Given the following properties, how much stretch would you expect to occur in a specimen of polyethylene to which a tensile load is applied?
l = change in length
Length = 10 cm
Cross-Sectional Area = 1 cm2
Young's Modulus = 200 GPa
Applied Load = 1000 kN
Find answer in cm.
The answer is 0.5 cm; however, I cannot figure out how to find this answer. Any help is appreciated.
Lengthchange= length*load/Y *1/area
= .1m*1000kN/200GN/m^2*1/1E-4m^2=
=.1*1E6/200E9 *1/E-4=.1/2 * 1E(6-11+4)
= .05E-1 m=.5cm
Where did you find this equation? I didn't find anything like that.
To find the stretch in a specimen of polyethylene, you can use Hooke's Law, which relates the applied force, the Young's modulus, and the change in length of the specimen.
Hooke's Law states: Stress = Young's Modulus x Strain
The stress applied to the specimen can be calculated using the formula: Stress = Force / Cross-Sectional Area
Given the properties:
Length = 10 cm
Cross-Sectional Area = 1 cm2
Young's Modulus = 200 GPa (1 GPa = 1 × 10^9 Pa)
Applied Load = 1000 kN (1 kN = 1 × 10^3 N)
First, convert the given values to the SI unit system:
Length = 10 cm = 0.1 m
Cross-Sectional Area = 1 cm2 = 0.0001 m2
Young's Modulus = 200 GPa = 200 × 10^9 Pa (1 GPa = 1 × 10^9 Pa)
Applied Load = 1000 kN = 1000 × 10^3 N (1 kN = 1 × 10^3 N)
Now, let's calculate the stress:
Stress = Force / Cross-Sectional Area
= (1000 × 10^3 N) / (0.0001 m2)
= 10 × 10^9 N/m2
= 10 GPa
Next, we can use Hooke's Law to calculate the strain:
Stress = Young's Modulus x Strain
10 GPa = (200 × 10^9 Pa) x Strain
Strain = 10 GPa / (200 × 10^9 Pa)
= 0.00005
Finally, we can calculate the change in length using the strain and the original length:
Change in Length = Strain x Length
= 0.00005 x 0.1 m
= 0.005 m
= 0.5 cm
Therefore, the stretch in the polyethylene specimen when a tensile load is applied is 0.5 cm.