
#1
Dec1413, 09:53 AM

P: 6

what is shear stress?what is its direction?




#2
Dec1413, 10:03 AM

P: 739

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!




#3
Dec1413, 10:16 AM

P: 6

moving parallel woudnt it mean sliding over it?can u explain with a diagram?




#4
Dec1413, 10:22 AM

P: 739

what is shear stress?
Take a look at here!




#5
Dec1413, 10:26 AM

P: 1

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.




#6
Dec1413, 02:48 PM

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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.




#7
Jan714, 10:29 PM

P: 6

wait what should i take as direction of surface?




#8
Jan714, 11:04 PM

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#9
Jan814, 07:04 AM

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#10
Jan814, 07:36 AM

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I'd like to understand better and correctly. Is this a misunderstanding of the information, or specifics of terminology, or something more? 



#11
Jan814, 09:26 AM

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[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 



#12
Jan814, 09:31 AM

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