Civil Engineering Reference
In-Depth Information
axes. Again combinations of stress corresponding to points inside the locus will not
cause failure, whereas combinations of
σ
I
and
σ
II
on or outside the locus will.
E
XAMPLE
14.10
×
500mmand is simply supported over a span of 4m. Determine themaximummid-span
concentrated load the beam can carry if the failure stress in simple tension of concrete
is 1
.
5N
/
mm
2
. Neglect the self-weight of the beam.
A concrete beam has a rectangular cross section 250mm
If the central concentrated load is
W
N the maximum bending moment occurs at
mid-span and is
4
W
4
=
W
Nm
(see Ex.3.6)
The maximum direct tensile stress due to bending occurs at the soffit of the beam
and is
10
3
W
×
×
250
×
12
10
−
5
N
/
mm
2
σ
=
=
W
×
9
.
6
×
(Eq. 9.9)
250
×
500
3
At this point the maximum principal stress is, from Eq. (14.8)
10
−
5
N
/
mm
2
σ
I
=
W
×
9
.
6
×
Thus from Eq. (14.57) the maximum value of
W
is given by
10
−
5
1
.
5N
/
mm
2
σ
I
=
W
×
9
.
6
×
=
σ
Y
(tension)
=
from which
W
=
15
.
6kN.
The maximum shear stress occurs at the horizontal axis of symmetry of the beam
section over each support and is, from Eq. (10.7)
3
2
×
W
/
2
250
τ
max
=
×
500
i.e.
10
−
5
N
/
mm
2
τ
max
=
W
×
0
.
6
×
Again, from Eq. (14.8), the maximum principal stress is
10
−
5
N
/
mm
2
1
.
5N
/
mm
2
σ
1
=
W
×
0
.
6
×
=
σ
Y
(tension)
=
from which
W
=
250 kN
Thus the maximum allowable value of
W
is 15
.
6 kN.
