Sample 1, Original Length 250 mm and 10 mm Diameter, Length after testing 250.23 mm the force used for the test was 15KN
Sample 2, Original Length 250 mm and 14 mm Diameter, Length after testing 250.21 mm the force used for the test was 15KN
1. Calculate the Stress and Strain for each of the Samples and from this and, assuming the material has remained within the elastic limit, determine their modulus of elasticity.
help, please.
To calculate the stress and strain for each sample and determine its modulus of elasticity, we can use the following formulas:
Stress (σ) = Force (F) / Area (A)
Strain (ε) = Change in Length (ΔL) / Original Length (L)
Modulus of Elasticity (E) = Stress (σ) / Strain (ε)
Let's calculate the values step by step for each sample:
Sample 1:
Original Length (L) = 250 mm
Diameter (d) = 10 mm
Original Area (A) = π(d/2)^2 = π(10/2)^2 = 78.54 mm^2
Length after Testing (Lt) = 250.23 mm
Change in Length (ΔL) = Lt - L = 250.23 - 250 = 0.23 mm
Force (F) = 15 KN = 15,000 N
Stress (σ) = F / A = 15000 / 78.54 = 191.07 N/mm^2
Strain (ε) = ΔL / L = 0.23 / 250 = 0.00092
Modulus of Elasticity (E) = σ / ε = 191.07 / 0.00092 = 207,989.13 N/mm^2
Therefore, the modulus of elasticity for Sample 1 is approximately 207,989.13 N/mm^2.
Sample 2:
Original Length (L) = 250 mm
Diameter (d) = 14 mm
Original Area (A) = π(d/2)^2 = π(14/2)^2 = 153.94 mm^2
Length after Testing (Lt) = 250.21 mm
Change in Length (ΔL) = Lt - L = 250.21 - 250 = 0.21 mm
Force (F) = 15 KN = 15,000 N
Stress (σ) = F / A = 15000 / 153.94 = 97.37 N/mm^2
Strain (ε) = ΔL / L = 0.21 / 250 = 0.00084
Modulus of Elasticity (E) = σ / ε = 97.37 / 0.00084 = 115,916.67 N/mm^2
Therefore, the modulus of elasticity for Sample 2 is approximately 115,916.67 N/mm^2.
To calculate the stress and strain for each sample, we can use the formulas:
Stress = Force / Area
Strain = Change in Length / Original Length
1. Sample 1:
Original Length = 250 mm
Diameter = 10 mm
Radius (r) = Diameter / 2 = 10 mm / 2 = 5 mm = 0.005 m (converting to meters)
Area = π * r^2
= π * (0.005 m)^2
= π * 0.000025 m^2
≈ 0.0000785 m^2
Change in Length = Length after testing - Original Length
= 250.23 mm - 250 mm
= 0.23 mm
= 0.00023 m (converting to meters)
Stress = Force / Area
= 15 kN / 0.0000785 m^2
≈ 191,082 kPa
Strain = Change in Length / Original Length
= 0.00023 m / 0.25 m
= 0.00092
Now, if we assume that the material has remained within the elastic limit, we can use the stress-strain relationship to determine the modulus of elasticity.
Modulus of Elasticity (E) = Stress / Strain
= 191,082 kPa / 0.00092
≈ 207,857 kPa (rounded to nearest whole number)
2. Sample 2:
Following the same steps as for Sample 1:
Original Length = 250 mm
Diameter = 14 mm
Radius (r) = Diameter / 2 = 14 mm / 2 = 7 mm = 0.007 m (converting to meters)
Area = π * r^2
= π * (0.007 m)^2
= π * 0.000049 m^2
≈ 0.0001539 m^2
Change in Length = Length after testing - Original Length
= 250.21 mm - 250 mm
= 0.21 mm
= 0.00021 m (converting to meters)
Stress = Force / Area
= 15 kN / 0.0001539 m^2
≈ 97,535 kPa
Strain = Change in Length / Original Length
= 0.00021 m / 0.25 m
= 0.00084
Modulus of Elasticity (E) = Stress / Strain
= 97,535 kPa / 0.00084
≈ 116,226 kPa (rounded to nearest whole number)
Therefore, the modulus of elasticity for Sample 1 is approximately 207,857 kPa, and for Sample 2 is approximately 116,226 kPa.