Determine the maximum shearing stress and elongation in a bronze helical spring composed of 20 turns of 1.0-in.-diameter wire on a mean radius of 4 in. when the spring is supporting a load of 500 lb. Use Eq. (3-10) and G = 6 x 10° psi.
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- , Solve the preceding problem using the numerical data: /) = 90mm, h = 280 mm, d = 210 mm, q = 14 kN/m, and L = L2 m.Two steel springs arranged in series support a load p the upper spring has 12 turns of 25 mm diameter wire on a mean radius of 100 mm the lower spring consist of 10 turns of 20 mm diameter wire on a mean radius of 75 mm use G=83GPA Compute A. the value of the total elongation of the assembly B.the most nearly gives the maximum value of pi if it the allowable shearing stress is 200 mpa c.which of the most nearly gives the equivalent spring constant assemblyTwo steel springs arranged in series supports a load P. The upper spring has 12 turns of 20-mm-diameter wire on a mean radius of 100 mm. The lower spring consists of 10 turns of 25-mmdiameter wire on a mean radius of 75 mm. If the maximum shearing stress in either spring must not exceed 300 MPa, compute the maximum value of P and the total elongation of the assembly. Use approximation formula and G = 83 GPa. Compute the equivalent spring constant by dividing the load by the total elongation.
- As shown below, a homogeneous 50-kg rigid block is suspended by the three springs whose lower ends were originally at the same level. Each steel spring has 24 turns of 10-mm-diameter on a mean diameter of 100 mm, and G = 83 GPa. The bronze spring has 48 turns of 20-mm-diameter wire on a mean diameter of 150 mm, and G = 42 GPa. Compute the maximum shearing stress in each spring using Wahl formula. Phosphor Bronze Steel Steel 2m 4m 50 kgThe diameter of and LcD = 240 mm. 1. The lengths LAB = LBC = 200 mm parts AB and CD of the bar is 25 mm and the diameter of part BC is 50 mm. The shear modulus of the material is G = 80 GPa. If the torque T = 2.2 kN- m, determine the magnitude of the angle of twist of the right end of the bar relative to the wall in degrees. 4 kN-m 8 kN-m LAB. LBC. 2. The lengths LAB = LBC = 200 mm and LCD = 240 mm. The diameter of parts AB and CD of the bar is 25 mm and the diameter of part BC is 50 mm. The shear modulus of the material is G = 80 GPa. What value of the torque T would cause the angle of twist of the right end of the bar relative to the wall to be zero? 4 kN-m 8 kN-m LAB. T LBCM (1) M 6 mm 2) P= 10 kN. The pin at point M has a radius R. If: 8 mm F.S.=2.5 a=250 mm & b=200 mm Angle between line of action of P and X-axis is 70⁰ Ultimate shearing stress at pin = 60 MPa Ultimate bearing stress at support posts= 150 Mpa Determine the minimum value of R 6 mm
- A steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rotate freely. If ends 4 and D are subjected to 85 Nm torques, determine the angle of twist of end B with respect to C. The tubes have an outer diameter of 30 mm and inner diameter of 20 mm. The solid section has a diameter of 40 mm. Steel has a shear modulus of elasticity G = 75 GPa. Provide your final answer in degrees.Two steel springs arranged in series supports a load P. The upper spring has 12 turns of 20-mm-diameterwire on a mean radius of 100 mm. The lower spring consists of 10 turns of 25-mm diameter wire on amean radius of 75 mm. If the maximum shearing stress in either spring must not exceed 300 MPa,compute the maximum value of P and the total elongation of the assembly. Use approximation formulaand G = 83 GPa. Compute the equivalent spring constant by dividing the load by the total elongation.Problem 5 A helical spring has a shearing stress of 120.6 Mpa with 20 turns. The spring has a deformation of 98.7 mm with mean radius at 80 mm. Determine minimum diameter of the wire to support a load of 2 kN. Use G = 83 Gpa.
- An aluminum alloy cylindrical roller with diameter 1.25 in and length 2 in rolls on the inside of a cast-iron ring having an inside radius of 6 in, which is 2 in thick. Find the maximum contact force F that can be used if the shear stress is not to exceed 4000 psi.Two forces, each of magnitude P, are applied to the wrench. The diameter of the steel shaft AB is 30mm. Determine the largest allowable value of P if the shear stress in the shaft is not to exceed 120MPa and it's angle of twist is limited to 7deg. Use G=83GPa for steel. -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.The mechanism below consists of a solid shaft with a diameter of 2 in between A and B and a hollow shaft with an outside diameter of 7 in and a thickness of 1/2 in between B and C. A torque T = 3 kip-in is applied as shown at A and two forces, each P = 322 lb, are applied as shown at the ends of the levers connected to Point B. Determine the shear stress that occurs on the exterior of the hollow shaft between B and C. Express answer in nearest whole psi.