Civil war sep 23-pwd

A plane element is subjected to shearing stresses of 50 MPa. The principal stresses existing in this element and the directions of the planes on which they occur would be _______ at __________ respectively.

• 50 MPa at 45°
• 50 MPa at 90°
• 7 MPa at 45°
• 7 MPa at 90°

Answer explanation

For a pure shear state of 50 MPa, the principal stresses are ±50 MPa, acting on planes oriented at 45° to the direction of shear.

If the principal stress at a point in a stressed body are 150 kN/m² tensile and 50 kN/m² compressive, then maximum shear stress at this point will be

• 100 kN/m²
• 150 kN/m²
• 200 kN/m²
• 250 kN/m²

Answer explanation

Maximum shear stress is (σ₁ - σ₂)/2 = (150 - (−50))/2 = 200/2 = 100 kN/m².

What is the value of max shear stress on a ductile material under action of uniform axial stress equal to ultimate tensile strength

• Ultimate tensile strength
• > Ultimate tensile strength
• 0.7 ultimate tensile strength
• 0.5 ultimate tensile strength

Answer explanation

For uniform axial stress σ = σut, the maximum shear stress is σ/2 = 0.5 σut.

The radius of Mohr circle is zero when the state of stress is

• Shear stress is zero
• Pure shear
• Hydrostatic stress
• There is no shear stress but equal and opposite normal stress

Answer explanation

Under hydrostatic stress, all principal stresses are equal, so the Mohr’s circle radius becomes zero.

Principal plane and maximum shear plane are at

• 90°
• 45°
• 0°
• 180°

Answer explanation

In Mohr’s circle, the maximum shear planes are oriented at 45° to the principal planes.

The shear stress on the principal plane is

• Maximum
• Minimum
• Zero
• Intermediate

Answer explanation

On principal planes, only normal stresses act and shear stress becomes zero.

On a plane, resultant stress is inclined at an angle of 45° to the plane. If the normal stress is 100 N/mm², the shear stress on the plane is:

• 71.5 N/mm²
• 100 N/mm²
• 86.6 N/mm²
• 120.8 N/mm²

Answer explanation

Formula:

tan θ = τ / σₙ

Substitute:

tan 45° = τ / 100

1 = τ / 100

τ = 100 N/mm²

Answer: 100 N/mm²

A 2D fluid element rotates like a rigid body. At a point within the element, the pressure is 1 unit. Radius of the Mohr's circle, for that point is

• 0.5
• 1
• 0
• 2

Answer explanation

In a fluid under pure hydrostatic pressure, all normal stresses are equal and shear stresses are zero, so the Mohr’s circle reduces to a point with radius 0.

A prismatic bar is subjected to axial tension. What is the aspect angle (φ) which defines an oblique section on which normal and shearing stresses are equal?

• 30°
• 45°
• 60°
• 90°

Answer explanation

A prismatic bar is subjected to axial tension.

Normal stress on an oblique plane:

σn = σ cos²φ

Shear stress on an oblique plane:

τ = σ sinφ cosφ

Condition: σn = τ

σ cos²φ = σ sinφ cosφ

Cancel σ cosφ:

cosφ = sinφ

So, tanφ = 1 → φ = 45°

Answer: 45°

Minor principal stress has minimum ________

• Value of shear stress acting on the plane
• Intensity of direct stress
• Both a and b
• None of the above

Answer explanation

The minor principal stress is the lowest value of direct (normal) stress acting on a plane.