LECTURE 6
Let us discuss few representative problems dealing with complex state of stress to be solved either analytically or graphically.
Therefore the required shear stress is 40.96 MN/m
For a given loading conditions the state of stress in the wall of a cylinder is expressed as follows: (a) 85 MN/m (b) 25 MN/m (c) Shear stresses of 60 MN/m Calculate the principal stresses and the planes on which they act. What would be the effect on these results if owing to a change of loading (a) becomes compressive while stresses (b) and (c) remain unchanged
The problem may be attempted both analytically as well as graphically. Let us first obtain the analytical solution The principle stresses are given by the formula
For finding out the planes on which the principle stresses act us the equation The solution of this equation will yeild two values q i.e they q (b) In this case only the loading (a) is changed i.e. its direction had been changed. While the other stresses remains unchanged hence now the block diagram becomes.
Again the principal stresses would be given by the equation.
Thus, the two principle stresses acting on the two mutually perpendicular planes i.e principle planes may be depicted on the element as shown below:
So this is the direction of one principle plane & the principle stresses acting on this would be s
Therefore the direction of other principal planes would be {-q + 90} since the angle -q is always less in magnitude then 90 hence the quantity ( -q + 90 ) would be positive therefore the Inclination of other plane with reference plane would be positive therefore if just complete the Block. It would appear as
If we just want to measure the angles from the reference plane, than rotate this block through 180
So whenever one of the angles comes negative to get the positive value, first Add 90 so q i.e q This is how we can show the angular position of these planes clearly.
Construct the graphical construction as per the steps given earlier.
Taking the measurements from the Mohr's stress circle, the various quantities computed are s
By taking the measurements, the various quantites computed are given as s
1. complementary shear stresses (on planes 90 2. The principal planes are orthogonal: points L and M are 180 3. There are no shear stresses on principal planes: point L and M lie on normal stress axis. 4. The planes of maximum shear are 45 5. The maximum shear stresses are equal in magnitude and given by points D and E 6. The normal stresses on the planes of maximum shear stress are equal i.e. points D and E both have normal stress co-ordinate which is equal to the two principal stresses.
As we know that the circle represents all possible states of normal and shear stress on any plane through a stresses point in a material. Further we have seen that the co-ordinates of the point ‘Q' are seen to be the same as those derived from equilibrium of the element. i.e. the normal and shear stress components on any plane passing through the point can be found using Mohr's circle. Worthy of note: 1. The sides AB and BC of the element ABCD, which are 90 2. It has been shown that Mohr's circle represents all possible states at a point. Thus, it can be seen at a point. Thus, it, can be seen that two planes LP and PM, 180 Thus , s s 3. The maximum shear stress in an element is given by the top and bottom points of the circle i.e by points J 4.The minimum normal stress is just as important as the maximum. The algebraic minimum stress could have a magnitude greater than that of the maximum principal stress if the state of stress were such that the centre of the circle is to the left of orgin. i.e. if s s Then t If should be noted that the principal stresses are considered a maximum or minimum mathematically e.g. a compressive or negative stress is less than a positive stress, irrespective or numerical value. 5. Since the stresses on perpendular faces of any element are given by the co-ordinates of two diametrically opposite points on the circle, thus, the sum of the two normal stresses for any and all orientations of the element is constant, i.e. Thus sum is an invariant for any particular state of stress. Sum of the two normal stress components acting on mutually perpendicular planes at a point in a state of plane stress is not affected by the orientation of these planes. This can be also understand from the circle Since AB and BC are diametrically opposite thus, what ever may be their orientation, they will always lie on the diametre or we can say that their sum won't change, it can also be seen from analytical relations We know on plane BC; q = 0 s on plane AB; q = 270 s Thus s 6. If s 7. If s |