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a_sh-v [17]
3 years ago
11

A three-point bending test is performed on a glass specimen having a rectangular cross section of height 5.3 mm and width 11.6 m

m; the distance between support points is 50 mm.
Required:
a. Calculate the flexural strength (in MPa) if the load at fracture is 283 N. (Round up the final number to integer-no decimal please.)
b. What is the deflection, Δy=__________ mm, for a load of 265 N if the elastic modulus if the glass is 66 GPa?
Engineering
1 answer:
wolverine [178]3 years ago
4 0

Answer:

a) 65.139 MPa

b) 17490 GPa

Explanation:

a) calculate the flexural strength

we now that for a three-point bending test the bending moment will occur at the midsection

Max stress ( flexural strength ) = MY / I

Max stress can be expressed as ;  3/2 * ( F*l / b*d^2 )  ------- ( 1 )

F ( load at fracture ) = 283

l (distance between support point ) = 50

b = 11.6

d = 5.3

substitute given data into equation 1 above

= 3/2 * (( 283 * 50) /( 11.6 * 5.3^2) )

= 3/2 * ( 14150 / 325.844 )

=  3/2 * 43.425  =  65.139 MPa

B ) Determine deflection ( Δ y )

   load = 265 N

   elastic modulus = 66 GPa

elastic modulus = tensile stress / tensile strain

Tensile stress = Δ y

 strain = load = 66 GPa

therefore Δ y = 265 * 66 = 17490 GPa

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Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

m_{Fe} = 97.5\,g

If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

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Answer:

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We need to take the input from the user

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Code

#include<stdio.h>

#include<math.h>

struct Point{

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int manhattan(Point A, Point B){

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float euclidean(Point A, Point B){

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int main(){

  struct Point A, B;

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