The state of stress at a point in an elastic material, with yield stress of 200 MPa in simple tension and Poisson's ratio 0.3, is as shown in the figure 1.40 20 20 1.40 The permissible value of o by maximum strain theory is
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- A body is subjected to stresses as shown in the figure given below. 90 MPa 35 MPa 145 MPa The FOS as per maximum principle strain theory if the yield stress of the material is 230 MPa and Poisson ratio of material is 0.38. (assume plane stress condition)The state of stress at a point in an elastic material, with yield stress of 200 MPa in simple tension and Poisson's ratio 0.3, is as shown in the figure 1.40+ 20 120 →1.40 The permissible value of o by maximum strain theory isa L₁ C 3 of 6 > Determine the orientation of the planes maximum and minimum normal stress, ₂1 and ₂2. respectively. Use the value for 8,2 that is negative. Then use this value to determine the value of 8,1. Express your answers to four significant figures separated by a comma. ► View Available Hint(s) VAΣo↓ vec ? 8p1, 8p2= Submit Part B - In-plane principal stresses Determine the in-plane principal stresses for the element (as oriented in Part A), and calculate 01 and 2 such that they correspond with 81 and 8,2. Express your answers to four significant figures separated by a comma. ► View Available Hint(s) IVE ΑΣΦ vec ? 5 O
- The major and minor principal strains in a 2D stressed element are 70μ and -25µ respectively. If the normal strain at a plane is given as 15µ, then what will be the normal strain on a plane perpendicular to it? A B d C D 40μ 30μ 25μ 20μ20 and σ are the principal stresses at a point in a strained material. Poisson ratio for the material is 0.25. Modulus of elasticity of the material is 2.20. Total strain energy stored per unit volume of the material is ko. Then the value of k is (Round off to two decimal places.)State of stress at a point is shown in figure. If the normal stress at inclined plane marked in the figure is 100 MPa tensile, the value of 'sigma' is
- A state of stress is specified in the figure shown. Determine the principal stress oy (mPa) if P =-27 mPa and v = 20 mPa P1) The state of plane stress at a point is shown on the element in Figure. Determine the maximum in-plane shear stress at this point. B 50 MPa ox Ox= -80 MPa = +50 MPa Oy Txy = -25 MPa 80 MPa 80 MPa A 25 MPa Ľ 50 MPaThe thin bar is subjected to an increase in temperature along its axis. Determine the end B of the bar due to the increase in temperature y; b) the average normal strain in the bar.The normal strain is 70X10^(-2) z^1/3, where z is expressed in meters. L = 300 mm.
- ox The state of stress at a given point in a material component is = 120 MPa; Oy = - 80 MPa and t = 60 MPa The loading on the component is increased so that stresses are increased to values which are k times the given values. Determine the maximum value of k if the material can withstand maximum normal and shear stresses of 300 MPa and 200 MPa respectively. 1 11The state of stress at a certain point in a stressed body is as shown in the figure. Normal stress in x-direction is 80 MPa (Tensile) and in y-direction is 40 MPa (Compressive). The radius of the Mohr's circle for this state of stress will be O, = 40 MPa O, = 80 MPa + Ox = 80 MPa Oy = 40 MPa (а) 60 MPа (b) 40 MPa (c) 20 MPa (d) 10 MPa14. For a linear elastic material with Young's modulus E=20 MPa: a) Calculate the amount of longitudinal strain ewhen a specimen made of this material is subjected to uniaxial tension at the stress level o =6 КРа. b) Draw the stress-strain diagrams corresponding to the respective loading protocols e1(t) and ɛ 2(t) given in the figure below Loading e (1) Loading e:(t) Ei (t) (1) 0.04 10 20 t(s)