and Strain, and FlexPDE Beam Solving
Introduction
Face names
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Consider a differential volume of material with size :
Faces of the cube can be referred to by referencing the direction of the outward normal vector. i.e., the “+x face” is the face whose normal vector that points out from the cube is in the +x direction.
Alternatively, you can refer to the faces by specifying their equation; i.e., the y = 0 face would be the -y face above, while the face would be the +y face.
Last time we saw that:
normal stress (defined as normal force per area) caused a normal deformation in an otherwise free object of ,
or more generally, considering coupling of stresses in other axes via Poisson’s ratio, of (where is the compliance matrix).
a differential cube with only normal forces acting on it required the force on the positive face to be equal to the force on the negative face to be in equilibrium:
Where , etc., and , the normal stress on the +y face, is the normal force per unit area on that face, which led to .
This time we’ll explore the effects of shear forces and stresses on objects (forces parallel to the faces) and determine:
1. What shear stress is, and how it relates to shear force
2. What shear strain is, and how it relates to shear stress
3. What “pure shear static equilibrium” means
4. Why shear stress is symmetric at any point, and what this means
How locally any object in static equilibrium requires not necessarily , but .
6. How to determine deformations and displacements due to both normal and shear stress in FlexPDE and Maple (as long as the object is loaded to have uniform shear and normal stress and strain).
Simple and Strain Example
In general, shear stress is shear force divided by the area of the face it’s applied on. This tends to deform the shape by sliding material planes relative to each other:
A +x-directed force on the +y-face produces a shear deformation (for the object to not accelerate to the right, there must be an equal and opposite shear force pointing to the left on the -y-face).
Specifically, shear force F produces a shear stress (“” here means on the +y-face in the +x-direction, or on the -y-face in the -x direction) where is the area of the y-face, and leads to a shear strain equal to the rate of change in the x location with respect to y coordinate:.
Shear strain is typically small, so this is approximately equal to the angle between the faces as well, since (where is in radians).
Shear stress and strain are related through the shear modulus G: , while the shear modulus itself is related to the Young’s modulus and the Poisson’s ratio through (at least for an isotropic material!)
The shear modulus is the ratio of shear stress to shear strain.
A large shear modulus tells us that a solid is very rigid and requires a large force in order to deform it. Contrarily, a small shear modulus indicates that a solid if flexible and requires little force to produce deformation.
Compared with axial (normal) stress & strain, shear has analogous stress and strain definitions & Hooke’s law expression using them:
Notice that the two forces above create a couple, causing the cube to rotate; this means it’s not in static equilibrium with just the forces shown. We could fix this by applying a moment on one of the surfaces,
but could also establish static equilibrium with a complementary pair of shear forces on the x-faces in the y-directions:
Thanks to the second couple, the shear on external faces is uniform and symmetric: . This loading situation is called pure shear static equilibrium.
Sign Convention
Here are the stresses and strains shown on the positive faces (the front faces):
While here are the stresses and strains acting on negative faces (the back faces):
These stresses don’t have exact directions unless you also know what face you’re talking about. This is because stress is the internal force per unit area, and like internal force before it points in one direction on the positive face and another direction on the negative face:
For the x-faces before, the V we were using was the shear force on the +x-face in the -y-direction. Since is the shear stress on the +x-face in the +y-direction, it is related to this V by . The overbar indicates that this is technically the average shear stress over the face.
Example: Pure Shear 10cm Cube
A 10 cm cube with one face glued to a table is put into pure shear by shear stresses on
the x- and y- faces given by , with directions shown below.
If the cube is made of plastic with and , determine:
a) The shear stress
b) The shear strain
c) The lateral distance the top face of the cube displaces relative to the bottom face
b) , where , so
c) This is definitely small enough to use the small angle approximation, where .
(compare with axial strain if the force was normal to the surface):
Shear Details and Proof of
Now consider an infinitesimal volume:
Let the back face actually be position (not necessarily the origin, so that the axes are there to show directions but not positions relative to ). As this is a differential volume of size , the front corner would then be :
Net Force = 0:
Now consider all the stresses acting on this which produce force in the + or -y-directions:
The positive y-face has a stress leading to a force in the +y direction of.
The negative y-face has a stress leading to a force in the +y direction of .
Together, they cause .
/****Remember that the linearization of when we subtract will lead to just as follows:
*********/
Similarly, the top surface (+z face) has a shear stress while the bottom surface has one of and combined they cause a force in +y of .
And the net force in +y from the shear stresses on the +&- x-faces is .
All together, the differential net force in the y-direction is .
Because is the differential volume, the net differential force in y per unit differential volume is
Similarly, the net force in x per unit volume is and in z is .
