How Do You Calculate the Modulus of Elasticity from Load and Extension Data?

[lordajm]

new to physics can some one please help me with b and c

Load (kN) 0 20 30 50 60 73 90 94
Extension (mm) 0 0.150 0.225 0.375 0.453 0.550 0.900 1.100

a) Draw a graph with load on the y-axis and extension on the x-axis (a load extension graph). Title the graph and Label the axis.
b) The diameter of the wire is 0.00735 m and its length is 36.75 mm. Determine the modulus of elasticity, E of the sample wire.
c) Select from the graph an appropriate point for the elastic limit of the wire and give the value of the:
(i) load,
(ii) stress if the diameter is 0.00735 mm.

 

[LightHero]

a) The graph should look something like this: Load (kN) | | * | * | * | * | * |_______*____ 0 20 30 50 60 73 90 94Extension (mm)Graph Title: Load Extension Graphx-axis Label: Extension (mm)y-axis Label: Load (kN)b) The modulus of elasticity, E, can be calculated using the following formula: E = (load x length) / (extension x area)E = (90 x 36.75) / (0.9 x 0.00735)E = 2.62 x 10^7 Pac) (i) The load at the elastic limit is 60 kN. (ii) The stress at the elastic limit is 8.16 x 10^7 Pa.

 

[astrokreng]

a) The graph should be titled "Load vs. Extension" and the x-axis should be labeled as "Extension (mm)" and the y-axis should be labeled as "Load (kN)". The points should be plotted and connected with a line to show the relationship between load and extension.

b) To determine the modulus of elasticity, we need to use the formula E = stress/strain. From the given data, we can calculate the strain (ε) using the formula ε = extension/length. Therefore, for the point where the extension is 0.900 mm, the strain would be 0.900/36.75 = 0.0245. To calculate the stress, we use the formula stress = load/area. The area can be calculated using the formula A = πr^2, where r is the radius of the wire. Since the diameter is given, we need to divide it by 2 to get the radius. Therefore, the area would be π(0.00735/2)^2 = 4.2379x10^-5 m^2. Now, we can calculate the stress for the point where the extension is 0.900 mm, which would be 90/4.2379x10^-5 = 2.123x10^6 Pa. Finally, we can calculate the modulus of elasticity using the formula E = stress/strain, which would be 2.123x10^6/0.0245 = 8.672x10^7 Pa or 86.72 GPa.

c) The elastic limit is the point where the material stops behaving elastically and starts to deform permanently. From the graph, we can see that the point of 0.550 mm extension and 50 kN load is where the line starts to curve, indicating the elastic limit. Therefore, the appropriate point for the elastic limit of the wire would be at 0.550 mm extension and 50 kN load. To calculate the stress at this point, we can use the same formula as in part b, which would be 50/4.2379x10^-5 = 1.180x10^6 Pa. To calculate the stress if the diameter is 0.00735 mm, we need to use the formula stress = load/area, where the area would be π(0.00735/2)^2 = 4.2379x