3rd Materials Flashcards
(24 cards)
- deformation (change in size & shape of body) is caused by …
-area under F-x graph reps…
defor caused by = tensile (stretch) or compressive (compress) forces
(forces and deformations will be assumed to be in one dimension only)
- area under the force–extension graph represents the work done
force-extension graphs.
-Every material has a unique force-extension graph depending on its properties
-contains a linear and then a non-linear region
-limit of P = point at which the force-extension graph changes from the linear to the non-linear region ;; right before.
area under = work done to stretch material
-work done in stretching a material is equal to the force applied multiplied by the extension created
-obeys Hooke’s Law => W=1/2Fx
-doesn’t => W = sum of the areas of the separate sections under the graph
eg triangle + trapezium areas, 3 sf data so ans to 3 sf.
worked egs
- rubber band that is stretched almost to its breaking point;; ans: A, shape ◞’ (more / at bottom)
bc - after the section of linear proportionality (the straight line), the gradient increases significantly, so, a large force is required to extend the rubber band by even a small amount
Graph B is incorrect as the gradient decreases, suggesting that less force is required to cause a small extension
Graph C is incorrect as this shows a material that does not break easily, such as a metal
Graph D is incorrect as the plateau suggests no extra force is required to extend the rubber as it has already been stretched
eg 2
STRETCHY MATERIAL - rubber: X, extends considerably then breaks with little force, low tensile strength.
DUCTILE MATERIAL - Y: steel, metal (crying stickman); elastic deformation, then plastic deformation, very high tensile strength.
BRITTLE MATERIAL - F: STRAIGHT DIAGONAL LINE, glass ;; breaks with little or no plastic deformation, elastic behaviour shown until breakpoint
1 load
2 extension
3 compression
4 limit of proportionality;; “upper end of proportional relationship”
load: applied force, attached to end of spring.
extension: increased length of spring due to applied force.
(extended) - (natural)
compression: decreased length of spring due to force
(natural) - (compressed)
limit of P: the point beyond which the spring is no longer able to return to its natural length after the load is removed and so longer obeys Hooke’s Law; force and extension are no longer proportional.
natural length of spring: length without any force applied.
Hooke’s law
extension is (directly) proportional to (applied) force
as long as limit of proportionality is not reached
formula for
spring constant
k = F / x
or F=kx.
F = N (newtons), k = N/m, x=extension (m).
k = spring constant
define?
[a measure of the stiffness of a spring. Stiffer spring will have a larger value of k.
k = the gradient of the linear part of a force-extension graph, but if Newtons on x-axis then 1/gradient]
;; force per unit extension up to the limit of proportionality
(tensile) stress
(applied) force / [or per unit] cross-sectional area of material.
σ Pa = F N / A m²
+ ultimate tensile stress: maximum force per original cross-sectional area a wire is able to support at the point it breaks
strain
extension / [or per unit] original length
+ -strain= deformation of a solid due to stress in the form of elongation or contraction
-strain, ε , is a dimensionless unit because it is the ratio of lengths given by the equation
the Young modulus
-define: stress/strain
FL / Ax or FL/Ae - 📌 ratio of tensile stress (σ) to tensile strain (ε). 📌
a measure of the ability of a material to withstand changes in length when a load is added ;; a measure of how stiff or elastic a material is
-units: Pa
-if material shows elastic behaviour, stress and strain, like force and extension, are also directly proportional to one another
-shown by drawing a stress-strain graph
-gradient of the 📍LINEAR 📍 section of a stress-strain graph is the Young modulus
A metal wire that is supported vertically from a fixed point has a load of 92 N applied to the lower end. Wire has a cross-sectional area of 0.04 mm2 and obeys Hooke’s law. The length of the wire increases by 0.50%.
What is the Young modulus of the metal wire?
ans: D) 4.6x10¹¹
-convert mm2 to m2 10-3 x 10-3 m = 0.04 x 10-6 m2
-find stress with F/A = 2.3x10^9
-find strain; 0.005L/L = 0.005.
-stress / strain = 2.3x10^9 / 0.005 = 4.6x10^11
describe an experiment to determine the Young modulus of a metal in the form of a wire
[4 or 5-marker] 📌
-wire clamped over a pulley
-tape or reference marker is needed on the wire to accurately measure the extension with the applied load
-iv load, dv extension
- Measure original length of wire using a metre ruler and mark this reference point with tape
- Measure the diameter of the wire with a micrometre screw gauge in several places and calculate an average
- Add the first mass and calculate the weight used to create the extension e.g. 300 g
- Record total length from the reference point on the metre ruler
- Add more masses and record the new total length from the metre ruler until at least 5 data sets obtained
- Subtract the original length from the new total length to obtain the extension
- Determine extension from final and initial readings
- Plot a graph of force against extension and draw a line of best fit
- Determine the gradient of the force-extension graph
- Calculate the cross-sectional area A of the wire from: A = πd²/4 [d=diameter]
- Calculate Young modulus. E = FL/Ax = gradient x L/A where gradient = of force-extension graph we plotted, L=initial length of wire, and A=average cross-sectional area
Improving experiment and reducing uncertainties
- Reduce uncertainty of the cross-sectional area by measuring the diameter d in several places along the wire and then calculating an average
+ - 2. After each reading remove the load and check that the wire returns to its original shape after each reading
- Take several readings with each load and find the average extension
- Use a Vernier scale to measure the extension of the wire
- use longer length of wire
Measurements required to determine the Young modulus
tables. DETAILS 📍
Mass/kg
Load/N
Initial length/mm
Final length/mm
Extension/× 10−³ m
+TABLE WITH ADDITIONAL DATA+
Length L /m
Diameter 1/mm
Diameter 2/mm
Diameter 3/mm
Average diameter/mm
elastic deformation
(returns)
when the load is removed, the object WILL return to its original natural length
plastic deformation
-is beyond elastic limit
permanently deformed; when the load is removed, the object will not return to its original length
elastic limit
-Below elastic limit, material exhibits elastic behaviour & returns to its original shape. Above elastic limit, material exhibits plastic behaviour.
📌maximum stress/force a material can withstand before it permanently deforms.📌
-point beyond which the object does not return to its original length when the load is removed
for a material deformed within its limit of proportionality [LINE IS STRAIGHT],
formula
potential energy =
1/2Fx
or 1/2kx²
Elastic potential energy
-energy stored within a material (e.g. in a spring) when it is stretched or compressed
-work done is also equal to the elastic potential energy stored in the material when it demonstrates elastic behaviour up to the limit of proportionality
brittle (glass & concrete)
FRACTURES WITH NO DEFORMATION
- have very little to no plastic deformation
-breaks with little elastic and insignificant plastic deformation
-is represented by a straight line through the origin with no or negligible curved region
ductile (eg rubber & copper)
- larger plastic region
-stretches into a new shape before breaking
-represented with a straight line through the origin then curving towards the x-axis
tips:
- total length of material after load placed + extended = original length + extension = final full length.
- suggest improvement for young mod EXP.
- elastic limit and limit of proportionality are not the same point on the graph
