Concept explainers
For the composite bar in Prob. 2.111, determine the residual stresses in the tempered-steel bars if P is gradually increased from zero until the deformation of the bar reaches a maximum value δm = 0.04 in. and is then decreased back to zero.
2.111 Two tempered-steel bars, each
Fig. P2.111
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
EBK MECHANICS OF MATERIALS
- 1. The normal strains of a solid under a uniaxial stress of 60MPa are and. When the solid is subject to a hydrostatic stress of 100MPa, determine (a) the bulk modulus (b) the dilatation of the solid.arrow_forwardTwo tempered-steel bars, each in. thick, are bonded to a 1/2 -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E= 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. Determine the residual stresses in the tempered-steel bars if the load P is gradually increased from zero until the deformation of the bar reaches a maximum value 5m = 0.04 in. and is then decreased back to zero. Take L = 17 in. 2.0 in. in. 33 in. 3 16 in. The residual stress in the tempered-steel bars is ksi.arrow_forwardA 50-kN force is subjected to a lap joint fastened by three 20-mm diameter rivets.(a) the shearing stress in each rivet.(b) the bearing stress in each plate.(c) the maximum average tensile stress in each plate. Assume that the axial load P is distributed equally amongthe three rivets. equally among the three rivets.arrow_forward
- solve i and ii please.arrow_forward2.36 A 250-mm bar of 150 x 30-mm rectangular cross section consists of two aluminum layers, 5 mm thick, brazed to a center brass layer of the same thickness. If it is subjected to centric forces of magni- tude P = 30 kN, and knowing that E, = 70 GPa and E, = 105 GPa, determine the normal stress (a) in the aluminum layers, (b) in the brass layer. P' 250 mm 5 mm 5 mm 5 mm Aluminum Brass Aluminum P 30 mm Fig. P2.36arrow_forward2. A steel wire 30 ft long, hanging vertically, supports a load of 500 lbf. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20,000 psi and the total elongation is not to exceed 0. in. Assume E = 29 × 106 psi.arrow_forward
- The rigid beam BC is supported by rods (1) and (2). The cross-sectional area of rod (1) is 9 mm². The cross-sectional area of rod (2) is 18 mm². For a uniformly distributed load of w = 4.5 kN/m, determine the length a so that the normal stress is the same in each rod. Assume L = 3.70 m. (1) (2) W a B Answer: a = i L Earrow_forwardProb.6: [2.53] A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (Es =105 GPa , ab= 20.9 x 106 /"C) and portion AB is made of aluminum (Ea =72 GPa , a.= 23.9 x 10-6 /C). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B. A 60-mm diameter 1.1 m - 40-mm diameter 1.3 m Carrow_forwardA spherical gas tank is fabricated by bolting together two thin-walled hemispherical shells with an inner diameter of 8 m. The gas is pressurized to 2.0 MPa. The shells have an allowable normal stress of 150 MPa, and the 25 mm-diameter bolts have an allowable normal stress of 250 MPa. (a) Determine the minimum thickness of the walls of the tank to the nearest mm. (b) Determine the minimum number of bolts to connect the hemispheres.arrow_forward
- 2.60 At room temperature (20°C) a 0.5-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 140°C, determine (a) the normal stress in the aluminum rod, (b) the change in length of the aluminum rod. 0.5 mm 300 mm 250 mm B Aluminum Stainless steel A = 2000 mm² E = 75 GPa a = 23 × 16-6/°C Fig. P2.60 A = 800 mm² E = 190 GPa a = 17.3 × 10-6/°Carrow_forwardThe rigid beam BC is supported by rods (1) and (2). The cross-sectional area of rod (1) is 11 mm². The cross-sectional area of rod (2) is 14 mm2. For a uniformly distributed load of w = 4.1 kN/m, determine the length a so that the normal stress is the same in each rod. Assume L = 5.35 m. A (1) (2) a L. Answer: a = i marrow_forward3) An eccentric force P is applied as shown in Fig. 2 to a steel bar of 25 x 90-mm cross section. The strains at A and B have been measured and found to be ƐA = + 400µ and be ЄB = - - 90μ. Knowing that E = 210 GPa, determine (a) the magnitude of force P, (b) the distance d, and (c) neatly draw the stress distribution diagrams of the system. 30 mm 25 mm- 90 mm A B Fig. 2 1 45 mm 15 mmarrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY