144
5 Two-Dimensional Problems in Cartesian Co-ordinate System
i.e. M = 2b
+h
−h
d 3 y
2 dy
= 2bd 3
+h
−h
y
2 dy
= 2bd 3
y
3
3
+h
−h
M = 4bd 3
h
3
3
Or
d 3 =
3M
4bh 3
d 3 =
M
I
where I =
4h
3 b
3
.
Therefore,
σ x =
M
I
y
(5.41)
From the above, it can be indicated that the simple theory of bending gives the
same solution.
5.7 Bending of a Narrow Cantilever Beam Subjected
to End Load
Consider a cantilever beam having a narrow rectangular cross section of unit width
and depth 2 h as shown in Fig. 5.9. Let P be a force applied at its free end as shown.
Let us a adopt fourth-degree polynomial function
i.e.,
φ =
a 4
24
x
4
+
b 4
6
x
3 y +
c 4
2
x
2 y
2
+
d 4
6
x y
3
+
e 4
24
y
4
Here,
5 Two-Dimensional Problems in Cartesian Co-ordinate System
i.e. M = 2b
+h
−h
d 3 y
2 dy
= 2bd 3
+h
−h
y
2 dy
= 2bd 3
y
3
3
+h
−h
M = 4bd 3
h
3
3
Or
d 3 =
3M
4bh 3
d 3 =
M
I
where I =
4h
3 b
3
.
Therefore,
σ x =
M
I
y
(5.41)
From the above, it can be indicated that the simple theory of bending gives the
same solution.
5.7 Bending of a Narrow Cantilever Beam Subjected
to End Load
Consider a cantilever beam having a narrow rectangular cross section of unit width
and depth 2 h as shown in Fig. 5.9. Let P be a force applied at its free end as shown.
Let us a adopt fourth-degree polynomial function
i.e.,
φ =
a 4
24
x
4
+
b 4
6
x
3 y +
c 4
2
x
2 y
2
+
d 4
6
x y
3
+
e 4
24
y
4
Here,