The truss lesson
for people that are pretty smart
For your reference throughout the lesson:
pin
member
wall
joint
roller
Force notation examples:
Direction symbols:
⊥ = perpendicular
FRB = reaction force exerted by joint B
(-) = negative
FAAC = force applied to member AC
(+) = positive
FAx = force applied to object in x direction
(⟳) = clockwise
FRy = reaction force exerted by object in y direction
(⟲) = counterclockwise
Force variables refer to the magnitude of the vector
(a positive number, not the signed force)
Problem:
Determine the forces on each joint and member.
C
y’
f lbs
θ˚
B
x’
A
Steps
① ∑m = 0 = mA + mR
② ∑Fx = 0 = FAx + FRx
∑Fy = 0 = FAy + FRy
For the entire structure
③ (Same as 2) for each joint
④ Determine tension or compression
Step 1: Σm = 0
(balance the moments on the structure)
m = Torque/Moment (Angular force)
Think of torque as a force trying to rotate an object about a point P.
Moment is measured in ft-lbs. To calculate angular force, use
m=r∙F
where r is the radius (distance to point P) ⊥ to F, the linear force
F
P
pretend it’s up here
P
r
F
r
the applied force
the applied force
To find r, imagine rotating F around P
so it is ⊥ to the object or member
OR Imagine rotating the object around P
so it is ⊥ to F
x’
C
B
f
A
Step 1.1 calculate applied torque/moment
To find mA (the applied moment), “rotate” AC
until the force is ⊥. r, or the distance to the
fulcrum C, is x feet.
mA = r ∙ F = x ∙ f = fx ft-lbs
Direction: (⟳) clockwise or (-)
This is the torque on the whole structure,
about the fixed pin at C.
Step 1.2 calculate reaction moment
So fx ft-lbs are acting on the structure.
To calculate the reaction to this,
Use ∑m = 0 = mA + mR
C
action reaction
0 = mR – fx
B
A
mR = fx
(⟲)
mA is made (-)
because it is (⟳)
Step 1.3 find the receiver of the reaction force
We know that the initial applied force, mA, is exerted on point A.
On what point is the reaction force exerted?
Consider the force on A. It is clockwise, or acting in the negative direction. The
only object in the negative direction of A linearly is the member AB, which is in
the (-) x direction. The reaction force cannot originate in AB, however, because
AB is not fixed. Because AB can only exert force in the x direction, we know the
mA force is then exerted straight in the negative x direction, onto joint B. Joint B
cannot push back, because it is not fixed. So point B continues to push in the
negative x direction into the wall. We know the wall is fixed, so it is the source of
the reaction force, and it exerts this force on the point B to counteract the applied
moment.
C
action
reaction
B
A
Step 1.4 convert reaction moment back to linear force
We now know that mR = fx ft-lbs is acting on joint B.
What is the linear reaction force exerted on B?
Start with the moment equation:
mR = r ∙ F
from step 1.2 we know mR = fx ft-lbs
fx ft ∙ 1 lbs = r ∙ F
To find r, remember that the moment forces are acting on the structure about the
fulcrum C. The reaction force is already ⊥ to the member BC, so the radius is
simply the measure of BC, which is y ft.
fx ft ∙ 1 lbs = y ft ∙ F
the ft units cancel
fx/y lbs = F
the linear force on B
(+), x
C
f lbs
fx/y lbs
B
A
Step 2: ∑Fx = 0 , ∑Fy = 0
(balance x and y forces on entire structure
using a global free body diagram)
Applied/action forces
(exerted on objects)
C
Reaction forces
(exerted by objects)
Member forces
B
A
Component forces
of angled vectors
Look at the whole structure and assign forces to the joints, drawing arrows.
(see next slide)
Rules for drawing arrows
① Applied Forces
Draw the force where it is applied initially (A). Then draw the mR reaction
force on the roller (this is considered an applied force because it acts on
B and is exerted by the wall rather than any object in the structure).
Because the size and position of the arrows don’t matter for now, you
can draw the applied forces so that they won’t overlap with members or
components of other vectors.
② Reaction Forces
The pin exerts the global reaction forces. (Mr. Kaluta spent a long time in
class explaining why this is. If you don’t get it, just trust me.) Draw in the
x and y directions, opposite the applied forces.
