
Materials Science and Engineering
9th Edition
ISBN: 9781118324578
Author: Jr. William D. Callister
Publisher: WILEY
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Chapter 6.12, Problem 24QP
To determine
To justify:
The materials according to the given criterion.
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Please find the torsional yield strength, the yield strength, the spring index, and the mean diameter. Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi·in, m = 0.190, and relative cost= 1.
my previous answers for a,b,d were wrong
a = 1050 b = 950 d=9.99
c was the only correct value i got previously
c = 100hz is correct
V₁(t)
ww
ZRI
ZLI
ZL2
ZTH
Zci
VTH
Zc21
Figure 8. Circuit diagram showing calculation approach for VTH and Z TH
we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference
ector circuit:
Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference)
Square wave with 2.4Vpp, with a frequency of 10 kHz (signal)
member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the
ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss
is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.
Chapter 6 Solutions
Materials Science and Engineering
Ch. 6.12 - Prob. 1QPCh. 6.12 - Prob. 2QPCh. 6.12 - Prob. 3QPCh. 6.12 - Prob. 4QPCh. 6.12 - Prob. 5QPCh. 6.12 - Prob. 6QPCh. 6.12 - Prob. 7QPCh. 6.12 - Prob. 8QPCh. 6.12 - Prob. 10QPCh. 6.12 - Prob. 11QP
Ch. 6.12 - Prob. 12QPCh. 6.12 - Prob. 13QPCh. 6.12 - Prob. 14QPCh. 6.12 - Prob. 15QPCh. 6.12 - Prob. 16QPCh. 6.12 - Prob. 17QPCh. 6.12 - Prob. 18QPCh. 6.12 - Prob. 19QPCh. 6.12 - Prob. 20QPCh. 6.12 - Prob. 21QPCh. 6.12 - Prob. 22QPCh. 6.12 - Prob. 23QPCh. 6.12 - Prob. 24QPCh. 6.12 - Prob. 25QPCh. 6.12 - Prob. 26QPCh. 6.12 - Prob. 27QPCh. 6.12 - Prob. 28QPCh. 6.12 - Prob. 29QPCh. 6.12 - Prob. 30QPCh. 6.12 - Prob. 31QPCh. 6.12 - Prob. 32QPCh. 6.12 - Prob. 37QPCh. 6.12 - Prob. 38QPCh. 6.12 - Prob. 39QPCh. 6.12 - Prob. 40QPCh. 6.12 - Prob. 41QPCh. 6.12 - Prob. 42QPCh. 6.12 - Prob. 43QPCh. 6.12 - Prob. 44QPCh. 6.12 - Prob. 45QPCh. 6.12 - Prob. 46QPCh. 6.12 - Prob. 47QPCh. 6.12 - Prob. 48QPCh. 6.12 - Prob. 49QPCh. 6.12 - Prob. 50QPCh. 6.12 - Prob. 51QPCh. 6.12 - Prob. 52QPCh. 6.12 - Prob. 53QPCh. 6.12 - Prob. 54QPCh. 6.12 - Prob. 55QPCh. 6.12 - Prob. 56QPCh. 6.12 - Prob. 57QPCh. 6.12 - Prob. 58QPCh. 6.12 - Prob. 59QPCh. 6.12 - Prob. 60QPCh. 6.12 - Prob. 61QPCh. 6.12 - Prob. 62QPCh. 6.12 - Prob. 63QPCh. 6.12 - Prob. 1DPCh. 6.12 - Prob. 2DPCh. 6.12 - Prob. 3DPCh. 6.12 - Prob. 4DPCh. 6.12 - Prob. 1FEQPCh. 6.12 - Prob. 2FEQPCh. 6.12 - Prob. 3FEQPCh. 6.12 - Prob. 4FEQPCh. 6.12 - Prob. 5FEQP
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- V₁(t) ww ZRI ZLI ZL2 ZTH Zci VTH Zc21 Figure 8. Circuit diagram showing calculation approach for VTH and Z TH we want to create a blackbox for the red region, we want to use the same input signal conditions as previously the design of your interference ector circuit: Sine wave with a 1 Vpp, with a frequency of 100 kHz (interference) Square wave with 2.4Vpp, with a frequency of 10 kHz (signal) member an AC Thevenin equivalent is only valid at one frequency. We have chosen to calculate the Thevenin equivalent circuit (and therefore the ackbox) at the interference frequency (i.e. 100 kHz), and the signal frequency (i.e. 10 kHz) as these are the key frequencies to analyse. Your boss is assured you that the waveform converter module has been pre-optimised to the DAB Receiver if you use the recommended circuit topology.arrow_forwardVs(t) + v(t) + vi(t) ZR ZL Figure 1: Second order RLC circuit Zc + ve(t) You are requested to design the circuit shown in Figure 1. The circuit is assumed to be operating at its resonant frequency when it is fed by a sinusoidal voltage source Vs (t) = 2sin(le6t). To help design your circuit you have been given the value of inductive reactance ZL = j1000. Assume that the amplitude of the current at resonance is Is (t) = 2 mA. Based on this information, answer the following to help design your circuit. Use cartesian notation for your answers, where required.arrow_forwardConsider the specifications for an asphaltic concrete mixture and the results of a sieve analysis below. Coarse aggregates: Fine aggregates: Filler: 60% 35% 5% Percent of Weight of Aggregate or Filler Passing Sieve Designation Retained on Sieve Designation Coarse Aggregate Fine Aggregate Mineral Filler 3/4 in. (19 mm) 1/2 in. 6 1/2 in. (12.5 mm) 3/8 in. 15 3/8 in. (9.5 mm) No. 4 50 - No. 4 (4.75 mm) No. 10 20 1 No. 10 (2 mm) No. 40 (0.425 mm) No. 40 9 35 - No. 80 31 40 No. 80 (0.180 mm) No. 200 (0.075 mm) Total No. 200 - 33 - - 25 35 100 100 100 Determine the proportion of different aggregates to obtain the required gradation. Percent of Total Weight of Mixture Passing Sieve Designation Retained on Sieve Designation Coarse Aggregate Fine Aggregate 3/4 in. (19 mm) 1/2 in. (12.5 mm) 3/8 in. (9.5 mm) 1/2 in. 3/8 in. No. 4 No. 4 (4.75 mm) No. 10 No. 10 (2 mm) No. 40 No. 40 (0.425 mm) No. 80 No. 80 (0.180 mm) No. 200 No. 200 (0.075 mm) Total Need Help? Read It Mineral Filler Total 100arrow_forward
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