5.11. Determining displacements by Mohr’s method

Consider now the general method of determining the displacements that is suitable for any linear deformed system under any load. The method was proposed by an outstanding German scientist O. Mohr.

For example, you have to determine the vertical displacement of the bar point B as presented in Fig. 5.13 a. Denote the given (load) state by f. Choose the auxiliary state of the same beam with the unit (dimensionless) force acting at the point B to the direction of the sought displacement. Denote the auxiliary state by k (Fig. 5.13 b).

Determine the work of the external and internal forces of the auxiliary state on the displacements caused by the force action of the load state.

The work of the external forces is equal to the product of the unit force and the sought displacement :

(5.34)

and the work of the internal force is equal to the integral:

(5.35)

But or (5.36)

Fig. 5.13.

This formula is in fact Mohr^’s formula (Mohr^’s integral) which gives the possibility to determine the displacement at any point of the linear deforming system.

The subintegral product M_kM_f in this formula is positive if both the bending moments have the same sign; and it is negative if M_k and M_f have different signs.

If we determine the angle displacement at the point B, we should apply at the point B a moment equal unit (dimensionless) for the state k.

Denoting by Δ any displacement (linear or angular) we write Mohr^’s formula (integral) in the following form

(5.37)

In a general case the expressions M_k, M_f can be different for different beam regions or for elastic system in general. Therefore a more general formula must be used.

(5.38)

If the system bars work in bending or tension, this formula must be used

(5.39)

In a particular case when the bars work only in tension or compression to determine the displacements, we have the formula

(5.40)

In this formula the product N_kN_f is positive if both stresses are elongated or compressed.

In the frame calculations when the bars work simultaneously both in bending and tension (compression) in usual cases – as the comparison calculations show – the displacement can be determined only by considering the bending moments because the shearing force influence is very small. For the same reason in usual cases the shearing force influence may not be taken into account.

If the states f and k are the same, we get:

(5.41)

6. Strengtn theory

6.1. The purpose of strength hypotheses

Up to now there have been studied the stress analyses for the cases when the material is in the uniaxial stress tension, compression or the simplest biaxial one, when the principal stresses at each point are equal to each other by the value and opposite by the sign (shear, torsion).

The composition of the strength conditions has not caused difficulties in these cases. To provide the material strength it was required that the maximum normal stress (in tension, compression) or the maximum shearing stress (in torsion) should not surpass the corresponding allowable working stress whose value was established experimentally by the received corresponding yield stress or the tensile strength (for brittle materials).

The general case of the triaxial stress state is represented in Fig. 6.1. The action plane of the maximum shearing stress is also shown in the figure. Remember that the following rule of the principal stress designation was accepted earlier: (with the sign consideration).

The question arises: what are the stress values ( ) so that the limit state of a material can take place, i.e. its failure or peasfic deformations take place.

Answering this question it would be necessary to solve another problem: how to determine the safe (allowable) values of the principal stresses .

The problem set is extremely difficult. The most reliable way of its solving would be to test the specimen under a given correlation of the principal stresses up to the failure or the yield beginning and thus to determine the limit principal stresses and then their allowable values.

However this way has to be refused as every new combination would require a new test.

Besides, these tests require very complicated machines and devices. Therefore it is necessary to have some hypothesis (theory) which would allow to appreciate the danger of passing a material in the limit state under the combined stress state, not resorting each time to the labor-consuming tests but using only the given simplest tests, i.e. the tests for the uniaxial stress.

Thus, the hypothesis construction is based on the assumption that any two stress states are considered as equally dangerous and equally strengthened, should they become with the proportional increasing of the principal stresses in the same number oftimes simultaneously.

Fig. 6.1.

In this case the factor of safety for two stress status under the given conditions will be equal.