STATICS+MECHANICS OF MATERIALS-ACCESS
2nd Edition
ISBN: 9781259748684
Author: BEER
Publisher: MCG
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An aluminum [E = 10000 ksi] control rod with a circular cross section must not stretch more than 0.21 in. when the tension in the rod is 2230 lb. If the maximum allowable normal stress in the rod is 11.2 ksi, determine:(a) the smallest diameter d that can be used for the rod.(b) the corresponding maximum length L of the rod.
The aluminum rod ABC (E = 70 GPa), which consists of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 200 GPa) of the same overall length.Determine the minimum required diameter d of the steel rod if its vertical deformation is not to exceed the deformation of the aluminum rod under the same load and if the allowable stress in the steel rod is not to exceed 165.5 MPa.
The rigid bar ABD is connected to the rod BD which consists of a single bar with a width of 25,0 mm and a thickness of 13 mm. Knowing that each pin has a diameter of 10 mm determine the value of the maximum average normal stress on rod BD considering the angle θ=90°. Calculate the deformation in the BD rod knowing that it is made of steel with an elastic modulus equal to 210GPa. Knowing that pins A, B and D are made of steel with a limit stress of 180Mpa, calculate the smallest allowable diameter so that they support the load applied to them considering an overall safety factor of 2,8.
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- Each of the four vertical Ilinks has an 8 x 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Take P= 19 kN. 0.4 m C 0.25 m 0.2 m B. P Determine the average bearing stress at Bin member ABC, knowing that this member has a 10 x 50-mm uniform rectangular cross section. MPa. The average bearing stress at Bin member ABC is.arrow_forward1. Determine the average normal stress in each of the 20-mm diameter bars of the truss. Set P = 40KN. If the average normal stress in each of the 20-mmdiameter bars is not allowed to exceed 150 MPa, determine the maximum force P that can be applied to joint C.arrow_forwardRigid triangle ABC is pinned at B and supported by a steel-reinforced aluminum shaft at A and an aluminum-copper at C. The steel-reinforced aluminum shaft is fixed at F while the aluminum-copper rod is fixed at E. Determine the maximum weight of triangle ABC if the allowable stress for steel, copper, and aluminum are 36 ksi, 11.4 ksi, and 60 ksi, respectively, and the allowable deformation for the aluminum-copper rod CDE is 0.1 in.arrow_forward
- Knowing that a 0.02-in. gap exists when the temperature is 75°F, determine (a) the temperature at which the normal stress in the alumi-num bar will be equal to –11 ksi, (b) the corresponding exact length of the aluminum bar.arrow_forwardDetermine the values of the stress in portions AC and CB of the steel bar shown Fig. when the temperature of the bar is 2508F, knowing that a close fit exists at both of the rigid supports when the temperature is 1758F. Use the values E=29 *106 psi and α=6.5 * 10–6/8F for steelarrow_forwardThe ABD rigid bar is connected to the BD rush which consists of a single bar with a width of 25.0 mm and a thickness of 13 mm. Knowing that each pin has a diameter of 10 mm determine the value of the maximum average normal stress on rod BD considering the angle θ= 90°. Calculate the deformation in the BD rod knowing that it is made of steel with an elastic modulus equal to 210 GPa. Knowing that pins A, B and D are made of steel with a limit stress of 180 Mpa, calculate the smallest allowable diameter so that they support the load applied to them considering an overall safety factor of 2.8.arrow_forward
- At a temperature of 32.98 °C a 0.41-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 131.06°C, determine the magnitude of the normal stress (in MPa) in the steel rod if L = 332.28 mm and M = 209 mm.arrow_forwardA14-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in length of the wire must not exceed 25 mm. The smallest diameter that can be used is ___mm?arrow_forwardStraight rods of 0.30-in. diameter and 200-ft length are sometimes used to clear underground conduits of obstructions or to thread wires through a new conduit. The rods are made of high-strength steel and, for storage and transportation, are wrapped on spools of 5-ft diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a rod, when the rod, which was initially straight, is wrapped on the spool, (b) the corresponding bending moment in the rod. Use E= 29 * 106 psi.arrow_forward
- A 5.3-m-long steel rod must not stretch more than 2.71 mm and the normal stress must not exceed 181 MPa when the rod is subjected to a 9.99-kNaxial load. Knowing that E = 199.3 GPa, determine the required radius of the rod in mm. Express your answer in four decimal places.arrow_forwardA 6.87-m-long steel rod must not stretch more than 2.5 mm and the normal stress must not exceed 181 MPa when the rod is subjected to a 9.54-kNaxial load. Knowing that E = 198.3 GPa, determine the required radius of the rod in mm.arrow_forward2 m long aluminium rod must not stretch more than 1 mm and the normal stress must not exceed 38 MPa when the rod is subjected to a 5 kN axial load. Knowing that Modulus of Elasticity for Aluminium (E)= 70 GPa, determine the required diameter of the rod. D= 13.49 mmarrow_forward
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EVERYTHING on Axial Loading Normal Stress in 10 MINUTES - Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=jQ-fNqZWrNg;License: Standard YouTube License, CC-BY