Normal Elasticity
Try out the following problem:
Suppose the green beams have negligible weight and are rigid (can’t bend or deform), and both springs are initially unstretched. If the spring constants are
Copyright By PowCoder代写 加微信 powcoder
1) Find the force provided by each spring, and the displacement of point E.
2) Draw an internal force and bending moment diagram for ABCDE.
Hint: Because the spring constants are so large, assume that the displacement will be so little that point E only moves vertically and angle angles stay basically the same.
Equilibrium for the frame means we’ll get 3 equations (no net force in x or y, and no net torque about any point), but we have 4 unknown forces:
However, the forces supplied by the springs aren’t really independent; they’re both uniquely determined by the deformation of the structure; i.e., how much the rigid beam rotates about point A determines each spring’s deformation, and that determines the spring’s force:
From similar triangles, the displacement that point E moves down determines the other two displacements:
(as a ratio, it’s the same with or without units in the denominator so they cancel here; this isn’t just being lazy!)
The spring forces are related to these extensions too; since the other sides of the springs are fixed points, the extension of spring BF is this -yB we’ve shown (the amount of downward displacement of point B), so
while the extension of spring CH is +yC (the amount point C moves up), meaning
These last two equations introduce only one new unknown () so together with the 2d net force and torque equations for ABCDE we’ll have 5 equations in 5 unknowns → unique solution possible!
> restart: kBF:=1e3/1e-3: kCH:=2e3/1e-3:
Ay+BF-CH-2e3,
.15*BF-.4*CH-200-.67*2e3,
BF=kBF*15/67*(-yE),
CH=kCH*40/67*(+yE)
Internal force and bending moment diagram:
> N:=Ax+piecewise(x>.55, -1e3):
V:=Ay+piecewise(x>.15, BF)+piecewise(x>.4, -CH)+piecewise(x>.67, -2e3):
M:=int(V, x=0..x)+piecewise(x>.55, 200):
plot([N,V,M], x=0..0.671);
Now, consider the following problem from a past EP2P04 Test:
Rigid member ABCDE is supported by three pin connections (at A, B, and C). All triangles are fixed in position. Members BF and CH are not rigid – each has a Young’s modulus of 1 GPa. They have cross sectional areas of 3 cm2 and 6 cm2, respectively. DG is rigid and connected to ABCDE via a fixed support at point D.
a) Determine the displacement of point E and the stress in members BF and CH. [5 marks]
b) Draw a shear force and bending moment diagram for ABCDE. [5 marks]
Hint: 1 cm2 is not 0.01 m2.
If we can determine the effective spring constant of members BF and CH, then this problem becomes identical to the first one. This section will develop relations for how solid objects deform in response to normal stresses and let us determine these spring constants in this and more complicated situations (e.g., the beams BF are not free to expand in one or both of the lateral directions, leading to higher effective spring constants via the Poisson’s ratio).
(Note: it will turn out for this problem that the spring constant is , where is BF’s cross-sectional area, is its effective elastic modulus in the extension direction, and is its length. We’ll see why in the following sections!)
The strategy to solve problems here will generally be:
1. Model the objects that can deform (like BF and CH above), using physics to determine their “effective spring constant” (how they’ll relate deformations to forces)
a. Sometimes this will involve using matrix methods in Maple or FEM simulations with FlexPDE to confirm these results, or deal with complicated boundary conditions.
2. Use an approach like the spring problem above to solve for the equilibrium of an extended object impacted by these “springs”.
Perspective: This week is only the first of 4 weeks on static elasticity (that’s right; not static electricity, but static elasticity):
1. Normal forces and stresses and response to them
a. Can analytically solve problems with normal forces
b. Can numerically solve problems with normal stresses
2. Shear forces and stresses and response to them
a. Can analytically also solve problems with normal forces, and pure shear only (no bending or torsion allowed)
b. Can numerically solve any problem
3. Bending moments
a. Can analytically also solve problems with bending moments
4. Torsion
a. Can analytically also solve problems with twisting moments
Intro to Elasticity
While for many systems it’s a good approximation, there’s no such thing as a perfectly rigid body – real materials have some degree of elasticity: they deform in response to the internal forces and moments we calculated in the last section. Since solids are made up of atoms chemically bonded to each other, and chemical bonding is governed by the electrostatic force, this means that:
1. Solids deform in response to force (they’re not perfectly rigid)
1. Solids exhibit a linear restoring force in response to deformation (like a spring), at least for small deformations
1. If the deformation is too large, the restoring force becomes nonlinear (the electrostatic attraction in chemical bonding depends on the arrangement of atoms, and if we start rearranging them or moving them far apart, we can cause permanent or “plastic” deformation).
