Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
Answer:
<u>Assistants</u><u> </u><u>works alongside and assists the engineers.</u>
Answer:
modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048
Explanation:
Modulus is the ratio of tensile stress to tensile strain
Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain within the direction of the stretching force
And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in
Its 0.001
0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m
Answer:
Explanation:
The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.
Round Robin:
we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A
C* = 8
D*=17
B*=23
E*=28
AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES
B) PRIORITY SCHEDULING:
1-6 7-14 15-24 25-26 27-30
B E A C D
AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.
C)FCFS
1-10 11-16 17-18 19-22 23-30
A B C D E
AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES
D)SJF
1-2 3-6 7-12 13-20 21-30
C D B E A
AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.
