The state of stress at a certain point in a stressed body is shown in figure. The magnitudes of normal stresses in x & y directions are 100 MPa& 100 MPa respectively. Determine the radius of Mohr's circle representing this state of stress (in MPa). Oxt Oy Ox
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- The stresses at a point in a bar are o1 = 28N/mm2 (tensile) and o2=12N/mm2 (compressive) as shown in Figure (a) Determine the normal, tangential and resultant stress in magnitude and direction on a plane inclined at 55° to the axis of the major stress. (b) Determine the maximum intensity of shear stress in the material at the point. (c) Also, solve the same problem using Mohr's circle method and compare the results. O2 Axis of Major Stress 55°1) 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 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 state of stress is specified in the figure shown. Determine the principal stress oy (mPa) if P =-20 mPa and y = 15 mPa P VA state of stress is specified in the figure shown below. Determine the Normal stress o (mPa) on the Principal Plane if P1 = -63 mPa; P2 = -80 mPa and v= 54 mPa Py PxThe stresses shown act at a point on the free surface of a machine component. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are 5x = 19.6 ksi, 5y = 4.3 ksi, and Sxy= 7.3 ksi. Assume ß = 38°. Determine the normal stresses ox and 6, and the shear stress Txy at the point. Say Sx Txy Answers: 0x = oy = Txy || ox ksi. ksi. ksi.
- A thin rectangular polymer plate PQRS of width b = 361 mmand height = 193 mm is shown in the figure. The plate is deformed so that corner Q is displaced upward by c = 3.1 mm and corner R is displaced leftward by the same amount. Determine the shear strain at corner P after deformation. a Y = b S a uradox 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 11Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on the horizontal and vertical planes at the point are shown in the figure. Note that the given stress values are absolute values and stress sign should be determined based on the figure. [o 107 MPa, o = 172 MPa, t= 149 MPa] Oy Try Determine the principal normal stress a A8T 7 MMP
- A state of stress is specified in the figure shown. Determine the principal stress oy (mPa) if P =-27 mPa and v = 20 mPa P2. 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 =2.23. An aluminum right-circular cylinder surrounds a steel cylinder as shown in Fig. 2-40. The axial compressive load of 200 kN is applied through the infinitely rigid cover plate shown. If the aluminum cylinder is originally 0.25 mm longer than the steel before any load is applied, find the normal stress in each when the temperature has dropped 20 K and the entire load is acting. For steel take E = 200 GPa: a = 12 × 10-*/°C, and for aluminum assume E = 70 GPa, a = 25 x 10° */°C. Ans. σ =9MPa, σ = 15.5 MPa 200 kN 80 mm- -85 mm 141 mm Fig. 2-40 500 mm