③ Member Forces
At each joint, draw an arrow for each of the connected members. As a
default, draw these arrows in tension (away from joints). This can be
changed later (if an answer comes out negative, it means the arrow’s
direction should be switched).
④ Component Forces
Draw x & y components for each angled arrow.
Step 3: ∑Fx = 0 , ∑Fy = 0
(balance x and y forces on each joint
using local free body diagrams)
3.1 Roller (B)
before
transfer arrows from Global FBD
Known: FAx = applied/action force = fx/y
Balance & satisfy the equilibrium equations:
∑Fx = 0 = FAx + FRx
0 = fx/y + FRx
FRx = –fx/y
B
FRx
negative answer;
change direction of FRx.
∑Fy = 0 = FBC
0
FAx
no force exists to counteract the y force
after
fx/y
fx/y
B
change the FRx direction and type (to a reactionary force)
Something to remember before we continue:
The arrows you draw are purely conceptual. If you think too hard about them,
you will realize that the source of the force doesn’t really matter, as objects can
exert both pushing and pulling forces, and any push or pull will have an
opposite and equal reaction force. However, because we are using the joint
method, we are interested in the forces involving the joints. For our purposes,
at the beginning of each step all previously calculated forces are considered
applied forces, acting on the joints. The reaction forces, which we consider to
be exerted by the joints, are then calculated to balance the applied forces.
Keeping that in mind…
3.15
B
FRB
FAAB
Put the new arrows in context. Every joint and member must
have a net force of zero, so AB must react to the force
exerted on it by B. Based on this you can determine
the
FRABdirection of the x force exerted by AB on A,
and its magnitude (equal to FRB).
A
Understand that FRAB could be calculated using equilibrium equations, but they
would be quite simple as a member can only push/pull in one direction.
This step is simply used to determine the x force on A. Do it in your head and
don’t over think it.
before
3.2 (A)
Known: FAx= FRAB = FRB = fx/y,
FAy= f
∑Fx = 0 = FAx + FRx
0 = fx/y – FRx
FRx = fx/y
∑Fy = 0 = FAy + FRy
0 = FRy – f
FRy = f
reaction force FRAC
and its components, FRx and FRy
θ˚
FAx
A
FAx direction found in step 3.15,
changed from the global FBD
FAy
f / sinθ
FRx made (-) because arrow points left
fx/y
FAy made (-) because arrow points down
after
f
Because both FRx and FRy came out positive, FRAC is in the right direction
Find FRAC by solving one of the following equations:
FRx = f = FRAC sinθ
(trig)
A
FRy = fx/y = FRAC cos θ
(trig)
FRAC = √ (FRy)2 + (FRx)2
(pythagorean)
Though the pythagorean method can be used in this case, you may have to
use one of the trig equations in another problem.
Note the result of this step, at right.
FRx
Note: the BC member force has been removed
because step 3.1 found it was 0.
FRy
3.3
(C)
C
FAx
FA
Known: FA = f/sinθ (found by one of 3 equations in 3.2)
Simple trig can be done to find that FAx and FAy
are equal to the components from joint A.
FAx = fx/y, FAy = f
FAy
∑Fx = 0 = FAx + FRx
0 = fx/y – FRx
FRx = fx/y
FRx made (-) because arrow points left
∑Fy = 0 = FAy + FRy
0 = FRy – f
FRy = f
FAy made (-) because arrow points down
Again, both are positive, and therefore these arrows are in the correct directions.
Step 4: Find tension & compression
Tension and Compression refer to the direction of the forces exerted by
members or joints (reaction forces).
Tension – the member or joint is exerting outwards from itself (push)
Compression – the member or joint is exerting inwards, towards itself (pull)
Put all non-zero member forces on the diagram.
B is pulling in along AB. Compression.
C
A is pulling in along AB. Compression.
AB reacts to these forces by pulling in. Compression.
BC is under 0 force.
C is pushing out along AC. Tension.
A is pushing out along AC. Tension.
AC reacts by pushing out. Tension.
A
B
Therefore AC is a string under tension and AB is a stick under compression.
ANSWER.
f
fx/y
C
f/sinθ
f/sinθ lbs of tension*
0 lbs
f/sinθ
f lbs
fx/y
fx/y
B
A
fx/y lbs of compression*
* I am not sure if these should be doubled. Anyone know?
Also, I used f, x, y, and θ just to make this a general case. I do not recommend
using these answers as formulas.
Questions, Comments, Corrections, Suggestions
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