Module 28 Solution Brittle Coulomb Mohr Theory [PDF]

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Mechanical Design of Machine Elements Complete after watching Module 27: Brittle Coulomb Mohr Theory Example

Module 27 Example: Brittle Coulomb Mohr Theory X The femur bone has an outside diameter of 24 mm. The tensile strength of the bone is Sut = 120 MPa, Suc = 170 MPa. The femur is subjected to a torque of 80 N-m, and a bending moment of 100 N-m. The positive z-axis comes out of the page, and therefore using the right-hand rule the bending moment will be negative. For this example we will assume the bone is acting in a brittle manner, is a solid cylinder, is isotropic and linear elastic.

Y

T

M

Find the factor of safety at point A. Brittle Coulomb Mohr Failure conditions: Case 1: σa ≥ σb ≥ 0 σa ≥ Sut n = Sut/σa Case 2: σa ≥ 0 ≥ σb Case 3: 0 ≥ σa ≥ σb

sa Sut

-

sb Suc

³1

σb ≤ -Suc

A

1 sa sb = n Sut Suc n = -Suc/σb

Assumptions: Isotropic, homogenous, linear elastic, brittle material, neglecting weight of bone. Thoughts: T results in torsional stress. At point A the torsional shear stress will be tyz. M results in bending stress that will be maximum at the edges of the bone, the neutral plane will be along the y-axis in the y-z plane. The stress will act in the y-direction as sy. Point A will be pulled into torsion. Transverse shear will be 0 at point A as it is on the edge of the bone, far from the neutral axis. Known: d = 0.024 m, Suc = 170 MPa, Sut = 120 MPa, T = 80 N*m, M = -100 N*m. Analysis: Determine stresses in X, Y, and Z directions.

sX = t xz =

Mc 32M (32)(-100N * m) === 73.7MPa 3 I pd p (0.024)3

Tr 16T (16)(80N * m) = = = 29.5MPa J p d3 p (0.024m)3

Note the positive value for tension, which is what we expected.

Mechanical Design of Machine Elements Complete after watching Module 27: Brittle Coulomb Mohr Theory Example

Module 27 Example: Brittle Coulomb Mohr Theory X The femur bone has an outside diameter of 24 mm. The tensile strength of the bone is Sut = 120 MPa, Suc = 170 MPa. The femur is subjected to a torque of 80 N-m, and a bending moment of 100 N-m. The positive z-axis comes out of the page, and therefore using the right-hand rule the bending moment will be negative. For this example we will assume the bone is acting in a brittle manner, is a solid cylinder, is isotropic and linear elastic.

Y

T

M

Find the factor of safety at point A. Brittle Coulomb Mohr Failure conditions: Case 1: σa ≥ σb ≥ 0 σa ≥ Sut n = Sut/σa Case 2: σa ≥ 0 ≥ σb Case 3: 0 ≥ σa ≥ σb

sa Sut

-

sb Suc

³1

σb ≤ -Suc

A

1 sa sb = n Sut Suc n = -Suc/σb

Analysis: Calculate principal stresses

s 1, s 2 =

sx +sz 2

73.7 æ 73.7 ö æs -sz ö ± ç x + t 2xz = ± ç + 29.4 2 = 84.0MPa, -10.3MPa ÷ ÷ è 2 ø è 2 ø 2 2

Apply the brittle coulomb mohr failure conditions:

Case 2: sa  0  sb

1 sa sb = n Sut Suc 1 n= = 1.3 84 -10.3 120 170

2