The state of stress at a certain point in a stressed body is as shown in the figure. Normal stress in x-direction is 80 MPa (Tensile) and in y-direction is 40 MPa (Compressive). The radius of the Mohr's circle for this state of stress will be

60 MPa

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40 MPa

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20 MPa

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10 MPa

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Solution

The correct option is A 60 MPa
Radius of Mohr's circle
$= \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + τ_{x y}^{2}}$

$= \sqrt{{(\frac{80 - (- 40)}{2})}^{2} + 0} = 60 M P a$

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The state of stress at a certain point in a stressed body is as shown in the figure. Normal stress in x-direction is 80 MPa (Tensile) and in y-direction is 40 MPa (Compressive). The radius of the Mohr's circle for this state of stress will be

The correct option is A 60 MPaRadius of Mohr's circle =√(σx−σy2)2+τ2xy = ⎷(80−(−40)2)2+0=60MPa

The correct option is A 60 MPa
Radius of Mohr's circle
$= \sqrt{{(\frac{σ_{x} - σ_{y}}{2})}^{2} + τ_{x y}^{2}}$

$= \sqrt{{(\frac{80 - (- 40)}{2})}^{2} + 0} = 60 M P a$