse of the
shaded triangle:
R = [( −60 MPa) − ( − 24MPa)] + (27 MPa − 0)
2 2
= 36 + 27 = 45 MPa
2 2
The angle between the x–y diameter and the σ axis is
2θ p , and it can be computed with the use of the tangent
function:
tan2θ
p
27
= ∴ 2θp
= 36.86°
36
Notice that this angle turns clockwise from point x to
the σ axis.
The principal stresses are determined from the
location of the center C and radius R of the circle:
σ
σ
p1
p2
= C + R =− 24 MPa + 45MPa = 21 MPa
= C − R =−24 MPa − 45MPa = −69 MPa
The maximum values of τ occur at points S 1 and
S 2 , located at the bottom and at the top of Mohr’s circle.
The shear stress magnitude at these points is simply
equal to the radius R of the circle, and the normal
stress σ at these points is equal to the center C.
The angle between points P 2 and S 2 is 90°. Since
the angle between point x and point P 2 is 36.86°, the
angle between point x and point S 1 must be 53.14°. By
inspection, this angle rotates in a counterclockwise
direction.
The angle between point x and point P 2 was
calculated as 2θ p = 36.86°. Thus, this principal plane
27 MPa
P 2
(–69, 0)
(–60, 27ccw) x
36.86°
Mohr’s circle point S2
corresponds to this face.
12 MPa
60 MPa
(–24, 45cw)
53.14°
C
S 2
(– 24, 45 ccw) S 1
26.57°
x
18.43°
24 MPa
Mohr’s circle point P1
corresponds to this face.
R = 45
(–24, 0)
S
τ
τ
45 MPa
P
21 MPa
y (12, 27 cw)
(21, 0)
P σ 1
24 MPa
Mohr’s circle
point S 1
corresponds
to this face.
69 MPa
Mohr’s circle
point P2
corresponds
to this face.
521