I think option c 12 is currect
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is ![8.621 \times 10^{-4} in/in](https://tex.z-dn.net/?f=8.621%20%5Ctimes%2010%5E%7B-4%7D%20in%2Fin)
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
![\frac{M}{I}=\frac{F}{y}](https://tex.z-dn.net/?f=%5Cfrac%7BM%7D%7BI%7D%3D%5Cfrac%7BF%7D%7By%7D)
Here
- M is the moment which is to be calculated
- I is the moment of inertia given as
![I=\frac{bd^3}{12}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bbd%5E3%7D%7B12%7D)
Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"
![I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bbd%5E3%7D%7B12%7D%5C%5CI%3D%5Cfrac%7B0.75%5Ctimes%208%5E3%7D%7B12%7D%5C%5CI%3D32%20in%5E4)
![y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bd%7D%7B2%7D%5C%5Cy%3D%5Cfrac%7B8%7D%7B2%7D%5C%5Cy%3D4%22%5C%5C)
![\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in](https://tex.z-dn.net/?f=%5Cfrac%7BM_y%7D%7BI%7D%3D%5Cfrac%7BF_y%7D%7By%7D%5C%5CM_y%3D%5Cfrac%7BF_y%7D%7By%7D%7BI%7D%5C%5CM_y%3D%5Cfrac%7B50%7D%7B4%7D%7B32%7D%5C%5CM_y%3D400%20k.%20in)
The yield moment is 400 k.in.
Part b:
The strain is given as
![Strain=\frac{Stress}{Elastic Modulus}](https://tex.z-dn.net/?f=Strain%3D%5Cfrac%7BStress%7D%7BElastic%20Modulus%7D)
The stress at the station 2" down from the top is estimated by ratio of triangles as
![F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi](https://tex.z-dn.net/?f=F_%7B2%22%7D%3D%5Cfrac%7BF_y%7D%7By%7D%5Ctimes%202%22%5C%5CF_%7B2%22%7D%3D%5Cfrac%7B50%20ksi%7D%7B4%22%7D%5Ctimes%202%22%5C%5CF_%7B2%22%7D%3D25%20ksi)
Now the steel has the elastic modulus of E=29000 ksi
![Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in](https://tex.z-dn.net/?f=Strain%3D%5Cfrac%7BStress%7D%7BElastic%20Modulus%7D%5C%5CStrain%3D%5Cfrac%7BF_%7B2%22%7D%7D%7BE%7D%5C%5CStrain%3D%5Cfrac%7B25%7D%7B29000%7D%5C%5CStrain%3D8.621%20%5Ctimes%2010%5E%7B-4%7D%20in%2Fin)
So the strain is ![8.621 \times 10^{-4} in/in](https://tex.z-dn.net/?f=8.621%20%5Ctimes%2010%5E%7B-4%7D%20in%2Fin)
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as
![shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi](https://tex.z-dn.net/?f=shape%5C%2C%20factor%3D%5Cfrac%7BPlastic%5C%2C%20Moment%7D%7BYield%5C%2C%20Moment%7D%5C%5C%7BPlastic%5C%2C%20Moment%7D%3Dshape%5C%2C%20factor%5Ctimes%20%7BYield%5C%2C%20Moment%7D%5C%5C%7BPlastic%5C%2C%20Moment%7D%3D1.5%5Ctimes400%20ksi%5C%5C%7BPlastic%5C%2C%20Moment%7D%3D600%20ksi)
The plastic moment is 600 ksi.
A material scientist investigates the characteristics of materials and their uses. Nature and human knowledge restrict the number of existing materials on Earth. This signifies that there is a limited amount of material available for use in the production of goods.
Given that they deal with the design and manufacturing of things, mechanical engineers sometimes lack the specialties seen in other branches of engineering. This necessitates a thorough understanding of all elements of engineering.
<h3>Why does Absolute Zero Exist?</h3>
Absolute zero, also known as zero kelvins, is equivalent to 273.15 degrees Celsius, or -459.67 degrees Fahrenheit, and represents the point on a thermometer where a system has the least amount of energy, or thermal motion. But there's a catch: absolute zero is unattainable, But exists as a measure of relativity.
Absolute zero is the temperature at which a thermodynamic system has the least amount of energy. It is equivalent to 273.15 degrees Celsius on the Celsius scale and 459.67 degrees Fahrenheit on the Fahrenheit scale.
Learn more about Absolute Zero:
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Answer:
Proper horizontal and vertical adjustments.
Explanation:
Proper horizontal and vertical adjustments will ensure that you can see far enough down the road to react to obstacles or avoid animals.