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

In the undeformed plate, the angle at P is π /2rad. After the plate is deformed, the angle
at P increases. Since the angle after deformation is equal to ( π /2) − γ , the shear strain
at P must be a negative value. A simple calculation shows that the shear strain at P is the
shear strain at P is
y
R
8 in.
γ
S
γ =− 0.00781 rad
Ans.
12 in.
P
π
– γ
2
V
Q
x
0.0625 in.
ExAmpLE 2.3
A thin rectangular plate is uniformly deformed as shown. Determine the shear strain γ xy at P.
y
0.50 mm
Plan the Solution
Shear strain is an angular measure. Determine the two angles created by the 0.25 mm
deflection and the 0.50 mm deflection. Add these two angles together to determine the
shear strain at P.
SOLUTION
Determine the angles created by each deformation. Note: The small-angle approximation
will be used here; therefore, sin γ ≈ tan γ ≈ γ . From the given data, we obtain
0.50 mm
γ 1 = = 0.000694 rad
720 mm
720 mm
P
y
R
480 mm
0.50 mm
S
0.25 mm
Q
x
γ
2
0.25 mm
= = 0.000521 rad
480 mm
The shear strain at P is simply the sum of these two angles:
R
S
γ = γ1 + γ 2 = 0.000694 rad + 0.000521 rad = 0.001215 rad
= 1, 215 µ rad
Ans.
Note: The angle at P in the deformed plate is less than π /2, as it should be for a positive
shear strain. Although not requested in the problem, the shear strain at corners Q and R
will be negative and will have the same magnitude as the shear strain at corner P.
720 mm
P
γ 1
π
– γ γ
2 2
480 mm
0.25 mm
Q
x
mecmovies
ExAmpLE
m2.5 A thin triangular plate is uniformly deformed. Determine
the shear strain at P after point P has been displaced
1 mm downward.
39