Internal Forces and Moments
Consider the following structure:
In week 3, we might have broken this into two separate segments AB and BC and determined the reaction forces and moments at B by viewing it as a fixed support holding up member BC:
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From the perspective of the combined frame ABC these forces and moments at B are internal. But, there’s not really anything special about point B – we could view beam BC as made up of two separate parts too, and then find the reaction forces and moments at that contact surface:
So, any object can be viewed as made up of a series of fixed supports that supply internal forces and moments to hold each part of the object in place.
This section will first cover how to describe these forces and moments, and how to classify them, defining a sign convention for positive and negative axial (normal) and shear force and as well as bending moment. Then the section will introduce the idea of internal force and moment diagrams and develop some important relations between the internal bending moment and shear force. Using these topics, we’ll be able to produce diagrams that help analyze and understand the internal forces and moments that arise in the types of problems we were solving last week, and pave the way for the next four weeks analyzing elasticity in response to them.
1. Able to produce internal force and moment diagrams given a structure
2. Able to read internal force and moment diagrams and understand where they may be faulty.
3. Able to relate internal shear force and bending moment
2D Internal Forces and Moments
Axial Force (Normal Force)
Consider a member in static equilibrium acted on by an axial force (one along the primary axis of the beam; aligned with its length). To be in equilibrium, the force on each end must be equal and opposite:
Now imagine the left and right sides of the beam are separate objects (they’re separate collections of atoms, after all):
For the individual pieces to be in static equilibrium too, there must be equal and opposite forces exerted on each face within the beam:
We’d get the same answer no matter where we sliced the beam, so the beam carries this same internal axial force all along its length.
The internal force can change along the length if the applied force also changes along the length. For example, a vertical beam subject to gravity. For this to balance, the ground must be pushing up with a normal force that matches gravity:
Now suppose we draw an FBD for a thin slice of the beam near the top. Since it has some small weight too the beam just under it must be exerting an internal normal force to hold it up:
If the beam has uniform density, then the weight of the object increases linearly as we look at lower faces, and therefore so does the internal normal force. The internal normal force applied just before the bottom is supporting its entire weight:
Example: Axial force variation
Find the internal axial force (i.e., normal force) as a function of position y from the bottom for one of the 1.5 m tall vertical supports on the sides of the bookcase shown. Assume the bottom shelf is supported 5 cm up from the bottom, the next shelf is 50 cm up and the top shelf is 1 m up. Assume the ends have a uniform density total mass of 7.5 kg, while each shelf has a mass of 4 kg.
The linear mass density of the ends is →
Each shelf has a weight of .
By symmetry, each shelf is supported equally on each side, so puts 20 N on the supports.
FBD for the support is then:
And also supports a weight of at each value y.
So, the total axial force as a function of y is
(negative indicates compressive internal force).
[Corrected] Video Solution:
> restart: g:=9.81:
WSideTotal:=7.5*g:
Wsupported:=(1-y/1.5)*WSideTotal;
Wshelf:=4*g:
Fy:=Wsupported+piecewise(y<.05, (Wshelf/2)*3, y<.5, (Wshelf/2)*2, y<1, Wshelf/2):
Wmax:=Wshelf/2*3+WSideTotal;
plot([Fy,Wmax], y=0..1.5);
& Bending moment
Consider a beam subjected to a bending moment at each end:
If we pick a cross section at some point along the beam and draw an FBD for the beam to the left and right of this cross section, then for each of those parts of the beam to be in equilibrium there needs there to be an internal moment to counteract the externally applied ones:
Now consider an L = 4 m-long beam supported by pins at each end with a point load P = 10 kN applied in the centre:
Externally, this beam has only forces applied in the y-direction and no torques. However, if we slice the beam as before, this time we find that we need a force and a moment to keep the slice in equilibrium (here V and M are drawn using the sign convention; note that V will be negative for the slice shown):
The force V along the face of the beam is called the internal shear force. If x is the distance from the left end of the beam, then for this piece to be in equilibrium we need and . Things change to the right of the support at the top though:
Now we find and .
This information is shown in statics using a and Bending Moment Diagram:
> restart:
P:=20e3: L:=4:
V:=piecewise(x
M:=piecewise(x
plot([V,M], x=0..L);
Important Observations
1. The units for this are different for each plot: N for the internal shear force and N-m for the internal bending moment.