If this object is in static equilibrium, we therefore have
This is the modification to the condition on force we had in FlexPDE last time; for an object with not just axial stress but shear stresses,
rather than , we require .
Net Torque = 0:
Unlike the differential volume where only axial stress is applied, this differential volume has a nontrivial torque equation because the shear stresses create couples. i.e., consider the stresses that produce force in the +&- y-direction again:
Assume that differential volume is small enough that the variation in a stress over a face is negligible (i.e., at the top of the +y face is the same as at the bottom of the +y face). This means the stresses shown create the following torques:
The forces from stresses together create a couple in the -x direction of
The forces from stresses together create a couple in the +z-direction of
The forces from stresses don’t form a couple.
So the net torque in the +x direction would be caused by the couples from and :
The net torque about x on this object is
This means that the net torque about the x-axis (or any point) in the x-direction per unit volume from these couples is , and therefore the only way to have no net torque on this object is if the shear stress is symmetric at any point: .
Similarly, having no net torque in the y-direction requires , and having no net torque in the z-direction requires .
Consider a macroscopic object subject to externally-applied shear forces only on the y-faces in the x-directions and held in equilibrium by an applied moment on the -y face:
Is for this object as well?
Yes – the derivation above showed that shear stress for any object in static equilibrium is symmetric at any point. This loading condition does mean that stress (including shear stress) is nonuniform: the top surface has an average but this doesn’t mean the stress on the sides needs to be equal to this as well. In fact, externally applying a moment requires nonuniform stress (for the direction shown, non-uniform axial stress). So, while the average stress on the top surface of is not equal to the average on the side surfaces, at any point on either surface or the interior . We’ll explore this with FlexPDE later.
Strain Details: True Strain & Engineering Strain
In FlexPDE in the last section we let u be the x-displacement and v be the y-displacement. Now consider an arbitrary displacement of a differential square of material with side lengths dx and dy and opposite corners at (x,y) and (x+dx, y+dy) as shown:
We’ve previously defined the axial strain as ; the change in the x-direction of the x-displacement.
True shear strain is defined the same as this but with the cross derivatives: ; change in the y-direction of the x-displacement and ; the change in the x-direction of the y-displacement.
Engineering shear strain (defined in the last section) is the one related to the shear stress though: . is the total angle that the faces have moved closer together so is actually the sum of the two angles , and each of these turn out to be true shear strains:
Hooke’s Law for of Hooke’s Law:
For normal or shear directions, [true] strains are related to displacements through (i.e., , where is the x-displacement of the material at wherever we’re looking).
For isotropic materials, axial stresses and strains are related through the compliance matrix:
Shear stress and engineering shear strain are related through the shear modulus:
Where the shear modulus is , and .
We can also write both via stiffness rather than compliance coefficients:
We’ve found that
1. locally Newton’s second law demands that net force and net torque are zero, which respectively lead to the relations:
Coupled PDEs for stresses: (from requiring net force on an infinitesimal volume to be zero)
Symmetric shear stresses: (from requiring no net torque on an infinitesimal volume)
2. Strains and displacements are related:
normal strain is defined as while
engineering shear strain is defined as ,
where is a point’s displacement vector due to deformation.
3. Hooke’s law relates stress and strain via stiffness and compliance matrices for normal directions, and shear moduli for shear directions:
, where the compliance matrix is (at least for an isotropic material)
, where the shear modulus is (at least for an isotropic material)
FlexPDE Complete Static Elasticity
Reading FlexPDE Code for Elasticity
The following sections of FlexPDE implement Hooke’s law, but don’t use the same variable names exactly. See if you can tell how it works.
DEFINITIONS
G=E/(2*(1+nu))
!Stiffness Matrix Elements
C11 = E*(1-nu)/(1+nu)/(1-2*nu)
C12 = E*nu/(1-nu-2*nu^2)
!Axial Strain
!Engineering Shear Strain
gxy=(dx(v)+dy(u))
gyz=(dy(w)+dz(v))
gxz=(dz(u)+dx(w))
!!Stress via Hooke’s law
!Axial Stress
sx = C11*ex+C12*ey+C13*ez
sy = C21*ex+C22*ey+C23*ez
sz = C31*ex+C32*ey+C33*ez
!Shear stress
!FNet = 0 on each differential element
u: dx(sx)+dy(sxy)+dz(sxz)=0
v: dx(sxy)+dy(sy)+dz(syz)=0
w: dx(sxz)+dy(syz)+dz(sz)=0
The first 3 equations under the !Hooke’s Law label express
The last 3 equations in the !Hooke’s Law section express .
