Flexural Elasticity Notes
The third elasticity mode is flexure (bending). Flexure results from moments with vectors parallel to a surface of the beam; i.e., an out-of-the-page moment applied on the right face of a beam:
Flexure is the dominant method of deformation for most beams, because most beams are long in one dimension which simultaneously strengthens the torque applied by forces at their ends (meaning they need higher internal moments to be in equilibrium) and decreases the beam’s resistance to them.
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Pure bending arises when a beam has equal and opposite bending moments applied to its faces on either end, e.g., equal and opposite z-moments applied to its opposite x-faces:
This loading causes the beam to bend (AKA flex, AKA curve) into a circular arc:
The inverse of the radius of curvature ( is the curvature (c):
In pure bending, a bending moment M causes a curvature c according to the bending elasticity relation (Hooke’s law for bending): ,
where EI is a beam parameter called its flexural rigidity and determines how difficult it is to bend the beam into a given curvature.
Specifically, E is the beam’s [effective] stiffness [in the length direction] and I is its “second moment of area” of its cross-section about its centroid: , where the stiffness-weighted centroid of the cross-section is defined as: (centroid weighted by Young’s modulus).
For a uniform material we can factor the elastic modulus out of the integral and this simplifies to: ( is the non-weighted centroid of the cross-section; i.e., average y position).
Averages and Weighted Averages
In general, the average value of some function of position over some area is the integral of that function over the area divided by the area; i.e., divided by the integral of 1 over the that area: .
e.g., Average y-position is just an example of this where : .
In 3d, we might also find the average over some volume: , while in 1d we might find it over some length .
Many times we’re interested in weighted averages, which are the same thing, but include some weighting function in both integrals: .
One example of where we’d need to do this is the y-coordinate of centre of mass when the object has a non-uniform density :
Many more areas of doing this exist in stats!
Notes on Beam Bending
1. Once the beam starts bending, its cross sections that were originally in the yz plane rotate about the z-axis, so they’re technically not in the “yz” plane anymore. This makes it especially necessary to distinguish between the original coordinates (x,y,z) of a point and its coordinates after deformation, (x+u, y+v, z+w). As long as the beam was initially un-curved, its cross sections along the length are always in the “yz” plane of the original coordinates.
2: The integral used to locate the y-coordinates of the cross-section’s centroid is also called the cross-section’s first moment of area about the z-axis,
3: The axis that runs through the stiffness-weighted centroid of each cross-section has no strain; as such, it’s called the neutral axis.
/*****Proof that
The axial stress in the x-direction at any point depends on the axial strain via . If the neutral axis has no stress, then it must have no strain either. Defining y = 0 at the neutral axis, we can write the radius of curvature at each y-coordinate relative to this point:
(Why? Because moving up means we’re closer to the origin of the circular arc so means a smaller radius.)
The strain is then where is the arc angle of the whole beam once it’s bent and we’ve used the arc length formula for the neutral axis: .
Substituting into Hooke’s law:
Now consider a differential piece of area of the cross section:
The tensile force on it from adjacent faces is
We can integrate this force to find the total torque that is put on this face about its centroid, which is the moment placed on the face:
(The negative sign is because y is upwards and dFx is right so the right hand rule gives Mz in the negative direction (into the page) for this coordinate system).
Example: Flexural rigidity of a rectangular beam
A rectangular beam with Young’s modulus E has length L (in x) thickness h (in y) and width B (in z). Suppose that the beam extends from (x,y,z) = (0,0,0) to (L,h,B). For bending in response to moments in the width (i.e., z) direction, determine the beam’s:
1. Cross-sectional area
1. Location of its stiffness weighted centroid
1. Second moment of area of its yz cross section about the z-parallel axis through the centroid
1. Flexural rigidity in response to this bending
1. Curvature
1. Radius of curvature
1. Bending angle
1. Maximum tensile and compressive stress
1. Cross-sectional area: A = Bh
Location of its stiffness weighted centroid: . Similarly, .
Second moment of area of its yz cross section about the z-parallel axis through the centroid:
Flexural rigidity:
Curvature:
Radius of curvature:
Bending angle:
Stress in the beam is , so (as long as M is positive and the beam bends up so that is positive) the max compressive stress is at the top while the max tensile stress is at the bottom:
Look at these equations and consider how things affect the answer to level up your physics intuition. e.g., because more bending moment means more bending angle. , because thicker beams are much tougher to bend.
Check your understanding: The following beam is called an “I-Beam”. It’s much more difficult to machine than just a rectangular beam, and yet it’s far more common to see in building construction.
1. Why is it called an “I-Beam”?
1. What advantages does this beam have over a rectangular beam?
Steel I-Beam from http://www.stlpipesupply.com/wp-content/uploads/2011/09/Steel-IBeam.jpg
Solutions:
1. Its cross section is in the shape of an I.
1. This cross section puts a lot of the area far from the centroid in both directions, giving it a high flexural rigidity for a given area, and thus a high strength-to-weight ratio.
Hooke’s Law
Normal (axial)
Flexural Rigidity for Multilayer Beams
If a beam has multiple layers with different material parameters, its average density and Young’s modulus are just weighted averages with respect to cross sectional area of each layer of the beam:
, , or, equivalently, , (this integral version is necessary if the material parameters change continuously rather than in a finite number of discrete steps. Fun fact: the III-V semiconductor Molecular Beam Epitaxy machine and Chemical Vapour Deposition machines in the Tandem Accelerator Building at McMaster are both examples of machines that can create continuously varying material parameters by gradually modifying growth conditions over the time of a deposition; and keep things all the same crystal grain too!).
(here is density not curvature like earlier, and is Young’s modulus).
While these are fine for calculating the mass of a beam per unit length and the axial stress-strain relation: , calculating bending is a bit more complicated if we bend about an axis that makes different materials compress and extend differently (i.e., about an axis out of the page or horizontal for the beam shown above).
Discrete Version:
The flexural rigidity of a multilayer beam is:
is the AE-weighted centroid of the beam: ,
Ai = Bti is the cross-sectional area of the ith layer and is its thickness
0. Ei is the Young’s modulus of the ith layer in the longitudinal direction
is the vertical centre of mass of the ith layer
is the moment of inertia of the ith section about its centroid;
i.e., if the section is a rectangle,
Note that still, .
Continuous Version:
Well, that’s how people without a computer algebra system might do it. A more general formula which allows for this or even continuously changing material parameters is:
(this time we didn’t even need the parallel axis theorem).
Check: show that and give the same result for the following example:
Example: Flexure of a Bimetallic Strip
A bimetallic strip is 30 cm long by 3 cm wide by 1 mm thick and is made of two layers:
the bottom is tungsten (E = 411 GPa, = 0.28, = 19.25 g/cm3) and
the top is aluminum (E = 70 GPa, = 0.35, = 2.7 g/cm3).
If the aluminum layer is 70% of the thickness,
a) Calculate linear mass density,
combined axial Young’s modulus,
curvature,
radius of curvature, and
maximum arc angle that results from a bending moment of M = 1 N-m (ignoring gravity).
b) Now suppose that in addition to being different materials, the bottom layer is only 1 cm wide. Redo part a) for this situation.
(not to scale)
Discrete version:
Cross sectional area is A = 3 cm * 1 mm = 30 mm2 = 30e-6 m2
Linear mass density is:
Combined axial Young’s modulus is:
The AE centroid is located at (relative to the bottom of the strip):
Flexural rigidity is then:
Therefore,
Continuous Version:
> restart:
Lx:=0.3: #Length
Ly:=0.001: #thickness
Lz:=0.03: #Width
E:=piecewise(y<0.3*Ly, 411e9, y>0.3*Ly,70e9): #Young’s Modulus
nu:=piecewise(y<0.3*Ly, 0.28, y>0.3*Ly,0.35): #Poisson’s Ratio
rho_mass:=piecewise(y<0.3*Ly, 19250, y>0.3*Ly,2700): #Density (kg/m^3)
M:=1: #Moment applied at the tip
A:=Ly*Lz: #Area
mu:=int(rho_mass*Lz,y=0..Ly); #LinearMassDensity
Ecombined:=int(E*Lz,y=0..Ly)/A; #Axial E
Ybar:=int(Lz*E*y,y=0..Ly)/int(Lz*E,y=0..Ly); #AE-centroid
EI:=int(Lz*E*(y-Ybar)^2,y=0..Ly); #Flexural Rigidity
c:=M/EI; #curvature;
rho:=1/c; #radius of curvature
phi:=c*Lx; #bending angle
Agrees! And is arguably a lot easier to see how it works!
