Problem 2 - An Introduction to the Mechanics of Solids in SI Units

Question

A hollow steel shaft \(2.5 \mathrm{~m}\) long must transmit a torque of \(25 \mathrm{kN} \cdot \mathrm{m}\). The total angle of twist over the length of the shaft is not to exceed \(2.0^{\circ}\) and the maximum allowable shearing stress is \(82 \mathrm{MN} / \mathrm{m}^{2}\). Find the dimensions of the shaft, i.e., its inside and outside diameters.

Step-by-Step Solution

Answer

To find the dimensions of the hollow steel shaft, key components of information must be plugging into the formula of the moment of inertia and the modulus of rigidity to define two equations with two unknowns; the outer diameter and the inner diameter of the shaft. Completing these calculations will accommodate optimal shearing stress levels and restrict the angle of twist.

Step 1: Calculation of polar second moment of area J

First, rearrange the torque equation to find the value of the moment of inertia (J). The equation becomes \(J=T* R / τ \). Given that \(T = 25 kN.m\), \(R\) is the outer diameter, and \(τ = 82 M N / m^2 \), substitute the known values to calculate \(J\). However, this calculation is dependent on the outer radius which has not been determined yet.

Step 2: Calculation of modulus of rigidity G

We can now use the equation for angle of twist to calculate the modulus of rigidity \(G\). Rearrange the equation to get, \(G = TL / (J∗Θ)\), where \(L = 2.5 m\) and \(Θ = 2−0 degrees = π/90 radians\). Substitute these values into the equation to calculate \(G\). This calculation will also yield an equation dependent on \(J\) which is dependent on the outer diameter \(R\).

Step 3: Calculating the outer diameter

We now have, two equations and two unknowns. By substituting the equation from step 1 into the equation found in step 2, we can solve for the outer diameter \(R\). Then, remember that the diameter \(D = 2R\) to find the outer diameter.

Step 4: Calculating the inner diameter

In a hollow-shaft, J = π/32 * (D_outer^4 - D_inner^4). We know from Step 1 that \(J = T* R / τ \), So now we can arrange for \(D_inner\), to get an equation in terms of \(D_outer\), \(T\) and \(τ\). After finding \(D_inner\), we can determine the specific dimensions of the shaft which will provide safe levels of shearing stress without exceeding the allowable twist angle.

Key Concepts

Polar moment of inertia

In the context of torsional mechanics, the polar moment of inertia, represented by the symbol \( J \), measures a shaft's ability to resist torsion. Imagine it as resistance against twisting forces around the axis of the shaft. The higher the polar moment of inertia, the better the shaft can handle applied torque without wrapping too much or breaking.

For a solid circular shaft, the formula to calculate \( J \) is quite straightforward: \( J = \frac{\pi}{32}D^4 \), where \( D \) is the diameter of the shaft. For hollow shafts, like the one in our example, the formula adapts to:

• \( J = \frac{\pi}{32}(D_{\text{outer}}^4 - D_{\text{inner}}^4) \)

Here, \( D_{\text{outer}} \) and \( D_{\text{inner}} \) are the outer and inner diameters, respectively.

Knowing \( J \) helps engineers in designing shafts that can withstand required operational constraints like torque without exceeding material stress limits.

Shearing stress

Shearing stress, denoted by \( \tau \), is a measure of the internal forces acting parallel to the surface of a material. In torsional mechanics, it's especially important because twisting is essentially a force trying to "slice" through the material by causing the layers to slide over each other.

In the exercise, the maximum allowable shearing stress is given as \( 82 \text{ MN/m}^2 \). This value is crucial as it determines how strong and thick the shaft's walls need to be to safely transmit the given torque without failing.

• To calculate the shearing stress in a circular shaft, the formula used is:
• \( \tau = \frac{T \cdot c}{J}\)

Where: - \( T \) is the applied torque - \( c \) is the distance from the center of the shaft to the outermost fiber (often the outer radius \( R \))This equation shows that shearing stress increases with higher torque or when the diameter, thus the polar moment of inertia, is smaller.

Modulus of rigidity

The modulus of rigidity, sometimes known as the shear modulus and denoted by \( G \), measures a material's stiffness or its resistance to shear deformation.Simply put, it quantifies how much a material resists changes in shape when a force is applied.

In the problem described, the modulus of rigidity helps calculate the angle of twist.This relationship is expressed in an equation where the torsional response of the shaft is tied to both the material properties and the physical dimensions through:

• \( \theta = \frac{T \cdot L}{J \cdot G} \)

Where:- \( \theta \) is the angle of twist in radians,- \( T \) is the applied torque,- \( L \) is the length of the shaft,- \( J \) is as previously mentioned the polar moment of inertia, - and \( G \) is the modulus of rigidity.By rearranging this formula, one can solve for \( G \), or use a known \( G \) to find permissible twisting angles or dimensions.

These principles ensure mechanical components remain within safe operational limits, optimizing both strength and functionality.