2. The shear force and bending moment diagrams are the same regardless of whether you consider the equilibrium of the object to the right side or left side of the slice. This is because the internal forces on the slice cancel out from the perspective of the entire beam.
3. When a beam fails, it fails at a specific spot, meaning you need to examine these internal forces and moments to determine whether, where, and how it will fail.
4. The shear force plot is the derivative of the bending moment plot: . Amazingly, this holds in general, see the proof below.
unless a point moment is applied at a poi along the beam, the beam has an increase in internal moment with each step dx given by the value of the shear force there: . To see why, consider a small section along the beam from x to x+dx:
Newton’s second law & a Approximation for gives:
Discussion:
V can only change if there are additional external forces on the beam added immediately in this sliver, either by the discontinuity of a point load being applied (like point load P for the beam above) or by a small amount of distributed load:
Likewise, the moment can change from new point moments, but also just from the couple caused by V:
(here you can neglect the direct moment effect of distributed loads: , because the term is infinitesimally small compared to the . term).
This means besides when point loads and point moments get involved, the slope of the shear force is the distributed load:
and the slope of the moment is the shear force: .
This gives a much faster way to find the bending moment diagram – integrate the shear force (and add any point moments if applicable):
> restart:
P:=20e3: L:=4:
V:=piecewise(x
M:=int(V, x=0..x):
plot([V,M], x=0..L);
Example: Internal force and moment diagram
Neglecting the mass of the members themselves, draw the shear force and bending moment diagram for member AB below. (Note: these diagrams are drawn to individual members. Thus, you must use the entire frame to calculate the reactions at pin G and roller C, then find the reaction forces that members FG and CD place on member AB, and use these to find the shear force and bending moment of AB.)
Enhanced Eng. Phys. Picture:
Draw an FBD for the whole structure:
Apply Newton’s 2nd law to find the reactions at the supports:
> restart:
Gy-2e3-8e3-2e3,
-.2*Gx-.3*Gy+.6*8e3-.2*Cx+1.2*2e3]);
FBD to find the forces and moments applied to ADEFB only:
> restart:
Gy-2e3-8e3-2e3,
-.2*Gx-.3*Gy+.6*8e3-.2*Cx+1.2*2e3]); assign(%):
MD-Cx*.2]); assign(%):
MF-Gx*.2]); assign(%):
FBD for internal forces and moments:
Solution and production of plots, as explained in the video:
> restart:
Gy-2e3-8e3-2e3,
-.2*Gx-.3*Gy+.6*8e3-.2*Cx+1.2*2e3]); assign(%):
MD-Cx*.2]); assign(%):
MF-Gx*.2]); assign(%):
N:=piecewise(x>.3, Dx)+piecewise(x>.9, -Fx):
V:=-2e3*piecewise(x>0, 1)+piecewise(x>.3, -Dy)+piecewise(x>.6,-8e3)+piecewise(x>.9, Fy)+piecewise(x>1.2, -2e3):
M:=-2e3*x+piecewise(x>.3, MD-Dy*(x-.3)) + piecewise(x>.6, -8e3*(x-.6))+
piecewise(x>.9, MF+Fy*(x-.9))+piecewise(x>1.2, (x-1.2)*(-2e3)):
Vcheck:=diff(M, x):
Mcheck:=int(V, x=0..x)+piecewise(x>.3, MD) +piecewise(x>.9, MF);
plot([N, V, M, Vcheck, Mcheck], x=-0.1..1.21);
Archive Solution writeup:
Force in x:
Force in y:
CCW Moment about G:
Now, consider members CD & FG to find out their effects on member AB.
Portion CD:
Portion FG:
With this information, we can write to forces on AB:
Notice the forces and moments drawn on AB are opposite in direction from the ones drawn on members CD and FG. This is because we are now drawing the forces acting on AB.
And now we can break up this object at different points to write the internal force and moment as a function of position within AB.
Sign Convention
When solving for internal forces and moments it is very important to follow a consistent sign convention. In the examples below we will use the following sign convention:
This diagram shows the positive vectors for shear and normal force as well as moments when they appear on either side of a member. So when analyzing a beam from left to right we say that downward shear forces are positive, normal forces to the right are positive and CCW moments are positive.
To do that, you section it at different points; e.g., between A & D, we’d have:
(The red force and orange moment vectors show our sign convention)
which, using Newton’s second law, gives that V = -2k, N = 0, M = -2kN*x, where x is how far from point A we’re slicing it.
The internal forces always oppose the external forces on a member.
