GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 50 of 65

Marks: +1, -0

The state of stress at a point, for a body in place stress, is shown in the figure below. If the minimum principal stress is

10 {kPa}

, then the normal stress

σ_{y}

(in

{kPa}

) is

Solution:

σ_{x}=100 k P a_{y}, τ_{x y}=50 {kPa}

Minimum principal stress

=\;{σ_{x}+σ_{y}}/{2}-√{(\;{σ_{x}-σ_{y}}/{2})^{2}+τ_{x y}^{2}}

10=\;{100+σ_{y}}/{2}-√{(\;{100-σ_{y}}/{2})^{2}+50^{2}}

∴ √{(\;{100-σ_{y}}/{2})^{2}+50^{2}}=50+\;{σ_{y}}/{2}-10

=40+\;{σ_{y}}/{2}

By squaring

2500+\;{σ_{y}^{2}}/{4}-50 σ_{y}+2500=1600+\;{σ_{y}^{2}}/{4}+40 σ_{y}

∴ \;\; 90 σ_{γ}=3400

σ_{y}=37.78 {MPa}

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