The value of stress produced due to 30 N is 44 N/mm2. Determine the cross sectional area of rod
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- Find the value of stress Force=2 N Area 2 mm2A steel rod with a cross sectional area of 120〖mm〗^2 is stretched between two fixed points. The tensile load at 20°C is 4200N. What will be the stress at-25°C? α=11.7x〖10〗^(-6)/°C and E = 200GPa. See figure 062. For the given state of stress shown in figure 2. 48 MPa Figure 2. a. Construct Mohr's circle for this plane stress. ||₁6 16 MPa 60 MPa Determine the principal stresses and principal planes.
- Ox = 45 Mpa, Oy = -15 Mpa, τxy(shear stress) = 20 Mpa, theta = 25 degreesConsider the state of stress as shown in figure. What will be the principal stresses? 02= 16Mpa oy= – 48 M pa, Try = 60 Mpa T 48 Mpa + 16 Mpa 60 MpaConsider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. Assume'stress magnitudes of 5,-80 MPa, S - 220 MPa. Sy For the given state of stress at a point in a structural member, determine the center Cand the radius R of Mohr's circle. Answers: C- MPa, R- i MPa. Save for Later Part 3 Determine the principal stresses (api > Op2 ) for this state of stress. Answers: Opl = MPa, o,2 = MPa Save for Later Part 4 For this Mohr's circle, determine the magnitude of the central angle between: (a) pointxand the point representing the principal stress opl- (b) point x and the point representing the principal stress Op2. (a) Angle - i (b) Angle - i
- (a) 128.5 GPa circular rod of diameter 30 mm and length 200 mm is subjected to a tensile force. The axtension of rod is 0.09 mm and change in diameter is 0.0045 mm. What is the Poisson's ratio of the material of the rod? (a) 0.35 (b) 0.33 (c) 0.32 (d) 0.302. For the state of stress shown, determine the range of value of for which the normal is equal to or less than 100 MPa and 50 MPa. stress x' + 90 MPa 60 MPa =۱۰:۱۲ ६६ docs.google.com The state of plane stress at a point is represented on the element shown in the left figure. The element oriented to the slope and direction shown in the right figure. Find the value of maximum principle stress o2 15 MPa - 5 MPa +17.81 Mpa Tension +17.81 Mpa Compression +2.19 Mpa Tension +2.19 Mpa Compression -2.19 Mpa Tension _17.81 Mpa Compression _2.19 Mpa Compression _17.81 Mpa Tension 6 MPa
- 10.Stress transformation. From the figure below, find the orientation of the maximum principal stresses, the magnitudes of the maximum principal stresses and the maximum shear stress. All formulae given at the end of the exam. 2/22 27200005 27200032 10 MPa 40 MPa 50 MPa3. The bar shown in Figure Q4 is subjected to tensile load of 160 kN. IIthe stress is the middle portion is limited to 150 N/mm', determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the har is to be (0.2 mm. Young's modulus is given as equal to 2. 1x 10 N/ mm. 160 EN 40 cm Figure 04A bronze bar 3 m long with a cross-sectional area of 320 mm² is placed between two rigid walls as shown in the figure below. At a temperature of -20°C, the gap A = 2.5 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use a = 18.0 x 10-6 m/(m-°C) and E = 80 GPa.
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