Answer:
MIS HIEVOSTES bbbbbbbbbbbb MIS HUEVOTES
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
