Problem 34 - An Introduction to the Mechanics of Solids in SI Units

Question

It is proposed to use torsion-spring suspensions for an automobile's front wheels. The conditions under which each steel torsion spring, consisting of a solid, circular shaft, must operate, are as follows: 1\. It must take up the static force at the wheel, \(5 \mathrm{kN}\). 2\. It must provide a spring constant at the wheel of \(25 \mathrm{kN} / \mathrm{m}\). 3\. Deflections of the wheel, up or down, of up to \(15 \mathrm{~cm}\) must be possible without a shear stress exceeding \(350 \mathrm{MPa}\) being set up in the shaft. 4\. The length \(L\) of the shaft should not exceed \(3 \mathrm{~m}\). 5\. The length \(x\) of the arm on the shaft cannot be larger than \(75 \mathrm{~cm}\). Derive a relation between \(x\) and \(L\) and see if you can design a suspension which meets these conditions. What will be the shaft diameter?

Step-by-Step Solution

Answer

After the steps are followed, the specific values for the torque and diameter can be calculated. Afterwards, it can be confirmed whether the torsion spring satisfies the given conditions. The needed relation between \(x\) and \(L\) is given by \(C = k \times x = G \times J / L\) or equivalently \(L = G \times J / (k \times x)\). However, the exact design may depend on the material's shear modulus \(G\), which was not provided.

Step 1: Calculating the Maximum Torque

First, the maximum torque (\(T\)) needs to be calculated that each spring must withstand. This can be determined using the static force at the wheel (\(F\)) and the length of the arm on the shaft (\(x\)). The direct relation is given by \(T = F \times x\). Given that \(F = 5 \mathrm{kN}\) and \(x = 75 \mathrm{cm} = 0.75 \mathrm{m}\), we substitute these values to find the maximum torque.

Step 2: Calculating the Maximum Shear Stress

With the maximum torque calculated, the maximum shear stress (\(τ\)) must be found that would result using the formula: \(τ = T \times r / J\) where \(J\) is the polar moment of inertia for the circular shaft (can be given by \(J = π \times d^4 / 32\)) and \(r\) is the radius of the circular shaft (which is \(d / 2\)). However, it's given that the shear stress should not exceed \(350 \mathrm{MPa}\). Thus, we can rearrange and solve for the diameter \(d\) using the inequality \(τ ≤ T \times r / J\).

Step 3: Satisfying the Spring Constant Condition

Given that the spring constant (\(k\)) at the wheel should be \(25 \mathrm{kN/m}\), one can realize that this is the same as stating the torsional stiffness (\(C\)) of the shaft should be \(C = k \times x\). Torsional stiffness itself is usually given by \(C = G \times J / L\), where \(G\) is the shear modulus of the material. However, \(G\) is not given in the problem, so this condition cannot be directly resolved. But now, one has the relation between \(x\) and \(L\) that was sought.

Step 4: Checking the Length of the Shaft

Lastly, it was stated that the length of the shaft should not exceed \(3 \mathrm{m}\). After calculating all the previous steps, the dimensions of \(L\) found should simply be checked that it does not exceed this limit.

Key Concepts

Shear Stress Calculation

Shear stress is an important factor to consider when dealing with torsion springs, as it reflects the internal forces within the material. Calculating shear stress helps us ensure that the spring can handle forces without failing. For a circular shaft like the one used in our suspension system, shear stress (\(\tau\)) can be derived from the torque (\(T\)) it experiences. The formula used is:
\[\tau = \frac{T \cdot r}{J}\]
where \(r\) is the radius of the shaft and \(J\) is the polar moment of inertia. Knowing the maximum allowable shear stress can help in choosing a suitable diameter for the shaft. In this exercise, the shear stress is limited to 350 \text{MPa}, meaning all stress calculations must adhere to this boundary to prevent mechanical failure.

Torsional Stiffness

Torsional stiffness defines a torsion spring's resistance to twisting under an applied torque, essentially measuring how rigid the spring is. It's denoted by \(C\) and expressed through the relationship:
\[C = \frac{G \cdot J}{L}\]
where \(G\) is the shear modulus of the spring's material, \(J\) is the polar moment of inertia, and \(L\) is the length of the shaft. This parameter is crucial when designing any mechanical system that relies on precise force management. In applications like vehicle suspensions, torsional stiffness must be carefully balanced to ensure the spring is neither too stiff, which can impact ride comfort, nor too soft, which could affect the vehicle's handling.

Polar Moment of Inertia

The polar moment of inertia (\(J\)) is a geometrical property of a shaft that affects its torsional stiffness and reflects how concentrated the material is about its axis of rotation. For solid, circular shafts, it is given by:
\[J = \frac{\pi \cdot d^4}{32}\]
where \(d\) is the diameter of the shaft. In the context of torsion springs, a larger polar moment of inertia signifies a stronger resistance to twisting, thus affecting the shaft's ability to transmit torque efficiently. Calculating \(J\) accurately allows us to predict how a shaft will perform under various loading conditions and is vital in ensuring that the system works as intended.

Spring Constant

The spring constant (\(k\)) is a measure of a spring's ability to resist deformation in response to an applied force, defined in terms of N/m (newtons per meter). In torsion spring applications, it equates to the torsional stiffness (\(C\)), linking the angular displacement with the torque applied. The formula used to establish this relation is:
\[k = \frac{C}{x}\]
where \(x\) is the arm length. This constant is central to ensuring that the torsion spring effectively absorbs shocks and maintains stability, especially in dynamic systems such as vehicle suspensions. A precise spring constant ensures that the system delivers consistent performance, adapting well to varying conditions while safeguarding mechanical components.