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Analysis and Design of Pratt Trusses

Aamer Haque

Abstract

Formulas for member forces are derived for deck and through Pratt trusses. Strength and deflection constraint designprocedures are developed for the trusses. The design procedures are applied to a hypothetical railroad bridge.

Preprint

Preprint submitted to Elsevier July 21, 2017

1. Formulation

1.1. Introduction

The optimal design of trusses is rarely discussed in struc-tural engineering texts. At most, one may find a commentthat states that truss depth should approximately equal thepanel length. Such a rule of thumb is said to result in anoptimal design. A proof is rarely offered and common designpractice is used to justify such remarks. The text by Troitsky[6] offers a formula for computing the optimal truss depth fora given panel length using strength constraints. A paramet-ric study of truss design was provided by Waling [7]. Thesesources provide guides for design practice but few details onthe derivation of the results.

This paper provides formulas for the member forces anddeflections for Pratt trusses. The method of sections and themethod of joints are used to compute member forces. De-flections are computed using the unit dummy load theorem.These concepts are explained in any standard text on struc-tural analysis such as Hibbeler [1], Kassimali [2], Timoshenkoand Young [5]. Historical information and terminology fortruss bridges can be found in Ketchum [3], Mallery [4], Troit-sky [6].

Computation of structural weight of a truss is detailed inthis paper. We assume a fixed span length L and knownuniform load w along the span. Specifying the number ofpanels and truss height determines the geometry of the truss.For a known geometry, the members are sized to satisfy ei-ther strength or deflection constraints. Members can be sizedindividually or according to their member type. The sizingcompletely determines the design of the truss. The weight ofthe design is given by the function W (n, h) where n is thenumber of panels for half the span and h is the truss height.

The optimal design of a truss is the design which minimizesW (n, h). For strength design we derive formulas for optimalvalues of h given a known n. For deflection constrained de-sign, we fix one of the variables and determine the optimalvalue of the other variable. It should be noted that strengthdesign usually results in a structure with excessive deflections.Members of such a design can be uniformly resized to satisfydeflection constraints.

1.2. Truss Geometry

In order to properly describe a truss, its geometry and load-ing must be specified. The deck and through Pratt trussesare shown in figures 1, 2, and 3. Let L be the span lengthand h be the truss height. Symmetry implies that we needonly consider half the span of the truss. n is the numberof complete panels for half the span. We shall assume thatn ≥ 1. The panel numbering depends on the type of truss.Index i = 1, . . . , n for deck trusses shown in figure 1. Indexi = 0, . . . , n for deck trusses of figure 2 through trusses offigure 3.

The panel length a is defined as:

a =

{

L2n Deck Truss

L2(n+1) Alternate Deck/Through Truss

(1.1)

Figure 1: Deck truss dimensions and loading

Figure 2: Alternate deck truss dimensions and loading

Figure 3: Through truss dimensions and loading

Define θ to be the angle between the diagonals and thechords. Some basic trigonometric functions are then easilydefined.

cos θ =1

√

1 +(

ha

)2

sin θ =1

√

1 +(

ah

)2

tan θ =h

a

cot θ =a

h

2

1.3. Loads

The maximum uniform live load is assumed constant, butthe dead load depends on the panel length. This is due tothe fact that more structural material is required for longerdeck and road lengths. We shall assume that the uniform(dead+live) load has the following simple form:

w = w0 + ω(a− a0)2 (1.2)

where w0, ω, and a0 are constants. a0 is the minimum feasiblepanel length. Structural weight of the truss is not includedas part of the dead load. This is a reasonable procedure forsteel structures where the structural weight is much less thanthe dead and live loads.

The uniform load w is transfered to the joints as a series ofpoint loads:

p = wa (1.3)

1.4. Weight

Suppose there are N structural members (i.e. chords, di-agonals, and posts). Let member k have cross-sectional areaAk and length Lk. The total structural weight of the truss iscomputed to be:

W = γ

N∑

k=1

AkLk (1.4)

where γ is the specific weight of the structural material.Clearly W depends on the number of members and their ge-ometry.

1.5. Strength Design

1.5.1. Strength Constraints

If the force Fk in member k is known, the member can besized according to strength criteria. All members are assumedto have yield strength σy in both tension and compression.The required cross-sectional area is thus:

Ak =|Fk|

σy

(1.5)

The weight function can be rewritten in terms of memberforces:

W =γ

σy

N∑

k=1

|Fk|Lk (1.6)

1.5.2. Strength Design Procedure

The following procedure is employed for a given truss ge-ometry and loading:

1. Calculate the member forces for the truss. Explicit ex-pressions for member forces are possible because thetrusses are statically determinant.

2. Use the strength constraint to size each member typeaccording to equation (1.5).

3. Compute the weight W using equation (1.6) for the de-sign determined by step 2.

1.5.3. Strength Design Optimization

The strength design procedure determines the member sizesand weight W (n, h). We wish to compute the truss heighth, for a given number of panels n, which minimizes weightW (n, h). This is written mathematically as:

optimal h for given n = arg minh∈R+

W (n, h) (1.7)

where R+ = (0,∞). The optimal h is computed analytically

by solving the following equation for h:

∂

∂hW (n, h) = 0 (1.8)

1.6. Deflection Design

1.6.1. Deflection Constraints

Sizing members according to strength constraints will usu-ally produce a structure which deflects excessively. It is nec-essary to limit deflections by increasing member size. Thefirst step is to compute the mid-span deflection using the unitdummy load theorem.

∆ =N∑

k=1

FkFk

(

Lk

EAk

)

(1.9)

Fk is the force in member k due to a unit load at the mid-span. The maximum allowable deflection ∆limit is typicallystated in terms of span length. The deflection constraint is:

∆ ≤ ∆limit =L

b(1.10)

where b is a constant. Assuming that ∆ ≥ ∆limit, the mem-bers can be uniformly resized using:

Ak :=

(

∆

∆limit

)

Ak (1.11)

This causes the deflection to be reduced to exactly ∆limit.

1.6.2. Deflection Design Procedure

The following procedure is employed for a given truss ge-ometry and loading:

1. Calculate the member forces for the truss. Explicit ex-pressions for member forces are possible because thetrusses are statically determinant.

2. Use the strength constraint to size each member typeaccording to equation (1.5).

3. Compute the weight W using equation (1.6) for the pre-liminary design determined by step 2.

4. Compute the deflection ∆ using the unit dummy loadtheorem formula (1.9).

5. Apply the deflection constraint (1.10) to the preliminarydesign. Equation (1.11) is used to resize the members.

6. Compute the weight W for the resized truss from step 5using equation (1.4).

3

1.6.3. Deflection Design Optimization

Fixing either n or h allows W be a function of only onevariable. We perform the following optimizations:

optimal h for given n ∈ N = arg minh∈R+

W (n, h) (1.12)

subject to ∆ = ∆limit

optimal n for given h ∈ H = arg minn∈N

W (n, h) (1.13)

subject to ∆ = ∆limit

where H ⊂ (0,∞) and N ⊂ N. These procedures are toocumbersome to perform analytically. Instead, we generateplots for the design example described next.

1.6.4. Design Optimization Example

Span L = 200 ft

Loading w0 = 9.5 kip/ft

ω = 3.125× 10−4 kip/ft3

a0 = 10 ft

Yield strength σy = 50 ksi

Young’s modulus E = 29, 000 ksi

Specific weight γ = 0.28 lb/in3

Deflection limit ∆limt = L/800

Table 1: Bridge parameters

A deflection design optimization study is applied to a dou-ble track steel railroad bridge with a truss on each side of thebridge. Pratt trusses in both the deck and through configura-tions are investigated. The fixed parameters for each truss aregiven in table 1. This example is for theoretical purposes andis not intended to satisfy the requirements of actual designcodes.

Cooper’s E80 loading is used to compute a uniform live loadof wlive = 8 kip/ft. Cooper’s loading is described in Mallery[4], Timoshenko and Young [5], Waling [7]. Dead loads varywith panel length a and are approximated by:

wdead = 1.5 kip/ft+(

3.125× 10−4 kip/ft3)

(a− 10 ft)2

The combined load w = wlive+wdead parameters for equation(1.2) are given in table 1. Note that the loads are unfactoredand strengths are not reduced.

The variable truss parameters are the numbers of panels nfor half the span and the truss depth h. The optimizationprocedure (1.12) uses N = {2, 3, 4, 5, 6, 7} for the deck trussand N = {1, 2, 3, 4, 5, 6} for the alternate deck and throughtruss. Procedure (1.13) uses H = {24 ft, 26 ft, 28 ft, 30 ft} forall truss types.

1.7. Sizing by Member Type

The procedures described in the previous sections willsize each member individually. For simplicity of fabrication,trusses are sometimes sized by member type. In other words,members of the same type all have identical cross-sectionalareas. We modify our design procedures to size members bytype.

We first categorize member forces by member type. Let Fbe the set of containing N member forces:

F = {F1, . . . , FN} = D + T + C + P

D are the diagonals, T are the bottom chords, C are the topchords, and P are the posts:

D = {D1, . . . , DND}

T = {T1, . . . , TNT}

C = {C1, . . . , CNC}

P = {P1, . . . , PNP}

Clearly we must have:

N = ND +NT +NC +NP

We size a member according to the largest force of thatmember type. Let Y ∈ {D,T,C, P} be a symbolic index anddefine:

Ymax = Yµ, µ = arg maxk=1,...,NY

|Yk| (1.14)

µ is the index to the largest absolute force in the member set Ywhere Y ∈ {D, T , C,P}. Note that Ymax may be negative, buthas the largest magnitude of all the forces in Y. All membersin set Y are then assigned a cross-sectional area of:

AY =|Ymax|

σy

(1.15)

1.8. Sizing Zero Force Members

Zero force members may occur in a truss. Such membersare still necessary to ensure stability of the truss. They willalso acquire non-zero forces under other loading conditions.A simple way to size these members is to assign them thesmallest size of the non-zero force members of the same type.Let KY be the set of indices to non-zero force members in setY and define:

Ymin = Yν , ν = arg mink∈KY

|Yk| (1.16)

Then we set the zero force member to have size:

AY 0 =|Ymin|

σy

(1.17)

This procedure is applied only to individual member sizing.Sizing by member type always assigns zero force members tonon-zero cross-sectional area.

4

2. Deck Pratt Truss

2.1. Member Forces

Figure 4: Deck Pratt truss

Figure 5: Free body diagram

Member Force Index

Vi =(

n− i+ 12

)

p i = 1, . . . , n

Di = Vi

√

1 +(

ah

)2i = 1, . . . , n

Ti = p(

ah

)

(i − 1)[

n− 12 (i − 1)

]

i = 1, . . . , n

Ci = −[

Ti + Vi

(

ah

)]

i = 1, . . . , n

P1 = −np i = 1

Pi = − (Vi + p) i = 2, . . . , n

Pn+1 = −p i = n+ 1

Table 2: Member forces

Maximum Member Force Index

Vmax =(

n− 12

)

p i = 1

Dmax = Vmax

√

1 +(

ah

)2i = 1

Tmax = p2

(

ah

) (

n2 − 1)

i = n

Cmax = − p2

(

ah

)

n2 i = n

Pmax = −np i = 1

Table 3: Maximum member forces

The shear that exists in the i-th panel is computed by con-sidering equilibrium in the vertical direction:

∑

Fy = 0

np−p

2− (i− 1)p− Vi = 0

np− ip+p

2− Vi = 0

(

n− i+1

2

)

p− Vi = 0

Solving for Vi and incorporating the value of p using equation(1.3) gives:

Vi =

(

n− i+1

2

)

p (2.1)

To locate the maximum shear, we compute the derivativeof 2.1 with respect to i. When computing the derivative, weassume that i is a continuous variable. However, we onlyevaluate the function Vi and its derivative at non-negativeinteger values of i. For the present case, the result is simply:

∂V

∂i= −p < 0

This implies that the shear decreases as i increases and thusthe maximum shear is located in the end panels.

