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

SolutioN
Plot point x as (ε x , −γ xy /2) = (435 µε, 321 µrad), and plot point y as (ε y , γ xy /2) = (−135 µε,
−321 µrad). Connect points x and y to define the diameter of Mohr’s circle for this strain
state. (See accompanying diagram.)
Since points x and y are always the same distance above
or below the ε axis, the center of Mohr’s circle can be found by
averaging the normal strains acting in the x and y directions:
εx
+ εy
C =
2
435 + ( −135)
=
= 150 µε
2
The center of Mohr’s circle always lies on the ε axis.
The geometry of the circle is used to calculate the radius. The
(ε, γ /2) coordinates of both point x and center C are known. Use
these coordinates with the Pythagorean theorem to calculate the
hypotenuse of the shaded triangle:
γ
2
(150, 0)
C
R = 429
48.4°
285
x (435, 321)
321
ε
R = (435 − 150) + (321 − 0)
2 2
(–135, –321) y
2 2
= 285 + 321 = 429 µ
Remember that the vertical coordinate used in plotting Mohr’s
circle is γ /2. The given shear strain is γ xy = −642 µrad; therefore, a
vertical coordinate of 321 µrad is used in plotting Mohr’s circle.
The angle between the x–y diameter and the ε axis is 2θ p , and its
magnitude can be computed with the tangent function:
tan2θ
p
321
= ∴ 2θp
= 48.4°
285
γ
2
S 2 (150, 429)
x (435, 321)
Note that this angle turns clockwise from point x to the ε axis.
The principal strains are determined from the location of the
center C and the radius R of the circle:
ε
ε
p1
p2
= C + R = 150 µε+ 429 µε = 579 µε
= C − R = 150 µε− 429 µε =− 279 µε
P 2
(–279, 0)
(150, 0)
C
R = 429
48.4°
(579, 0)
ε
P 1
The maximum values of γ occur at points S 1 and S 2 , located at
the bottom and at the top of Mohr’s circle. The shear strain magnitude
at these points is equal to the circle radius R times 2;
therefore, the maximum in-plane shear strain is
γ = 2R
= 2(429 µ ) = 858 µ rad
max
(–135, –321) y
S 1 (150, –429)
The normal strain associated with the maximum in-plane shear
strain is given by the center C of the circle:
ε = C = 150 µε
avg
The problem states that this is a plane strain condition. Therefore, the out-of-plane normal
strain ε z = 0 is the third principal strain ε p3 . Since ε p1 is positive and ε p2 is negative,
the absolute maximum shear strain equals the maximum in-plane shear strain. Therefore,
the magnitude of the absolute maximum shear strain (see Table 13.2) is
γ = ε − ε = 858 rad
Ans.
absmax p1 p2 µ
553