Verify the equality of the following shearing stresses: (2) σεν = σνε (b) σzz =σ22
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- The distribution of stress in an isotropic aluminium machine component is given (in MPa) as: σ₂ = y +2z² - 6 σ₁ = x+z-6 oy σ₂ = 3x+y-13 6: T =3z² -11 xy (i) (ii) =x²-14 Tyz Txz = y² XZ x, y and z are coordinates of a point within the machine component. By taking Young's modulus, E = 70 GPa, Poisson ratio, v= 0.3 and yield stress, Y = 5 MPa, do the following for a point P located at (4, 1, 2): a) Provide the stress and strain tensors. b) Determine all the principal stresses and principal strains. c) Determine if the machine component will fail based on the failure criteria below: Tresca criterion Von Mises criterionThe shear stress is maximum on the principal plane Select one: O True O FalseFollowing plate is connected to the wall using 7 bolts. Cross sectional area of each bolt is A. Force F is applied to the plate. The magnitude of shear stress experienced by bolt B is, 100.00 25.00 25.00 25.00 O 1. Less than F/(7A) O 2. F/(7A) O3. More than F/(7A) 80.00 20.00 20.00 20.00 (40 00)
- For a steel rod with a circular cross section with a diameter of D = 20 mm, the following is required: 1. Draw a diagram of the longitudinal force 2. Draw a diagram of normal stresses 3. Determine the total elongation of the rod if Е = 2∙10^5 MPa When calculating, take: а = 2 m, b =1.2 m, F=10 кN - The point of application of force! The work must be done on one sheet of A4 paper, which must show: - Using the method of sections to determine the longitudinal forces in the rod. - Draw a diagram of longitudinal forces N - Determination of normal stresses based on the constructed plot of longitudinal forces. - Draw diagram a normal stresses - Determination of the full extension of the rod8 x = 17 MPa y = 41 MPa T = 56 MPa 5 T σχ X At a point on a structural member subjected to plane stress, normal and shear stresses exist on horizontal and vertical planes through a point as shown. Use the stress transformation equations to determine the normal stress to the nearest 0.01 MPa on the indicated plane. Be sure to use - as appropriate. 277Determine the resulting maximum value of the normal stress. Specify the orientation of the plane on which these maximum values occur. **The answer is tensile stress is 0 ksi at 90 degrees. **The answer is compressive stress is 7 ksi at 0 degrees. Can you explain how that is? This was my thought process: I know that tensile would be zero because the force P is actually going inwards and not outwards. I know that means that there would be a compressive force. I am confused on the angles, how is a tensile force going 90 degress if there technically is no force in the tensile direction. And how is there a compressive force at 90 degrees if there is a stress? thank you!
- At a point on the surface of a pressurized cylinder, the material element is subjected to stresses; σx = 60 MPa, σy = - 40 MPa, and τxy = - 30 MPa. Construct Mohr’s circle, then use it to determine the following: Do not use the equations of transformations.1- The principal stresses and the maximum shear stress.2- Show these stresses on sketches of properly stress elements.3- A point on the circle where the element is only subjected to pure shear stress. Calculate the value of that pure shear stress.A carbon steel ball with 27-mm diameter is pressed together with an aluminum ball with a 35-mm diameter by a force of 11 N. Determine the maximum shear stress and the depth at which it will occur for the aluminum ball. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. Ratio of stress to Pma σ. T 1.0 0.8 0.6 0.4 0.2 0 J₂, J₂ 0.5a σ₂ Tmax a 2a Distance from contact surface 1.5a 2.5a 3a The maximum shear stress is determined to be 134.6 Z MPa. The depth in the aluminum ball at which the maximum shear stress will occur is determined to be 0.0519 mm.2. determine the maximum normal stress and the maximum shear stress at point O in dia. B 15 in 7 in F = 15 lbf
- Consider the given state of stress. Take X = 35 MPa and Y = 50 MPa Determine the orientation of the planes of maximum in-plane shearing stress in the first and third quadrantsAt a point in the cross-section of an engineering component an element is subject to the following stresses: σx =100MPa σy =45MPa τxy = -50MPa Construct a Mohr’s Stress Circle and hence determine: (a) The Principal Stresses (σ1 & σ2) and the Directions of the Principal Planes (φ1 & φ2) (b) The Maximum Shear Stress (τmax) (c) Find the Normal Stress (σn) acting on a plane at 40 degs anticlockwise from the y-axis2. A prismatic bar under an axial load produces a tensile stress of 75MPA on a plane at an angle of 0 = 20°, as seen below. Determine the normal and shearing stress on all faces of an element at an angle of e = 50°. 75 MPa 20 deg