A brittle material has the properties Sut= 30 kpsi and Suc= 90 kpsi. Using the brittle-Coulomb-Mohr...

Question

Question

A brittle material has the properties Sut= 30 kpsi and Suc= 90 kpsi. Using the brittle-Coulomb-Mohr and modified-Mohr theorie

Answer 1

Answer #1

Please Thumbs Up.

Add a comment

Answer 2

Similar Homework Help Questions

A brittle material has the properties Sut = 31 kpsi and Suc = 109 kpsi. Using...

A brittle material has the properties Sut = 31 kpsi and Suc = 109 kpsi. Using the brittle Coulomb-Mohr (BCM) and Modified-Mohr (MM) theories, estimate the factors of safety from the two theories for the following state of plane stress: Ox = - 35 kpsi, Oy = 13 kpsi, Txy = -10 kpsi.
A brittle material has the properties Sat 19 kpsi and Suc 95 kpsi. Using the brittle...

A brittle material has the properties Sat 19 kpsi and Suc 95 kpsi. Using the brittle Coulomb Mohr and modified Mohr theories, determine the factor of safety for the following 0x = 18 kpsi, Gy = 14 kpsi, Txy 0 Hint: Coulomb-Mohr Theory: Oxt Oy 2 + + If A 2 0820 A = Sut If 0 202 03 OA Sut Suc Suc If 0 2 042 OB ов a. n-1.22 b. n - 2.72 Ocn- 3.14 Od.n - 1.52...
A brittle material has the properties su kpsi and S 125 kps using the bitte Coulomb...

A brittle material has the properties su kpsi and S 125 kps using the bitte Coulomb Mohr and modified Mortheories, determine the factor of safety for the following x = -20 kpsi, Gy = 18 kpsi, Txy = -13 kpsi Hint Coulomb-Mohr Theory: Oox + Oy + 04.08 1 2 3* + z3 2 If on 2020 OA Suit 71 If on 20 203 GA Sut Sue If 0 201208 OR a. n-2.65 Ob.n-1.26 On-1.68 d. n-3.14 en - 1.08

A brittle material has the properties Sut = 30kpsi and Sue = 90kpsi. Using the Maximum...

A brittle material has the properties Sut = 30kpsi and Sue = 90kpsi. Using the Maximum Normal Stress theory, determine the factor of safety for the following states of plane stress. (a) Ox = 20kpsi, Oy = 10kpsi, Txy = -15kpsi (b) Ox = 15kpsi, oy = -25kpsi (c) Ox = 10kpsi, oy = -40kpsi (d) Ox = -20kpsi, oy = -20kpsi, Txy = -40 (1) Draw the stress element Plot the failure diagram in the o-plane Estimate the factor...
A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a...

A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of & = 0.55. Calculate the factor of safety (n) for the following stress states by using both MSS and DE theory: (a) Ox = -50 kpsi, oy = -50 kpsi, Txy = -30 kpsi (b) Ox = -50 kpsi, Oy = 0 kpsi, Txy = -30 kpsi (c) Ox = 0 kpsi, oy = 50 kpsi, Txy = 30...
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt...

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using the ductile Coulomb-Mohr theory, determine the factor of safety for the states of plane stress. 0x = -21.8 kpsi, oy" -21.8 kpsi, and Txy=-16 kpsi The factor of safety is 2.455

Please help with this mechanical engineering machine design homework. Please show all work! Thanks 1) Using...

Please help with this mechanical engineering machine design homework. Please show all work! Thanks 1) Using typical values for strength of ASTM No. 40 cast iron, find the factor of safety corresponding to fracture by the maximum normal stress theory, the Coulomb-Mohr theory and the modified Mohr theory for each of the following stress states: (a) 0x = 10 ksi, oy = -4 ksi (b) Ox = 10 ksi, Txy = 4 ksi (c) Ox = -2 ksi, oy =...
2. Figure below shows the state of stress at a critical point on a component of a lawn mower. Mat...

2. Figure below shows the state of stress at a critical point on a component of a lawn mower. Material properties 0.005 are Sut# 170 MPa; Suc-650MPa; Sus-240 MPa;Ep a. Identify the type of material (ductile or brittle) and justify it. 2 marks) 13 marks c. Formulate the solution and determine if the component will fail or not fail. b. Identify the appropriate failure theories. [10 marksl d. Solve for the factor of safety from appropriate failure theories using appropriate...