Put you're finger on the table and push it parallel to its surface.The pressure you're exerting on the table is called a shear stress in contrast to normal stress.It doesn't have direction because its a scalar!
Imagine pushing down on a table at an angle with your finger. The component of the force perpendicular to the table gives the normal stress, and the component parallel to the table gives the shear stress.
Suppose you have a surface, and an force acting obliquely on that surface. The stress vector is the force per unit area of the surface. The shear stress is the component of the stress vector in the direction tangent to the surface. This is also the same as the stress vector minus the normal component of the stress vector. Clearly, the shear stress has direction.
The force vector has a direction. Shear stress is a scalar as Shyan indicated - the force per unit area - measured for example in psi. Just as with normal stress, there's no direction on shear stress.
I don't know where you and Shyan learned mechanics, but based on a PhD in fluid mechanics and over 50 years of real world working experience in solid- and fluid mechanics, I can assure you that both normal stress and shear stress acting on a surface have direction. Of course, as with any vector, the magnitude of a vector is a scalar. Maybe that's what you are thinking of.
Chester, I use topics at PF to continue learning and try to help, too. I'm thankful for more experienced members like yourself who keep the information in check. The information I wrote was based on my understanding of the shear stress wiki that Shyan linked to in post #4. I've searched further and found this: I clipped the above from http://www.continuummechanics.org/cm/tractionvector.html I'd like to understand better and correctly. Is this a misunderstanding of the information, or specifics of terminology, or something more?
I can see why you have been confused. The write-up in Wiki is very confusing, and this does a disservice to students like yourself who are seeking clarity. The quantities σ and τ that they refer to are the magnitudes of the normal stress and shear stress components of the traction vector. In terms of the unit normal vector [itex]\vec{n}[/itex] and the unit vector in the tangential direction [itex]\vec{s}[/itex] (whatever direction that happens to be) to the surface, the traction vector [itex]\vec{T}[/itex] is represented as: [tex]\vec{T}=σ\vec{n}+τ\vec{s}[/tex] The normal stress (vector) is the component of [itex]\vec{T}[/itex] in the direction perpendicular to the surface [itex]σ\vec{n}[/itex], and the shear stress (vector) is the component of [itex]\vec{T}[/itex] in the direction of the [itex]\vec{s}[/itex] unit vector [itex]τ\vec{s}[/itex]. The magnitude of the normal stress component is obtained by dotting the traction vector with the unit normal: [tex]σ=\vec{T}\centerdot \vec{n}=(σ\vec{n}+τ\vec{s})\centerdot\vec{n}[/tex] The magnitude of the shear stress component is obtained by dotting the traction vector with the tangential unit vector: [tex]τ=\vec{T}\centerdot \vec{s}=(σ\vec{n}+τ\vec{s})\centerdot\vec{s}[/tex] Initially, you may know the traction vector [itex]\vec{T}[/itex] and the unit normal vector [itex]\vec{n}[/itex], but you will probably not know the direction of the shear stress vector [itex]\vec{s}[/itex]. You can determine the direction of the shear stress vector [itex]\vec{s}[/itex] as follows: [tex]\vec{T}=σ\vec{n}+τ\vec{s}=(\vec{T}\centerdot \vec{n})\vec{n}+τ\vec{s}[/tex] From this, if follows that: [tex]τ\vec{s}=\vec{T}-(\vec{T}\centerdot \vec{n})\vec{n}[/tex] If we take the dot product of this equation with itself, we obtain: [tex]τ^2=\vec{T}-(\vec{T}\centerdot \vec{n})\vec{n}=T^2-(\vec{T}\centerdot \vec{n})^2[/tex] where T is the magnitude of the traction vector. Therefore, the magnitude of the shear stress component is given by: [tex]τ=\sqrt{T^2-(\vec{T}\centerdot \vec{n})^2}[/tex] The unit vector in the shear stress direction is then given by: [tex]\vec{s}=\frac{(\vec{T}-(\vec{T}\centerdot \vec{n})\vec{n})}{\sqrt{T^2-(\vec{T}\centerdot \vec{n})^2}}[/tex] So, from knowledge of the traction vector and the unit normal vector, one can determine both the magnitude and direction of the shear stress vector. I hope this helps and provides greater clarity. Chet