A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te
1 answer:
You might be interested in
Answer:
The shear plane angle and shear strain are 28.21° and 2.155 respectively.
Explanation:
(a)
Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.
Given:
Rake angle is 12°.
Chip thickness before cut is 0.32 mm.
Chip thickness is 0.65 mm.
Calculation:
Step1
Chip reduction ratio is calculated as follows:
r = 0.4923
Step2
Shear angle is calculated as follows:
Here, is shear plane angle, r is chip reduction ratio and
is rake angle.
Substitute all the values in the above equation as follows:
Thus, the shear plane angle is 28.21°.
(b)
Step3
Shears train is calculated as follows:
.
Thus, the shear strain rate is 2.155.
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