In the EQUATIONS section, we have the expression:
These come from the force balance we did on the infinitesimal cube Net Force = 0
Reminder: Boundary Conditions
Consider the code below:
surface ‘top’
value(u) = 0 !Fixed position (i.e., zero deformation in the x-direction) at this boundary
load(w) = -50e9 !Uniform stress of 50 GPa compressive applied to this surface
The value boundary condition specifies a fixed value for the variable, so value(u) = 0 means u is 0 (i.e. make the x displacement 0 and apply whatever load is necessary to make that happen).
The second boundary condition (load(w)= -50e9) needs explaining. Using the load condition in FlexPDE sets the flux of the variable through the boundary. Since this is a top surface, its perpendicular direction is z. Therefore, load(w)=-50e9 means that the equations
w: dx(sxz)+dy(syz)+dz(sz)=0
when integrated across the boundary will know that it needs to use a constant of -50e9 in the z-direction; i.e., as if there’s a stress of 50 GPa inwards on that surface. Even though this FlexPDE behaviour (using load to set a constant of integration of the PDE) is convenient for writing stresses, it’s a bit surprising, because it means that what a given BC means technically depends on seemingly irrelevant constants in the PDE. For example, the following equation will give different results for the boundary condition load(w)=-50e9:
w: dx(sxz/10)+dy(syz/10)+dz(sz/10)=0
Complications aside, as long as you stick with the standard form of the PDEs,
u: dx(sx)=0
v: dy(sy)=0
w: dz(sz)=0
load(u), load(v) and load(w), set the stress, respectively, in the x, y and z-directions.
Example: Beam thinning by pulling on it with FlexPDE
Suppose you take a 3 m by 1 m by 1 m steel (E = 200 GPa, = 0.3) beam and apply a 200 MN force to each end of the long section of the beam.
a) Calculate the deformation of the beam, if the beam is otherwise free and verify the results in FlexPDE.
b) Rerun the program after changing your extension in x- equation to:
u: dx(sx/10)+dy(sxy/10)+dz(sxz/10)=0
c) Returning to the original DEs, try to model the more realistic situation where the x = 0 surface is totally fixed and the y & z surfaces are all free. Compare the deformed shape and displacement of the centre of the tip here to the ones you got in part a).
We did this previously analytically and found:
where , so that , and.
Similarly, .
Therefore, the beam deforms to 3 mm longer in x, and 0.3 mm narrower in y & z.
We can solve this problem in FlexPDE using this code:
‘Lecture19BeamExtension’
errlim=1e-4
spectral_colors
COORDINATES
cartesian3
u !Displacement in x
v !Displacement in y
w !Displacement in z
DEFINITIONS
G=E/(2*(1+nu))
s_applied = 200e6 !applied stress
C11 =E*(1-nu)/(1+nu)/(1-2*nu)
C12 = E*nu/(1+nu)/(1-2*nu)
!Axial Strain
!Engineering Shear Strain
gxy=(dx(v)+dy(u))
gyz=(dy(w)+dz(v))
gxz=(dz(u)+dx(w))
!!Stress via Hooke’s law
!Axial Stress
sx = C11*ex+C12*ey+C13*ez
sy = C21*ex+C22*ey+C23*ez
sz = C31*ex+C32*ey+C33*ez
!Shear stress
u: dx(sx)+dy(sxy)+dz(sxz)=0
v: dx(sxy)+dy(sy)+dz(syz)=0
w: dx(sxz)+dy(syz)+dz(sz)=0
surface ‘bottom’ z=0
surface ‘top’ z=Lz
BOUNDARIES
surface ‘bottom’
value(w)=0
surface ‘top’
START(0,0) !y=0 surface:
value(v)=0
LINE TO (Lx,0) !x=Lx surface
load(u)=s_applied
LINE TO (Lx,Ly) !y=Ly surface
LINE TO (0,Ly) !x=0 surface
value(u)=0
LINE TO CLOSE
contour(u) painted on x=0
contour(v) painted on x=0
contour(w) painted on x=0
grid(x+u,y+v,z+w)
grid(x+u, y+v, z+w)
contour(u) on surface z=0
elevation(sx,sy,sz) from (0,0,0) to (0,0,Lz)
report val(u,Lx,Ly,Lz)
report val(v,Lx,Ly,Lz)
report val(w,Lx,Ly,Lz)
Running this produces the same deformation (3e-3).
b) After changing the u: equation from:
dx(sx)+dy(sxy)+dz(sxz)=0
dx(sx/10)+dy(sxy/10)+dz(sxz/10)=0
The deformation becomes 3e-2. This is because the load BC sets up the constant that will be assumed for the integral of sx/10 in this case, not sx, meaning that even though the new differential equation is identical to the previous one, because of how the load boundary condition works they’re not the same in FlexPDE.
