Re: How do you efficiently design and rehabilitate a bridge?

Alice:

I'm pleased to see another young woman interested in Engineering.  Also, 
congratulations on a clear layout of the problem.

First:  Most engineering is compromise.  The Engineer must balance several 
constraints, such as span, material strengths, clearances, stream flows, 
loadings, aesthetics, and others, when designing a bridge.  In the 
following discussion, I'm going to deal with the most important ones 
regarding your proposed project.  Some other considerations will be left 
by the wayside either because they do not enter into what you are doing or 
because the analysis is complicated.  

The bridge you describe appears to be a simple span, non-composite steel 
bridge with a concrete deck.  "Simple" span bridges are supported at each 
end, like a plank across two sawhorses, as opposed to a "Continuous" 
bridge which spans in one piece over three or more supports -- a plank 
with a sawhorse in the middle as well as at the ends.  "Non-Composite" 
means that the concrete road deck is resting on and completely supported 
by the beams beneath, while "composite" construction utilizes the deck as 
part of the structure.  Composite bridges are by far the most common, but 
the analysis is tricky.

Three things must be investigated when analyzing a bridge:  the stress at 
mid-span due to bending, the reaction at the end of the span, and the 
deflection, or how much the bridge sags downward.  These stresses are 
caused by the weight of the bridge itself;  the weight of the live load 
rolling across the bridge;  and the "impact" load, a little extra that's 
included. 

A lot of the formulas and data I'm giving here appear in the American 
Institute of Steel Construction "Steel Design Manual".  A copy of this may 
be in your local library -- but probably not. You might try a local 
engineering office.  Another resource that might be easier to find is the 
Engineer's Manual by Hudson. 

The village would normally specify what kind of load they want the bridge 
to support.  I've included a picture of the American Association of State 
Highway and Transportation Officials (A.A.S.H.T.O.) H20-44 standard truck, 
which is a good candidate for a small secondary road as you have 
described.  As the load rolls over the bridge, the maximum forces in the 
structure change also.  The maximum end reaction takes place when the 
heaviest axle has just come onto the end of the bridge, while the maximum 
bending occurs under the heaviest axle when that load is as far from the 
support as the center of gravity of all the wheels on the beam is from the 
other support, that is, when the heavy axle is the same distance from the 
middle of the span as the center of gravity.  The end reaction is 
calculated by the formula R = (w * d) / L, where w is each wheel load, d 
is the distance from the wheel to the support bearing, and L is the length 
of the span.  This is done for every wheel on the bridge.  The maximum 
bending stress (remember the truck is in a different place!)  is the 
reaction at the end of the beam caused by the truck, times the distance 
from the heavy axle to the nearest support.

You will have to decide how many beams to place under the deck.  Using 
many beams close together means that the deck can be lighter (it doesn't 
have to span as far between the beams).  But, although the beams 
themselves can be smaller, there are more of them and the weight adds up.  
Using fewer beams requires a heavier deck, and each beam must be bigger 
since it shoulders more load.  But fewer, heavier beams tend to be more 
efficient.  

(An aside:  There's also a question of redundancy.  If there are only two 
beams and one breaks, the bridge will collapse.  If there are lots of 
beams, however, and one breaks, the others are more likely to carry the 
load until somebody notices.)

A.A.S.H.T.O. and the State of Illinois use a standard deck thickness of 7-
1/2 inches for beam spacings from 5'-6" to 8'-9", and 8 inch thick deck 
for spacings of  9 feet to 14 feet.  They adjust the quantity of 
reinforcing steel in the deck to handle the increased stress.

The truck load is distributed to the beams by the following formula:  For 
a two lane bridge, the wheel load (i.e. 1/2 a truck) is placed on the beam 
at a ratio of  the beam spacing (BS) / 5.5, up to 14 feet.   For beams 
spaced 14 feet or greater, the road is assumed to act like a simple beam 
spanning between the supporting beams.  

"Dead Loads" are the weight of the bridge itself.  The weight of the deck 
per beam is the width of the deck between beams, times its thickness, 
times 150 pounds per cubic foot for concrete.  The weight of the beams 
themselves is the cross-section area times 3.4 pounds per square inch per 
foot of length for steel. 

Now it's time to design a bridge.  The span length is set by the distance 
between the abutments.  You need to select the number of beams, and how 
far apart they'll be spaced.  From this, you calculate the bridge deck 
load and the truck load.  The load of the beams themselves is estimated at 
first.  In these formulas, "*" means multiply, "/" is divide, "^" is raise 
to the power.  This is the convention used in most spreadsheet programs.

Let's assume that your bridge spans 42 feet from bearing to bearing.  
Start with four beams supporting two lanes, about 6 feet apart, with a 7-
1/2 inch thick concrete deck:

Dead load of deck (per beam) = 6 feet wide by .625 ft. thick by 150 lb / 
cubic foot = 562 lb. per foot of bridge.
Dead load of beam (guessed) say 88 lb. per running foot, to make an even 
650 pounds per running foot per beam.

Truck load:  Distribution factor is 6 ft divided by 5.5, or about 1.1.

