Question

The tensile stresses at a point across two mutually perpendicular planes are 150 N/mm² and 75 N/mm². What is the normal stress on the plane inclined at 35° to the axis of the major stresses?

1. 128.64 N/mm2
2. 125.33 N/mm2
3. 115 N/mm2
4. 120.50 N/mm2

Answer 1

Correct Answer - Option 2 : 125.33 N/mm2

Concept:

For an element under the effect of bi-axial state of normal stress, the normal stresses on the plane inclined at θ° to the axis of the major stresses is given by

\({{\rm{σ }}_{{\rm{n}}1}} = \left( {\dfrac{{{{\rm{σ }}_1} + {{\rm{σ }}_2}}}{2}} \right) + \left( {\dfrac{{{{\rm{σ }}_1} - {{\rm{σ }}_2}}}{2}} \right){\rm{cos}} \ 2{\rm{θ }}\)

Where, 

σ₁ and σ₂ are stresses at two mutually perpendicular planes.

Calculation:

Given,

σ₁ = 150 N/mm²

σ₂ = 75 N/mm2

θ = 35°

\({{\rm{σ }}_{{\rm{n}}1}} = \left( {\dfrac{{{{\rm{σ }}_1} + {{\rm{σ }}_2}}}{2}} \right) + \left( {\dfrac{{{{\rm{σ }}_1} - {{\rm{σ }}_2}}}{2}} \right){\rm{cos}} \ 2{\rm{θ }}\)

\({σ _{{\rm{n}}1}} = \left( {\dfrac{{150 + 75}}{2}} \right) + \left( {\dfrac{{150 - 75}}{2}} \right){\rm{cos}}\ 2 \times 35 = 112.5 + 12.826 = 125.33\rm \ N/m{m^2}\)