- Home
- questions and answers
- 2 download the concept questions page for understanding car crashes wh

Strength of Material

2) Download the Concept Questions Page for Understanding Car Crashes: When Physics Meets Biology, watch the video and complete the concept questions.

Verified

Strength of Material

For the support of Prob. 1.47, knowing that the diameter of the pins d = 16 mm and that the magnitude of the load is P 20 kN,determine (a) the factor of safety for the pin, (b) the required values of b and e if the factor of safety for the wooden member is the same-as that found in part a for the pin.

Read More

Strength of Material

17. Iron has a BCC crystal structure and an atomic radius of 0.1241 nm. Determine the interplanar spacing corresponding to (111) and (211) set of planes. Please look at the previous question to understand what (111) and (211) means.[5 points]

Read More

Strength of Material

16. Determine the expected diffraction angle for the first-order reflection (i.e., n=1) from the (310)set of planes for BCC chromium when monochromatic radiation of wavelength 0.0711 nm is-used. (311) means h=3, k=1, and l=1. Please look at the determination of dnki in the slide set corresponding to Lecture 2. Also note that the diffraction angle means 20. You will determine"0" from the equation, and will have to multiply it by 2.[5 points]

Read More

Strength of Material

14. Explain the differences between ionic and covalent bonding.

Read More

Strength of Material

13. Molybdenum has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute its theoretical density.[3 points]

Read More

Strength of Material

12. Calculate the radius of a Palladium atom if it has a FCC crystal structure, a density of 12.0g/cm3 and an atomic weight of 106.4 g/mol.[3 points]

Read More

Strength of Material

10. Derive the value for the atomic packing factor of a FCC structure.

Read More

Strength of Material

9. In Question 1, it was noted that the net bonding energy En between two isolated positive and negative ions is a function of interionic distance r as follows:

E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}

where A, B, and n are constants for the particular ion pair. This equation is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity "E" is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation that is is E =kx(df/dr)

Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion system) using the following procedure:

\text { a. Establish a relationship for the force } \mathrm{F} \text { as a function of } \mathrm{r} \text {, realizing that } F=\frac{d E_{N}}{d \mathbf{r}}

b. Now take the derivative dF/dr.

c. Substitute the value of ro obtained in Question 1 in the above expression and determine the expression for the modulus of elasticity “E". [Assume k = 2]

Read More

Strength of Material

8. The net potential energy between two adjacent ions, En, may be represented as:

E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}

Calculate the bonding energy Eo in terms of the parameters A, B, and n using the following procedure:

a. Differentiate En with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at Eo.

b. Solve for r in terms of A, B, and n, which yields ro, the equilibrium interionic spacing.

c. Determine the expression for Eo by substitution of ro into the above equation.

Read More

Strength of Material

7. Show that volumetric strain is the sum of strains in three orthogonal directions.

Read More

Getting answers to your urgent problems is simple. Submit your query in the given box and get answers Instantly.

Success

Assignment is successfully created

2017-2022 TutorBin. All Rights Reserved*The amount will be in form of wallet points that you can redeem to pay upto 10% of the price for any assignment.

**Use of solution provided by us for unfair practice like cheating will result in action from our end which may include permanent termination of the defaulter’s account.Disclaimer:The website contains certain images which are not owned by the company/ website. Such images are used for indicative purposes only and is a third-party content. All credits go to its rightful owner including its copyright owner. It is also clarified that the use of any photograph on the website including the use of any photograph of any educational institute/ university is not intended to suggest any association, relationship, or sponsorship whatsoever between the company and the said educational institute/ university. Any such use is for representative purposes only and all intellectual property rights belong to the respective owners.

**Use of solution provided by us for unfair practice like cheating will result in action from our end which may include permanent termination of the defaulter’s account.Disclaimer:The website contains certain images which are not owned by the company/ website. Such images are used for indicative purposes only and is a third-party content. All credits go to its rightful owner including its copyright owner. It is also clarified that the use of any photograph on the website including the use of any photograph of any educational institute/ university is not intended to suggest any association, relationship, or sponsorship whatsoever between the company and the said educational institute/ university. Any such use is for representative purposes only and all intellectual property rights belong to the respective owners.

Get Homework Help Now!

+

a

+

Success

Assignment is successfully created

Error

Please add files or description to proceed