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sveticcg [70]
3 years ago
11

The steel bar has a 20 x 10 mm rectangular cross section and is welded along section a-a. The weld material has a tensile yield

strength of 325 MPa and a shear yield strength of 200 MPa, and the bar material has a tensile yield strength of 350 MPa. An overall factor of safety of at least 2.0 is required. Find the largest load P that can be applied, to satisfy all criteria.

Engineering
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

Explanation:

Find attached the solution

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Construction support involves mostly what kind of work?
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Outdoors

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Construction workers perform outdoors.

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A vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c
Ann [662]

Answer:

Amount of air left in the cylinder=m_{2}=0.357 Kg

The amount of heat transfer=Q=0

Explanation:

Given

Initial pressure=P1=300 KPa

Initial volume=V1=0.2m^{3}

Initial temperature=T_{1}=20 C

Final Volume=V_{2}=0.1 m^{3}

Using gas equation

m_{1}=((P_{1}*V_{1})/(R*T_{1}))

m1==(300*0.2)/(.287*293)

m1=0.714 Kg

Similarly

m2=(P2*V2)/R*T2

m2=(300*0.1)/(0.287*293)

m2=0.357 Kg

Now calculate mass of air left,where me is the mass of air left.

me=m2-m1

me=0.715-0.357

mass of air left=me=0.357 Kg

To find heat transfer we need to apply energy balance equation.

Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})

Where me=m1-m2

And as the temperature remains constant,hence the enthalpy also remains constant.

h1=h2=he=h

Q=(me-(m1-m2))*h

me=m1-me

Thus heat transfer=Q=0

6 0
3 years ago
Five batch jobs A through E arrive at a computer center in the order A to E at almost the same time. They have estimated running
Nikolay [14]

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.

3 0
3 years ago
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