Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section?
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Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section?
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- Determine the shape factor f for a cross section in the shape of a double trapezoid having the dimensions shown in the figure. Also, check your result for the special cases of a rhombus (b1= 0) and a rectangle (b1= b2).Consider a two-component rod fixed at the left end. The first component (on the left, positioned from z=0 to z=L/2) is a solid rod with a shear modulus of G and radius c1. The second component (on the right, positioned from z=L/2 to z=L) is a hollow pipe fixed to component 1 at z=L/2 with the same shear modulus and outer radius c1, but with an inner radius of c2. A torque of 6To is applied at z=L/2, and a torque of -2To is applied at z=L. What value of c2 is required to make the maximum shear stress equivalent in each component? What is the angle of twist at the free end at z=L?Use the graphical method to construct the shear-force diagram and identify the magnitude of the largest shear force (consider both positive and negative). The ground reactions at the wall of the cantilever are provided. L₁= 14.00 ft L2= 7.75 ft Vc = 87.00 kips Mc = 621.875 kip-ft 20 kips 6 kips/ft L₁ 64.00 kips 99.00 kips 110.00 kips 87.00 kips O 75.00 kips 70 kips B 12 kips/ft L2 Vc Mc
- Use the graphical method to construct the shear-force diagram and identify the magnitude of the largest shear force (consider both positive and negative). The ground reactions at the wall of the cantilever are provided. L₁= 14.75 ft L₂ = 7.75 ft Vc = 91.50 kips Mc = 706.4375 kip-ft 20 kips 6 kips/ft L₁ 103.50 kips 68.50 kips 91.50 kips O 79.50 kips O 114.50 kips 70 kips BO 12 kips/ft L2 O Vc McIf the internal torque experienced by a hollow brass shaft (G = 39 GPa) at a section with diameter d = 100 mm with 3 mm thick walls is T = 12 kNm, which of the following is closest to the magnitude of the minimum shear strain developed on that section? Select one: O A. 0.0072 rad O B. 0.0137 rad O C. 0.0128 rad O D. 0.0067 radDetermine the torsional shear stress in the shaft section AB and BC. There is a fixed support at C that can resist a moment. (Image taken from Statics and Mechanics of Materials by Beer, Johnson, DeWolf and Mazurek) T₁ = 300 N-m 30 mm TB = 400 N-m 46 mm 055 Shear Stress in AB = Shear Stress in BC= B 0.9 m 0.75 m units: units:
- 1. A circular shaft with a keyway can be approximated by the section shown in Fig. r=b Ay r r=2acose Figure 1: Circular shaft with a keyway. The keyway is represented by the boundary equation r = b and the shaft is represented by the boundary equation r = 2a cos 0. Show that using a Prandtl stress function of the form ²) (₁ (1-2a cos) X v = K (b² − ²) (1 – : will solve the problem of torsion on this shaft and find the constant K (assume an applied torque T). Compute the shear stress components Txz and Tyz. (Hint: start by converting the stress function to Cartesian coordinates).The copper pipe [G = 6,000 ksi] has an outside diameter of 3.375 in. and a wall thickness of 0.250 in. The pipe is subjected to a uniformly distributed torque of t = 85 lb-ft/ft along its entire length. Using a = 2.5 ft, b = 3.5 ft, and c =7.0 ft, determine (a) the maximum shear stress in the pipe. (b) the rotation angle pc at the free end of the shaft. b B C Answers: (a) Tmax psi i rad (b) ¤c = iA 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. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: d. Modulus of elasticity for the brass.2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: a. Strain in the steel rod2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: b. Load carried by the steel and brass rods