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

The forces acting on the vertical and horizontal faces of the wedge can be resolved
into components acting in the n direction (i.e., the direction normal to the inclined plane)
and the t direction (i.e., the direction parallel, or tangential, to the inclined plane).
From these force components, the sum of forces acting in the direction perpendicular
to the inclined plane is
∑ F = σ dA + (30 MPa)( dA sin53.13 ° )sin53.13°
n
n
− (150 MPa)( dA cos53.13 ° )cos53.13°=
0
Notice that the area dA appears in each term; consequently, it will cancel out of the equation.
From this equilibrium equation, the normal stress acting in the n direction is found
to be
σ = 34.80 MPa (T)
Ans.
When forces are summed in the t direction, the equilibrium equation is
n
∑ F = τ dA + (30 MPa)( dA sin53.13 ° )cos53.13°
t
nt
+ (150 MPa)( dA cos53.13 ° )sin53.13°=
0
Therefore, the shear stress on the n face of the wedge acting in the t direction is
τ =− 86.4 MPa
Ans.
nt
The negative sign indicates that the shear stress really acts in the negative t direction on
the positive n face. Note that the normal stress should be designated as tension or
compression. The presence of shear stresses on the horizontal and vertical planes, had
there been any, would merely have required two more forces on the free-body diagram:
one parallel to the vertical face and one parallel to the horizontal face. Note, however,
that the magnitude of the shear stresses (not the forces) must be the same on any two
orthogonal planes.
n
(150 MPa)(dA cos 53.13°)
t
τ
σn
dA
nt dA (30 MPa)(dA sin 53.13°)
53.13°
(150 MPa)(dA cos 53.13°) sin 53.13°
53.13°
(150 MPa)(dA cos 53.13°) cos 53.13°
(30 MPa)(dA sin 53.13°) sin 53.13°
53.13°
(30 MPa)(dA sin 53.13°) cos 53.13°
490