Method of Joints and Free Body Diagrams

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    We need to find the tension in the bottom chord  of this 3-member truss-the string that connects

 the 2 legs of the truss. If the force is too great, the string might break, so we'd better find

 exactly what that force is.


We know the string is under Tension right?

The 2 legs of the truss are hinged at the top

and they're being pulled down by the lead

weight (the grey cylinder) . The only thing

keeping the bottoms of the 2 legs from getting

farther apart is the string-so the string is being

stretched. Engineers say it's IN TENSION- but we

have to prove that mathematically. 

We'll need to find the angle BAC. Each leg is 22" long and the string is 38".

Since the truss is symetrical, we can erect a perpendicular line from line AC to point B,

which will bisect line AC  


In this right triangle, the hypotenuse is 22" and the adjacent side is 19" so we'll use cosine.

  cos BAC = 19/ 22 = .8636   to  simplify let's call it .8660 which makes the angle 30 degrees.



I encourage my students to use the trig tables instead of just punching numbers

 into their calculators. Looking at the tables reminds them of how the trig values

go up or down as the angle changes. 

First we have to find the EXTERNAL FORCES acting on the truss.

The truss is in equilibrium- it's not moving or rotating so we can use the

3 basic equilibrium equations-


The Sum of the forces in the X-direction must equal zero
The Sum of the forces in the Y-direction must equal zero and
The Sum of the moments must equal zero.

The Greek letter sigma stand for "the sum of" 


These 3 formulas tell us that the truss as a whole and as well as every part
or member of the truss is in equilibrium- not moving.


                    Now let's find the EXTERNAL FORCES. 

There are 2 reaction forces   at  A and C.    We have a pin connecton at C so there will be  vertical and horizontal forces there.

           At A there's a roller connection ( you can see the wheel) so we only have a vertical force there.



1-the sum of all forces in the horizontal direction ...sigma Fx = 0 
 All we have is  the horizontal force at C    R/cx    so    Fx=0= R/cx which makes R/cx=0

2-the sum of all forces in the vertical direction.....sigma Fy = 0
     We've got 2 external vertical forces-
-the force (weight) of the lead cylinder which weighs 109 ounces and
-the weight of the truss -which weighs just over 5 pounds. Half of that weight is carried
by the bolts at points A and C, so we'll just use the other 2 1/2 lbs. or 41 ounces.




That gives us a total external force of 150 oz. (109 + 41) pulling down
and along the Y-axis, down is a negative value.
This action force is countered by a REACTION force at joints A and C.   Ra  Rc 


You remember Newton's 3rd law about action and reaction?
Well the 150 oz. load pushes down (action) on the table at points A and C
and the table pushes back in reaction.
 I always had trouble understanding how the table actually pushes back.
You can't see or feel it pushing back. That's because it all happens at the atomic level.
The atoms making up the surface of the table are compressed by the 150 oz. force.
Obviously we can't see this tiny compression force.  The bonds holding the atoms
together are so strong that when the atoms are displaced or moved, they(the atoms)
want to go back to their original position-this movement is the table pushing back( reaction).

  So back to our equation Sigma Fy = 0

  The 150oz. force pushes  down..........................-150
and the 2 reactions Ra and Rc push up..................+Ra and +Rc
And the sum of all these forces must be 0 so
        +Ra +Rc -150 =0  and to simplify we get      +Ra +Rc =150
Since everything about the truss is symmetrical we could assume the 150oz. load
is carried equally by the 2 members  so the reactions are probably the same too.
But we have to prove it mathematically and right now we have 1 equation with
2 unknowns, so we'll have to go to the 3rd equilibrium equation-  


                 SUM OF MOMENTS
The sum of the moments is equal to zero    Sigma M =0
If we use point A as the pivot, we can compute all the forces 

that try to rotate the truss around point A.
First let's look at the reaction force at A. This force goes right through A
so its distance from the pivot point is zero which makes the moment zero too.
Why?    Remember M = Force x Distance
and the Distance is measured from the line of force to the pivot, so 
                          if D = 0 then M = 0
Next is the 150 oz. force that's pulling down on the truss. Imagine if we could
stick a pin through point A and let the truss rotate around A. Then the 150 oz. force
would try to rotate the truss in a clockwise (CW) direction. 



