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Question

If the diameter of a solid steel bar is to be designed for torque T, then the permissible shear stress is considered as:

The correct answer is \(\frac{\text { Yield stress }}{(2 \times \text { Factor of safety })}\)

When designing engineering components, such as a solid steel bar intended to withstand a specific torque T, it is essential to determine the maximum allowable stress the material can safely endure. This is known as the permissible stress or allowable stress. This calculated value ensures the component functions reliably without failure or excessive deformation under operational loads.

Permissible Stress Fundamentals

The concept of permissible stress is a cornerstone in mechanical design. It is derived by considering the material's inherent strength properties and applying a factor of safety. The choice of which material strength property to use (e.g., ultimate stress or yield stress) largely depends on whether the material is ductile or brittle.

Yield Stress vs. Ultimate Stress

  • Yield Stress (\(\sigma_y\)): For ductile materials like steel, the yield stress represents the point at which the material begins to deform plastically. Beyond this point, the material will not return to its original shape after the load is removed. In design, yielding is often considered a failure criterion for ductile materials because permanent deformation can render a component unusable.
  • Ultimate Stress (\(\sigma_{ult}\)): This is the maximum stress a material can withstand before it begins to neck down and eventually fractures. While it represents the material's absolute maximum strength, designing ductile components based on ultimate stress is generally not practical. Significant permanent deformation would occur before reaching this stress, which is unacceptable for most functional designs.

Factor of Safety (FoS)

The factor of safety is a critical design parameter that provides a margin against failure. It is typically defined as the ratio of a material's strength to the actual or expected maximum stress applied during service. For design purposes, it's used to calculate the permissible stress:

\[ \text{Permissible Stress} = \frac{\text{Material Strength}}{\text{Factor of Safety}} \]

A higher factor of safety indicates a more conservative and typically safer design, although it may lead to larger or heavier components.

Permissible Shear Stress Calculation for Steel

For ductile materials such as a solid steel bar, especially when subjected to shear stress caused by torque T, the design calculation for permissible stress is based on the material's yield stress. This is because ductile materials show significant plastic deformation beyond the yield point, which for most engineering applications is considered a failure.

When dealing with shear stress, it's important to understand the relationship between the material's tensile yield stress (\(\sigma_y\)) and its shear yield stress (\(\tau_y\)). According to widely accepted theories of failure for ductile materials (like the Tresca or Von Mises criterion), the shear yield stress is often taken as approximately half of the tensile yield stress:

\[ \tau_y \approx \frac{\sigma_y}{2} \]

Therefore, to determine the permissible shear stress (\(\tau_{perm}\)) for a solid steel bar designed for torque T, we apply the factor of safety to the shear yield stress:

\[ \tau_{perm} = \frac{\text{Shear Yield Stress}}{\text{Factor of Safety}} \]

By substituting the relationship between shear yield stress and tensile yield stress into the formula, we get:

\[ \tau_{perm} = \frac{\left(\frac{\text{Yield Stress}}{2}\right)}{\text{Factor of Safety}} \]

This simplifies to the formula for permissible shear stress:

\[ \tau_{perm} = \frac{\text{Yield Stress}}{(2 \times \text{Factor of safety})} \]

This formula ensures that the design is safe, accounting for both the material's yielding behavior under shear and the necessary safety margin.

Detailed Option Analysis

Let's evaluate each given option against the principles of mechanical design for a solid steel bar under torque T:

  • Option 1: \(\frac{\text { Ultimate stress }}{\text { Factor of safety }}\)
    This option uses ultimate stress. While it represents the material's maximum load-carrying capacity, for ductile materials like steel, designing based on ultimate stress is generally not preferred for service conditions because it would involve significant permanent deformation, which is usually unacceptable.
  • Option 2: \(\frac{\text { Ultimate stress }}{(2 \times \text { Factor of safety) }}\)
    Similar to Option 1, this option also bases the calculation on ultimate stress, which is typically not the chosen design criterion for ductile materials in the context of permissible stress. The factor of 2 is also applied, but the primary basis remains less appropriate.
  • Option 3: \(\frac{\text { Yield stress }}{(2 \times \text { Factor of safety })}\)
    This option correctly uses yield stress as the basis for design, which is appropriate for ductile materials like steel. The factor of '2' in the denominator accurately reflects the approximate relationship between tensile yield stress and shear yield stress (\(\tau_y \approx 0.5 \sigma_y\)), which is essential when calculating permissible shear stress. This formula is consistent with standard engineering practices for torsion design.
  • Option 4: \(\frac{\text { Yield stress }}{\text { Factor of safety }}\)
    While this option correctly uses yield stress, it lacks the factor of '2' in the denominator. This formula would typically be used for calculating permissible normal stress (tensile or compressive), not permissible shear stress, as shear yield strength is generally lower than tensile yield strength for ductile materials.

Considering the material (steel, which is ductile) and the type of stress (shear stress due to torque), the most appropriate formula for permissible shear stress for a solid steel bar designed for torque T is the one that uses yield stress and includes the factor of '2' in the denominator to account for the shear strength relationship.

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Important Questions from Equation of Torsion

  1. If two aluminum bar have a different length (L1 = 2L2) and diameter (d1 = 2d2) with an identical angle of a twist then, find torque value for bar 1, If bar 2 torque value is 50 N-m
  2. A tubular shaft, having an inner diameter of 30 mm and an outer diameter of 40 mm, is to be used to transmit 80 kW of power. The speed of rotation of the shaft so that the shear stress will not exceed 50 MPa is

  3. The magnitude of shear stress induced in a shaft due to applied torque varies from:

  4. Torsional rigidity of a circular section for unit twist and unit length will be:
  5. What is the maximum torque transmitted by a hollow shaft of external radius ‘R’, internal radius ‘r’ and maximum allowable shear stress τ?

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