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Engineering Exams RRB-JE Civil and Allied Engineering Technical Abilities 2019 19-Sept-2019 Paper-2 Shift-1 Medium +1 -0.33
At a point in a steel member, major and minor principal stress in 1000 kg/cm2, and minor principal stress is compressive. If uniaxial tensile yield stress is 1500kg/cm2,then magnitude of minor principal stress at which yielding will commence, according to maximum shearing stress theory is-
Correct Answer: B. 500
Explanation: According to the Maximum Shear Stress Theory (also known as Tresca's Criterion), yielding starts when the maximum shear stress in a material reaches the maximum shear stress observed at the yield point in a simple tension test. For a 2D stress state with principal stresses σ1 and σ2, the maximum shear stress is calculated as: τmax = (σ1 - σ2) / 2 In a uniaxial tension test at the point of yielding, the maximum shear stress is: τyield = σy / 2 Where σy is the uniaxial tensile yield stress. Equating these two: (σ1 - σ2) / 2 = σy / 2 σ1 - σ2 = σy Given: Major principal stress (σ1) = 1000 kg/cm2 Uniaxial tensile yield stress (σy) = 1500 kg/cm2 Minor principal stress (σ2) is compressive, meaning its value will be negative in the equation. Plugging in the values: 1000 - (-σ2) = 1500 1000 + σ2 = 1500 σ2 = 1500 - 1000 σ2 = 500 kg/cm2 Thus, the magnitude of the minor principal stress at which yielding will commence is 500 kg/cm2.
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