For most of the materials why is shear always lesser than tensile strength?
I agree with Miquel. According to two failure criteria for ductile materials, the failure of a pure shear specimen occurs as 0.5 to 0.577 of a unixial loaded specimen.chapter 7 of
Beer, Johnston & Dewolf (2004). Mechanics Of Materials, McGraw-Hill Education
explains it well.
Immediate idea - if you have a material that fails in a ductile manner in a tensile test, this should happen in the plane that has the maximum shear stress, and that's at a 45° angle with respect to the direction of the force you apply. Nevertheless you will determine your tensile strength by dividing the force along the specimen axis by the specimen cross section perpendicular to it, while in fact your specimen fails (in shear) due to a smaller force component associated with a larger area (45° angle).
As far as I know, shear strength and tensile strength are independent and it can't be said that shear strength is always less than tensile strength.
Just a little more. A uniaxial tensile stress state [Sxx, 0, 0; 0, 0, 0; 0, 0,0] is what we obtain at the cross section A0 in the central part of the specimen at a tensile test machine applying a axial force F.
As previous answer mentioned, for ductile material the specimen will fail due the shear stress (in planes at 45º angle to the specimen axis). These shear stresses Sx'y' and Sx'z are caused by the same traction force which is decomposed in normal N and shear V force. Sx'y'=V/A'=F.cos(theta))/(A0/sin(theta) but the resisting section that support this shear is A=A0/sin(theta0) and so Sx',y'= F.cos(theta)sin(theta) /A0 =1/2*Sxx.
It means that Sxy_yield=0.5*Sxx_yield which corresponds to the Tresca criteria (seen also with Mohr circles). But, test shown that this is a conservative assumption, and usually a better approach is given for ductile material by a Von Mises criteria getting Sxy_yield=0.577*Sxx (it comes from strain energy considerations).
These considerations explain why shear strength have to be less than tensile strength, and the value is between 0.5 to 0.577 of the tensile one.
I agree with Miquel. According to two failure criteria for ductile materials, the failure of a pure shear specimen occurs as 0.5 to 0.577 of a unixial loaded specimen.chapter 7 of
Beer, Johnston & Dewolf (2004). Mechanics Of Materials, McGraw-Hill Education
explains it well.
I don't know why you downvoted my comment... You guys are talking about an energy criterion to estimate shear strength in terms of axial strength. But this is not a way to obtain the shear strength of a given material. You could only find the shear strength of a material using the axial test if the specimen fails by shear (45 degree plane), then you can set the shear strength to be 0.5 of the normal stress applied. If this is not the case, you should perform a shear test.
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