91Ó°ÊÓ

To understand axial force, think of it as the straightforward stretching or compressing force acting along an object's length. In this exercise, the solid circular bar undergoes an axial force, which results in a strain detected by strain gauge. Strain, denoted by \( \varepsilon \), measures how much an object stretches per unit length.The axial stress, which is the force per unit area, can be calculated using Hooke's Law, linking stress and strain via Young’s modulus, \( E \). Young's modulus is a measure of material stiffness, here provided as 210 GPa for steel. The equation relating these is:\[ \sigma = E \cdot \varepsilon \]Our problem gives two strain readings from gauges: \( \varepsilon_A = 140 \times 10^{-6} \) and \( \varepsilon_B = -60 \times 10^{-6} \). Using these, calculate axial stress assuming symmetry: \( \sigma_x = \sigma_y = \sigma \), so:\[ \varepsilon_A - v \varepsilon_B = \frac{\sigma}{E} \]Solving for \( \sigma \) and substituting into the equation for axial force:\[ P = \sigma \cdot A \]where \( A \) is the cross-sectional area, found using the equation for a circle: \( A = \frac{\pi d^2}{4} \). You substitute \( E \), \( v = 0.29 \); solve for \( \sigma \) and then \( P \). With these, identify the force stretching or compressing the bar.

Torque in a physical system, like the bar in our exercise, depicts a rotational force causing twisting. This can be a bit challenging, as it manifests as shear stress rather than direct length changes.To counsel torque, relate it to shear strain, \( \varepsilon_B \), via the polar moment of inertia, \( J \), and the shear modulus, \( G \). The shear modulus is a ratio similar to Young's modulus but concerning shear stresses and strains:\[ G = \frac{E}{2(1+v)} \]For our exercise, the torque equation is:\[ T = -JG\varepsilon_B \]To assess \( J \), use the equation for a solid cylinder:\[ J = \frac{\pi d^4}{32} \]Substitute the strain value \( \varepsilon_B = -60 \times 10^{-6} \), using the given material properties to find \( G \), and compute \( J \). Plug these into the torque equation to find the twisting force exerted on the bar, expressed in Newton-meters (Nm). This determined torque tells us how much the bar twists under the given conditions, crucial in ensuring the bar’s structural integrity during rotational motion.

Shear stress represents the internal forces that cause layers within a material to slide past one another. In the context of this exercise, the maximum shear stress is pivotal to understanding the bar's response to combined loading conditions.Maximum shear stress, \( \tau_{\max} \), correlates directly to torque \( T \) and is calculated using the formula:\[ \tau_{\max} = \frac{T \cdot c}{J} \]where \( c \) is the radius of the bar (half the diameter). This formula considers torque's tendency to cause twisting at the outermost surface of the bar, where shear is most pronounced.The computation begins with torque \( T \) already derived from previous steps. Remember the polar moment of inertia \( J \) which accounts for the circular cross-section’s resistance to torsion. Insert the calculated torque \( T \) and \( J \) alongside known geometric parameters, like the radius \( c = \frac{d}{2} = 0.016 \text{ m} \). From here, solve to find \( \tau_{\max} \).Understanding maximum shear stress is essential for predicting material failure and ensuring the component operates safely under expected loads, helping design engineers and materials scientists gauge the strength of components to twist or rotational forces.