英文弹性力学课件课程.ppt

文档介绍：STRESS ON OBLIQUE PLANE THROUGH A POINT
Now it is possible to determine the stresses on any other oblique plane passing through the point using the direction cosines.
Now it is possible to determine the stresses on any other oblique plane passing through the point using the direction cosines.
In practice, the state of stress at a point with reference to the
general Cartesian co-ordinate system is not very significant.
 
Because of that:
 
The failure of a structure or a body is due to fracture or plastic yielding.
The failure may occur on a different plane.
The failure plane is always inclined to the three co-ordinate axes.
 
Therefore, in this section the method of determining the
stresses on an oblique plane will be described.
Considering the equilibrium of the tetrahedron OABC in
Let sx , txy , txz be stresses on plane OBC
Let sy , tyx , tyz be stresses on plane OAC
Let sz , tzx , tzy be stresses on plane OAB
and snx , sny , snz be stresses on plane ABC
()
where dA is the area of ABC
Dividing Eq.() by dA,
()

Similar expressions for sny and snz can be
obtained. Expressing these relations in matrix
form:
()
()
Eq.() are gives the ponents of stress on an inclined plane.
To obtain normal and shear stresses on this inclined plane,
consider a set of axes x', y', z'. Let x' axis coincide with
the normal n to the plane ABC in Fig. .

The direction cosines can be
written as follows:
Using these direction cosines,
()
Substituting for snx, sny and snz from Eq.()
()
Similarly,
()
or
()
and
()
EXAMPLE
The state of stress at a point is given as follows: sx=800kPa, sy = 1200 kPa, sz = -400 kPa, txy=400 kPa, tyz=-600kPa and tzx = 500 kPa. Determine (a) the stresses on a plane whose normal has direction cosines l1 = 1/4 and l2 = 1/2, and (b) the normal and shearing stresses on that plane.
SOLLUTION:
From the relation:
get:
;from Eq.():
4