Learning Goal: To calculate the principal stresses and maximum in-plane shear stress for a plane state of stress. The normal and shear stresses for a state of stress depend on the orientation of the axes. For a general plane state of stress, there is one orientation where the two normal stresses are a maximum and minimum and the shear stresses on the faces of the element are zero. The two normal stresses 01 and 2 are known as principal stresses. The directions of the faces on which the principal stresses act are known as the principal planes. The principal stresses can be calculated using the following formula: 01,2 = σI + Oy 2 Tmax= ± OT The maximum in-plane shear stress occurs on a plane that is 45° away from the principal planes. The average normal stress on this plane is O₂ + Oy generally not zero, but is avg = . The 2 maximum shear stress is given by the following equation: 2 JI Ty 2 2 2 tiểu + Ty Once avg and Tmax are calculated, the principal stresses can be found very directly: 1,2 = avg Tmax. The state of stress at a point can be described by |0₂| = 22 MPa, oy = 52 MPa, and Try 9 MPa, acting in the directions shown (Figure 1). Part A - In-plane shear stress What is the maximum in-plane shear stress for the given state of stress? Express your answer with appropriate units to three significant figures. ► View Available Hint(s) Tmax in-plane= Submit UA Value Units Part B - Maximum normal stress magnitude uA What is the principal normal stress with the largest magnitude? Express your answer with appropriate units to three significant figures. ► View Available Hint(s) ? ?

1- Water vapor at a flow rate of 20000 kg / hour enters a steam boiler with a specific volume of 150 kPa 0.001053 m3 / kg. With the energy it takes from the fuel in the combustion chamber of the boiler, it rises at 150˚C at 1200 kPa. If the boiler has poor insulation and loses 100 kW of heat to the environment, calculate the time required to increase 1 ton of water to 150 ° C in the boiler. A. 347 sB. 3752 sC. 1651 sD. 648 sE. 2494 s

Solve fast and accurate,it needs attention

When a beam section is subjected to a shear load, a shear stress distribution is developed on the section. The distribution of the shear stress is not linear. Elasticity theory can be used to calculate the shear stress at any point. However, a simpler method can be used to calculate the average shear stress across the width of the section, a distance y above or below the neutral axis. The average VQ shear stress is given by T = Here V is the shear It force on the section, I is the moment of inertia of the entire section about the neutral axis, and t is the width of the section at the distance y where the shear stress is being calculated. Q is the product of the area of the section above (or below) y and the distance from the neutral axis to the centroid of that area (Figure 1). In short, Q is the moment of the area about the neutral axis. I Figure 1 of 1 An I-beam has a flange width b = 250 mm, height h = 250 mm, web thickness tw = 9 mm, and flange thickness tf = 14 mm. Use the…