Original BCs give this displacement (this is magnified by a factor of 500):
Totally fixed x = 0 surface instead gives slightly less:
Code to do this:
‘Lecture19BeamExtension’
errlim=1e-4
spectral_colors
COORDINATES
cartesian3
u !Displacement in x
v !Displacement in y
w !Displacement in z
DEFINITIONS
mag = 500 !magnification of displacement
G=E/(2*(1+nu))
s_applied = 200e6 !applied stress
C11 =E*(1-nu)/(1+nu)/(1-2*nu)
C12 = E*nu/(1+nu)/(1-2*nu)
!Axial Strain
!Engineering Shear Strain
gxy=(dx(v)+dy(u))
gyz=(dy(w)+dz(v))
gxz=(dz(u)+dx(w))
!!Stress via Hooke’s law
!Axial Stress
sx = C11*ex+C12*ey+C13*ez
sy = C21*ex+C22*ey+C23*ez
sz = C31*ex+C32*ey+C33*ez
!Shear stress
u: dx(sx)+dy(sxy)+dz(sxz)=0
v: dx(sxy)+dy(sy)+dz(syz)=0
w: dx(sxz)+dy(syz)+dz(sz)=0
surface ‘bottom’ z=0
surface ‘top’ z=Lz
BOUNDARIES
surface ‘bottom’
surface ‘top’
START(0,0) !y=0 surface:
LINE TO (Lx,0) !x=Lx surface
load(u)=s_applied
LINE TO (Lx,Ly) !y=Ly surface
LINE TO (0,Ly) !x=0 surface
value(u)=0
value(v)=0
value(w)=0
LINE TO CLOSE
contour(u) painted on x=0
contour(v) painted on x=0
contour(w) painted on x=0
grid(x+u,y+v,z+w)
grid(x+u*mag, y+v*mag, z+w*mag)
contour(u) on surface z=0
elevation(sx,sy,sz) from (0,0,0) to (0,0,Lz)
report val(u,Lx,Ly/2,Lz/2)
report val(v,Lx,Ly/2,Lz/2)
report val(w,Lx,Ly/2,Lz/2)
Example: Pure Shear 10 cm Cube in FlexPDE
Modify the previous FlexPDE program to check the results from the Pure Shear 10 cm Cube example.
Example reminder: we found that for a plastic ( and ) cube with one face glued to a table put into pure shear by shear stresses on the xy faces given by , with directions shown below, that:
First part of DEFINITIONS changed to the new numbers
G=E/(2*(1+nu))
s_applied = 10/(Lx*Ly) !applied stress
The boundary conditions updated to the new setup as well:
START(0,0) !y=0 surface:
value(u)=0
value(v)=0
value(w)=0
LINE TO (Lx,0) !x=Lx surface
load(v)=s_applied
LINE TO (Lx,Ly) !y=Ly surface
load(u)=s_applied
LINE TO (0,Ly) !x=0 surface
load(v)=-s_applied
LINE TO CLOSE
The program was already outputting a plot of u at the edge of the cube, but we need to tell it to output s_applied and exy as well. Let’s try gxy at a couple of different places (it should be constant if the pure shear BC worked.)
report val(s_applied, Lx/2,Ly/2,Lz/2)
report val(gxy, Lx/2,Ly/2,Lz/2)
report val(gxy, Lx,Ly,Lz)
report val(u,Lx,Ly,Lz)
One final point: in the student version, you’ll need to reduce the grid density so the program will run. I used ngrid = 5
‘Lecture19BeamExtension’
errlim=1e-4
spectral_colors
COORDINATES
cartesian3
u !Displacement in x
v !Displacement in y
w !Displacement in z
DEFINITIONS
mag = .4*globalmax(magnitude(x,y,z))/globalmax(magnitude(u,v,w))!magnification of displacement
G=E/(2*(1+nu))
s_applied = 10/(Lx*Ly) !applied stress
C11 =E*(1-nu)/(1+nu)/(1-2*nu)
C12 = E*nu/(1+nu)/(1-2*nu)
!Axial Strain
!Engineering Shear Strain
gxy=(dx(v)+dy(u))
gyz=(dy(w)+dz(v))
gxz=(dz(u)+dx(w))
!!Stress via Hooke’s law
!Axial Stress
sx = C11*ex+C12*ey+C13*ez
sy = C21*ex+C22*ey+C23*ez
sz = C31*ex+C32*ey+C33*ez
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