Now what would happen if we lower the boom to 30 degrees?
The crane body is still
135 inch pounds, and the brick still weighs 4.7 pounds, but now it's farther away from the fulcrum. Again draw a line through
the CG of the brick and then another line at right angles to the fulcrum.
The new rt. triangle has a base angle of 30 deg., the hyp (the boom)
is still 34" but the formula now is
Cos 30 = Dist/ hyp
Cos 30 = .866 and hyp. = 34" so .866 = Dist./34 and
Dist = 29.44" ---call it 30"
and if W X D
, then 4.7 lbs X 30" = 141 inch pounds. Now for the boom- weight 2 pounds but the distance has
Can you do
the math now? Same rt triangle with a 30 deg base angle, but now the hyp (the boom) is 17".
Use the cosine to find the Dist. and multiply by the Weight of the boom (2pounds)--and
then add this to the 141 inch pounds of the brick. Did you get 189.1 inch pounds?? What would happen now???
What to do?? We could add a brick as counter-weight, all the way
at the far end of the crane.
W x D
4.7 lbs X 20" = 94 inch pounds
Which we add to the 135 of the crane to get 229 inch pounds-- a
lot more than the 190 . Now the crane is stable and won't tip.
#2 --- a 3 member truss is used to illustrate the Method of Joints
or how to prevent the bridge from collapsing.
Here's a simple 3 part truss, made of 2 legs (wood) and a tie (string) along the bottom.
It's a bare bones version of a King Post truss bridge like this one.
of the cars, people etc. (the live load) on the deck of the bridge, pulls down on the king post (the vertical part that connects
the deck with the top of the truss), and this force is transfered to the 2 legs. In our bridge the live
load is the brick which exerts the same force on the 2 legs.
force causes the legs to try and slide apart, which they would if they weren't held together by the string.
The string is in TENSION-that's the term engineers use to mean it's being stretched.
How much tension can the string support before it breaks? So we'll need a little geometry and the Method
of Joints to figure it out.
We'll first need to find the angle BAC
Each leg is 25" and the string is 36" . The truss is symmetric
and we can erect a perpendicular line from AC to point B which will bisect line AC.
Now we've got 2 rt. triangles- the hypotenuse is 25" and the adjacent side
is 18". To find the base angle BAC, we'll use the cosine function. Cos BAC = 18/25
So the base angle is 44 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.
step is to find the EXTERNAL FORCES acting on the truss.
Now we can introduce
the concept of equilibrium. The truss is not moving or rotating-it's
so we can use the 3 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 Torques/moments must equal zero.
The greek letter sigma means "sum of"
3 formulas tell us that the truss as a whole, and every member (part) of the truss is in equilibrium -not moving or rotating.
No horizontal force ( X) , no vertical force ( Y) and no Moments ( M) . Engineers prefer the term
Moment instead of torque---same thing.
there are no horizontal ( X) forces, so we go to
forces in the Y direction.
sigma Fy= 0
We've got 2 external forces in the Y direction---
the force (weight) of the brick
4.6 lbs and
the force (weight) of the truss which weighs 3 lbs.
Half this weight is carried by the small wood dowel that goes through A and C, so we'll just use 1 1/2 lbs.
gives us a total force Fy in the vertical direction of 4.6 + 1.5 = 6.1 lbs.
This force is pulling down along
the Y axis so that gives us a negative -6.1 lbs.
This force is countered by the REACTIONS at A and C Ra and Rc
You remember Newton's 3rd law about action and reaction? When your teacher
blew up a balloon and then released it the balloon took off like a jet.
air rushes out -that's the action, and the balloon is propelled in the opposite direction-that's the reaction.
Well the same thing happens with something as simple as the truss sitting on the table. The combined force of
the 6.1 lbs pushes down on the table at points A and C ---action--- and the table pushes back with equal
force- 6.1 lbs.
I always had trouble understanding
how the table actually pushes back. You can't see it 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 6.1 lb force.Obviously
we can't see this tiny compressive force. The bonds holding the atoms together are so strong, that when the atoms
are compressed or moved, they (the atoms) want to go back to their original position --this movement is the table pushing
back (the reaction).
