A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield str
2 answers:
Answer:
The critical length of a surface flaw is 6.17 mm
Explanation:
according to the exercise:
flat plate size=25 mm
σ=545 MPa
plane strain fracture toughness=29.6 MPa-m1/2
Y=1.3
Design stress = 0.3*σ = 163.5 MPa
For plain structure fracture toughness is:
Where ac is the maximum allowable flat size. Clearing ac:
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Answer:
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answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)
Answer:
the net work per cycle Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
the mass in air ( lb) can be determined by using the formula:
where;
R = 53.3533 ft.lbf/lb.R°
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
At state of volume 2; the relative volume can be determined as:
The specific energy at
is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
To determine the relative volume at state 4; we have:
The specific energy at
is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
where ;
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp