Trusses, Machines, and Distributed Loads
Last week we covered static equilibrium of extended objects in 3d subject to applied forces and moments. This week we will introduce simplifications to this analysis in 2d which make it easier to handle and understand larger problems quickly (especially problems that require analyzing many connected objects forming a superstructure). We’ll also look at simplifications for 2-force members and how to analyze structures made up of them (trusses), define and analyze simple machines, and learn how to analyze problems with distributed loads.
1. Can use the force component method to quickly analyze 2d static equilibrium problems with many interconnected frames
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2. Can recognize 2-force members, show how they lead to pure tension or compression reactions, and use this method to simplify solving static equilibrium problems
3. Can recognize and explain zero force members
4. Can solve pure truss problems with the method of joints or method of sections, understand why these work, and apply their lessons to better understand and solve frame problems
5. Can analyze simple machines
6. Can analyze problems with distributed loads
2D Static equilibrium simplifications
Consider the previous example problem:
A beam is fixed into a wall that extends from to , centred on z = 0. Forces and moments are placed on the beam also at z = 0 as shown:
where , , . In addition, a moment of is applied by a force distribution at the point of application of F2.
Find the equivalent net reaction force and reaction moment of the wall on the beam, if the point of application of this force is the:
a) the midpoint of the wall’s contact surface,
b) the origin
To solve this problem, we solved two vector equations in 3d, equivalent to 6 scalar equations; 3 for forces and 3 for torques in each direction:
However, since all the forces are in the xy-plane, we don’t get a reaction force in z, and only the first two force equations are needed to solve for the reaction force components:
Similarly, since all of the forces are also applied at positions in the xy-plane, when we find the torque they produce about a point in that plane we’ll be crossing a moment arm in that plane with a force in that plane so are guaranteed to get only torques perpendicular to it:
For every force applied, about every point in the xy plane, we’ll find: .
Therefore, this is entirely a 2d problem, so rather than doing it with arbitrary 3d vectors we can use a different solution method that will help understand the problem on a deeper level (especially if you mentally link it back to the 3d vector version we’ve been doing).
Start drawing a free body diagram and assigning directions of the reaction forces and moments in advance as follows:
Note: it doesn’t matter which way you decide the reaction forces are as long as you’re consistent; e.g., we’ve arbitrarily drawn as left, as downward, and as clockwise, meaning the vectors would be and . When we solved it before we found , therefore (left and down). If we had instead drawn going to the right we would have found (right) – that’s the same thing! Don’t reassign it because a value is negative; it’s unnecessary, time consuming, and could lead to errors if you miss a spot.
The two vector equations reduce to 3 nontrivial scalar equations:
(where we can take the torque equation about any point in the xy plane that’s convenient).
Solving these equations,
For the torque equation, we’ll find the torque in the +z direction about the centre point of the wall:
This is easiest to do using components; e.g., for , we’ll have a CCW (+z) torque given from its y-component times how much it’s displaced to the right and a clockwise (-z) component from its x-component times how much it’s displaced upwards: . You should confirm this using the right hand rule, and by taking the cross product algebraically:
So, using either method of thinking about it, we have:
Continuing this way for the other torques, the net torque in the +z direction about the centre point of the wall is:
This method of analysis is especially helpful if the beams are relatively thin compared to their lengths so that we can ignore the displacements forces have from their axes.
Example: Frame Problem with Thin Beams
Consider the following setup made of light materials except for mass M.
Mass M has a weight of 1 kN and is attached at the midpoint between B and D and you can ignore the width of the members when considering coordinates of points (i.e., the pins are small compared to the members). Coordinates are (in m):
Determine the reactions at all pins and at the fixed support A.
First, draw a FBD and label all of the force & moment components:
(It doesn’t matter which direction we pick as long as we follow Newton’s third law and stay consistent.)
Generate equations for each frame by declaring that each is in static equilibrium and therefore has no force in x, no force in y, and no torque in z about any point.
> restart: W:=1e3:
-Ay+By+Cy,
MA-2*Bx-4*Cx,
-1*W+2*Dy,
2*Cx-2*Dy]);
2-Force Members
In the previous example note that member CD had exactly two forces on it (the force of pin D on it and the force of pin C on it) but no moments applied to it (because the pins can’t apply moments). A member like this is called a 2-force member:
Notice also that the reaction forces from pins C and D on member CD needed to be equal and opposite for the net force on that member to be zero, and needed to be collinear for the net torque on it about any point to be zero. This will be true for any member that has only two forces (and no moments) applied on it and is in static equilibrium.