So now back
to our equation sigma Fy = 0 We've got a downward force of 6.1lbs. and since along
the Y axis, down is negative that gives us -6.1 lbs. The 2 reactions at A and
C are pushing up (remember opposite ) so Ra and Rc are positive. And if the sum of all the forces in the vertical
direction are equal to 0 then
+ Rc - 6.1 = 0 With a little algebra we get Ra + Rc = 6.1 Since the truss
is symmetric, we could assume that the 6.1 lb. load is carried equally by the 2 legs and therefore the reactions would also
be equal, but we have to prove it mathematically. And we have 2 unknowns with only 1 equation, so we'll have to
go to the 3rd equation.
sum of the Moments is 0 sigma M = 0 No the bridge isn't rotating, but we can use this
formula as if it were.
Imagine that we put a pin through
the truss at A so that the truss could in fact rotate around point A. Scroll back up and review the concept of Torque/
moment. Since we're working with bridges ( and structural engineering ) we'll use the term Moment instead of torque.
there's a pin stuck through the truss at pont A. The force of the brick (4.7 lbs.)and the truss (1.5lbs)
combined to 6.1 lbs. would try to rotate the truss in a clock wise (CW) direction. The convention among engineers
is that a CW rotation would give a negative value.
to compute the moment we multiply the force times the distance.
is the 6.1 lbs. and the distance is from the line of force to the fulcrum --Db is 18"
so 6.1lbs X 18" = 109.8" lbs
Next is the
Reaction force at C. Remember this force is pushing up, which would cause the truss to rotate around the point A in a CCW
direction- and CCW rotation is positive.
to find the Moment at point C it's Force X distance
the force is Rc and the distance is 36" . The sum of all the moments at
points A and C
is sigma M = 0 or 0= Rc (36") - 109.8"lbs (remember the 109
is negative-CW rotation)
so some algebra and we get Rc(36") = 109.8" lbs. and
= 3.05lbs. Scroll back up to Ra + Rc = 6.1 When we substitute 3.05
for Rc we find that Ra is also 3.05
so we've proved that the reactions
at A and C are in fact equal.
We cut through the truss to isolate 1 joint and the attached members.
We're making a free body diagram at joint A
the joint and the forces attached to it and free it from the rest of the truss. We need to find how much force is in the left
hand leg - from joint A to B or member Fab, and how much tension (remember why we started this) is in
the tie between A and C or member Fac.
We drew the
arrows pointing away from the joint (A) which indicates the forces are pulling away from A -in other words they're in
tension. We don't know that for a fact but the convention is to asssume the forces are in tension- a positive value.If when
we solve the equation, the answer is negative , then the force was in Compression and not tension. We'll use the
summation of forces in the horizontal direction ( X ) and vertical direction ( Y ) again - no Moments.
First we look at the force in member AB Fab. This force is pulling
up and away to the right, not really vertical or horizontal. We can convert this diagonal force into vertical and horizontal
components using trig.
rt. triangle, sine is opposite side divided by the hypotenuse or in our example--vertical/ hyp.
so sine 44 = vert./hyp. or sine 44 = vert./Fab and then
to get the vertical component of Fab,
we multiply Fab by the sine of 44 degrees.
VERTICAL COMPONENT = Fab
What about the horizontal
component? cosine is adjacent side over hyp. so cos 44 = adj./hyp or cos 44=horiz./Fab and then we
= Fab (cos 44)
back to our truss. First the forces in the X (horizontal) direction.
Sigma Fx = 0 we've got 2 forces in the X direction. Fac and the horizontal component of Fab
so sigmaFx = 0 = Fac + Fab cos44=0 Both Fac and
Fab cos44 are acting to the right and along the X axis so they've both positive forces.
0= Fac + Fab cos44 We have 1 equation with 2 unknows
so let's go to the forces in the Y direction.
are 2 forces acting in the Y direction. Ra and the vertical component of Fab---Fab sin44 so
sigma Fy = 0 = Ra + Fab sin 44 Ra is pushing up and Fab is
pulling up and to the right, so both forces are positive. We know that Ra is 3.05, plug that in 0
= 3.05 + Fab sin 44 and sin 44 = .6947 so
0 = 3.05 + Fab (.6947) and now
Fab = - 4.39 A negative value which means the leg AC is under Compression not
That actually makes sense. The brick is pulling down on the
top of the truss, and the bottom of the truss is fixed in place by the tie (string), so the leg AB is being compressed.