Vmax = V1 =

(

n−1

2

)

p (2.2)

The tensile force in the diagonal is:

Di =Vi

sin θ= Vi

√

1 +(a

h

)2

(2.3)

Since sin θ is constant for fixed n, the largest force in adiagonal member occurs at the end panels:

Dmax = D1 = Vmax

√

1 +(a

h

)2

(2.4)

The tensile force in the bottom chord is computed by sum-ming the moments about the joint connecting the post, diag-onal, and top chords:

∑

M = 0

Tih+ (i − 2)p(i− 1)a

2−

(

n−1

2

)

(i− 1)pa = 0

Tih+

[

1

2(i − 2)−

(

n−1

2

)]

(i− 1)pa = 0

Tih+

[

i

2− 1− n+

1

2

]

(i− 1)pa = 0

Tih+

[

1

2(i − 1)− n

]

(i− 1)pa = 0

5

Solving for Ti produces:

Ti = p(a

h

)

(i− 1)

[

n−1

2(i− 1)

]

(2.5)

Notice that T1 = 0.Computing the derivative to locate the location of maxi-

mum tensile force:

∂Ti

∂i= p

(a

h

)

[

n−1

2(i− 1) + (i− 1)

(

1

2

)]

∂Ti

∂i= p

(a

h

)

n

Clearly we have:∂Ti

∂i> 0

Thus the maximum tensile force for bottom chords is locatedat the n-th panel:

Tmax = Tn =p

2

(a

h

)

(

n2 − 1)

(2.6)

The force in the top chord is computed by considering equi-librium in the horizontal direction:

∑

Fx = 0

Ti +Di cos θ + Ci = 0

Ti + Vi cot θ + Ci = 0

Ci is given by the equation:

Ci = −[

Ti + Vi

(a

h

)]

(2.7)

The negative sign indicates that this force is compressive in-stead of tensile.

Since Ci is always negative, we compute the derivative of itsabsolute value to locate the force with the largest magnitude.

∂ |Ci|

∂i=

∂Ti

∂i+

∂Vi

∂i

(a

h

)

∂ |Ci|

∂i= p

(a

h

)

n− p(a

h

)

∂ |Ci|

∂i= p

(a

h

)

(n− 1)

Obviously:∂ |Ci|

∂i≥ 0

Equality holds only if n = 1. In any case, the maximum islocated at panel n:

Cmax = −p

2

(a

h

)

n2 (2.8)

The force in the post is not computed using the methodof sections. Instead, consider equilibrium in the vertical di-rection at the joint connecting the post, diagonal, and topchords. For i = 2, . . . n we compute vertical equilibrium asfollows:

∑

Fy = 0

−Pi −Di sin θ − p = 0

The formula for Pi is:

Pi = − (Vi + p) , i = 2, . . . , n (2.9)

The negative sign again indicates that this is a compressiveforce.

For the end post i = 1 is:

P1 = −np

The force in the center center post i = n+ 1 is:

Pn+1 = −p

Hence the maximum post force is the end post:

Pmax = P1 (2.10)

6

2.2. Unit Dummy Load

Figure 6: Unit dummy load


Member Force Index

Vi =12 i = 1, . . . , n

Di =1

2 sin θi = 1, . . . , n

Ti =12 (i − 1)

(

ah

)

i = 1, . . . , n

Ci = − i2

(

ah

)

i = 1, . . . , n

Pi = − 12 i = 1, . . . , n

Pn+1 = −1 i = n+ 1


The unit dummy load theorem is used to compute the mid-span deflection. The loading and free body diagrams for theunit dummy load are given in figures 6 and 7. The mem-ber forces for the unit dummy load are computed using themethod of sections. The shear in each panel is given by:

∑

Fy = 0

1

2− Vi = 0

Vi =1

2(2.11)

The force in the diagonal is then computed to be:

Di =1

2 sin θ(2.12)

To compute the tension in the bottom chords, we sum themoments about the joint connecting the post, diagonal, andtop chords:

∑

M = 0

Tih−1

2(i− 1)a = 0

Ti =1

2(i − 1)

(a

h

)

(2.13)

The compressive force in the top chord is computed by con-sidering equilibrium in the horizontal direction:

∑

Fx = 0

Ti + Di cos θ + Ci = 0


Ci = −i

2

(a

h

)

(2.14)

The compressive force in the posts are easily calculated:

Pi = −1

2, i = 1, . . . , n (2.15)

The force in the center post is:

Pn+1 = −1

7

2.3. Strength Design Optimization

2.3.1. Individual Member Sizing

In order to facilitate the computation of total weight, weorganize the calculations according to member type. Equa-tion (1.6) is applied to each member in order to satisfy thestrength constraints. We also wish to clearly write the weightsas functions of h. This will allow us to compute the optimalvalue of h.

The total weight of the diagonals is:

WD =2γ

σy

n∑

i=1

Di

√

a2 + h2 (2.16)

This formula is rewritten using (2.3):

WD =2γ

σy

n∑

i=1

Di

√

a2 + h2

WD =2γ

σy

n∑

i=1

Dih

√

1 +(a

h

)2

WD =2γ

σy

n∑

i=1

Vih

[

1 +(a

h

)2]

WD =2γh

σy

[

1 +(a

h

)2] n∑

i=1

Vi

The desired form is:

WD =2γ

σy

(

h+a2

h

) n∑

i=1

Vi (2.17)

When computing the combined weight of the bottomchords, we must size the zero force member. The smallestnon-zero size is given by T2. Remembering that T1 = 0, wecan write:

WT =2γ

σy

[

T2 +n∑

i=1

Ti

]

a (2.18)

We rewrite this in the more convenient form:

WT =2γ

σy

(

a2

h

)

[

T2 +n∑

i=1

Ti

]

(2.19)

where Ti is defined using equation (2.5):

Ti =(a

h

)

Ti, Ti = p(i− 1)

[

n−1

2(i − 1)

]

For top chords, the weight function is:

WC =2γ

σy

n∑

i=1

|Ci| a (2.20)

Factoring out the appropriate terms with h in equation (2.7),we have:

WC =2γ

σy

(

a2

h

) n∑

i=1

Ci (2.21)

using the definitions:

|Ci| =(a

h

)

Ci, Ci = Ti + Vi

The weight of the posts must be handled more carefullysince the number of posts is not equal to the number of trusspanels.