This course will be concerned with “small” deformations, meaning linear elasticity holds. We’ll look at some problems that your friends in Engineer 2P04 are working on which don’t need computers to solve (but you still get to use them to save time), and then tackle some more elasticity problems using Maple and FlexPDE.
Axial Stress & Strain
Recall that a spring exerts a restoring force in response to a change in its length given by the spring “constant” ; The negative sign means that the force opposes the change in length:
1. if you pull a spring into tension it pulls back on you to try to compress, and
1. if you push a spring into compression it pushes back on you to try to extend.
Notice that the net force on the spring is in the same direction as its extension:
Any real material has some degree of springiness and so applied forces cause a length change proportional to the force, (at least until it reaches its “elastic limit” where the material starts to deform in other ways depending on the type of material). The spring constant for a beam with length L and uniform cross-sectional area A is:
where E is a material parameter called the Young’s modulus, elastic modulus, or elasticity [measured in Pa, typically MPa or GPa]. k is also called the stiffness of the beam.
Substituting the spring constant and rearranging:
(the [normal] force per unit area) is the [normal] stress [in Pa] and
(the change in length per unit length = relative length change) is the [normal] strain [dimensionless].
1. In this context, “normal” means perpendicular, just like in the good ol’ “normal force”
Unlike the force-extension relationship ( ) the stress-strain relationship is only dependent on the material parameter E (and not on the geometry parameters A & L).
3. Normal stress & strain come in two flavours:
a. tensile (positive); stress is pulling away from the surface; object has been stretched out and is pulling back, so needs an outward stress to stop from moving.
b. compressive (negative); stress is pushing into the surface; object has been compressed and is trying to expand back out, so needs an inward stress to stop from expanding.
1 Pa is a very small amount of stress in most contexts; (1 N is about the weight of a chicken egg, so distributed over a contact surface of about 1 cm2 in your hand, holding up the weight of a chicken egg in the palm of your hand puts an extra compressive stress on your skin beneath the egg of about ).
5. Because strain is fractional length change, a strain of +1 means the object doubles in length. Strain of -0.5 means it shrinks to half its original length. Strain of -1 means it’s folded in on itself and is now just as long in the other direction (not possible for uniform solid objects like steel beams, but with some imagination could be applicable to some non-uniformly solid object like a sweater).
Wait a minute: Which length should you use in the denominator of if the length changes?
Generally, you use the length before the change in the denominator: .
However, for most (but not all) situations the strain is very small; e.g., we can estimate the strain limit by dividing the ultimate strength by the elastic modulus (Young’s modulus):
Properties of Annealed Materials:
yield strength (Mpa)
Strength (Mpa)
“Ultimate Strain”
(using upper range of ultimate strength)
Zinc (wrought)
(Start of table is from https://en.wikipedia.org/wiki/Ultimate_tensile_strength )
Note that for most metals the strain limit when they break apart is around 0.1%, and that silicon is a remarkable mechanical material which can be stretched to 80 times the amount of most metals. In fact, the numbers in the table above are just to give you an idea; in reality after the elastic limit materials have a lower effective marginal Young’s modulus as strain increases in ductile plastic deformation. Still, these give you an idea of the order of magnitude of strain.
That said, materials like rubber may have a strain limits of over 100%, so this is highly dependent on the type of material considered. Strain limits may be very different in tension and compression for some materials. e.g., the same piece of concrete (much stronger in compression than tension) may have a limit of 0.7% in compression but only 0.01% in tension.
Example 6.1: Compound Cylinder with Loading
Two solid cylindrical rods AC & CD are both made of the same aluminum alloy with E = 70 GPa, and are welded together at C.
a) Determine the axial stress in regions AB, BC, and CD
b) Determine the total deformation of the composite rod ACD, and the deflection of point C.
The area changes along the bar. Define
This problem is 1d – the bars are free and unloaded in y & z, so we can say that at any point, , where and E is a constant 70 GPa.
Because the loading is changing, we need to be careful about what the axial force is as a function of distance.
At the support, RA = -90+65*2-40 = 0.
Therefore,
From B to C we have a compressive axial force of 90 kN, so . From C to D, we have tensile of 40 kN, and again .
The strain in each section is , and since strain is changing with position, rather than saying , we need to add up the length changes of each individual region; i.e., , where u is the total x-position change of the piece at x, and so .
(This is admittedly overkill for this problem. You could just find for both sections and add them up, but this approach is closer to how we’ll be doing things eventually with FlexPDE).
A1:=Pi*(.06)^2/4:
A2:=Pi*(.045)^2/4:
A:=piecewise(x<.5, A1, x<1, A2): #final point doesn't matter.
F:=piecewise(x<.3, 0, x< .5, -90e3, x<1, 40e3):
sigma:=F/A;
plot(sigma, x=0..1);
ep:=sigma/E:
plot(ep, x=0..1);
DeltaL:=int(ep, x=0..0.9);
DeltaLC:=int(ep, x=0..0.5);
Intuitively, it's hard to guess at the start whether the bar should get shorter or longer in this problem. It's clear that C moves left, but because D moves right, we need to know what moves more. BC has more force but also larger area and shorter length than CD. Looking a bit closer though, BC has about double the force, but also about double the area, so the stress is around the same magnitude in each section. Since it has half the length, CD extends more than BC compresses.
Note that these numbers are pretty reasonable for small displacement.
A strain of 0.2% is a reasonably large strain and often the start of plastic deformation for a metal, where Hooke's law no longer applies (or at least, the spring constant changes).
Example 6.2: ENG 2P04 A9 P10
We can make these problems pretty hard, but the same principles apply.
In the figure shown, the RIGID bar ABC is pin-supported at B, and is connected to steel rods (E = 200 GPa) AD and CE. The rods are 8-mm diameter and are threaded at their upper end with a pitch of 2 mm. After being snugly fitted the nut at A is tightened two full turns (that is, lowered by 4 mm on the rods).
a) Determine the force in each rod
b) Determine the deflection of point A of the rigid bar ABC
Since AC is a rigid member, it doesn't bend. Instead, we have from the torque equation about B:
The length changes are related through AC as well. Calling V the change in y-position,
(used small angle approximation, since the thread pitch is small compared to 150 & 200 mm)
And finally, we know that in total, we know that the deformation of A and C are related to the length change of their rods, minus the amount each nut moved:
And the deformations are related to the force produced through the area and stress:
A:=Pi*(4e-3)^2:
FA=3/4*FC,
VC=-3/4*VA,
VA=2*epA-4e-3,
VC=2.5*epC,
FA=A*E*epA,
FC=A*E*epC]);
So, A moves down and C moves up. This makes sense because A's nut was tightened. What if we were to tighten both nuts?
A:=Pi*(4e-3)^2:
FA=3/4*FC,
VC=-3/4*VA,
VA=2*epA-4e-3,
VC=2.5*epC-4e-3,
FA=A*E*epA,
FC=A*E*epC]);
A still moves down, but not by as much. Why does it win?
1. It's on a shorter rod, so it produces more force per displacement -> won’t be lengthened as much from its ideal length.
1. It’s on the longer end of the pivot arm, so its forces have more effect -> won’t need to produce as much force to balance.
Example 6.3: Hanging Rod
1. The rigid rod ABC is hung by three identical wires. Knowing that x= (2/3) L, determine the tension in each wire due to the force P (hint: assume each wire extension is small compared to L.)
