Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter [tex]d_{o}[/tex] = 12.8 mm
Final diameter [tex]d_{f}[/tex] = 10.7
Engineering stress [tex]\alpha _{E}[/tex] = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4([tex]d_{o}[/tex] )² ; Ag = π/4([tex]d_{f}[/tex] )²
% = π/4 [ ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ( π/4 ([tex]d_{o}[/tex] )²) ]
= ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ([tex]d_{o}[/tex] )² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress [tex]\alpha _{T}[/tex] = [tex]\alpha _{E}[/tex] ( 1 + [tex]E_{E}[/tex] )
[tex]E_{E}[/tex] is engineering strain
[tex]E_{E}[/tex] = dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
[tex]E_{E}[/tex] = 0.431
so we substitute the value of [tex]E_{E}[/tex] into our initial equation;
True stress [tex]\alpha _{T}[/tex] = 460 ( 1 + 0.431)
True stress [tex]\alpha _{T}[/tex] = 460 (1.431)
True stress [tex]\alpha _{T}[/tex] = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa
A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length
Answer:
a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b) length of the specimen is 66.97 mm
Explanation:
Given the data in the question;
a) Determine the maximum stress that can be applied without plastic deformation
when know that; maximum stress σ[tex]_{max}[/tex] = F / A
where F is the force in the rod ( 39872 N )
A is the cross-sectional area of the rod ( 100 mm² )
so we substitute;
σ[tex]_{max}[/tex] = 39872 N / 100 mm²
σ[tex]_{max}[/tex] = 398.72 N/mm²
Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b)
strain in the members can be calculated using the expression
ε = σ / E
where σ is the stress in the rod
E is the module of elasticity ( 110 GPa = 110000 N/mm² )
(Sl-L) / L = σ/E
where Sl-L is the change in length of the member
L is the original length of the specimen
so we substitute
(67.21 - L) / L = 398.72 / 110000
110000( 67.21 - L) = 398.72L
7393100 - 110000L = 398.72L
7393100 = 398.72L+ 110000L
7393100 = 110398.72L
L = 7393100 / 110398.72
L = 66.97 mm
Therefore; length of the specimen is 66.97 mm
Axial forces in a column due to service loads are as follows (assume the live load used in these calculationsis less than 100 psf):Dead:150 kcompressionLive:280 kcompressionRoof Live:40 kcompressionSnow:50 kcompressionWind:120 kcompression or tensionEarthquake:200 kcompression or tension1. Compute the required axial strength,TuandPu, for this column in tension and compression usingLRFD load combinations. (Neglect self-weight.)2. Describe the loading scenario that represents the worst case tension and compression loading for thecolumn (remember, wind and earthquake loading can act in either direction).
Answer:
a) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
Explanation:
Given data :
D = 150k , L = 280k , Lr = 40k , s = 50k , w = ± 120k
E = ± 200k
attached below is a detailed solution to the given problem ( problem 1 )
A) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
3.) Technician A says that a scan tool can be used to verify engine operating temperature,
Technician B says that a refractometer can be used to verify engine operating temperature.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B
Engineer drawing:
How can i draw this? Any simple way?
PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%
Answer:
Attached below
Explanation:
PWM signal source has 1 KHz base frequency
Analog filter : with time constant = 0.01 s
low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]
PWM duty cycle is a constant block
Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
Perform the following unit conversions. Please do not use an on-line unit converter since this problem is given to you as practice in preparation for what you need to be proficient in:
a. 180 in^3 to L
b. 750 ft-lbf to kJ
c. 75.0 hp to kW
d. 2500.0 lb/h to kg/s
e. 120 psia to kPa
f. 120 psig to kPa
g. 300 ft/min to m/s
h. 125 km/h to miles/h
i. 6000 N to Ibf
j. 6000 N to ton
Answer:
The answers are below
Explanation:
a. 180 in^3 to L
1 in³ = 0.0164L
180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]
b. 750 ft-lbf to kJ
1 ft-lbf = 0.00136 kJ
750 ft-lbf = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]
c. 75.0 hp to kW
1 hp = 0.746 kW
75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]
d. 2500.0 lb/h to kg/s
1 lb/h = 0.000126 kg/s
2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]
e. 120 psia to kPa
1 psia = 6.89 kPa
120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]
f. 120 psig to kPa
1 psig = 6.89 kPa
120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]
g. 300 ft/min to m/s
1 ft/min = 0.005 m/s
300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]
h. 125 km/h to miles/h
1 km/h = 0.62 mph
125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]
i) 6000 N to Ibf
1 N = 0.2248 lbf
6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]
j. 6000 N to ton
1 N = 0.000102 Ton-force
6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]
2.) Technician A says that milky colored ATF could indicate a leaking transmission cooler in the radiator.
Technician B says that milky colored ATF could indicate the presence of leak detection dye.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B
