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snow_lady [41]
2 years ago
15

While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently

seal sidewall area punctures. Technician B says that most of these tires have reinforced sidewalls that are able to support the vehicle at any speed without air in the tire. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Engineering
1 answer:
musickatia [10]2 years ago
4 0

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

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