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:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
Answer: l = 2142.8575 ft
v = 193.99 ft/min.
Explanation:
Given data:
Thickness of the slab = 3in
Length of the slab = 15ft
Width of the slab = 10in
Speed of the slab = 40ft/min
Solution:
a. After three phase
three phase = (0.2)(0.2)(0.2)(3.0)
= 0.024in.
wf = (1.03)(1.03)(1.03)(10.0)
= 10.927 in.
Using constant volume formula
= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf
Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)
= 6750 /0.2625
= 25714.28in = 2142.8575 ft
b.
vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)
= (0.12)(424.36)/0.2625
= 50.9232/0.2625
= 193.99 ft/min.
Answer:
shoe box covered in aluminum foil
