## Shear stress Difficulty

Hi all,

I'm having problems with visualising shearforce for a body.Let us take a rectangular prism and fix one of its ends.Let the other end be subjected to a tensile force T Newtons.The normal stress acting on the cross section is the tensile force /area of cross section perpendicular to it.The net effect will be to produce elongation in the bar.I know that the shear loads are applied tangentially to the cross section of the body,In that case how does a body loaded by pure tensile load fail by shear ? how are shear stresses induced in this condition?.And why is shear strength for a material less than the tensile strength ?

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 Quote by sriram123 Hi all, I'm having problems with visualising shearforce for a body.Let us take a rectangular prism and fix one of its ends.Let the other end be subjected to a tensile force T Newtons.The normal stress acting on the cross section is the tensile force /area of cross section perpendicular to it.The net effect will be to produce elongation in the bar.I know that the shear loads are applied tangentially to the cross section of the body,In that case how does a body loaded by pure tensile load fail by shear ? how are shear stresses induced in this condition?.And why is shear strength for a material less than the tensile strength ? Thanks in advance
This is a very good question. Think of a plane crossing through your rectangular prism at an arbitrary angle. The stress in your body is not just at the ends, but it is also present at all locations throughout your body. So there is stress acting on both faces of the plane. The material on one side of the plane exerts a stress on the material on the other side of the plane, and the material on the other side of the plane exerts an equal but opposite stress on the material on the first side of the plane. But, because the plane is oriented at an angle to the axis of the prism, the stress on the plane is not perpendicular to the face of the plane. There is a component of stress tangent to the plane. This is the shear stress on the plane. In general, if you take an arbitrary plane within a body and ask what the stress is that is acting on this plane, unless the plane is perpendicular to one of the principal directions of stress, there will be a shear stress on the plane.

 @ chestermiller Thanks for you Reply When we are plotting a stress strain curve for a material in tension ,How do we ensure that the material will fail by pure tension (Necking for ductile materials)?.If shear strength is less than tensile strength won't the material fail by shear first?.Then how can we find yield strength and Ultimate tensile strength ?.I learnt that care will be taken that the material will not fail by bending but the shear failure ?How do we ensure that it won't happen?

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## Shear stress Difficulty

 Quote by sriram123 @ chestermiller Thanks for you Reply When we are plotting a stress strain curve for a material in tension ,How do we ensure that the material will fail by pure tension (Necking for ductile materials)?.If shear strength is less than tensile strength won't the material fail by shear first?.Then how can we find yield strength and Ultimate tensile strength ?.I learnt that care will be taken that the material will not fail by bending but the shear failure ?How do we ensure that it won't happen?
Another good question. Usually, the failure criterion is expressed in terms of all the components of the stress tensor within a single tensorially correct scalar relationship. When a certain function of the stress components exceeds a critical value (for the material), the material will fail. For simple loadings like tension on a prism, the function takes on a simple form when the axes of the coordinate system are aligned with the principal stress direction. However, the same function will give the correct answer if the axes of the coordinate system are rotated so there are shear stresses. In short, if the failure criterion is properly expressed mathematically, it doesn't matter whether there are only normal stresses present, or whether shear stresses are also present.

 "For simple loadings like tension on a prism, the function takes on a simple form when the axes of the coordinate system are aligned with the principal stress direction. However, the same function will give the correct answer if the axes of the coordinate system are rotated so there are shear stresses" I could not follow that can you please explain in more detail? Thanks in Advance
 Recognitions: Gold Member This is not easy to explain in detail without the mathematics. To get your feet wet, check out the following web page in wiki: http://en.wikipedia.org/wiki/Von_Mises_yield_criterion