Write the extension of each wire as . Since ABC is a rigid rod, we must have that the extensions are related via:
i.e., by similar triangles:
Also, each wire puts a force on the rod given by (where k is the spring constant), so in terms of the tension, this expression reads:
And we have the net force in y is zero, = 0
And the net moment about, say, A, is zero:
TB-TC=(TA-TC)/2,
TA+TB+TC=P,
TB+2*TC-2/3*P=0],
[TA, TB, TC]);
Extension: thinking about this as a moving line
We can also think more broadly about the geometry equation by imagining the beam moves up along a line with offset at point A of b, and slope of m:
> restart:
TA+TB+TC-P,
L*TB+2*L*TC-x*P,
yC=b+2*L*m,
TA=k*(-yA),
TB=k*(-yB),
TC=k*(-yC)], [TA, TB, TC, yA, yB, yC, b, m]);
This line information incorporates the same information as what we did before, expressing that the displacement of point B is the average of that of points A & C:
> restart:
TA+TB+TC-P,
L*TB+2*L*TC-x*P,
(yB-yA)/L =(yC-yB)/L,
TA=k*(-yA),
TB=k*(-yB),
TC=k*(-yC)], [TA, TB, TC, yA, yB, yC]);
We can also redo the first example in this section using this method. Rather than this:
> restart: kBF:=1e3/1e-3: kCH:=2e3/1e-3:
Ay+BF-CH-2e3,
.15*BF-.4*CH-200-.67*2e3,
BF=kBF*15/67*(-yE),
CH=kCH*40/67*(+yE)
…the same information (including the fact that point A is fixed) looks like this:
> restart: kBF:=1e3/1e-3: kCH:=2e3/1e-3:
Ay+BF-CH-2e3,
.15*BF-.4*CH-200-.67*2e3,
yB = b+.15*m,
yC = b+.4*m,
yE = b+.67*m,
BF=kBF*(-yB),
CH=kCH*(yC)
Poisson’s Ratio: Axial coupling in Stress and Strain
If you compress a material in the x-direction, it tends to expand in the y- and z- directions:
Figure 62 [3]
Specifically, stress and strain along all three axes are in general coupled together:
where (lower case Greek letter “nu”) is a dimensionless factor called the Poisson’s ratio. Poisson’s ratio represents the degree of coupling between the axes of the material i.e. how strongly stresses in one axis result in strains in the other.
In vector form, this coupled-axis stress-strain relation (i.e., Hooke’s law) is
Matrix Multiplication Review
To multiply a matrix (e.g., 3×3 matrix ) by a vector (e.g., 3×1 column-vector ), you dot each row of the matrix by the vector to form the new vector:
In Maple, you can do this operation using the following commands:
> restart:
with(LinearAlgebra):
M:=Matrix([[1,2,3],[0,1,0],[0,0,1]]); #a matrix
r:=
M.r; #r left-multiplied by M
Columns of the Poisson’s ratio matrix map from the x, y and z stresses, and all the entries in the same row of the matrix map to the x, y and z components of the strain in 1 direction (i.e. the first, second and third entries in the first row of the matrix, respectively, contribute to the x, y and z components of the strain in the x direction).
Table 2: Poisson’s ratios for some common materials [4]
Poisson’s ratio
0.4999 [4]
saturated clay
0.252-0.289
0.265-0.34
aluminium-alloy
stainless steel
Poisson’s ratio for almost all materials is between 0 and 0.5.
Steels and polymers are typically around 0.3.
Cork has a Poisson’s ratio of close to 0 (meaning it doesn’t actually expand in the y and z when compressed in x or in the x & y when compressed in z, etc.)
Rubber has a Poisson’s ratio close to 0.5 (meaning it expands in y and z to counteract the compression in x and keep volume constant).
Crystal materials in general have different Young’s moduli and Poisson’s ratios for different directions. For now, we’ll assume all materials are isotropic (same E and for all axes) to make the calculations simpler.
Isotropic materials have identical values of a property in all directions
Check your understanding:
For a material with a typical Poisson’s ratio of 0.3, since stress and strain in other axes can affect the stress and strain in x through Poisson’s ratio, is it ever actually fair to say that like we did earlier?
Answer: Yes – As long as the beam is free in y & z so that , because then:
Check your understanding:
What does a negative Poisson’s ratio mean?
That the material tries to extend in one dimension due to extending in the other, like this folding chair:
(Flash Furniture Folding Camping Chair with Drink Holder, Blue (TY1410BL) from www.staples.ca)
Example 6.4: Tensile Cube Shape & Volume Change
Suppose you apply a tensile stress in the x-direction to an otherwise free cube of side length L, isotropic Young’s modulus E, and Poisson’s ratio . Calculate:
a) its new dimensions and
b) its change in volume.
Figure 63: Elongating in x leads to shrinking in y & z via the Poisson’s ratio; the red cube is the shape before displacement and the yellow prism is the shape afterwards (likely exaggerated!).
The strains are:
and are related to the changes in length via , so the new dimensions are
The new volume is:
(the notation means “+ some other terms that all have or to a higher power in them”, which we can ignore because )
So, the change in volume is:
Considering this change in volume result we can get more insight into Poisson’s ratio: is the degree to which an isotropic material object resists changes in its volume more than changes in its shape. e.g., for rubber with, the object is much more resistant to changes in volume: you can easily deform it, but it takes up nearly
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com