Question 1.3: Figure 1.7, where the dimensions are in mm., shows part of t......
Figure 1.7, where the dimensions are in mm., shows part of the spring suspension unit of a small road vehicle. It consists of a circular-section torsion bar, and rigid lever. The effective length of the torsion bar, L_{1}, is 900 mm, and that of the lever, L_{2}, is 300 mm. The bar is made of steel, having elastic shear modulus G=90×10^{9} N/m² ( = 90 GN/m² or 90 GPa), and its diameter, d, is 20 mm. Find the vertical stiffness, for small displacements, as measured at point B.
From Eq. (1.28),
k_{\varphi }=\frac{\delta T}{\delta \varphi }=\frac{GJ}{L} (1.28)
the torsional stiffness, k_{\varphi }, of the bar between point A and the fixed end, is
k_{\varphi }=\frac{\delta T}{\delta \varphi }=\frac{GJ}{L_{1}} (A)
where \delta T is a small change in torque applied to the bar at A and \delta \varphi the corresponding twist angle. G is the shear modulus of the material, J the polar moment of inertia of the bar cross-section and L_{1} the length of the torsion bar. However, we require the linear stiffness k_{x} or \delta F/\delta x, as seen at point B. Using Eq. (A), and the relationships, valid for small angles, that \delta T = \delta F . L_{2}, and \delta \varphi = \delta x / L_{2}, it is easily shown that
k_{\varphi }=\frac{\delta F}{\delta x}=\frac{\delta T}{\delta \varphi } \cdot \frac{1}{L^{2}_{2} }=\frac{GJ}{L_{1}L^{2}_{2} } = k_{\varphi }\cdot \frac{1}{L^{2}_{2}} (B)
Numerically, G = 90 × 10^{9} N/m² ; J = \pi d^{4}/32 = \pi (0.02)^{4}/32 = 15.7 \times 10^{-9} m^{4} ; L_{1} = 0.90 m ; L_{2} = 0.30 m giving k_{x} = 17 400 N/m or 17.4 k N/m.