Now we can
plug in the values of Fab- -4.39 lbs. and cos 44 (.7193 into our previous equation
sigma Fx= 0= Fac + Fab cos 44 0= Fac
+ (-4.39) .7193 and we get 3.15 lbs. of force
I used this
fish scale/ luggage scale to measure the tension in the tie.
reading of 3.13 lbs.is pretty close to the computed figure of 3.15, considering that all the measurements of weight and distance
were not very accurate. The idea is to illustrate how the tension in the tie will increase as the 2 legs get farther apart
and the angle decreases
LESSON #3- a deck truss is used to illustrate the Method of Sections
This 3ft long truss bridge is supporting a live load of 80lbs. Notice the string
and turn-buckle connecting the bottom chord of the truss. To find out how much tension is in the string ( and prevent the
string from failing and dropping the bridge) we use the Method of Sections.
The 80lb. live load is concentrated on joints B and E. If we use
sigma Fy =0, then all the downward forces (80lbs.) must equal all the upward forces, that's where the reactions come
in. Since the truss is symmetric , both reactions (40lbs.) are equal.
In this diagram of the truss, the string is member BE. To use
the Method of Sections, we can draw a line through the truss, making sure the line goes through the member we want (
BE). Now we can work with just that section to find the forces in the members CD BD and BE.
the truss isn't moving or rotating we can sy it's in EQUILIBRIUM and
then we can apply the 3 equilibrium equations:
1- the sum of the
forces in the X direction are zero ---Sigma Fx =0
2-the sum of the forces in the Y are zero
sigma Fy=0 and
3-the sum of the moments sigma M=0
To find the force in member BE we'll only
need the last equation. We'll compute the sum of the moments about joint D.
Why D? Remember that moment is the product
of Force X Distance, and
Dist. is the shortest distance from the line of force to the joint in question (joint
Notice that the arrows are drawn pointing away from the joints- this
indicates the forces are pulling away and are in tension. We don't know that for certain yet, but the convention among engineers
is to assume they're in tension =-a positive value. If when we solve the equations we get a negative value, then the
force was in Compression- a negative value.
Member CD has
its line of force going through D, so the Dist.from this line to joint D is zero, and
since Moment = Force X Distance if
the D is zero, then M is zero . Same thing for member BD, the line of force goes through joint D, so no Moment.
Member BE is 10" from D Now
imagine the section has a pin stuck through joint D, so the whole section
can rotate around D. Now the force in member BE (Fbe) goes to the right , so that if we
were pulling on the end of BE, the section would rotate counter-clock wise (CCW) around joint D. and CCW
rotation gives a positive value.
= Force x Distance so our first equation is
M = Fbe x 10"
is the force of the 40lb load pushing down through CB. The distance from this line of force to joint D is 9.5" so
M = F xD M = 40lbs. x 9.5"
or M = 380"lbs.
Finally we have the force of the reaction at A which pushes and would
cause the truss to rotate CW which gives a negative value. M = F x D -40lbs x 12.5+9.5= 22"
so M = -40lbs. x 22" = -880" lbs.
Finally the sum of all the moments at Joint D sigma Md = Fbe(10") + 380"lbs
- 880"lbs , do a little algebra and we get Fbe = 50 lbs of force. Now there are lines on both sides of the truss
so we have 25lbs. of force in each line. If we used thread instead of string , the thread would have broken and the bridge
would have collapsed.
Engineers have to do calculations like this
when designing a bridge so that doesn't happen.
Tower cranes need counter-weights to keep from tipping over- that's what the
red brick is for.
is a deck truss- the load (cars, people etc.) pass on top of the truss. I used single wide beams, connected at the joints
8-32 machine screws. The bottom middle member is a string
tensioned with a turn-buckle.
The live load is approx. 100lbs. This puts tension on the line (string) on
the bottom. I'll use the Method of Sections to compute the amount of tension on the string.
In a through truss, the live load passes through the truss. I only put around
50 lbs. on the deck.
In this lift bridge, the line is attached to the end of the deck, then goes
over a sheave (pulley) high up on the tower, and then down to the windlass.
Here the line goes over a pulley mounted lower down on the tower.
The lower angle means that it will take more force to raise the deck of the bridge.
In a cable-stayed bridge, the deck is supported by cables (string in this
case) that go in straight lines from the towers to the deck.The main span (the distance between the 2 towers) is almost 6ft.
Suspension bridges differ from cable-stayed bridges. The 2 main
cables rest on top of the towers, curve down between the towers and then attach to the anchorages. The deck is held
up by the "suspenders" the short black lines that hang down from the main cables and connect to the deck.