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I've recently switched to using 1/4" plywood strips for my bridge/crane models. So far I've designed a number of new structures--- 2 cranes, 3 truss bridges, 2 cable-stayed bridges, a suspension bridge, a tower crane and a lift bridge. 

Some of the topics I cover with  these models include:   

Center of Gravity    Equilibrium equations   Mechanical Advantage    Torque    engineering principles    Applied trig     How bridges work   useful knots 

                            LESSON # 1-

I use the short boom crane to explain static equilibrium, torque,

basic engineering and Mechanical Advantage.

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This crane has a 3ft boom and 2 windlasses (to control the boom and the load). Also, there are outriggers at each corner to keep the crane level.


I used the windlass to raise the boom to  60 degrees. 


The weight of the load (the brick) tipped the crane over.   Let's do the numbers to find out why.     The crane weighs 15lbs. and the distance from the pivot point to the center of gravity ( CG) is 9".    So what??          Well a crane operates like a see saw.    




How can 1  100lb guy balance 2 100lb. guys??   If the green guy sits farther away from the  pivot point that'll work.  If you were ever on a see saw as a kid you'd know this instinctively.   

 Archimedes came up with a formula that can be adapted to show that --- 

Weight   X  Distance on 1 side must equal  weight  x  distance on the oppposite side for everything to balance.

The numbers on both sides are 1000 foot pounds so the see saw is balanced.


Foot pounds....what's that??   The force measured in pounds times the distance measured in feet/inches/meters..When the see saw or a crane moves, it pivots around a fulcrum, so it's not a straight line movement, but a circular one.  That's why we need the distance from the fulcrum as part of the equation.

What we've been talking about here is TORQUE.  The torque of the 2 kids is their force (weight) times the distance from the pivot point.....1000 foot pounds which equals the torque of the kid on the other side--so the see saw is balanced.



Now imagine the crane is on one side of the pivot point (fulcrum). And the other side is the boom and the load hanging from the boom. The triangle shows the fulcrum. 


Remember we need weight times distance for our formula.   The crane weighs 15 lbs and the distance from the Center of Gravity ( CG) of the crane to the fulcrum is 9".   What's the CG? 

Imagine the guy on the see saw is laying down. Maybe he's 6ft tall or in this case 6ft long. To get the Distance would you measure from his head or his foot? No we measure from his CG which is probably near his belly button. 

 Same with the crane - we measure from the CG to the fulcrum. 

So now we've got  Weight  X Distance   or  15 lbs.  X   9"   =  135 inch pounds of TORQUE

What about the other side of the equation?   We've got the weight of the brick  (4.7 lbs) but what's the distance?   Draw a line through the CG of the brick , and make sure that line goes straight down.  That's the line of force---what force??? 

 Gravity. Gravity pulls everything straight down right?   To find the Dist. we need to measure from this line to the fulcrum, but we need it to be the shortest distance, and it turns out the shortest Dist.  is the line that's at a right angle to the line of force.--  in this case that's   D/L.    Looks like we've got a rt. triangle, with a 60 degree base angle and a 34" hyp.(the boom)


So we need to find the Dist. and that's the adjacent side of the triangle.  Remember your trig?   To find the adj. side we use cosine. 

Cos  = adj/hyp       Cos 60  = adj/ 34"   and Cos 60 = .5   so substituting we get     .5 = adj/34   or  adj= 17" so that's our Dist.    Now back to our basic formula of   Weight X Dist.     4.7lbs. X  17"   =  79.9 let's call it 80 inch pounds of TORQUE. 

Don't forget the boom. It weighs 2lbs. and the distance is ????  The CG of the boom is halfway along the boom (34" divided by 2 is 17"). 


So now we've got another rt triangle, same 60 degree base angle but now the hyp. is 17"     

Cos 60=  adj/hyp      Cos 60 =.5 and hyp. is now 17" so    .5 = adj/17 so the adj. or Dist is 8.5" 

and back to  W x D   we get  2lbs. x 8.5 "  - 17inch pounds.   almost done---

Now to get the total we add the 17 inch pounds to the 80 and get  97 inch pounds of TORQUE

Remember we found that the other side of the equation (the crane body) was 135 which is way more than 97 so the crane is stable, it won't tip over.   UNLESS of course if it's a windy day and the load starts to swing. Then the crane could tip over.  That's why we need to discuss the safety factor--more about that later.  

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 changed. 


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???  TIP OVER.  

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.      


                     LESSON #2 --- a 3 member truss is used to illustrate the Method of Joints

TORQUE 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.


The weight 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. 

This 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   = .72 


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.

The next 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 in equilibrium,

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"




These 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.       

            EXTERNAL FORCES   

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. 




This 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. 


The 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     

Ra + 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. 

The 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. 


So imagine 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.   

Now to compute the moment we multiply the force times the distance.


The force 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.    





Remember 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

then   Rc = 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.


                                         INTERNAL FORCES   

We cut through the truss to isolate 1 joint and the attached members.



We're making a free body diagram at joint A


We isolate 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. 


In a 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  (sin44)


What about the horizontal component?  cosine is adjacent side over hyp. so  cos 44 = adj./hyp  or  cos 44=horiz./Fab and then we get                   

                           HORIZONTAL COMPONENT =   Fab (cos 44)

Now 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  (Fab cos44)

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.  



There 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 tension

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. 

The 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. 


Since 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 D)-



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.


 Moment  = Force  x Distance    so   our first equation  is 

                                   M = Fbe x 10"  

Next 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. 


Almost done.   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. 



This is a deck truss- the load (cars, people etc.) pass on top of  the truss. I used single wide beams, connected at the joints with

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.


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