91Ó°ÊÓ

An aluminum bar of solid circular cross section is twisted by torques \(T\) acting at the ends (see figure). The dimensions and shear modulus of elasticity are as follows: \(L=1.4 \mathrm{m}, d=32 \mathrm{mm},\) and \(G=28 \mathrm{GPa}\) (a) Determine the torsional stiffness of the bar. (b) If the angle of twist of the bar is \(5^{\circ},\) what is the maximum shear stress? What is the maximum shear strain (in radians)? (c) If a hole of diameter \(d / 2\) is drilled longitudinally through the bar, what is the ratio of the torsional stiff- nesses of the hollow and solid bars? What is the ratio of their maximum shear stresses if both are acted on by the same torque? (d) If the hole diameter remains at \(d / 2,\) what new outside diameter \(d_{2}\) will result in equal stiffnesses of the hollow and solid bars?

Step by step solution

01

Torsional Stiffness for Solid Bar

The torsional stiffness \(k_t\) of a solid circular shaft is given by the formula \[ k_t = \frac{GJ}{L} \]where:\( G \) is the shear modulus of elasticity,\( J \) is the polar moment of inertia,\( L \) is the length.For a solid circular cross-section, the polar moment of inertia \( J \) is:\[ J = \frac{\pi d^4}{32} \]Substitute \( d = 32 \text{ mm} = 0.032 \text{ m} \), \( G = 28 \text{ GPa} = 28 \times 10^9 \text{ Pa} \), and \( L = 1.4 \text{ m} \) into these formulas to find \( J \) and then \( k_t \).

02

Stress and Strain for Solid Bar

The angle of twist \( \theta \) is given as \( 5^\circ \), which needs to be converted to radians.1 \( \text{degree} = \frac{\pi}{180} \text{ radians} \), so:\[ \theta = 5 \times \frac{\pi}{180} \text{ radians} \] Maximum shear stress \( \tau_{max} \) in the shaft is given by:\[ \tau_{max} = \frac{T c}{J} \]The torque \( T \) can be calculated using:\[ T = \frac{k_t \theta}{L} \]The maximum shear strain \( \gamma_{max} \) is:\[ \gamma_{max} = \frac{\tau_{max}}{G} \]Substitute the values to find both \( \tau_{max} \) and \( \gamma_{max} \).

03

Torsional Stiffness Comparison for Hollow and Solid Bars

For a hollow bar with an outer diameter \( d \) and inner diameter \( d/2 \), the polar moment of inertia \( J_h \) is:\[ J_h = \frac{\pi}{32} \left( d^4 - \left(\frac{d}{2}\right)^4 \right) \]Calculate \( J_h \), then the torsional stiffness of the hollow bar \( k_{t,h} \) is:\[ k_{t,h} = \frac{GJ_h}{L} \]Now compute the ratio \( \frac{k_{t,h}}{k_t} \) to find the stiffness ratio between the hollow and solid bars.

04

Shear Stress Ratio for Same Torque

When both bars are subjected to the same torque \( T \), the maximum shear stress ratio can be found by:For the hollow bar:\[ \tau_{max,h} = \frac{T \cdot \frac{d}{2}}{J_h} \]And similarly for the solid bar:\[ \tau_{max,s} = \frac{T \cdot \frac{d}{2}}{J} \]Calculate the ratio \( \frac{\tau_{max,h}}{\tau_{max,s}} \).

05

Outer Diameter for Equal Stiffnesses

To find the new outer diameter \( d_2 \) such that the torsional stiffnesses of hollow and solid bars are equal, equate the expressions of torsional stiffness \( k_{t,h} = k_t \).This implies:\[ \frac{\pi}{32} \left( d_2^4 - \left( \frac{d}{2} \right)^4 \right) = \frac{\pi d^4}{32} \]Solve for \( d_2 \) from this equation. Use algebraic manipulation to isolate and solve for \( d_2 \) in terms of \( d \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Moment of Inertia

The polar moment of inertia, denoted as \( J \), is a crucial factor in understanding the torsional properties of a shaft. It represents the shaft's resistance to twisting and significantly influences its torsional stiffness. For a solid circular cross-section, the formula is given by:\[ J = \frac{\pi d^4}{32} \]where \( d \) is the diameter of the shaft. Substituting the diameter into the formula lets us calculate \( J \), indicating how uniformly or variably the shaft will twist under applied torque.

• The larger the value of \( J \), the more resistant the shaft is to twisting.
• Hence, a shaft with a larger diameter will have a higher polar moment of inertia.

Understanding the polar moment of inertia helps in designing shafts that can endure higher torques without excessive twisting or failure.

Shear Modulus

The shear modulus, represented by \( G \), measures a material's rigidity and ability to endure shear forces. It's analogous to the elastic modulus, but specifically for shear. For aluminum, as in the exercise, \( G = 28 \text{ GPa} = 28 \times 10^9 \text{ Pa} \).

• Higher shear modulus values indicate stiffer materials that deform less under shear stress.
• When coupled with the polar moment of inertia, \( G \) helps calculate the torsional stiffness \( k_t \) using the equation:\[ k_t = \frac{GJ}{L} \]

By using this modulus, we deploy materials effectively, ensuring they can handle expected stresses without reaching their limits.

Maximum Shear Stress

Maximum shear stress, symbolized by \( \tau_{max} \), represents the highest amount of shear force a material can withstand across its cross-section. In our problem, it's expressed as:\[ \tau_{max} = \frac{T c}{J} \]Here, \( T \) is the torque, \( c \) is the radius of the shaft, and \( J \) is the polar moment of inertia. Properly computing this ensures that the material doesn't succumb to excessive stress, leading to potential failure.

• This parameter is critical for ensuring that designs can withstand operational conditions.Using the correct \( \tau_{max} \) helps assess whether the design will hold or require adjustments to material choice or dimensions.

It's vital when designing components to ensure the material's strength is not exceeded when maximum operational stress conditions are applied.

Shear Strain

Shear strain, indicated by \( \gamma \), describes the amount of deformation a material undergoes when subjected to shear stress. It measures how much the material shape changes without changing the volume, usually resulting from twisting or sliding forces.

• In the context of the exercise, maximum shear strain \( \gamma_{max} \) can be calculated as:\[ \gamma_{max} = \frac{\tau_{max}}{G} \]
• A lower shear strain indicates that the material maintains its shape better under stress.

This measurement is crucial in applications where maintaining structural integrity and minimum tolerance changes are important for operational efficiency and safety.