A simply supported nonmetallic beam of 0.25 m height, 0.1 m –width, and
1.5 m span is subjected to a uniform loading of 6KN/m. Determine the factor of safety for
this loading according to (a)the maximum distortion energy theory, and (b)the maximum
shearing stress theory. Use
MPa
28
=
s
yp
.
Solution:
This beam is being subjected to two different types of stresses: bending and shear
The bending stress for a beam subjected to a bending moment "M", and moment of
inertia "I" is:
I
My
b
=
s
Where, M= bending moment
y= distance of the fiber form the neutral axis
I = moment of inertia
The bending moment at any crosssection distance "x" from the left end is,
2
)
2
)(
.
(
2
x
x
x
M
w
w
=
=
where,
w
is load per unit length of the beam.
And, the moment of inertia of the crosssection is:
12
3
h
b
I
=
Where b= width of the beam
h= height of the beam
12
.
2
3
2
h
b
y
x
b
w
s
=
∴
…………………….(1)
Now, we evaluate the shear stress on the crosssection,
A
sf
=
t
Here, "sf" is the shear force , and “A” is crosssectional area.
h
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 Fall '09
 Law
 Force, Shear Stress, Second moment of area, Shear strength, distortion energy theory, σv

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