Any 2-force member in static equilibrium must have:
1. Both forces equal and opposite (to have no net force on it):
2. Both force lines of action collinear (to then have no net torque on it):
Note that often the beam is designed so that this line of action passes through the centre of the beam, and that does useful things like making the internal stress reasonably uniform throughout. This proof shows that even for other shapes of beam what’s required is that the forces have collinear lines of action:
If you can recognize a 2-force member like this exists in the problem from the start, then you can simplify it substantially. For example, here’s that same problem re-solved with CD recognized as a 2-force member:
> restart: W:=1e3:
Ax+Bx+CD/sqrt(2),
-Ay-By-CD/sqrt(2),
MA-2*Bx-4*CD/sqrt(2),
-Bx-CD/sqrt(2),
By+CD/sqrt(2) – W,
-1*W+2*CD/sqrt(2)]);
Alternate solution:
Only 2 frames –> 6 equations and 6 unknowns!
> restart:
Ax+Bx+CD/sqrt(2),
Ay+By-CD/sqrt(2),
MA-2*Bx-4*CD/sqrt(2),
-Bx-CD/sqrt(2),
-By-1e3+CD/sqrt(2),
2*By+1e3]);
(in Newtons)
A truss is a structure built by arranging 2-force members in a triangular pattern using pins at their ends:
Truss members are assumed to have negligible weight compared to the forces from the pins.
Conventions for truss forces:
1. in tension (positive force on the member); a truss in tension wants to shrink back down so is pulling on things around it and things around it are pulling on it with a “tensile force”
2. in compression (negative force on the member) it wants to stretch back out so is pushing on things around it and things around it are pushing on it with a “compressive force”
By default, we write the truss member force as tensile (exactly like cables, except that truss members can have negative tension in them; i.e., compression).
Example: Simple Truss
Phys1D03 Exam Question: The truss ABCD is made of members with negligible mass forming equilateral triangles, and supports a weight of 600 N at pin C. The truss is fixed at pin A, and supported at pin D. Find the internal force in member BD, and state whether this is a compressive or tensile force.
Solution: write the force balance equations at each pin drawing the truss force as tensile by default. Don’t actually draw the FBD for the truss members (since we already analyzed 2-force members in general and used their equilibrium equation to determine that their reaction forces are equal & opposite and collinear, i.e., tension or compression). We instead draw the FBD for the pins (if the pins aren’t part of a frame we can draw the FBD for instead):
Now, each pin must be in equilibrium, so we can apply Newton’s second law at them. We only get force equations though, because the forces on each pin are applied at zero moment arm so the torque equations are trivial (i.e., 0=0).
The net force on Pin A in the x-direction must be zero:
where the 60o comes from this being an equilateral triangle, AD is the tension in member AB (just like if it were a cable, except with a beam connected like this the “tension” may end up being negative, meaning it’s pushing on the pins its attached to), and is the x-component of the reaction at A (the force the ground connected to the pin joint at A “reacts” with when you try to move it.) Because A is a pin joint, it can’t exert a torque (about z) to resist any attempted rotation in that direction, but can exert a force (in x and in y) to resist translations.
The net force on pin A in the y-direction must also be zero:
Similarly, the net force in each direction on Pin B are:
Net force on Pin D:
(D can only exert a reaction force in the positive y direction; it doesn’t resist x-motions or z-torques. So if we later find Dy needed to be negative, we either made a math error or the structure will not actually be in static equilibrium!)
Net force on pin C:
In 2P04 since we have Maple the hard part is generating these equations rather than solving them. We have 8 independent linear equations and 8 unknowns. Therefore, typing them in Maple and using the solve command will give a unique solution for all of the unknowns.
> restart: q:=60*Pi/180:
BC+BD*cos(q)-AB*cos(q),
-AB*sin(q)-BD*sin(q),
CD*cos(q)-BD*cos(q)-AD,
CD*sin(q)+BD*sin(q)+RD,
-BC-CD*cos(q),
-CD*sin(q)-600]): evalf(%);
This solution method where we analyze a truss piece-by-piece by applying static equilibrium at each of the connections is called the method of joints.
