Stress Components on an Arbitrary Plane:

You have already learnt in previous Section in Unit 1, as how to determine the normal and shear stress components on any arbitrary plane whose inclination is described by the aspect angle q and Eqs. (30) and (31)of unit 1 furnish expressions for these components. Let us have an instance of practical application.

Example

Given Figure shows the projection of a rectangular prism ABCD, composed by adhesive bonding of two triangular prisms ABC and ACD. The state of stress in the prism is given by the components σ _x = 40 N/mm2, σ _y = 0 and t = 0.

If the tensile and shear strengths of the adhesive are 10 N/mm² and 12 N/mm², check the safety of the joints and find out the value of σ _x at which the joint shall fail.

Solution

The aspect angle  q of the bonding plane AC

=90o +tan^-1 (50/75) = 90^o  +33.69^o  + 123.69^o

Figure

Known stress components are follows such as :

σ x = 40 N/mm2, σ y = 0 and txy = 0

Stress components on plane AC, Normal stress, σ n  , σ x  cos2 q

= 40 cos² 123.69^o

= 12.3076 N/mm2 > 10 N/mm2

t_nt  =-   σ _x / 2 sin 2q

= - (40 /2 )sin (2 ´ 123.69^o )

= 18.4615 N/mm² > 12 N/mm²

The tensile stress on plane AC is not within the tensile strength of the bond. Also, the shear stress on the plane exceeds the shear strength of the bond and therefore, the bond shall fail both in tensile and shear.

Let us discover the normal stress σ _x that might be safely applied. Shear strength of the bond = 12 N/mm²

 Shear stress on bonding plane = - σ _x/2 sin 2q

∴          12 = - σ _x /2 sin 247.38o

σ _x  = 12 ´ 2 / sin 247.38^o =  26 N/mm²

So the maximum stress we might apply on the plane CB is 26 N/mm². Here, you might note that in strength analysis the sign of the shear stress has no significance, while the sign of the normal stress is significant, as the tensile and compressive strengths might differ considerably.