Civil Engineering Reference
In-Depth Information
Element in the vertical plane of symmetry
on the upper surface of the beam
D
Point load
A
C
D
s
B
A
C
F
IGURE
14.1
Representation of
stress at a point in
a structural
member
B
s
(a)
(b)
D
Q
s
x
s
n
s
n
u
Q
τ
A
C
u
s
x
s
x
P
F
IGURE
14.2
Determination of
stresses on an
inclined plane
R
τ
s
x
P
R
B
(a)
(b)
of a thin, very small element; such an element is known as a two-dimensional element
and the stress system is a plane stress system as we saw in Section 7.11.
14.2 D
ETERMINATION OF
S
TRESSES ON
I
NCLINED
P
LANES
Suppose that we wish to determine the direct and shear stresses at the same point in
the cantilever beam of Fig. 14.1 but on a plane PQ inclined at an angle to the axis of
the beam as shown in Fig. 14.2(a). The direct stress on the sides AD and BC of the
element ABCD is
σ
x
in accordance with the sign convention adopted previously.
Consider the triangular portion PQR of the element ABCD where QR is parallel to
the sides AD and BC. On QR there is a direct stress which must also be
σ
x
since there
is no variation of direct stress on planes parallel to QR between the opposite sides of
the element. On the side PQ of the triangular element let
σ
n
be the direct stress and
τ
the shear stress. Although the stresses are uniformly distributed along the sides of the
elements it is convenient to represent them by single arrows as shown in Fig. 14.2(b).
The triangular element PQR is in equilibriumunder the action of forces corresponding
to the stresses
σ
x
,
σ
n
and
τ
. Thus, resolving forces in a direction perpendicular to PQ
and assuming that the element is of unit thickness we have
σ
n
PQ
=
σ
x
QR cos
θ