Instead of applying Newton’s second law to each pin individually you can generate different equations by viewing the entire structure as a rigid body. E.g., the net torque on the structure about point A is . However, these equations, while often convenient to quickly solve for a variable of interest, are not linearly independent from the internal force equation set we generated from the pins, so don’t give us any new information:
> RD=(1+cos(q))*600;
…but we already knew that.
Without Maple though, solving all 8 equations is not the fastest way to get the solution. In this particular question, the Phys1D03 exam only asked you to solve for , which you can get by solving equation 8 () for and then substituting into equation 6 () after you use the external torque-about-A equation to find RD.
Method of Joints and Method of Sections
The method of joints (analyzing a truss piece-by-piece by applying static equilibrium at each of the connections) is typically the fastest way to totally solve a truss system. But it’s not the most straightforward way to get a target piece of information, e.g., determine the force in CD, DK, and KJ in the following system from the old ENGINEER 2P04:
Solving this with method of joints would mean applying Newton’s second law at each of the 12 joints, for 24 equations in 24 unknowns!
Instead of that, we can more conveniently solve it in two steps
1. Assert that the entire truss as one object is in static equilibrium and solve for the support reactions at A
2. Mentally “section” (i.e., split) it along a line between the members of interest, then reason that the left side of it must be in static equilibrium too, and solve for the three unknown truss member tensions:
> restart:
Ay+Fy-30e3,
-2*15e3-6*15e3+8*Fy]); assign(%):
-KJ-CD/sqrt(2)-DK*1/sqrt(1^2+3^2),
Fy-15e3+CD/sqrt(2)-DK*3/sqrt(10),
-1*15e3+3*Fy-3*KJ]);
Note: knew the angle of DK by looking at the locations of the pins
> DKx=DK*cos(q)=DK*1/(sqrt(10))
Trickier torque equation, about point F:
> restart:
Ay+Fy-30e3,
-2*15e3-6*15e3+8*Fy]); assign(%):
-KJ-CD/sqrt(2)-DK*1/sqrt(1^2+3^2),
Fy-15e3+CD/sqrt(2)-DK*3/sqrt(10),
2*15e3+3*DK*3/sqrt(10)+3*DK*1/sqrt(10)
Archive solution:
This approach (sectioning the object to get directly the equations that will let us solve for the variables we want) is called the method of sections.
Solution by method of sections in full:
We’re best off finding the force at A first. There’s no horizontal force applied to the structure, so the reaction force at A is purely vertical:
Viewing the entire structure, we can say:
→ RF solved.
→ RA solved.
Method of sections to determine the forces in members CD, DK, and KJ.
Having determined RA from the entire truss’ static equilibrium condition, we can take a section through all 3 members of interest and then apply static equilibrium to the truss section on the left side:
and from the original truss picture, .
Solving with maple:
C45:=cos(Pi/4);
evalf(solve([
theta=arctan(3),
0=2*15-4*TCD*C45-4*15,
0=TCD*C45+TKD*cos(theta)+TKJ,
0=15-15+TKD*sin(theta)-TCD*C45
So, member CD is in compression (10.6 kN), member KD is in compression (7.9 kN), and member KJ is in tension (10 kN).
0-Force Members
Another trick to watch out for is a 0-force member. Consider member BL in that last truss problem:
Looking at joint L immediately shows that member BL must have 0 force in it:
Sometimes this effect cascades through the truss. e.g., if not for the 15 kN applied force, it would then also require BK to be a 0-force member.
Another way to recognize it is as follows:
In this system, member CD and DE are both zero-force members, because they are each 2-force members so must have a reaction at point D along the lines to their other pins:
But that non-180° connection between then makes this only work if both forces are 0.
Truss and Frame Members Together
Sometimes after doing a bunch of truss problems you can forget about the assumptions that led to something being a truss member in the first place; that it only has 2 forces acting on it and no torques. e.g., in this next problem, CF is not a 2-force member:
Example: Truss and Frame Combination Problem
For the following Frame system:
a) For each member, state whether it is a frame, 2-force member (truss), or zero-force member.
b) Draw an FBD for any and all frames
c) Write the system of equations you can use to solve for the force in all truss elements, and the reaction at fixed support A, then solve them.