WP =γ

σy

[

|Pn+1|+ 2 |P1|+ 2n∑

i=2

|Pi|

]

h (2.22)

The term in the brackets can be simplified further:

WP =γ

σy

[

|Pn+1|+ 2 |P1|+ 2

n∑

i=2

|Pi|

]

h

WP =γ

σy

[

p+ 2np+ 2

n∑

i=2

(Vi + p)

]

h

WP =γ

σy

[

p+ 2np+ 2n∑

i=2

(Vi + p)

]

h

WP =γ

σy

[

p+ 2np+ 2

n∑

i=2

[(

n− i+1

2

)

p+ p

]

]

h

WP =γ

σy

[

1 + 2n+ 2

n∑

i=2

(

n− i+3

2

)

]

ph

WP =γ

σy

[1 + 2n+ 2n(n− 1)

−n(n+ 1) + 2 + 3(n− 1)] ph

WP =γ

σy

[

n2 + 2n]

ph

Thus we have:

WP =

(

γ

σy

)

n(n+ 2)ph (2.23)

The total weight of the truss is the sum of the memberweights:

W = WD +WT +WC +WP (2.24)

8

Assume that we fix the number of panels n. Then h, whichminimizes total weight W , is computed by setting:

∂W

∂h= 0 (2.25)

and solving for h. We notice that each term of the resultingexpression will contain a factor of 2γ/σy and this term isdivided on the each side of the equation. We thus have:

(

1−a2

h2

) n∑

i=1

Vi−a2

h2

[

T2 +

n∑

i=1

Ti

]

−a2

h2

n∑

i=1

Ci+1

2n(n+2)p = 0

This equation is easily solved for h2/a2:

a2

h2

[

T2 +

n∑

i=1

(

Vi + Ti + Ci

)

]

=1

2n(n+ 2)p+

n∑

i=1

Vi

a2

h2

[

T2 +n∑

i=1

(

Vi + Ti +(

Ti + Vi

))

]

=1

2n(n+ 2)p+

n∑

i=1

Vi

a2

h2

[

T2 + 2

n∑

i=1

(

Vi + Ti

)

]

=1

2n(n+ 2)p+

n∑

i=1

Vi

a2

h2=

12n(n+ 2)p+

∑n

i=1 Vi

T2 + 2∑n

i=1

(

Vi + Ti

)

h2

a2=

T2 + 2∑n

i=1

(

Vi + Ti

)

12n(n+ 2)p+

∑n

i=1 Vi

Taking the square root of both sides gives the final result:

h

a=

√

√

√

√

T2 + 2∑n

i=1

(

Vi + Ti

)

12n(n+ 2)p+

∑ni=1 Vi

(2.26)

We note that every term inside the square root has a factor ofp. The division ensures that (2.26) does not depend on p. Weprefer to rewrite this equation the formula in terms of h/L:

h

L=

1

2n

√

√

√

√

T2 + 2∑n

i=1

(

Vi + Ti

)

12n(n+ 2)p+

∑ni=1 Vi

(2.27)

Preprint

9

2.3.2. Sizing by Member Type

The total weights for each type of member is given by:

WD =2γ

σy

[

h+a2

h

]

nVmax

WT =2γ

σy

(

a2

h

)

nTmax

WC =2γ

σy

(

a2

h

)

nCmax

Wp =

(

γ

σy

)

(2n+ 1) |Pmax|h

where Vmax =(

n− 12

)

p, Tmax = p2 (n

2 − 1), Cmax = 12pn

2,and |Pmax| = np. The total truss weight is again:

W = WD +WT +WC +WP

Setting the derivative of W , with respect to h, to zero:

(

1−a2

h2

)

nVmax −

(

a2

h2

)

nTmax

−

(

a2

h2

)

nCmax +

(

n+1

2

)

|Pmax| = 0

Solving for h2/a2 produces:

a2n

h2

(

Vmax + Tmax + Cmax

)

= nVmax +

(

n+1

2

)

|Pmax|

a2

h2=

nVmax +(

n+ 12

)

|Pmax|

n(

Vmax + Tmax + Cmax

)

h2

a2=

n(

Vmax + Tmax + Cmax

)

nVmax +(

n+ 12

)

|Pmax|

We factor out n on the numerator and denominator, and thentake the square roots of both sides:

h

a=

√

Vmax + Tmax + Cmax

Vmax +(

1 + 12n

)

|Pmax|

Further simplification is possible by substituting the definedvalues and recognizing that p occurs in every term:

h

a=

√

(

n− 12

)

+ 12 (n

2 − 1) + 12n

2

(

n− 12

)

+(

1 + 12n

)

n

The final formula is:

h

a=

√

n2 + n− 1

2n(2.28)

Rewriting equation (2.28) in terms of h/L results in:

h

L=

1

2

√

n2 + n− 1

2n3(2.29)

Preprint

10

2.4. Deflection Design Optimization

2.4.1. Optimal h for given n

0.1

0.15

0.2

0.25

0.3

2 3 4 5 6 7

h/L

n

Deck Pratt

Strength Design - IndividualStrength Design - By type

Deflection Design - IndividualDeflection Design - By type

Figure 8: Optimal h/L vs. n

1

1.2

1.4

1.6

1.8

2

2.2

2 3 4 5 6 7

h/a

n

Deck Pratt



Figure 9: Optimal h/a vs. n

Figures 8 and 9 show how truss height h should be chosenfor a given number of panels n.

2.4.2. Optimal n for given h

240

260

280

300

320

340

360

380

2 3 4 5 6 7

W [k

ip]

n

Deck Pratt - Individual

h=24 fth=26 fth=28 fth=30 ft

Figure 10: Individual member sizing: minimum W vs. n

260

280

300

320

340

360

380

400

420

2 3 4 5 6 7

W [k

ip]

n

Deck Pratt - By type


Figure 11: Sizing members by type: minimum W vs. n

Figure 10 demonstrates that 3 panels for half the span areoptimal for individual member sizing. Figure 11 shows that 3panels are optimal when sizing members by type. The weightincreases considerably when using only 2 panels. This is dueto the increased deck weight caused by long panel length.

11

3. Alternate Deck Pratt Truss

3.1. Member Forces

Figure 12: Alternate deck Pratt truss


Member Force Index

Vi =(

n− i + 12

)

p i = 0, . . . , n

Di = Vi

√

1 +(

ah

)2i = 0, . . . , n

Ti = p(

ah

) [

n+ 1− 12 i]

i i = 1, . . . , n

C0 = −V0

(

ah

)

i = 0

Ci = −[

Ti + Vi

(

ah

)]

i = 1, . . . , n

Pi = − (Vi + p) i = 1, . . . , n

Pn+1 = −p i = n+ 1



Vmax =(

n+ 12

)

p i = 0

Dmax = Vmax

√

1 +(

ah

)2i = 0

Tmax = p2

(

ah

) (

n2 + 2n)

i = n

Cmax = − p2

(

ah

)

(n+ 1)2 i = n

Pmax = −(

n+ 12

)

p i = 1


The deck truss of the previous section had end panels withlower chords as a zero force member. Also, a very large com-pressive load is taken by the end post. These members canbe eliminated to produce lighter truss. Figure 12 shows thealternative deck truss.

The formulas for member forces in tables 2 and 3 remainvalid for this truss. However, the indexing must be adjustedsince we now index from i = 0 and have 2(n + 1) total pan-els. The following variable replacements are made in thoseformulas:

i → i+ 1, n → n+ 1 (3.1)

The new formulas are given in tables 5 and 6.

12




Member Force Index

Vi =12 i = 0, . . . , n

Di =1

2 sin θi = 0, . . . , n

Ti =12

(

ah

)

i i = 1, . . . , n

Ci = − 12

(

ah

)

(i+ 1) i = 0, . . . , n

Pi = − 12 i = 1, . . . , n

Pn+1 = −1 i = n+ 1


Preprint

13

3.3. Strength Design


We first compute weight according member type. Thestrength constraints (1.6) are used to compute the weight ofeach member. We write weights as functions of h in order toperform the optimization.

The total weight of the diagonals is:

WD =2γ

σy

n∑

i=0

|Di|√

a2 + h2 (3.2)

Using the value for Di in table 5, we compute weight as:

WD =2γ

σy

n∑

i=0

|Di|√

a2 + h2

WD =2γ

σy

n∑

i=0

|Di|h

√

1 +(a

h

)2

WD =2γ

σy

n∑

i=0

Vih

[

1 +(a

h

)2]

WD =2γh

σy

[

1 +(a

h

)2] n∑

i=0

Vi

The expression for WD is:

WD =2γ

σy

(

h+a2

h

) n∑

i=0

Vi (3.3)

The total weight of the bottom chords is:

WT =2γ

σy

n∑

i=1

Tia (3.4)

This is rewritten as:

WT =2γ

σy

(

a2

h

) n∑

i=1

Ti (3.5)

where Ti is defined using:

Ti =(a

h

)

Ti, Ti = p(a

h

)

[

n+ 1−1

2i

]

i, i = 1, . . . , n


WC =2γ

σy

n∑

i=0

|Ci| a (3.6)

Or equivalently:

WC =2γ

σy

(

a2

h

)

[

C0 +

n∑

i=1

Ci

]

(3.7)


|Ci| =(a

h

)

Ci, Ci = Ti + Vi, i = 1, . . . , n

|C0| =(a

h

)

C0 , C0 = V0

The weight of the posts is computes as:

WP =γ

σy

[

|Pn+1|+ 2

n∑

i=1

|Pi|

]

h (3.8)

The term in the brackets can be simplified further:

WP =γ

σy

[

p+ 2n∑

i=1

(Vi + p)

]

h

WP =γ

σy

[

p+ 2np+ 2

n∑

i=1

Vi

]

h

WP =γ

σy

[

(2n+ 1)p+ 2

n∑

i=1

Vi

]

h

WP =γ

σy

[

(2n+ 1)p+ 2

n∑

i=1

(

n− i+1

2

)

p

]

h

WP =γ

σy

[

(2n+ 1) + 2n∑

i=1

(

n− i+1

2

)

]

ph

WP =γ

σy

[

(2n+ 1) + 2n2 − n(n+ 1) + n]

ph

Thus we have:

WP =

(

γ

σy

)

(n+ 1)2ph (3.9)


W = WD +WT +WC +WP (3.10)

14

Setting the derivative of W with respect to h equal to zero:

∂W

∂h= 0 (3.11)

Solving this equation for h/a gives the optimal value of h.Diving both sides of the equation by 2γ/σy produces:

(

1−a2

h2

) n∑

i=0

Vi−a2

h2

n∑

i=1

Ti−a2

h2

[

C0 +

n∑

i=1

Ci

]

+1

2(n+ 1)

2p = 0

This equation is easily solved for h2/a2:

a2

h2

[

V0 + C0 +

n∑

i=1

(

Vi + Ti + Ci

)

]

=1

2(n+ 1)

2p+

n∑

i=0

Vi

a2

h2

[

V0 + V0 +n∑

i=1

(

Vi + Ti +(

Ti + Vi

))

]

=1

2(n+ 1)2 p+

n∑

i=0

Vi

2a2

h2

[

V0 +

n∑

i=1

(

Vi + Ti

)

]

=1

2(n+ 1)

2p+

n∑

i=0

Vi

a2

h2=

12 (n+ 1)

2p+

∑n

i=0 Vi

2[

V0 +∑n

i=1

(

Vi + Ti

)]

h2

a2=

2[

V0 +∑n

i=1

(

Vi + Ti

)]

12 (n+ 1)

2p+

∑n

i=0 Vi

The final result is:

h

a=

√

√

√

√

2[

V0 +∑n

i=1

(

Vi + Ti

)]

12 (n+ 1)

2p+

∑n

i=0 Vi

(3.12)

We observe that every term inside the root has a factor ofp. The division ensures that (3.12) does not depend on p.Rewriting the equation the formula in terms of h/L:

h

L=

1

2n

√

√

√

√

2[

V0 +∑n

i=1

(

Vi + Ti

)]

12 (n+ 1)

2p+

∑n

i=0 Vi

(3.13)

Preprint

15


The total weights of each member type are:

WD =2γ

σy

[

h+a2

h

]

(n+ 1)Vmax

WT =2γ

σy

(

a2

h

)

nTmax

WC =2γ

σy

(

a2

h

)

(n+ 1)Cmax

Wp =

(

γ

σy

)

(2n+ 1) |Pmax|h

where Vmax =(

n+ 12

)

p, Tmax = 12 (n

2 + 2n)p, Cmax =p2 (n+ 1)

2. |Pmax| =

(

n+ 12

)

p. The total truss weight isagain:

W = WD +WT +WC +WP

Setting the derivative of W with respect to h to zero anddividing by 2γ/σy:

(

1−a2

h2

)

(n+ 1)Vmax −

(

a2

h2

)

nTmax

−

(

a2

h2

)

(n+ 1)Cmax +

(

n+1

2

)

|Pmax| = 0


a2

h2

[

(n+ 1)(

Vmax + Cmax

)

+ nTmax

]

= (n+ 1)Vmax +

(

n+1

2

)

|Pmax|

a2

h2=

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

(n+ 1)(

Vmax + Cmax

)

+ nTmax

h2

a2=

(n+ 1)(

Vmax + Cmax

)

+ nTmax

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

Taking the square roots of both sides:

h

a=

√

√

√

√

(n+ 1)(

Vmax + Cmax

)

+ nTmax

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

Every term inside the root again contains a factor of p andthus it can be cancelled. The maximum values are substitutedand the algebraic simplification is performed on the next page.

Preprint

16

h

a=

√

√

√

√

(n+ 1)(

Vmax + Cmax

)

+ nTmax

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

h

a=

√

(n+ 1)[(

n+ 12

)

+ 12 (n+ 1)2

]

+ 12n(n

2 + 2n)

(n+ 1)(

n+ 12

)

+(

n+ 12

) (

n+ 12

)

h

a=

√

(n+ 1)[(

n+ 12

)

+ 12 (n

2 + 2n+ 1)]

+ 12 (n

3 + 2n2)(

n+ 12

) [

(n+ 1) +(

n+ 12

)]

h

a=

√

(n+ 1)[

12n

2 + 2n+ 1]

+ 12n

3 + n2

(

n+ 12

) (

2n+ 32

)

h

a=

√

12n

3 + 2n2 + n+ 12n

2 + 2n+ 1 + 12n

3 + n2

(

n+ 12

) (

2n+ 32

)

The optimal height is given by:

h

a=

√

n3 + 72n

2 + 3n+ 1(

n+ 12

) (

2n+ 32

) (3.14)

In terms of h/L:

h

L=

1

2(n+ 1)

√

n3 + 72n

2 + 3n+ 1(

n+ 12

) (

2n+ 32

) (3.15)

Preprint

17



0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6

h/L

n

Alt Deck Pratt




1.2

1.4

1.6

1.8

2

2.2

2.4

1 2 3 4 5 6

h/a

n

Alt Deck Pratt




Figures 16 and 17 show the optimal truss height h for agiven number of panels n.


200

220

240

260

280

300

320

340

1 2 3 4 5 6

W [k

ip]

n

Alt Deck Pratt - Individual



200

220

240

260

280

300

320

340

360

380

1 2 3 4 5 6

W [k

ip]

n

Alt Deck Pratt - By type



Figures 18 and 19 imply that 2 panels produce the trusswith the lowest weight. This is true for both individual mem-ber sizing and sizing by member type.

18

4. Through Pratt Truss

4.1. Member Forces

Figure 20: Through Pratt truss


Member Force Index

Vi =(

n− i+ 12

)

p i = 0, . . . , n

D0 = −V0

√

1 +(

ah

)2i = 0

Di = Vi

√

1 +(

ah

)2i = 1, . . . , n

T0 = T1 i = 0

Ti = p(

ah

) (

n+ 1− 12 i)

i i = 1, . . . , n

Ci = −[

Ti + Vi

(

ah

)]

i = 1, . . . , n

P1 = p i = 1

Pi = −Vi i = 2, . . . , n

Pn+1 = 0 i = n+ 1



Vmax =(

n+ 12

)

p i = 0

Dmax = −Vmax

√

1 +(

ah

)2i = 0

Tmax = p(

ah

) (

12n+ 1

)

n i = n

Cmax = − p2

(

ah

)

(n+ 1)2

i = n

Pmax =

{

P1 = p if 1 ≤ n ≤ 2

P2 = −(

n− 32

)

p if n > 2

i = 1i = 2


When computing member forces, we need to consider thefirst panel i = 0 as a special case. Considering force equilib-rium in the vertical direction gives an expression for shear inpanel i:

∑

Fy = 0

(n+ 1)p−p

2− ip− Vi = 0

np− ip+p

2− Vi = 0

(

n− i+1

2

)

p− Vi = 0

The formula for shear is identical to that of the deck truss.

Vi =

(

n− i+1

2

)

p (4.1)

This formula is also valid for i = 0.The maximum shear occurs at the end panel i = 0:

Vmax = V0 =

(

n+1

2

)

p (4.2)

The force in the diagonals are:

Di =Vi

sin θ= Vi

√

1 +(a

h

)2

(4.3)

Di is a tensile force when i > 0, but is compressive wheni = 0.

The maximum force in the diagonals occurs at the end pan-els:

Dmax = D0 = −Vmax

√

1 +(a

h

)2

(4.4)

The bottom chords have tensile forces which are computedby summing the moments about the joint connecting the post,diagonal, and top chords. For i = 1, . . . , n we have:

∑

M = 0

Tih+ (i − 1)pia

2−

(

n+1

2

)

ipa = 0

Tih+

[

1

2(i− 1)−

(

n+1

2

)]

ipa = 0

Tih+

[

i

2−

1

2− n−

1

2

]

ipa = 0

Tih+

[

i

2− 1− n

]

ipa = 0

19

Ti is thus given by:

Ti = p(a

h

)

(

n+ 1−i

2

)

i (4.5)

We must still compute the force in the end panel’s bottomchord. It is easily computed by looking at the forces at thefirst joint to connect a post with two bottom chords:

T0 = T1

We compute the derivative of Ti to locate the maximumforce in the bottom chords:

∂Ti

∂i= p

(a

h

)

(n+ 1− i) > 0

The derivative positive for i = 1, . . . , n and thus the maximumoccurs at panel n.

