Question

The figure below shows force (N) versus distance (mm) plots of tensile test data for nylon and LDPE films (30 um thickness: 2.5 cm width). The initial length of the test strips was 8 cm. Labels 1 and 2 indicate specimens cut from the same web in either machine direction or transverse direction. Both graphs show the same data but in different axis scales. 

a) Calculate ultimate tensile strength (in MPa), elongation at break (in %), apparent modulus (in MPa) and toughness (in J/m3)values for all four films. Hint: toughness is equivalent to the area under the stress-strain curve, which can be estimated from the number of squares between the curve and X-axis or using basic geometries.

b) Which film, nylon or LDPE, is likely to be uniaxial oriented? Justify your answers.

c) Between Nylon and LDPE which film is tougher, and which is stiffer? Justify your answer.

Answer 1

Ultimate tensile strength = Maximum load / Area of cross section

Elongation at break = (increase in distance at break / Initial length) * 100

Apparent modulus = stress / strain (within elastic limit and considering the curve is linear)

Toughness = area under stress strain curve = No. of boxes * Area of Each box

stress = applied force / cross sectional area

strain = change in length(distance in graph) / initial length of film

Given: (for all the 4 films)

Initial length of film = 8cm = 80 mm

Thickness = 30 micrometer = 0.03 mm

width = 2.5 cm = 25 mm

Cross sectional area of the film = width * Thickness = 0.03 * 25 = 0.75 mm²

Area of Each box in graph 2 = 10 *100 = 1000 N mm

convering force into stress and distance into strain 1000 / (0.75mm² * 80mm)

Area of Each box = 16.7 N mm / m= 16.7 N m / (m³  *1000000) = 16.7 * 10⁶ N m / m³ = 16.7 * 10⁶ J / m³

Nylon 1:

From Graph:( values are taken approximatly by seeing the graph)

Maximum Force = 75 N => Ultimate strength = 75 / 0.75 = 100 N/mm² = 100 MPa

Increase in distance at break = 295 mm => Elongation at break = (295/80) * 100 = 369.75%

Within Elastic limit, ( chossing a point near the end of elastic limit (straicht line portion) )

Force = 14 N => stress = 14/0.75 = 18.7 Mpa

Displacement (when F = 14 N) = 4 mm => strain = 4/80 = 0.05 mm/mm

Therefore, Apparant modulus = stres / strain = 18.7/0.05 = 374 MPa

No of boxes below the graph = 10.75 (Approx.)

=> Toughness = 10.75 * 16.7 * 10⁶ J / m³ = 180 * 10⁶ J / m³  (Approx.)

Nylon 2:

From Graph:( values are taken approximatly by seeing the graph)

Maximum Force = 68 N => Ultimate strength = 68 / 0.75 = 90.7 N/mm² = 90.7 MPa

Increase in distance at break = 290 mm => Elongation at break = (290/80) * 100 = 362.5%

Within Elastic limit, ( chossing a point near the end of elastic limit (straicht line portion) )

Force = 15 N => stress = 15/0.75 = 20 Mpa

Displacement (when F = 15 N) = 4.4 mm => strain = 4/80 = 0.055 mm/mm

Therefore, Apparant modulus = stres / strain = 20/0.055 = 363.6 MPa

No of boxes below the graph = 10.5 (Approx.)

=> Toughness = 10.5 * 16.7 * 10⁶ J / m³ = 175.3 * 10⁶ J / m³  (Approx.)

LDPE 1:

From Graph:( values are taken approximatly by seeing the graph)

Maximum Force = 46 N => Ultimate strength = 46 / 0.75 = 61.3 N/mm² = 61.3 MPa

Increase in distance at break = 440 mm => Elongation at break = (440/80) * 100 = 550%

Within Elastic limit, ( chossing a point near the end of elastic limit (straicht line portion) )

Force = 10 N => stress = 10/0.75 = 13.3 Mpa

Displacement (when F = 10 N) = 5.2 mm => strain = 5.2/80 =0.065 mm/mm

Therefore, Apparant modulus = stres / strain = 13.3/0.065 = 204.6 MPa

No of boxes below the graph = 9 (Approx.)

=> Toughness = 9 * 16.7 * 10⁶ J / m³ = 150.3 * 10⁶ J / m³  (Approx.)

LDPE 2:

From Graph:( values are taken approximatly by seeing the graph)

Maximum Force = 38 N => Ultimate strength = 38 / 0.75 = 50.7 N/mm² = 50.7 MPa

Increase in distance at break = 520 mm => Elongation at break = (520/80) * 100 = 650%

Within Elastic limit, ( choosing a point near the end of elastic limit (straight line portion) )

Force = 11 N => stress = 11/0.75 = 14.7Mpa

Displacement (when F = 11 N) = 6 mm => strain = 6/80 = 0.075 mm/mm

Therefore, Apparant modulus = stres / strain = 11/0.075 = 146.7 MPa

No of boxes below the graph = 8 (Approx.)

=> Toughness = 8 * 16.7 * 10⁶ J / m³ = 133.6 * 10⁶ J / m³  (Approx.)

b)

LDPE is likely to be Uniaxially oriented.

Justification:

Uniaxially oriented film exhibits ductile yielding in one direction and brittle yielding in another. Ductile yielding will result in a large area below the stress-strain graph and thus result in higher toughness. Comparing the difference in the toughness of films 1 and 2 for both nylon and LDPE fiber, LDPE fiber has a more significant difference in its toughness. So it is likely to be Uniaxially oriented.

c)

Nylon is both Tougher and Stiffer than LDPE.

Justification:

As Nylon has a more excellent toughness value than LDPE from our calculations.

Stiffness is defined as the force required to produce uint displacement.

From the graph, it seems that to create a displacement of 300mm, Nylon needs Nearly 70N, and LDPE needs only 40N, so Nylon is stiffer.