Coordinates (in m):
A(0,0) (centre of where the beam is connected to the wall)
D(0.1,1.5)
F(2.4,1.5)
The force is in the downwards direction and is applied to member CF 2/3 of the way from pin C to pin F.
d) This very similar looking problem to problem 2 will not produce a unique solution set for the reactions at the supports in Maple. Why?
Deep Dive into Problem 16 Background: truss analysis of DE – Phantom Menace.
Could deal with DE (or any 2-force member) like this:
(Assigned arbitrary force magnitude and direction at each end.)
But that’s inconvenient when using Newton’s second law:
Now, for = 0 with those constraints on the angles, and also knowing E and D are both positive need one cos to be negative and the other positive for it to work; therefore , and E = D.
–> forces are equal and opposite and collinear.
This is a bit simpler with x & y components though:
So we can skip all that and go with this FBD for the truss:
Next step is to look at the pins around the truss:
(you don’t need to draw the truss FBDs to jump right to this step)
OK, so we can jump right to the frame and know that EF is just putting that tension force right on the frame CF through the pin (because there’s no other forces on the pin, the pin transmits the truss force to the frame; this means you only really need FBDs for pins (or anything!) when the pins have 3+ forces on them).
similarly,
Note that when we jump to the force of CF at C on ABCD, we draw it in the opposite direction right away. Technically, these forces are transmitted through the pin at C:
The FBD for ABCD, CF, and pin E give us 8 equations in 8 unknowns which uniquely determine the reaction forces and moments. Once the background concepts of how to transmit forces through pins, which things to use FBDs and apply Newton’s Second Law to, and how to use Newton’s 3rd law to set them up are clear you can skip many of these steps and solve the problem more like the following (Fast mode) solution.
Solution (Fast mode):
a) ADBC and CF are frames since force acts on more than their ends (and ADBC has torques applied at the wall).
BE is a zero-force member since pin E shows that it can’t supply any force
DE & EF are trusses.
> restart:
MA-1.5*DE+1.8*Cy,
-1.5*EF+2/3*0.6*2e3]);
(In Newtons)
d) It’s over-constrained; pin joint D gives two additional reaction forces (along with the two forces and torque unknowns from point A) but the external body being in equilibrium still only gives 3 equations in these 5 unknowns.
Simple machine: mechanical device for transmitting forces. Used to obtain some desired result.
Analysis is the same as frames.
Engineer 2P04 Problem Set 3 #10
1. Draw FBDs for each member
1. Write the static equilibrium equations for each member
1. Solve the equations
Here’s where the pins are:
Split into the frames:
Find the net torque on the three frames BGH, FCE, and GD about points B, C, and A to solve for x:
> restart:
100*FG=(100+300+75+350-x)*.5,
75*DE = 300*FG,
50*100=375*DE]);
Historical Solution:
Using the torque equations to initially ignore the pin forces, we can quickly solve for ED, FG, and then x. With those known, we can apply the force equations to each member to solve for the pin forces.
> restart;
50*98.1=375*ED,
75*ED=300*GF,
100*GF=(100+300+75+350-x)*0.49
Now the 6 pin equations are each one equation in one unknown:
Distributed Loads
Often, a load isn’t applied at a single point but instead is distributed across a surface (e.g., hydrostatic pressure on the wall of a dam is a distributed load along that wall).
https://images.theconversation.com/files/295772/original/file-20191007-121088-1ftaw7b.jpg?ixlib=rb-1.1.0&q=45&auto=format&w=496&fit=clip
If the distributed force per unit length of wall is , then the infinitesimal force it produces on a segment would be :
The total force in the x direction would be .
The torque this force dF creates about the point at location would be , so the total torque created by the distributed load is
This method works for forces in the x and y directions, as well as non-linear force distributions. We’ll explore this with several examples.
Distributed load & Piecewise Functions – Engineer 2P04 Problem Set 2 #7A
Setup equations to determine the reactions at the supports for the beam and loading as shown:
Interpreting Distributed Load Diagrams
This diagram convention tells you the size of the max distributed load force densities; it reaches 4.8 kN/m at the left and 6 kN/m at the right, but goes down to zero linearly from A to B and C to B.
B is a pin support, so we have (assign positive as to the right and up, respectively), while C is a roller joint, so gives only a reaction force in y ().
Pin supports resist translational forces but do not resist moments
Roller joints are free to rotate and move horizontally. They only apply reaction forces in the vertical direction.
where is the distributed load (in force/length). From the picture, you can work out the inte
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