Tmax = Tn = p(a

h

)(n

2+ 1

)

n (4.6)

The derivation for the force in the top chords is identicalto that of the deck truss and the results are:

Ci = −[

Ti + Vi

(a

h

)]

(4.7)

Taking the derivative of the absolute value of Ci to locatethe maximum:

∂ |Ci|

∂i=

∂Ti

∂i+

∂Vi

∂i

(a

h

)

∂ |Ci|

∂i= p

(a

h

)

(n+ 1− i)− p(a

h

)

∂ |Ci|

∂i= p

(a

h

)

(n− i)

The critical point is i = n and the second derivative confirmsthat this is the location of the maximum:

∂2 |Ci|

∂i2= −p

(a

h

)

< 0

The maximum value of force in the top chord is:

Cmax = Cn =p

2

(a

h

)

(n+ 1)2

(4.8)

Equilibrium of vertical forces at each joint provides theforces in the posts. Consider vertical equilibrium of forcesat the joint connecting the post, diagonal, and top chord.

∑

Fy = 0

−Pi −Di sin θ = 0

Pi = −Vi, i = 2, . . . , n (4.9)

The tensile force in the first post is easily determined by in-spection:

P1 = p

The center post i = n+ 1 is a zero force member:

Pn+1 = 0

The maximum post force depends on the number of panels:

Pmax =

{

P1 = p if 1 ≤ n ≤ 2

P2 = −(

n− 32

)

p if n > 2(4.10)

20




Member Force Index

Vi =12 i = 0, . . . , n

D0 = −D1 i = 0

Di =1

2 sin θ, i = 2, . . . , n i = 1, . . . , n

T0 = T1 i = 0

Ti =12

(

ah

)

i i = 1, . . . , n

Ci = − 12 (i+ 1)

(

ah

)

i = 1, . . . , n

P1 = 0 i = 1

Pi = − 12 i = 2, . . . , n

Pn+1 = 0 i = n+ 1


The unit dummy load theorem is again used to computethe mid-span deflection. The derivation proceeds in the exactsame manner as for the deck truss. The shears are identicalto that of the deck truss:

Vi =1

2(4.11)

The forces in the diagonals are:

Di =1

2 sin θ(4.12)

Except for the end panel which has a compressive load in thediagonal:

D0 = −1

2 sin θ

To compute the tension in the bottom chords, we sum themoments about the joint connecting the post, diagonal, andtop chords:

∑

M = 0

Tih−1

2ia = 0

Ti =i

2

(a

h

)

(4.13)

The compressive force in the top chord is computed by con-sidering equilibrium in the horizontal direction:

∑

Fx = 0

Ti + Di cos θ + Ci = 0


Ci = −1

2(i+ 1)

(a

h

)

(4.14)

The compressive force in the posts are easily computed:

Pi = −1

2, i = 2, . . . , n (4.15)

The end post and center post are zero force members:

P1 = Pn+1 = 0

21

4.3. Strength Design Optimization


The total weight of the diagonals is again:

WD =2γ

σy

n∑

i=0

|Di|√

a2 + h2 (4.16)

The algebra is identical to the previous trusses and the ex-pression for WD is:

WD =2γ

σy

(

h+a2

h

) n∑

i=0

Vi (4.17)

The total weight of the bottom chords is:

WT =2γ

σy

n∑

i=0

Tia (4.18)

This is rewritten as:

WT =2γ

σy

(

a2

h

) n∑

i=0

Ti (4.19)

where Ti is defined using equation (4.5):

Ti =(a

h

)

Ti, Ti = p

(

n+ 1−i

2

)

i, i = 1, . . . , n

T0 = T1


WC =2γ

σy

n∑

i=1

|Ci| a (4.20)

Or equivalently:

WC =2γ

σy

(

a2

h

) n∑

i=1

Ci (4.21)


|Ci| =(a

h

)

Ci, Ci = Ti + Vi

The weight of the posts is complicated by the fact that thereare special formulas for P1 and Pn+1. Recall that Pn+1 = 0and this post must be sized to non-zero cross-sectional area.We use Pn to size that post.

WP =γ

σy

[

|Pn|+ 2 |P1|+ 2

n∑

i=2

|Pi|

]

h (4.22)

The term in the brackets can be simplified:

WP =γ

σy

[

|Pn|+ 2p+ 2

n∑

i=2

|Pi|

]

h

WP =γ

σy

[

1

2p+ 2p+ 2

n∑

i=2

Vi

]

h

WP =γ

σy

[

5

2p+ 2

n∑

i=2

Vi

]

h

WP =γ

σy

[

5

2p+ 2

n∑

i=2

(

n− i+1

2

)

p

]

h

WP =γ

σy

[

5

2+ 2

n∑

i=2

(

n− i+1

2

)

]

ph

WP =γ

σy

[

5

2+ 2n(n− 1)− n(n+ 1) + 2

+(n− 1)] ph

Thus we have:

WP =

(

γ

σy

)(

n2 − 2n+7

2

)

ph (4.23)


W = WD +WT +WC +WP (4.24)

22

Setting the derivative of W with respect to h equal to zero:

∂W

∂h= 0 (4.25)

Solving this equation for h gives the optimal value of h. Div-ing both sides of the equation by 2γ/σy produces:

(

1−a2

h2

) n∑

i=0

Vi−a2

h2

n∑

i=0

Ti−a2

h2

n∑

i=1

Ci+1

2

(

n2 − 2n+7

2

)

p = 0

This equation is solved for h2/a2:

a2

h2

[

V0 + T0 +

n∑

i=1

(

Vi + Ti + Ci

)

]

=1

2

(

n2 − 2n+7

2

)

p+

n∑

i=0

Vi

a2

h2

[

V0 + T0 +

n∑

i=1

(

Vi + Ti +(

Ti + Vi

))

]

=1

2

(

n2 − 2n+7

2

)

p+

n∑

i=0

Vi

a2

h2

[

V0 + T0 + 2n∑

i=1

(

Vi + Ti

)

]

=1

2

(

n2 − 2n+7

2

)

p+n∑

i=0

Vi

a2

h2=

12

(

n2 − 2n+ 72

)

p+∑n

i=0 Vi

V0 + T0 + 2∑n

i=1

(

Vi + Ti

)

h2

a2=

V0 + T0 + 2∑n

i=1

(

Vi + Ti

)

