Answer:
the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Explanation:
From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
The Critical Stress for a maximum internal crack can be expressed by the formula:
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
where;
[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation
[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa
Y = dimensionless parameter
a = length of the internal crack
given that ;
the maximum internal crack length is 8.6 mm
half length of the internal crack will be 8.6 mm/2 = 4.3mm
half length of the internal crack a = 4.3 × 10⁻³ m
From :
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]
[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]
[tex]Y= \dfrac{26}{13.01748802}[/tex]
[tex]Y=1.99731315[/tex]
[tex]Y \approx 1.997[/tex]
For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
when the length of the internal crack a = 3mm
half length of the internal crack will be 3.0 mm / 2 = 1.5 mm
half length of the internal crack a =1.5 × 10⁻³ m
From;
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]
[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]
[tex]\sigma_c =189.6595506[/tex]
[tex]\sigma_c =[/tex] 189.66 MPa
Thus; the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:
[tex]\sigma_b = \frac{MC}{I}[/tex]
Where,
[tex]\sigma_b[/tex] is the compressive stress or tensile stress[tex]M[/tex] is the B.M [tex]C[/tex] is the N.A distance[tex]I[/tex] is the moment of interiorSo making this formula for the max, we have:
[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]
With all this information we can put the formula as:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
See more about stress in the beam at brainly.com/question/23637191
. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 rpm and the mixer must operate as close to 250 rpm as possible. Select an appropriate belt size, commercially available sheaves, and a belt for this application. Also calculate the actual belt speed and the center distance.
Answer:
Hello the table which is part of the question is missing and below are the table values
For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1
Answers:
belt size = 140 in with diameter of 20.1n
actual speed of belt = 288.49 in/s
actual center distance = 49.345 in
Explanation:
Given data :
Electric motor (driver sheave) speed (w1) = 950 rpm
Driven sheave speed (w2) = 250 rpm
pick D1 ( diameter of driver sheave) = 5.8 in ( from table )
To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first
VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250
also since the speed of belt would be constant then ;
Vb = w1r1 = w2r2 ------- equation 1
r = d/2
substituting the value of r into equation 1
equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex] = VR
Appropriate belt size ( d2) can be calculated as
d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04
From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value
next we have to determine the belt length /size
[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]
inputting all the values into the above equation including the value of C as calculated below
L ≈ 140 in
Calculating the center distance
we use this equation to get the ideal center distance
[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]
22.04 < c < 3 ( 5.8 + 20.1 )
22.04 < c < 77.7
the center distance is between 22.04 and 77.7 but taking an average value
ideal center distance would be ≈ 48 in
To calculate the actual center distance we use
[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3
B = [tex]4L -2\pi (d2 + d1 )[/tex]
inputting all the values into (B)
B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )
B ≈ 399.15 in
inputting all the values gotten Back to equation 3 to get the actual center distance
C = 49.345 in ( actual center distance )
Calculating the actual belt speed
w1 = 950 rpm = 99.48 rad/s
belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]
= 99.48 * 5.8 / 2 = 288.49 in/s
Describe the meaning of the different symbols and abbreviations found on the drawings/documents that they use (such as BS8888, surface finish to be achieved, linear and geometric tolerances, electronic components, weld symbols and profiles, pressure and flow characteristics, torque values, imperial and metric systems of measurement, tolerancing and fixed reference points)
Answer:
Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.
There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.
Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.
Explanation:
An inventor claims to have developed a heat pump that produces a 200-kW heating effect for a 293 K heatedzone while only using 75 kW of power and a heat source at 273 K. Justify the validity of this claim.
Answer:
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
Explanation:
Heat generated Q = 200 kW
power input W = 75 kW
Temperature of heated region [tex]T_{h}[/tex] = 293 K
Temperature of heat source [tex]T_{c}[/tex] = 273 K
For this engine,
coefficient of performance COP = Q/W = 200/75 = 2.67
The maximum theoretical COP obtainable for a heat pump is given as
COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] = [tex]\frac{293 }{293 - 273 }[/tex] = 14.65
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
Who plays a role in the financial activities of a company?
O A. Just employees
O B. Just managers
O C. Only members of the finance and accounting department
O D. Everyone at the company, including managers and employees
Hey,
Who plays a role in the financial activities of a company?
O D. Everyone at the company, including managers and employees
Answer:
Everyone at the company, including managers and employees
Explanation: