Popular posts from this blog
Solved Numerical Problems from Boiler Operation Engineer Exams
Boiler Mechanics Problem (Jharkhand BOE 2024) A solid round bar 3 m long 5cm in diameter is used as a strut with both ends hinged. Determine the crippling load. Take E=2 x 10 5 N/mm 2 . Solution Moment of inertia \(I = \frac{\pi }{{64}}x{5^4} = 30.68\,c{m^4} = 30.68x{10^4}\,m{m^4}\) Both ends of the bar are hinged. Crippling load for this condition is \(P = \frac{{{\pi ^2}EI}}{{{l^2}}} = \frac{{{\pi ^2}x2x{{10}^5}x30.68x{{10}^4}}}{{{{3000}^2}}}\) = 67288 N = 67.288 kN These questions are taken from study material for Boiler Operation Engineer Exam by amiestudycircle.com . Problem (Jharkhand BOE 2022, 3 marks) A hollow shaft is to transmit 300 kW at 80rpm. If shear stress is not to exceed 60 N/mm 2 a diameter is 0.6 of the external diameter, find external and internal ...
Geotechnical & Foundation Engineering (Solved Numerical Problems)
Numerical A 1000 cc core cutter weighing 946.80 g was used to find out the in-situ unit weight of an embankment. The weight of the core cutter filled with soil was noted to be 2770.60 g. Laboratory tests on the sample indicated a water content of 10.45 % and specific gravity of solids of 2.65. Determine the bulk unit weight, dry unit weight, void ratio, and degree of saturation of the sample. (AMIE Summer 2023, 8 marks) Solution Weight of soil in core cutter = 2770.60 - 946.80 = 1823.8 g Bulk unit weight, γ t = W/V = 1823.8/1000 = 1.82 g/ccDry unit weight, γ d = γ t /(1 + w) = 1.82/(1 + 0.1045) = 1.65 g/cc Void ratio, e = (G s γ w /γ d ) - 1 = (2.65 x 1.0/1.65) - 1 = 0.61 Degree of saturation, S = wG s /e = (0.1045 x 2.65)/0.61 = 0.4540 = 45.4% Numerical What is the theoretical height of the capillary rise and the capillary pressure in fine-grained soil with an effective size (D 10 ) of 0.002 mm? (AMIE Summer 2023, 4 marks) Solution D 10 = 0.002 mm; Using the assumption that th...
Comments