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From See. 3-12, the maximum shear stress in a solid round bar of diameter. d, due to an applied torque. T, is given by
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Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
- Three round, copper alloy bars having the same length L but different shapes are shown, in the figure. The first bar has a diameter d over its entire length, the second has a diameter d over one-fifth of its length, and the third has a diameter d over one-fifteenth of its length. Elsewhere, the second and third bars have a diameter Id. All three bars are subjected to the same axial load P. Use the following numerical data: P = 1400 kN, L = 5m,d= 80 mm, E= 110 GPa. and v = 0.33. (a) Find the change in length of each bar. (b) Find the change in volume of each bar.arrow_forwardThe principal stresses at a point in a critical section of a machine component are sigma_{1} = 60MPa, sigma_{2} = 5MPa and sigma_{3} = - 40MPa For the material of the component, the tensile yield strength is sigma_{y} = 200 MPa According to the maximum shear stress theory,what is the factor of safety.arrow_forwardA test piece is cut from a brass bar and subjected to a tensile test. With a load of 6.4 kN the test piece, of diameter 11.28 mm, extends by 0.04 mm over a gauge length of 50 mm. Determine: (i) the stress, (ii) the strain, (iii) the modulus of elasticity. (b) A spacer is turned from the same bar. The spacer has a diameter of 28 mm and a length of 250 mm, both measurements being made at 20~ The temperature of the spacer is then increased to 100~ the natural expansion being entirely prevented. Taking the coefficient of linear expansion to be 18 x 10-6/~ determine: (i) the stress in the spacer, (ii) the compressive load on the spacer. rC.G.] [64 MN/m 2, 0.0008, 80 GN/m 2, 115.2 MN/m 2, 71 kN.] Could you please answer thisquestion fully? If you use * could you please explain what you mean by it! Thank You :)arrow_forward
- Determine the yield strength of a material required such that the component would not fail when subject to the following stresses sigma1 = 2 MPa and sigma 2= -15 MPa sigma 3 = 10 MPa). Use a yield criterion that assumes that yield failure will occur when the maximum shear stress in the complex system becomes equal to the limiting shear strength in a simple tensile test.arrow_forwardCalculate the values of Poisson’s ratio of a rod under uniaxial tensile test. Initially it has 15.5 mm diameter and 62 mm gauge length, then it changes to 15.415 mm diameter and 62.102 mm gauge length under a load of 20 kN.a) 0.186 b) 0.242 c) 0.259 d) 0.342arrow_forwardThe material of a certain transmission shaft uses AISI 1030 HR, its yield strength is 260 MPa, the transmission torque of this shaft is 160 N·m, and it is subjected to a uniform axial tension of 0.5 kN. Please calculate according to the Distortion energy theory. The minimum shaft diameter with a safety factor n of 2.5 (to 2 decimal places).arrow_forward
- A hard-drawn steel wire extension spring has a wire diameter of 4mm, an outside coil diameter of 28 mm, and initial tension of 1200 N is applied to the spring. The spring shows an extension of 85 mm. The approximate number of turns is 25. From the given information: Determine the physical parameters of the spring such as mean diameter, spring index, maximum shear stress, total number of turns, Wahl’s stress factor, free length of the spring, and pitch of the coilarrow_forwardA bar of dimension 120 mm x 50 mm x 40 mm, is subjected to an axial pull of 50 kN. The measured extension is 0.25 mm and change in depth is 0.0036 mm. Calculate: (i) Young’s Modulus, (ii) Poisson’s ratio (iii) Shear Modulus. and (iv) Bulk Modulus (i) The Young's modulus is (N/mm2) = _________________ (ii) Poisson's ratio = ___________________ (iii) The shear modulus is (unit in N/mm2) = ________________ (iv) The Bulk modulus is (unit in N/mm2) =arrow_forwardDescribe the stress analysis process and the concept of the following failure theories for mechanical design. Maximum Normal Stress Theory (MNST) Distortion Energy Theory (DET) Coulumb-Mohr Theory (CMT) Modified Mohr Theory (MMT)arrow_forward
- The armature shaft of a 50 kW, 580 rpm electric motor, mounted on two bearings A and B, is shown in Fig. The total magnetic pull on the armature is 5 kN and it can be assumed to be uniformly distributed over a length of 400 mm midway between the bearings. The shaft is made of steel with an ultimate tensile strength of 770 N/mm2 and yield strength of 580 N/mm2. Determine the shaft diameter using the ASME code if, kb=1.5 and kt=1.0 Assume that the pulley is keyed to the shaft.arrow_forwardHow many 1/2 inch diameter hole that can be punch in one motion of a 1/8 inch thick plate using a force of 50 tons. The ultimate shear stress is 52 ksi and factor of safety of 3. A. 7 B. 9 C. 8 D. 10 Refer Machine Elements and Stresses Equations from the figure. Please solve the Problem elaborately. Your solution will be use as reference for my studies. Thank you so much your work will be appreciated much!arrow_forwardAt a temperature of 60°F, a 0.05-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10-6/°F] bar with a width of 2.9 in. and a thickness of 0.80 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.9 in. and a thickness of 0.80 in. The supports at A and C are rigid. Assume h1=2.9 in., h2=1.9 in., L1=28 in., L2=42 in., and Δ=Δ= 0.05 in. Determine(a) the lowest temperature at which the two bars contact each other.(b) the normal stress in the two bars at a temperature of 255°F.(c) the normal strain in the two bars at 255°F.(d) the change in width of the aluminum bar at a temperature of 255°F. Determine the lowest temperature, Tcontact, at which the two bars contact each other.arrow_forward
- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning