Answer:
a) 0.697*10³ lb.in
b) 6.352 ksi
Explanation:
a)
For cylinder AB:
Let Length of AB = 12 in


For cylinder BC:
Let Length of BC = 18 in




b) Maximum shear stress in BC

Maximum shear stress in AB

Answer:
Explanation:
a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.
b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.
c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.
d. lead can be used as a container for strong acids because of its anti-corrosive properties
e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.
f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.
Answer:
b. Technician B only
Explanation:
A watchdog timer is a circuit that automatically monitors the MCU (Microcontroller Unit) for any anomaly, detects it and helps the MCU to recover from the malfunction it has detected.
If the input signal turn-on time is too fast for the input circuit, that is a malfunction and this activates the watchdog timer circuit to correct this malfunction immediately. So Technician B only is correct as the watchdog timer is activated immediately once there is a malfunction.
Answer:
P=361.91 KN
Explanation:
given data:
brackets and head of the screw are made of material with T_fail=120 Mpa
safety factor is F.S=2.5
maximum value of force P=??
<em>solution:</em>
to find the shear stress
T_allow=T_fail/F.S
=120 Mpa/2.5
=48 Mpa
we know that,
V=P
<u>Area for shear head:</u>
A(head)=π×d×t
=π×0.04×0.075
=0.003×πm^2
<u>Area for plate:</u>
A(plate)=π×d×t
=π×0.08×0.03
=0.0024×πm^2
now we have to find shear stress for both head and plate
<u>For head:</u>
T_allow=V/A(head)
48 Mpa=P/0.003×π ..(V=P)
P =48 Mpa×0.003×π
=452.16 KN
<u>For plate:</u>
T_allow=V/A(plate)
48 Mpa=P/0.0024×π ..(V=P)
P =48 Mpa×0.0024×π
=361.91 KN
the boundary load is obtained as the minimum value of force P for all three cases. so the solution is
P=361.91 KN
note:
find the attached pic