What is the minimum yield strength required for the


1. Test data are given for siltstone from Virginia under simple tension, simple compression, and compression with lateral pressure. The values of σ3 correspond to fracture.

a) Fit the data to Eq. 7.50 in the book to obtain value of m and τi that describes the Coulomb-Mohr failure envelope line. Also calculate µ, ?, and θc

b) Plot the resulting failure envelope line, along with the largest Mohr's circles, for each test. Does the line reasonably represent the test data?

c) Also, calculate the ultimate strengths in compression and tension, σ'uc and σ'ut, that correspond to the fitted C-M failure envelope, and compare these with the actual values from the tests

Note: Use a software for the drawings (excel, Matlab, etc.), do not use our hands.

σ3

MPa

σ3 = σ2

MPa

21.9 0
-185.4 0
-278 -7.1
-291 -10.49
-343 -14.34
-345 -19.65
-392 -23.1

2. Consider a 50 mm diameter shaft of the gray cast iron of Table 7.1. If a safety factor of 3.0 against fracture is required, what is the largest torque that can be applied along with a compressive axial force of 250 kN? (Use M-M criterion)

3. A steel joint in building construction is supposed to be designed to withstand a multiaxial state of stress with the following stress values:

σx = 150,          σy = -60,          σz = 210,          τxy = 50,           τyz = τzx  = 0 MPa

What is the minimum yield strength required for the material if a safety factor of 2.0 against yielding is required?

a) Employ Maximum Shear Stress Criterion

b) Employ Octahedral Shear Stress Criterion

c) Based on the results from (a) and (b), list all possible options from tables 4.2, 4.3, and 4.4. (DO NO list options for (a) then options for (b), you need to decide if option (a) or option

(b) is safer, then list its options)

Attachment:- corresponding-tables.rar

Request for Solution File

Ask an Expert for Answer!!
Mechanical Engineering: What is the minimum yield strength required for the
Reference No:- TGS01282260

Expected delivery within 24 Hours