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QUESTION: 1

A certain steel has proportionality limit of 300 N/mm^{2} in simple tension. It is subjected to principal stress of 120 N/mm^{2} (tensile), 60 N/mm^{2} (tensile) and 30 N/mm^{2} (compressive). The factor of safety according to maximum shear stress theory is

Solution:

Proportionality limit shear stress

= 300/2 = 150N/mm^{2}

Maximum shear stress

∴ Factor of safety = 150/75 = 2.0

QUESTION: 2

All the theories of failure, will give nearly the same result when

Solution:

When one of the principal stresses at a point is large in comparison to the other, the situation resembles uniaxial tension test. Therefore all theories give nearly the same results.

QUESTION: 3

The principal stresses developed at a point are +60, -60, 0 MPa. Using shear strain energy theory, factor of safety obtained is √3. What is yield stress of the material?

Solution:

QUESTION: 4

For ductile material the suitable theory of failure is

Solution:

For ductile material the most suitable theory is maximum shear stress theory.

Other theories for ductile material, Maximum Strain Energy Theory and Maximum Shear Stress Theory (Most Conservative Theory)

For brittle material the most suitable theory is Maximum Principal Stress Theory.

Other theories for brittle material, Maximum Principal Stress Theory.

QUESTION: 5

Which of the following theories of failure is most appropriate for a brittle material?

Solution:

Maximum principal stress theory (Rankine theory) is suitable for brittle materials.

QUESTION: 6

In a structural member, there are perpendicular tensile stresses of 100 N/mm^{2} and 50 N/mm^{2}. What is the equivalent stress in simple tension, according to the maximum principal strain theory? (Poisson’s ratio = 0.25)

Solution:

Equivalent stress

= σ_{1} - μσ_{2}

= 100 - 0.25 x 50 = 87.5 N/mm^{2}

QUESTION: 7

Permissible bending moment in a circular shaft under pure bending is M, according to maximum principal stress theory of failure. According to maximum shear theory of failure, the permissible bending moment in the shaft is

Solution:

According to maximum principal stress theory,

σ_{1} = σ_{y}

According to maximum shear stress theory,

σ_{1} - σ_{2} = σ_{y}

Under pure bending,

Therefore, in both the cases, permissible bending moment is M,

QUESTION: 8

A shaft subjected to pure torsion is to be designed which of the following theories gives the largest diameter of shaft?

Solution:

It is clear from the relation T/J = (Shear Stress)/ (Radial Distance) that a shaft subjected to pure torsion is to be designed for maximum shear stress theory.

QUESTION: 9

If maximum principal stress σ_{1} = 60 N/mm^{2}, σ_{2} and σ_{3} of value zero act on a cube of unit dimensions, then the maximum shear stress energy stored in it would be

Solution:

Shear strain energy

QUESTION: 10

In a 2D stress system, the two principal stress are p_{1} = 180 N/mm^{2} (tensile) and p_{2} (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm^{2} and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p_{2} shall yielding commence?

Solution:

Yielding may occur either in tension or compression.

For yielding in tension,

According to normal strain theory,

For yielding in compression,

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