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- An oil pipeline and a 1.200 m^3 rigid air tank are connected to each other by a manometer, as shown in the figure. The tank contains 15 kg of air at 80°C. Assume the pressure in the oil pipeline to remain constant and the air volume in the manometer to be negligible relative to the volume of the tank. Determine the change in Δh when the temperature in the tank drops to 20°C.arrow_forwardCalculate the collapse load (P) for the two fixed ended beam shown below. Use virtual work method P 2 m 4 m L= 6 marrow_forwardFind the collapse load (Wu) for the one-end continuous beam shown below. Use virtual work method Wu 6 marrow_forward
- Find the maximum distributed load can be applied to the two fixed ends beam shown below. Use Virtual work method Wu L=6marrow_forwardCalculate the collapse load (P) for the two fixed ended beam shown below. Use virtual work method P 2 m 4 m L=6marrow_forwardQuestion 1 (Approximate Method - Superposition). Using Superposition determine the displacement at C of this beam. El is constant. (Note - you must use the PE Handbook Shears, Moments and Deflection Tables. The FE handbook does not have one of these conditions) (On an exam I will make sure it is found in both the FE and the PE handbook). 60 kN 30 kN/m A C 3 m 3 m B Question 2 (Slope and Deflection - virtual work - statically determinate beam) Using virtual work determine the slope at A and the displacement at C of this beam. El is constant. Same beam as question #1arrow_forward
- Question 4 (Force Method). Determine the reaction at the supports. Assume A is a pin and B and C are rollers. El is constant. 1.5 k A A 10 ft 10 ft B C - 20 ftarrow_forwardFind the maximum load (collapse load) that can be carried by the simply supported beam shown below. P ↓ 3 m 3 marrow_forwardFind the maximum distributed load can be applied to the two fixed ends beam shown below. Wu L=6marrow_forward
- In excavation for a wall footing, the water table level was lowered from a depth of 1.0 m to a depth of 3.0 m in a clayey soil deposit. Considering that the soil has a water content of 28% when it is fully saturated, and above the water table the (dry) unit weight of the soil is 17 kN/m³. Assuming initially that all of the soil above the water table is dry, then compute the following: 1. The effective stress at a depth of 4.0 m after the lowering of the water table. Take Gs = 2.68. (Hints: w*Gs=Sr*e) 2. The increase in effective stress at a depth of 5 m. (You also need to plot the values of total vertical stress and effective vertical stress against depth before and after lowering the water table.)arrow_forwardCalculate the collapse load (P) for the two fixed ended beam shown below. P 2 m 4 m L=6marrow_forwardThe vertical stress at a point in soil is σx =400 kN/m², Txz = 50kN/m² while the horizontal stress at the same point is σ =100 kN/m², Tzx = -50kN/m². (a) Draw the Mohr circle that describes the 2D stress state at the point. (b) Find the maximum shear stress that acts at the point and its orientation angle from the horizontal plane. (c) Find the principal stress (σ₁ and σ3) that acts at the point and locate the major principal stress plane and its orientation angle from the horizontal plane (Use the pole method). (d) Determine both the normal and shear stress at a plane that orientates from the major principal stress plane with an angle of 30° (counterclockwise direction) and verify your results with the stress transformation equations.arrow_forward
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