The convention is that a CW rotation will give a negative value so we'd have a force of  -150

  Next we have the reaction force at C. It's pushing the truss up and would make the truss
rotate CCW, which by convention gives a positive value so we have   +Rc


So we have the 2 forces, now to compute the Distance in the formula
                  M = F x D
Remember the D is the shortest distance between the line of force and the pivot,
and the LINE OF FORCE is the line going through the center of gravity of the force/load/weight
straight down to the center of the earth- that's what gravity does- it pulls everything down.
 The shortest distance  for the 150 oz. force is the perpendicular distance from the line
of force to the point A.---19"


So we have a moment of .....M = F x D or  M= -150oz. x 19" = -2850 inch ounces
negative because we have a CW rotation. 
Next is the Reaction force at C...Rc which is pushing up and would rotate the
truss CCW which gives us a positive value. The D from the line of force
to point A is 38" so              M = Rc x 38

Now we can sum the moments and solve for Rc.

  Sigma M = 0   -2850 inch ounces + Rc (38) = 0   or
                     Rc = 2850/38 = 75
and now we can plug the value of Rc into the previous equation
    Ra + Rc = 150  and get Ra + 75 =150 so
                            Ra = 75
So we've proved that both reactions are the same.

Now to find the INTERNAL FORCES  in the members.
We cut through the truss to isolate 1 joint and the attached members.
  Here's a FREE BODY DIAGRAM  at joint A      


We draw just the joint A and the 2 forces attached to A. The arrows are pointing
away from the joint which indicates the forces are in Tension. We don't know that
for a fact but again the convention is to assume these forces are in Tension-a positive value.
If when we solve our equations the answer is negative, then the force was in Compression.

We'll use the summation of forces in the X and Y direction again- no Moments. 
However we have to convert any angled forces into their vertical and horizontal components.
Take the force acting along AB...Fab. It pulls up and away to the right. So some of this force
is acting up and some to the right.
To find the horizontal component we multiply the angle force Fab, times the cosine of angle A
and for the vertical component, we use the sine of the angle.  
                           It's basic trig.


In a right triangle the sine of an angle is opposite side / hypotenuse, so
       sin 30 = vertical component / Fab  or
            vert. component = Fab (sin 30)
Now for the horizontal component we use the cosine function.
  cos 30 = horiz. / Fab     or   horiz. = Fab (cos 30)
 So back to our truss. We've got 2 forces in the horizontal (X- axis) direction
             Fac and the horiz. part of Fab 


Both forces . Fac and Fab(cos30) are acting to the right and along the
X-axis, to the right is positive so
 Sigma Fx(at joint A) = 0    or
    +Fac + Fab(cos 30) = 0 
Again we have an equation with 2 unknowns so we have to go to 
                  Sigma Fy = 0
 We have 2 forces acting in the Y direction- Ra and the vert. component of Fab- Fab (sin 30).
Ra is pushing up and Fab(sin30) is pulling away (and up) from point A so both are positive.


            +Ra + Fab sin 30 = 0  and we know that Ra = 75 so

             75 + Fab sin30 =0    and sin 30 = .5  so
                   Fab = -75/.5 ......Fab = -150
and a negative value means member AB is in Compression, not Tension.
That makes sense since the 150 oz. load is pushing down on the 2 legs.
                       Now back to finding the force in member AC.
 We had the Sigma Fx =0 formula as
                 Fac + Fab cos30 = 0  so we can plug in -150 for Fab and .866 for cos 30

                 Fac +(-150) .866 = 0               Fac = 150 (.866) = 129.9
 which we'll call 130 oz. or 8.1 pounds of Tension in the string.

The load cell read 8 lbs.4oz.-not too far off from the computed value.
No matter how accurately we measure the weights and distances in these
exercises the readings on the gauge will only approximate the calculated value.
The idea is to show (roughly) the correlation between the base angle, the load
and the Tension in the bottom chord. 

   What result would we get if we doubled the load to 300 oz. and decreased
the base angle to 15 degrees? Figure it out!
The string might be able to withstand an increased Tension, but we know
students like to see something dramatic.

 Why not replace the string with a thin thread? Common sewing thread is
surprisingly strong but with enough tension it will break and the
truss will collapse with a satisfying noise.