The minimum diameter. The angle of twist.

Question

2: A rectangular aluminum block is loaded uniformly in three directions. The loadings are as follows:a 50 kN total resulting compressive load in the x-direction, a 200 kPa uniformly distributed tensile load in they-direction, and a 0.03 MN total resulting tensile load in the z-direction. The block has the following dimensions:L = 1 m, b = 20 cm, h = 350 mm. Use E = 70 GPa and ν = 0.25.Determine the strain in the x and y axes respectively. For the strain in the y-direction to be equal to 0, how much uniformly distributed load inthe surface of y-direction should be added? (+ for tensile, - for compressive) Answers: 5 -1.122 x10-5 / 3 decimals 6 5.102 x10-6 / 3 decimals 7 -0.357 MPa / 3 decimals

Answer 1

Question

Chapter 3, Problem 3.4.7P

To determine

The minimum diameter.

The angle of twist.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

2: A rectangular aluminum block is loaded uniformly in three directions. The loadings are as follows:a 50 kN total resulting compressive load in the x-direction, a 200 kPa uniformly distributed tensile load in they-direction, and a 0.03 MN total resulting tensile load in the z-direction. The block has the following dimensions:L = 1 m, b = 20 cm, h = 350 mm. Use E = 70 GPa and ν = 0.25.Determine the strain in the x and y axes respectively. For the strain in the y-direction to be equal to 0, how much uniformly distributed load inthe surface of y-direction should be added? (+ for tensile, - for compressive) Answers: 5 -1.122 x10-5 / 3 decimals 6 5.102 x10-6 / 3 decimals 7 -0.357 MPa / 3 decimals

A spherical balloon with a diameter of 9 m is filled with water vapour at 200˚C and 200 kPa. Determine the mass of water in the balloon. The R value for water is 0.4615 kJ/kg∙K.

Correct answers are written below. Detailed and correct solution only with fbd. I will upvote. 1: A 3 m alloy shaft fixed at one end has a torsional shearing stress capacity of 55 MPa. Due to improper fabrication, its cross-sectionalarea has become irregularly shaped. Its effective polar moment of inertia has become 2 x10-7 m4, and the maximum torque stress acts at 7.5 cm fromthe center of the shaft.[1]: If the shaft is to be replaced by a properly manufactured solid circular shaft that has a maximumshearing stress capacity of 70 MN/m2, what is the minimum diameter required so it can withstand the sameload? [2]: Calculate the thickness of a hollow circular shaft with the same outside diameter calculated initem [1] that can carry the same load. Limit the maximum shearing stress of the hollow circular shaft to0.09 GPa.Determine the angle of twist on the free end of the shaft. Use G = 150 x103 GPa. [3]: Use the solidcircular shaft from [1] and use the hollow circular shaft from [2].…