Now the area integrals involved in and would need to consider the different width at different values of y. Keeping the coordinate system where y goes from 0 to Ly = 1 mm over the beam, the width would be
This should make the average values closer to that of Al than Tungsten, and generally make the beam less stiff:
> restart:
Lx:=0.3: #Length
Ly:=0.001: #thickness
w:=piecewise(y<0.3*Ly, 0.01, y>0.3*Ly, 0.03): #Width
E:=piecewise(y<0.3*Ly, 411e9, y>0.3*Ly,70e9): #Young’s Modulus
nu:=piecewise(y<0.3*Ly, 0.28, y>0.3*Ly,0.35): #Poisson’s Ratio
rho_mass:=piecewise(y<0.3*Ly, 19250, y>0.3*Ly,2700): #Density (kg/m^3)
M:=1: #Moment applied at the tip
A:=int(w,y=0..Ly); #Area
mu:=int(rho_mass*w,y=0..Ly); #LinearMassDensity
Ecombined:=int(E*w,y=0..Ly)/A; #Axial E
Ybar:=int(w*E*y,y=0..Ly)/int(w*E,y=0..Ly); #AE-centroid
EI:=int(w*E*(y-Ybar)^2,y=0..Ly); #Flexural Rigidity
c:=M/EI; #curvature;
rho:=1/c; #radius of curvature
phi:=c*Lx; #bending angle
The Beam Equation (Small Deformations Only): Displacements from Arbitrary Bending
We’ve seen for beams that locally (i.e., at any point) and learned how to calculate EI and M, and how the beam bends into a circular arc for constant (a beam in pure bending with the same cross section along its length). However, in general M and aren’t constant so it would be helpful to learn how local curvature leads to overall beam displacement in general. This section will derive The Beam Equation which is a way to determine this overall beam displacement from curvature caused by non-uniform moments (as long as the displacement stays small enough, which is typically true).
For small deformation bending, the beam cross-sections stay approximately in the y-z plane, and curvature is .
/*******To see this, consider the equation of a circle of radius r centred at (0,r):
When the radius of the circle is much larger than y (i.e., we’re close to the origin and r is large), , so we have:
where is the curvature. Notice that in this case , and so while really the displacement v vs. x for pure bending is in the shape of a circle of curvature c, for small deformation bending: (If you take v as not necessarily the y-displacement but the perpendicular (transverse) displacement, then this is also true locally even for large deformations from straight beams since as y goes to 0).
*********/
Small deformation bending lets us solve beam problems using differential equations as follows:
Consider a simply-supported beam with a uniform distributed load per unit length on the top, :
Solving the static equilibrium equation gives the reaction force at each support as .
Using our previous sign convention for internal shear force and moment,
the internal shear force is then ,
and the internal bending moment is . (As before, this is also the integral of the shear force: )
Now we know how moment leads to displacement: (for small displacements)
So, we can integrate this equation twice to find , the transverse displacement of the beam vs. position:
Integrating again:
Therefore:
We also know that . Incorporating these conditions,
So, in total, the beam’s transverse deformation is given by:
e.g., suppose this beam is 1 m long, 10 cm thick, and 20 cm wide (meaning ), with E = 100 GPa and the distributed force has a magnitude of (i.e., a stress of ). Then the transverse displacement vs x is:
I0:=0.1*0.2*0.1^2/12;
v:=x-> -p0*L^3/(24*EI)*x+1/EI*(L*p0*x^3/12-p0*x^4/24);
plot(v(x), x=0..L);
Another way to solve this problem which is a bit more general is to differentiate the moment equation twice to arrive at the beam equation:
This is a 4th order DE in position, which we can solve by integrating 4 times:
To determine the constants, we’d need to know what our boundary conditions are. Because the beam equation is 4th-order in x, we need 4 boundary conditions in x. In the simplest version of the beam problem, we get these by constraining two things at each end of the beam. For example,
Type of boundary
Clamped (Fixed)
Pivot joint
Vertical Slider
e.g., applying the pivot joint boundary conditions at each end:
Why No in The Beam Equation?
We learned earlier that shear stress leads to shear deformation through , but we didn’t have to consider this for the beam above, even though it had shear force on it. Why was that?
Answer: Because the dimensions of a long thin beam make it negligible. Inside the beam:
(at the end of the beam)
Now for simple loading at each end:
But we also said:
Therefore, when L/h is 10 and the object is in the combined shear and flexure that results from transverse loading, the shear bending is about 100x smaller than the flexure bending.
The internal shear force and bending moment are related through (for the appropriate sign convention:
The moment causes a curvature in the beam via flexural rigidity:
If the curvature caused by the moment is all of the strain, and the strain is reasonably small, then the curvature is equal to the second derivative of the deformation caused by bending:
This therefore ignores the deformation caused by shear, which is:
Thus, while it looks like we’re simultaneously saying that and , the first is based only on the v caused by flexure while the second is only on the v caused by shear.
Example: Cantilever boundary conditions
Now suppose we have a 1 m long 10 cm thick and 20 cm wide beam () with E = 100 GPa, , and a distributed load on the top surface with magnitude of (i.e., a stress of ) the left edge is a fixed support.
Find the deformation in the y direction vs. position relative to the wall.
Since the distributed load is a constant, we don’t need to re-solve the beam equation () and can jump directly to its solution:
as before, but this time the boundary conditions are a bit different:
which lead to
I0:=0.1*0.2*0.1^2/12;
v:=x-> -p0/EI*x^4/24+a*x^3+b*x^2+c*x+d;
subs(x=0, diff(v(x), x))=0,
subs(x=L, diff(v(x), x,x))=0,
subs(x=L, diff(v(x), x,x,x))=0]);
assign(%):
plot(v(x),x=0..L);
v(1) = -75 μm.
