A long bone is subjected to a torsion test. Assume that the inner diameter is 0.375 in. and the outer diameter is 1.25 in., both
for a circular cross section. If E = 16 GPa and v = 0.325, find the largest torque that can be applied prior to yield, where σ yield = 1.25 ksi (i.e., a maximum normal stress). Humphrey, Jay D.; O’Rourke, Sherry L.. An Introduction to Biomechanics (Kindle Locations 5346-5349). Springer New York. Kindle Edition.
1 answer:
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Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for
Multiplying both sides of equation (1) by
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by we got:
(4)
Replacing the values into equation (4) we got:
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.
Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
Substituting figures in the expression above, we have:
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection