2. Linear stress state

Linear stress state is experienced by some points of the rod working for bending or complex resistance, but in most cases this type of stress state is experienced by points of the rod working in tension or compression.

Consider a rod experiencing simple tension (Fig. 9.5, a). We calculate the stresses acting on any inclined section. We cut the rod by section making an angle with cross section perpendicular to the axis of the rod. The same angle is made up of the normals to the transverse and inclined sections. For the positive direction of the reference angle take the direction counterclockwise. The normal OA directed outward with respect to the cut off part of the rod will be called the external normal to the cross section . Cross-sectional area denote , cross-sectional area denote .

Mentally discard the upper part of the rod and replace its action with the lower part by stresses (Fig. 9.5b).

Fig.9.5

Accepting the hypothesis of flat sections, we assume that the stress evenly distributed over the area :

. (9.1)

Given that and substituting in (9.1), we obtain:

, (9.2)

Where  normal site voltage perpendicular to tensile force.

Designing to normal and on the section plane, we obtain the expressions for the normal and tangential stresses on the inclined platform:

;

or

, (9.3)

. (9.4)

As can be seen from formulas (9.3)  (9.4), for voltage , ; at voltage and are equal to zero (Fig. 9.6).

Figure 9.6

Thus, with simple tension and compression at each point of the body, the main areas are perpendicular and parallel to its axis, and the main stresses in it are respectively equal:

;  tensile

;  in compression.

Maximum tangential stresses act in areas that are inclined to the main areas at an angle . Wherein

(9.5)

Example 9.1 Determine the normal and tangential stresses on inclined platforms for the elements shown in Fig. 9.7.

Decision:

1. For the element in Fig. 9.7, a: MPa; ; .

From: MPa; MPa

1. For the element in Fig. 9.7, b: MPa; ; .

From: MPa; MPa

3. For the element in Fig. 9.7, c: ; MPa; .

From: MPa; MPa