Note: rather than substituting into the solution to the beam equation, we could also use dsolve like before:
> restart;
I0:=0.1*0.2*0.1^2/12;
BeamEquation:=-p0=EI*diff(v(x),x,x,x,x);
dsolve([BeamEquation,v(0)=0,D(v)(0)=0,D[1,1](v)(L)=0,D[1,1,1](v)(L)=0]);
assign(%):
plot(v(x),x=0..L);
Check using direct statics
Because the cantilever is statically determinate without knowing how the beam is deforming, we could also have solved for the reaction shear force and moment at the left wall and the internal shear force and bending moment using our statics knowledge, then could determine the local curvature and use it to find the displacement by integrating :
> restart:
I0:=0.1*0.2*0.1^2/12;
Ry:=int(p0,x=0..L);
RM:=int(p0*x,x=0..L);
V:=Ry-p0*x;
M:=-RM+int(V,x=0..x);
plot([V,M],x=0..L);
> c:=M/EI:
dsolve([diff(v(x),x,x)=c, v(0)=0,D(v)(0)=0]);
v:=evalf(rhs(%)):
plot(v,x=0..L);
vtip:=subs(x=L, v);
which confirms what we got using the beam equation.
Reflection: what are the pros and cons of using the beam equation rather than this direct statics approach?
Pro: The beam equation can easily deal with more constraints; e.g., fixed at both ends, or roller supported at one side; the direct statics approach requires the beam to be statically determinate without considering deformations.
Con: The beam equation is less flexible to complicated loading along the length; we’d need to re-derive a different version to deal with a point load applied somewhere or a non-uniform distributed load p0. (you can still use but won’t be constant so you’ll need to integrate).
Minor Con: The beam equation doesn’t directly find the displacement due to shear, as it ignores it – You can still find the displacement due to shear alone from the beam equation solution by using to get the local shear force then saying where and .
Check the check using the beam equation:
We can re-produce a shear force and bending moment diagram using the beam equation and compare to the one we got using direct statics:
> restart;
I0:=0.1*0.2*0.1^2/12;
BeamEquation:=-p0=EI*diff(v(x),x,x,x,x);
dsolve([BeamEquation,v(0)=0,D(v)(0)=0,D[1,1](v)(L)=0,D[1,1,1](v)(L)=0]);
v:=rhs(%):
#plot(v,x=0..L);
M:=EI*diff(v,x,x);
V:=diff(M,x);
plot([V,M],x=0..L);
This agrees with our earlier method using direct statics.
Example 8.3b: Bridge Boundary Conditions
Repeat example 8.3 but for bridge boundary conditions instead of cantilever ones. Draw a shear force and bending moment diagram, and find the displacement at the centre from moment alone and from shear alone.
This time the direct statics method doesn’t work because we can’t solve for the reaction force and moment independent of the beam bending. You can assume by symmetry that the reaction force and moment at each support wall are symmetric, and this lets you find the force, but makes the reaction moment a free parameter. Using trial and error you can zero that a reaction moment of 83 at each end is pretty close:
> restart:
I0:=0.1*0.2*0.1^2/12;
Ry:=int(p0,x=0..L)/2;
V:=Ry-p0*x;
M:=-RM+int(V,x=0..x);
plot([V,M],x=0..L):
dsolve([diff(v(x),x,x)=c, v(0)=0,D(v)(0)=0]);
v:=evalf(rhs(%)):
plot(v,x=0..L);
vtip:=subs(x=L, v);
But with the beam equation we can just change the boundary conditions to bridge ones and get the solution very quickly
Then we can solve for the v produced by shear alone by where and :
> restart;
I0:=0.1*0.2*0.1^2/12;
BeamEquation:=-p0=EI*diff(v(x),x,x,x,x);
dsolve([BeamEquation,v(0)=0,D(v)(0)=0,v(L)=0,D(v)(L)=0]);
v:=rhs(%):
plot(v,x=0..L);
vMomentCent:=subs(x=L/2,v);
M:=EI*diff(v,x,x); #bending moment
V:=diff(M,x); #Shear force
plot([V,M],x=0..L);
nu:=0.2: G:=E/(2*(1+nu)): tau:=-V/(0.2*0.1):
vShear:=int(tau/G, x=0..x);
vShearCent:=subs(x=L/2, vShear);
vTotalCent:=vMomentCent+vShearCent;
For this beam, shear is a relatively large (10%) correction at the centre of the bridge compared to moment.
FlexPDE Flexure
Since FlexPDE
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