148
5 Two-Dimensional Problems in Cartesian Co-ordinate System
−P =
+h
−h
−b 2 dy +
+h
−h
−2
b 2
h 2
y
2
2
dy
Simplifying, we get
b 2 =
3P
4h
Therefore, expressions for σ x , σ y and τ xy can be written as
σ x = −
2b 2
h 2 x y =
−
3
2
·
P
h 3
x y
or
σ x = −
P
I
x y
Here I = Moment of inertia of c/s of beam =
2
3
h
3
σ y = 0
and
τ xy = −
3P
4h
+
3P
4h 3 · y
2
or
τ xy = −
3P
4h
1 −
y
2
h 2
= −
3P
4h 3
h
2
− y
2
τ xy = −
P
2I
(h
2
− y
2
)
This distribution of bending stress and shear stress is shown in Fig. 5.9c, d
respectively.
5 Two-Dimensional Problems in Cartesian Co-ordinate System
−P =
+h
−h
−b 2 dy +
+h
−h
−2
b 2
h 2
y
2
2
dy
Simplifying, we get
b 2 =
3P
4h
Therefore, expressions for σ x , σ y and τ xy can be written as
σ x = −
2b 2
h 2 x y =
−
3
2
·
P
h 3
x y
or
σ x = −
P
I
x y
Here I = Moment of inertia of c/s of beam =
2
3
h
3
σ y = 0
and
τ xy = −
3P
4h
+
3P
4h 3 · y
2
or
τ xy = −
3P
4h
1 −
y
2
h 2
= −
3P
4h 3
h
2
− y
2
τ xy = −
P
2I
(h
2
− y
2
)
This distribution of bending stress and shear stress is shown in Fig. 5.9c, d
respectively.