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

ExAmpLE 12.3
y
8
42 MPa
5
50 MPa
16 MPa
x
At a point on a structural member subjected to plane stress, normal and shear
stresses exist on horizontal and vertical planes through the point as shown. Use
the stress transformation equations to determine the normal and shear stress on
the indicated plane surface.
Plan the Solution
Problems of this type are straightforward; however, the sign conventions used in
deriving the stress transformation equations must be rigorously followed for a
successful result. Particular attention should be given to identifying the proper
value of θ, which is required to designate the inclination of the plane surface.
SolutioN
The normal stress acting on the x face creates tension in the element; therefore, it
is considered a positive normal stress (σ x = 16 MPa) in the stress transformation
equations. Likewise, the normal stress on the y face has a positive value of σ y =
42 MPa.
The 50 MPa shear stress on the positive x face acts in the negative y direction; therefore,
this shear stress is considered negative when used in the stress transformation equations (τ xy =
−50 MPa). Note that the shear stress on the horizontal face is also negative. On the positive
y face, the shear stress acts in the negative x direction; hence, τ yx = −50 MPa = τ xy .
In this example, normal and shear stresses are to be calculated for a plane surface
that has a slope of −5 (vertical) to 8 (horizontal). This slope information must be converted
to the proper value of θ for use in the stress transformation equations.
A convenient way to determine θ is to find the angle between a vertical plane and the
inclined surface. This angle will always be the same as the angle between the x axis and
the n axis. For the surface specified here, the magnitude of the angle between a vertical
plane and the inclined surface is
t
y
58°
n
θ =+58°
x
8
tanθ
= ∴ θ = 58°
5
Notice that the preceding calculation determines only the magnitude of the angle. The
proper sign for θ is determined by inspection. If the angle from the vertical plane to the
inclined plane turns in a counterclockwise direction, the value of θ is positive. Therefore,
θ = 58° for this example.
With the proper values for σ x , σ y , τ xy , and θ now established, the normal and shear
stresses acting on the inclined surface can be calculated. The normal stress in the n direction
is found from Equation (12.3):
2 2
σ = σ cos θ + σ sin θ + 2τ sinθcosθ
n x y xy
2 2
= (16MPa) cos 58 ° + (42 MPa) sin 58° + 2( − 50 MPa) sin58° cos58°
=−10.24 MPa
Note that Equation (12.5) could also be used to obtain the same result:
σ
n
σx + σ y σx − σ y
= + cos2θ + τxy
sin2θ
2 2
(16MPa) + (42 MPa) (16MPa) − (42 MPa)
=
+
cos2(58 ° ) + ( − 50MPa)sin 2(58 ° )
2
2
=−10.24 MPa
494