From the stress transformations, an equation in the form of a circle can be derived as follows: $$ [\sigma _ \theta - \frac{\sigma _x + \sigma _y}{2}]^2 + \tau _ \theta ^2 = ( \frac {\sigma _x - \sigma _y}{2}) + \tau _{xy} ^2 $$ Where the centre of the circle is \( ( \frac{ \sigma _x

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Uniaxial Stress A solid beam under tensile stress will eventually fracture due to a shearing effect in an internal plane. Here, \( \sigma _x \) is the tensile stress acting in the same axis as the beam, \( \tau _ \theta \) is the resultant shear stress acting at an angle \( \theta \) to the vertical

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To calculate the shear stress in the beam at point \( P \), given that the beam is subject to a resultant internal vertical shear force of \( V = 3 kN \). The second moment of area of the cross-section of the beam about the neutral axis: $$ I = \frac{bd^3}{12} = \frac{1}

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Stress ( \( \sigma \) ): Internal force per cross-sectional area $$ \sigma = \frac{F}{A} $$ Strain ( \( \epsilon \) ): Ratio of elongation to original $$ \epsilon = \frac{\Delta L}{L} $$ Thermal Strain ( \( \epsilon _T \) ), Thermal Stress ( \( \sigma _T \) ): $$ \epsilon _T = \alpha \Delta L $$ $$ \sigma _T = E \alpha \Delta L $$ Where \( \alpha \) is the linear coefficient of thermal

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