The end reaction is the weight at one end of the bridge.  Imagine a plank 
spanning two sawhorses, with a row of bricks laid on it.  The reaction is 
how much you'd have to lift if you picked up one end.  The reaction caused 
by the bridge's weight  is the weight per ft times half the span length, 
or 

R = (W * L/2).  

In this example, it would be 650 * 42 / 2, 0r about 13,650 pounds.  (for 
convenience, engineers usually work in units of 1,000 pounds, 
called "kips".  Thus, 13.6 kips).  The maximum bending at the middle of 
the span is the bridge weight times the span length squared, divided by 
eight, or:

M = (W * L^2 / 8)               In our example, this is M = 650 * 42 ^2 / 
8, or   143.3 kips. 

The maximum end reaction caused by the truck occurs right when the heavy 
axle rolls onto the bridge.  This is calculated by 

Rmax = (4,000 lb. wheel * 28 ft + 16,000 lb. wheel * 42 ft) / 42 ft. * 1.1 
distribution factor.  or 20,533 lb.  

The maximum bending occurs when the rear axle of the truck is 20 ft. past 
the end.  The end reaction then is 

(4000 * 8 ft. + 16000 * 22ft.) / 42 ft. * 1.1 , or 10,050 lb.  The maximum 
bending stress is 

10,050 lb * 20 ft, or 201,000 foot-pounds.  

Engineers always like to put in a little extra.  This is called "impact" 
and it's given as percentage of the live load.  The formula is I = 50 / 
length + 125.  For our 42 foot bridge, the I = 50 / (42 + 125), 0r .299.  
The maximum impact is set by A.A.S.H.T.O. at 30%, so we're almost at that.

So our loading conditions are:

                                     end reaction                  bending
bridge load:                    13.6 kips                      143.3 kips
Truck load:                     20.5 kips                      201.0 kips
Impact of truck                6.1 kips                        60.1 kips

TOTALS:                       40.2 kips                       404.4 
kips    

The formulas for determining the size of beams you need to carry these 
loads are f = M * S for bending and f = r / (area of web) for the end 
reaction.  "f" is the allowable stress in the steel, "M" is the bending in 
the beam, "S" is the Section Modulus of the structural shape.

"S" is a property of the area, shape and depth of a section.  The units 
are inches to the third power ( in ^3 ).It takes a little getting used 
to.  The Section Modulus is actually a property of a section called the 
Moment of Inertia, divided by the distance from the centroid of the shape 
to the outermost fiber.  I wanted to avoid a long discussion of moment of 
inertia, though, so I am limiting this article to section modulus.   

For a rectangular shape, the formula for "I" is:  I = b * h ^3 / 12 .  

The formula for "S" is:   S = b * h^2 /6.  

  The A.I.S.C. Manual I mentioned earlier has tables giving the Section 
Modulus for common sizes of beams.  If you can't get a copy, you can 
calculate the "S"  of a symmetrical section by adding the area of the 
flanges times their distance from the middle, squared to b * h ^3/12 of 
the rectangular web plate, then dividing this number by 1/2 the total 
depth of the section.

[2 * (width * thickness of flange ) * y ^ 2] + ( b * h ^3 /12) / (1/2 the 
depth of the section. 

A.A.S.H.T.O. says that the allowable stress in tension for the most 
commonly used structural steel -- Type M270 Grade 36 -- is 20,000 pounds 
per square inch in tension, and 12,000 pounds per square inch for "end 
shear".  Multiply 404.4 foot-kips by 12 to convert it to inch-kips, and 
divide by 20 kips per square inch.  This gives a "Section Modulus" of 
242.6 inches ^3.   By doing several trial-and-error iterations, I came up 
with a beam consisting of a 1/2" by 21 inch web plate with two 10 inch by 
1 inch flange plates.  

The section modulus of this beam is:   2 * (1" * 10") * 11"^2 + .5" * 
21"^3 / 12, divided by 11.5 inches = 244.0 inches ^3. Check this by 
dividing  the bending of ( 404.4 foot kips times 12 inches per foot) by 
244 to get 19.9 kips per square inch, less than the allowable stress.  

The end reaction is carried by the web alone:  40.2 kips divided by ( 21 
* .5") =3.8  kips per square inch, which is less than 12 kips / sq in.

The only way to find out if you have the most efficient beam section is to 
try different beam configurations (deeper beams have a higher moment of 
inertia for their weight) and different numbers of beams --three, five, 
six, whatever.  Continue to do trials until you have a set of beams that 
can carry the loads with the minimum amount of steel in the beams.  HINT:  
constructing a spreadsheet to compute the beam properties and determine 
the loads for a given layout of beams is a tremendous time saver.

As for computing the cost estimates:

 For the steel, the weight of beams is found by multiplying the area of 
the beam times 3.4 pounds per square inch per foot, times the number of 
beams, times the span length.  We estimate our bridges based on $1.60 a 
pound.  

Concrete is estimated by taking the deck thickness times the width, times 
the length.  Add sidewalks if any.    Divide this by 27 to convert to 
cubic yards, and multiply by cost per cubic yard -- we use $400.   Paint 
is estimated by calculating the perimeter of the steel beams, less the top 
which is in the concrete, times the length, times the price per square 
foot for paint -- I don't have one handy.

I trust that you find this information useful in your project.

Chas.