If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mmmm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12.0 MPaMPa .
Answer: hello some data related to your question is missing attached below is the missing data
answer:
T2 = 265°C
Explanation:
First step : calculate sum of vertical forces
∑ y = 0
Fmg - 2(0.5 Fst ) = 0
∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )
= 1884.96 N
Also determine the Compatibility equation in order to determine the change in Temperature
ΔT = 250°C
therefore Temperature at which average normal stress becomes 12.0 MPa
ΔT = T2 - T1
250°C = T2 - 15°C
T2 = 250 + 15 = 265°C
attached below is the detailed solution
The worst case signal-to-noise ratio at the output of an FM detector occurs when: ________
a. the desired signal is 90 degrees out of phase with the intelligence signal.
b. the desired signal is 90 degrees out of phase with the noise signal.
c. the noise signal is 90 degrees out of phase with the resultant signal of adding the signal to the noise.
d. the desired signal is 90 degrees out of phase with the resultant signal of adding the signal to the noise.
Answer:
a. the desired signal is 90 degrees out of phase with the intelligence signal.
Explanation:
The signal to noise ratio of FM detector is defined as function of modulation index for SSB FM signal plus narrow band Gaussian noise at input. The ratio is usually higher than 1:1 which indicates more signals than noise.
What method is most likely to be used to measure the
perature of a liquid contained in an open vessel?
Answer:
Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure and display pressure in an integral unit are called pressure meters or pressure gauges or vacuum gauges.
a. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant equations corresponding to the mechanisms.
b. Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.
Answer:
a) Solid solution strengthening and alloying, Precipitation hardening, work hardening
b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle
Explanation:
A) strengthening mechanism
i) Solid solution strengthening and alloying:
In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal
also the shear stress required can be represented as: Δz = Gb√Ce^3/2
where : C = solute concentration , e = strain on material
ii) Precipitation hardening:
During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments
particle bowing cab be written as : Δz = Gb / L-2x
iii) work hardening :
Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism
dislocation can be represented as ; Gb √ p
where : G = shear modulus , b = Burgess vector, p = dislocation density
B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle