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algol13
3 years ago
5

When a rubber is stretched during a tensile test, its elongation is initially proportional to the applied force, but as it reach

es about twice its original length, the force required to stretch the rubber increases rapidly. The force, as a function of elongation, that was required to stretch a rubber specimen that was initially 3cm long.
(a) Curve-fit the data with a forth order polynomial. Make a plot of the data points and the polynomial. Use the polynomial to estimate the force when the rubber specimen was 11.5cm long.
(b) Fit the data with spline interpolation (use MATLAB built-in function interp1). Make a plot that shows the data points and a curve made by interpolation. Use interpolation to estimate the force when the rubber specimen was 11.5cm long

Table 1

Elongation vs. force for rubber during tensile test.

Elongation (cm) Force (KN)
0 0
1.2 0.6
2.4 0.9
3.6 1.16
4.8 1.18
6.0 1.19
7.2 1.24
8.4 1.48
9.6 1.92
10.8 3.12
12.0 4.14
13.2 5.34
14.4 6.22
15.6 7.12
16.8 7.86
18.0 8.42
Engineering
1 answer:
Softa [21]3 years ago
4 0
I don’t feel like reading
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1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used
bazaltina [42]

Answer:

F_x = 100,200 N

x' = 21.321 cm ... Length

y' =0.7825 cm

z' = 1.565 cm

A' = ( 0.783 x 1.565 ) cm  

Explanation:

Given:

- The Modulus of Elasticity E = 645 MPa

- The poisson ratio v = 0.28

- The Yield Strength Y = 501 MPa

- The Length along x-direction x = 12 cm

- The length along y-direction y = 1 cm

- The length along z--direction z = 2 cm

Find:

The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

Solution:

- The Tensile forces within the limit of proportionality is given as:

                                   F_i = б_i*A_jk

- A maximum tensile Force F_x along x direction can be given as:

                                   F_x = Y*A_yz

                                   F_x = 501*( 0.01*0.02)*10^6

                                  F_x = 100,200 N

- The corresponding strains in x, y and z direction due to F_x are:

                                    ξ_x = Y / E

                                    ξ_x = 501 / 645 = 0.7767

                                    ξ_y = ξ_z = -v*Y / E

                                    ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175

- The corresponding change in lengths at tensile elastic stress are:

                                    Δx = x*ξ_x = 12*0.7767 = 9.321 cm

                                    Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm

                                    Δz = z*ξ_z = - 2*0.2175 = -0.435 cm

- The final lengths are:

                                    x' = x + Δx = 12 + 9.321 = 21.321 cm

                                    y' = y + Δy = 1 - 0.2175 = 0.7825 cm

                                    z' = z + Δz = 2 - 0.435 = 1.565 cm

                                       

3 0
4 years ago
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