Two bolts are used to form a joint connecting rectangular bars in tension as shown. If the bolts have a diameter of 9.5mm and the load is P = 100kN, determine the average shear stress (in MPa) on the bolt surfaces that are subjected to direct shear. Note: Do NOT include units in your answer 25 P2 P2 Answer: X w/2 P The temperature at which Nitinol remembers its high temperature form/shape can be adjusted by slight changes in Q Search IPPODZIL hp ENG GA

An aluminum bar has length L = 6 ft and diameter d = 1.375 in. The stress-strain curse for the aluminum is shown in Fig. 1.34. The initial straight, line part of the curve has a slope (modulus of elasticity) of 10.6 × 106 psi. The bar is loaded by tensile forces P = 44.6 k and then unloaded. (a) That is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional limit? hint: Use the concepts illustrated in Figs. l.39b and 1.40.

A 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.

In a sheet metal of 2 mm thickness, a hole of 10 mm diame- ter needs to be punched. The yield strength in tension of the sheet material is 100 MPa and its ultimate shear strength is 80 MPa. The force required to punch the hole (in kN) is

With the force P appliced on a member, the stress distributions at different cross sections are different as shown below. However, at locations far enough away from the support and the applied load, the stress distribution becomes uniform. This is due to:                ​     Inelastic property     Poisson’s effect     Saint Venant’s Principle     Principle of superposition

A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram is shown in Fig.a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Fig. b, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?

A specimen of copper having a rectangular cross section of 15.2mm x 19.1mm is pulled with a force of 44.5kN. Calculate the stress in MPa. Provide your answer to 2 decimal places. Answer: 0 0 Vũ Vũ (DE) + TI a W

You are in charge of designing a new fixture for a “universal testing machine” that will attach a tensile specimen to the machine using a clevis—a U-shaped piece with holes drilled through the two arms—and a cylindrical pin that passes through the clevis and the specimen. If the maximum load exerted by the machine is 30 kN and the pin is to be made of some sort of steel, what is the minimum pin diameter needed to ensure that the shear stress in the pin does not exceed 600 MPa? Assume that the steel has similar elastic properties to pure Fe.

Sketch a tensile member with (a) a rectangular crosssection, (b) a solid circular cross section, and (c) a circulartubecross section, and label the dimensions symbolically (e.g.,label the radius for the solid circular case). For each member, write out the definition of engineering stress in terms of the actual dimensions of the component. If therectangular member has dimensions of width and thicknessequal to 1 cm 0.3 cm, what would be the radius of a solidcircular member such that the stress is equal for an equaltensile load? If a tube has an outer radius equal to that ofthis same solid cylinder, what is the maximum inner radiussuch that the stress does not exceed 200% of the stress inthe solid cylinder?

Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025