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A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum normal stress will be
A
400 MPa
B
500 MPa
C
900 MPa
D
1400 MPa
Correct Answer:
1400 MPa
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be
A
400 MPa
B
500 MPa
C
900 MPa
D
1400 MPa
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The minimum normal stress will be
A
400 MPa
B
500 MPa
C
900 MPa
D
1400 MPa
A body is subjected to a direct tensile stress of 300 MPa in one plane accompanied by a simple shear stress of 200 MPa. The maximum shear stress will be
A
-100 MPa
B
250 MPa
C
300 MPa
D
400 MPa
A body is subjected to a direct tensile stress of 300 MPa in one plane accompanied by a simple shear stress of 200 MPa. The maximum normal stress will be
A
-100 MPa
B
250 MPa
C
300 MPa
D
400 MPa
A body is subjected to a direct tensile stress of 300 MPa in one plane accompanied by a simple shear stress of 200 MPa. The minimum normal stress will be
A
-100 MPa
B
250 MPa
C
300 MPa
D
400 MPa
When a body is subjected to biaxial stress i.e. direct stresses $$\left( {{\sigma _{\text{x}}}} \right)$$ and $$\left( {{\sigma _{\text{y}}}} \right)$$ in two mutually perpendicular planes accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ then maximum normal stress is
A
$$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
B
$$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
C
$$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
D
$$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
When a body is subjected to a direct tensile stress $$\left( {{\sigma _{\text{x}}}} \right)$$ in one plane accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ the maximum normal stress is
A
$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
B
$$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
C
$$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$
D
$$\frac{1}{2} \times \sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm²Find the maximum amount of shear stress to which the body is subjected.
A
22.4mm
B
25mm
C
26.3mm
D
27.2mm
When a body is subjected to biaxial stress i.e. direct stresses $$\left( {{\sigma _{\text{x}}}} \right)$$ and $$\left( {{\sigma _{\text{y}}}} \right)$$ in two mutually perpendicular planes accompanied by a simple shear stress $$\left( {{\tau _{{\text{xy}}}}} \right),$$ then minimum normal stress is
A
$$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
B
$$\frac{{{\sigma _{\text{x}}} + {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
C
$$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} + \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
D
$$\frac{{{\sigma _{\text{x}}} - {\sigma _{\text{y}}}}}{2} - \frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
Tresca or maximum-shear stress criteria assumes that yielding occurs when the maximum shear stress reaches a value of the shear stress in the uniaxial tension test. Assume the principal stress being σ1, σ2, σ3 where σ1 is largest, and σ3 is the smallest principal stresses. Find the value of minimum shear stress to cause yielding, given that yield stress in tension is equal to σo?
A
τ = σo
B
τ = σo/2
C
τ = σo/3
D
τ = σo/4