If we section a bit further we’d have to include the forces and moments applied at point D,
now we’d find
(That’s using the torque about the point we’re slicing it. Note that we could instead have summed the torque about another point; e.g., about point A we’d have:
(which is the same)
Then at point E,
For this to be in equilibrium, we’d have new things added to the previous expressions to account for the new 8 kN force:
Then to the right of point F, you’d have:
Using piecewise functions, we can actually use just this diagram and write an expression that works for all points along the member:
ND:=0: VD:=9e3: MD:=1.8e3:
NF:=12e3: VF:=9e3: MF:=1.8e3:
N:=piecewise(x>.3, VD)+piecewise(x>.9, -VF);
#Same as this, but not as scalable: N:=piecewise(x>.3, -VD, x>0.9, -VD-VF);
V:=-2e3+piecewise(x>.3, ND)+piecewise(x>.6,-8e3)+piecewise(x>.9,+NF);
M:=-2e3*x+piecewise(x>.3, MD+ND*(x-.3))+piecewise(x>.6,-8e3*(x-.6))+piecewise(x>.9,MF+NF*(x-.9));
plot([N,V,M], x=0..1.2, legend=[‘N’,’V’,’M’]);
Massive time-saving level-2 spell
Remember that as long as you use the same sign convention as before,
then so after you write the shear equation you can get the moment one by saying
ND:=0: VD:=9e3: MD:=1.8e3:
NF:=12e3: VF:=9e3: MF:=1.8e3:
N:=piecewise(x>.3, VD)+piecewise(x>.9, -VF);
V:=-2e3+piecewise(x>.3, ND)+piecewise(x>.6,-8e3)+piecewise(x>.9,+NF);
M:=int(V, x=0..x)+piecewise(x>.3, MD)+piecewise(x>.9, MF);
plot([N,V,M], x=0..1.2);
(where the shear plot is in kN and the moment plot is kN-m)
Meanwhile, Over in ENGINEER 2P04…
In a purely-paper course they might show this info all in one diagram like this:
Note that at any point you could slice it and look at the other side instead and should get the same answer:
Example: Distributed load Internal force and Moment Diagram
Consider the loading in Figure 3A:
a) Find the equivalent force to the distributed load and its location
b) Solve for the reactions at the supports
c) Draw the internal force and moment diagrams for the actual and the equivalent loading
d) What can you conclude about using equivalent forces?
First, replace each block section and the triangle as follows:
Each rectangle is equivalent to 400N/m*2.5m = 1000N acting in its center, while the triangle is equivalent to 500N acting 2/3 of the way towards its right side.
All together we have:
F:=1e3, 1e3, 500, 2e3;
x:=1.25, 2.5+1.25, 2.5+2*2.5/3, 5+1.25;
FNet:=sum(F[i], i=1..4);
tauNet:=sum(F[i]*x[i], i=1..4);
xEquiv:=tauNet/FNet;
If we figure out how to write an equation for the distributed force vs. position then we can integrate all at once to find the net force and equivalent position:
dF:=piecewise(x<2.5, 400, x<5, 400*x/2.5, x<7.5, 800);
plot(dF, x=0..7.5);
FNet:=int(dF, x=0..7.5);
tauNet:=int(dF*x, x=0..7.5);
xEquiv:=tauNet/FNet;
Either way, the entire distributed load is equivalent to an FNet = 4500 N load placed at xequiv = 4.35 m from the left end.
b) Taking the reaction forces to be up and to the right in y and x respectively, the support equations are:
Therefore,
0=FAy+FB-FNet,
0=7.5*FB-tauNet};
solve(Eqns);
c) Since there are no forces in x acting on this beam, it has no normal internal force. The shear force at any point must balance the total force to the left (or right) of that point, while the moment is the integral of the internal shear force plus any externally applied moments (in this example there aren't any):
dF:=piecewise(x<2.5, 400, x<5, 400*x/2.5, x<7.5, 800):
plot(dF, x=0..7.5):
FNet:=int(dF, x=0..7.5):
tauNet:=int(dF*x, x=0..7.5):
xEquiv:=tauNet/FNet:
0=FAy+FB-FNet,
0=7.5*FB-tauNet}:
solve(Eqns);
assign(%):
VActual:=FAy-int(dF, x=0..x);
MActual:=int(VActual, x=0..x);
plot([VActual,MActual], x=0..7.5);
VEquiv:=FAy-piecewise(x