Civil Engineering Reference
In-Depth Information
The maximum shear stress
τ
max
occurs at the horizontal axis of symmetry of the beam
section and is, from Eq. (10.7)
S
y
3
2
×
τ
max
=
(i)
60
×
100
Thus from Eqs (14.8) and (14.9)
10
2
10
2
10
2
1
2
10
2
1
2
4
τ
max
4
τ
max
σ
I
=
+
+
σ
II
=
−
+
or
25
25
σ
I
=
5
+
+
τ
max
σ
II
=
5
−
+
τ
max
(ii)
25
τ
max
|
It is clear from the second of Eq. (ii) that
σ
II
is negative since
|
+
>
5. Thus
in the Tresca theory Eq. (14.42) applies and
2
25
150N
/
mm
2
σ
I
−
σ
II
=
+
τ
max
=
from which
74
.
8N
/
mm
2
τ
max
=
Thus from Eq. (i)
S
y
=
299
.
3kN
Now substituting for
σ
I
and
σ
II
in Eq. (14.55) we have
5
τ
max
2
5
τ
max
2
5
τ
max
5
τ
max
25
25
25
25
150
2
+
+
+
−
+
−
+
+
−
+
=
which gives
86
.
4N
/
mm
2
τ
max
=
Again from Eq. (i)
S
y
=
345
.
6kN
BRITTLE MATERIALS
When subjected to tensile stresses brittle materials such as cast iron, concrete and
ceramics fracture at a value of stress very close to the elastic limit with little or no
permanent yielding on the planes of maximum shear stress. In fact the failure plane is
generally flat and perpendicular to the axis of loading, unlike ductile materials which
have failure planes inclined at approximately 45
◦
to the axis of loading; in the latter
case failure occurs on planes of maximum shear stress (see Sections 8.3 and 14.2). This
