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Stress Under General Loading Conditions and Components of Stress

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Most structural components are under more involved loading conditions other than axial loading and transverse loading.

Fig1.

Fig2.

Fig3.

Consider a body subjected to several loads P₁, P₂, etc. such as in Fig1. To understand the stress condition created by these loads at some point Q within the body, a section is passed throughout Q, using a plane parallel to the yz plane. The portion of the body to the left of the section is subjected to some of the original loads, and to normal and shearing forces distributed over the section. The normal and shearing forces acting on a small area ΔA surrounding point Q are denoted by ΔF^x and ΔV^x, respectively, as shown in Fig2. The superscript x is used to indicate that the forces ΔF^x and ΔV^x act on a surface perpendicular to the x axis. While the normal force ΔF^x has a well-defined direction, the shearing force ΔV^x may have any direction in the plane of the section. Therefore ΔV^x is resolved into two component forces, ΔV_y^x and ΔV_z^x, in directions parallel to the y and z axes, respectively, as shown in Fig3. Now dividing the magnitude of each force by the area ΔA, and letting ΔA approach zero, as in Fig4, the three stress components are defined:

(Eq1)

σx =    lim ΔA → 0 ΔFx ΔA

(Eq2)

τxy =    lim ΔA → 0 ΔVyx ΔA

(Eq3)

τxz =    lim ΔA → 0 ΔVzx ΔA

Note that the first subscript in σ_x, τ_xy, and τ_xz is used to indicate that the stresses under consideration are exerted on a surface perpendicular to the x axis. The second subscript in τ_xy and τ_xz identifies the direction of the component. The normal stress σ_x is positive if the corresponding arrow points in the positive x direction, i.e., if the body is in tension, and negative otherwise. Similarly, the shearing stress components τ_xy and τ_xz are positive if the corresponding arrows point, respectively, in the positive y and z directions.

The above analysis may also be carried out by considering the portion of the body located to the right of the vertical plane through Q as shown in Fig5.

Fig4.

Fig5.

The same magnitudes, but opposite directions, are obtained for the normal and shearing forces ΔF^x, ΔV_y^x, and ΔV_z^x. Therefore, the same values are also obtained for the corresponding stress components, but since the section in Fig5 now faces the negative x axis, a positive sign for σ_x will indicate the corresponding arrow points in the negative x direction. Similarly, positive signs for τ_xy and τ_xz will indicate that the corresponding arrows point, respectively, in the negative y and z directions, as in Fig5. Passing a section through Q parallel to the zx plane, the stress components σ_y, τ_yz, and τ_yx are defined in the same manner. Finally, a section through Q parallel to the xy plane yields the components σ_z, τ_zx, and τ_zy

To facilitate the visualization of the stress condition at point Q, a small cube of side a centered at Q and the stresses exerted on each of the six faces of the cube are considered, as shown in Fig6.

Fig6.

Fig7.

The stress components shown in the figure are σ_x, σ_y, and σ_z, which represent the normal stress on faces respectively perpendicular to the x, y, and z axes, and the six shearing stress components τ_xy, τ_xz, τ_yx, τ_yz, τ_zx, and τ_zy. Recall from above that according to the definition of the shearing stress components, τ_xy represents the y component of the shearing stress exerted on the face perpendicular to the x axis, while τ_yx represents the x component of the shearing stress exerted on the face perpendicular to the y axis. Only three faces of the cube are actually visible in Fig6, and that equal and opposite stress components act on the hidden faces. While the stresses acting on the faces differ slightly from the stresses at Q, the error involved is small and vanishes as side a of the cube approaches zero.

Important relations among the shearing stress components will now be derived. Consider the free-body diagram of the small cube centered at point Q as shown in Fig7. The normal and shearing forces acting on the various faces of the cube are obtained by multiplying the corresponding stress components by the area ΔA of each face. The three equilibrium equations are:

(Eq2a)

∑Fx = 0

(Eq2b)

∑Fy = 0

(Eq2c)

∑Fz = 0

Since forces equal and opposite to the forces actually shown Fig7 are acting on hidden faces of the cube, Eq2a, Eq2b, and Eq2c are satisfied. Considering now the moments of the forces about axes Qx', Qy', and Qz' drawn from Q in directions respectively parallel to the x, y, and z axes, the three additional equations are:

(Eq3a)

∑Mx' = 0

(Eq3b)

∑My' = 0

(Eq3c)

∑Mz' = 0

Using a projection on the x'y' plane as shown in Fig8.

Fig8.

it is noted that the only forces with moments about the z-axis different from zero are the shearing forces. These forces form two couples, one of counterclockwise (positive) moment (τ_xyΔA)a, the other of clockwise (negative) moment −(τ_yxΔA)a. Eq3c yields, therefore:

∑M_z = 0:      (τ_xyΔA)_a − (τ_yxΔA)_a = 0

from which it is concluded that:

(Eq4a)

τxy = τyx

The relation obtained shows that the y-component of the shearing stress exerted on a face perpendicular to the x-axis is equal to the x-component of the shearing stress exerted on a face perpendicular to the y-axis. From the remaining two equations, Eq3b and Eq3c, the following relations can be derived in a similar manner:

(Eq4b)

τyz = τzy

and:

(Eq4c)

τzx = τxz

It can then be concluded that from Eq4a, Eq4b, and Eq4c, that only six stress components are required to define the condition of stress at a given point Q, instead of nine as originally assumed. These six components are σ_x, σ_y, σ_z, τ_x, τ_y, and τ_z. It is also noted that, at a given point, shear cannot take place in one plane only; an equal shearing stress must be exerted on another plane perpendicular to the first one. For example considering a bolt and a small cube at the center Q of the bolt as shown:



It is found that the shearing stresses of equal magnitude must be exerted on the two horizontal faces of the cube and on the two faces which are perpendicular to the forces P and P'.

Lastly, consider again the case of a member under axial loading. If a small cube with faces respectively parallel to the faces of the member is considered,