12

(

n2 − 2n+ 72

)

p+∑n

i=0 Vi

This gives the equation:

h

a=

√

√

√

√

V0 + T0 + 2∑n

i=1

(

Vi + Ti

)

12

(

n2 − 2n+ 72

)

p+∑n

i=0 Vi

(4.26)

Rewriting the equation the formula in terms of h/L:

h

L=

1

2(n+ 1)

√

√

√

√

V0 + T0 + 2∑n

i=1

(

Vi + Ti

)

12

(

n2 − 2n+ 72

)

p+∑n

i=0 Vi

(4.27)

Preprint

23


The total weights of each member type are:

WD =2γ

σy

[

h+a2

h

]

(n+ 1)Vmax

WT =2γ

σy

(

a2

h

)

(n+ 1)Tmax

WC =2γ

σy

(

a2

h

)

nCmax

Wp =

(

γ

σy

)

(2n+ 1) |Pmax|h

where Vmax =(

n+ 12

)

p, Tmax =(

n2 + 1

)

np, Cmax =p2 (n+ 1)

2. |Pmax| = p if 1 ≤ n ≤ 2 and |Pmax| =

(

n− 32

)

p ifn > 2. The total truss weight is again:

W = WD +WT +WC +WP

Setting the derivative of W , with respect to h, to zero:

(

1−a2

h2

)

(n+ 1)Vmax −

(

a2

h2

)

(n+ 1)Tmax

−

(

a2

h2

)

nCmax +

(

n+1

2

)

|Pmax| = 0


a2

h2

[

(n+ 1)(

Vmax + Tmax

)

+ nCmax

]

= (n+ 1)Vmax +

(

n+1

2

)

|Pmax|

a2

h2=

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

(n+ 1)(

Vmax + Tmax

)

+ nCmax

h2

a2=

(n+ 1)(

Vmax + Tmax

)

+ nCmax

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

Taking the square roots of both sides:

h

a=

√

√

√

√

(n+ 1)(

Vmax + Tmax

)

+ nCmax

(n+ 1)Vmax +(

n+ 12

)

|Pmax|

The maximum values are substituted and further simplifica-tion is performed on the next page.

Preprint

24

h

a=

√

√

√

√

(n+ 1)(

Vmax + Tmax

)

+ nCmax

(n+ 1)Vmax + 1p

(

n+ 12

)

|Pmax|

h

a=

√

(n+ 1)[(

n+ 12

)

+(

12n+ 1

)

n]

+ 12n(n+ 1)2

(n+ 1)(

n+ 12

)

+ 1p

(

n+ 12

)

|Pmax|

h

a=

√

√

√

√

(n+ 1)[(

n+ 12

)

+(

12n+ 1

)

n+ 12n(n+ 1)

]

(

n+ 12

)

[

n+ 1 + 1p|Pmax|

]

h

a=

√

√

√

√

(n+ 1)(

n2 + 52n+ 1

2

)

(

n+ 12

)

[

n+ 1 + 1p|Pmax|

]

For 1 ≤ n ≤ 2:

h

a=

√

(n+ 1)(

n2 + 52n+ 1

2

)

(

n+ 12

)

(n+ 2)(4.28)

For n > 2:

h

a=

√

(n+ 1)(

n2 + 52n+ 1

2

)

(

n+ 12

) (

2n− 12

) (4.29)

These equations are equivalently given in terms of h/L as:For 1 ≤ n ≤ 2:

h

L=

1

2(n+ 1)

√

(n+ 1)(

n2 + 52n+ 1

2

)

(

n+ 12

)

(n+ 2)(4.30)

For n > 2:

h

L=

1

2(n+ 1)

√

(n+ 1)(

n2 + 52n+ 1

2

)

(

n+ 12

) (

2n− 12

) (4.31)

Preprint

25



0.1

0.15

0.2

0.25

0.3

0.35

0.4

1 2 3 4 5 6

h/L

n

Thru Pratt



Linear Fit by Waling


1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

1 2 3 4 5 6

h/a

n

Thru Pratt




Figures 24 and 25 show the optimal truss height h for agiven number of panels n. Figure 24 also shows a linear fit(4.32) by Waling [7] to actual truss data.

h

L= 0.381− 0.0185N, N = 2(n+ 1) (4.32)


160

180

200

220

240

260

280

300

1 2 3 4 5 6

W [k

ip]

n

Thru Pratt - Individual



160

180

200

220

240

260

280

300

320

340

1 2 3 4 5 6

W [k

ip]

n

Thru Pratt - By type



Figures 26 and 27 imply that 2 panels produce the trusswith the lowest weight. This is true for both individual mem-ber sizing and sizing by member type.

26

5. Conclusions

Explicit formulas were developed for strength design of thePratt truss. Unfortunately, strength design usually producestrusses which do not satisfy deflection requirements. Mem-bers can be uniformly resized to satisfy deflection constraints.The deflection design procedure is performed by numericallyevaluating the resulting algebraic expressions.

It is seen that increasing truss height generally reducesweight. However, there are limits to such an approach. Theprimary concern is that deep trusses have long members incompression. The buckling of such members requires a heav-ier member or additional bracing. For our design example, welimited members to lengths within reasonable design practice.

The trusses were design by assuming a uniform load. Mov-ing loads must be considered when designing bridges. Suchloads may alter the design in the following ways:

• Forces in certain members may be larger for partial load-ing than for uniform loading. Influence lines must beused to determine these conditions.

• Stress reversals occur for moving loads. For example, di-agonals will be in compression and thus must be designedfor buckling.

• Horizontal and other loads will be induced when a trainenters the bridge.

Self-weight of the truss was ignored in this article, but it needsto be considered in the analysis of the final design. Further-more, unfactored loads and unreduced strengths were used forthe design example. Thus the design procedures described inthis paper should be viewed as a means to approximatinglower bounds on the truss weight. The total weight of thebridge includes two trusses, lateral bracing, deck weight, gus-set plates, and other connection details.

References

[1] R.C. Hibbeler. Structural Analysis. Prentice Hall, UpperSaddle River, New Jersey, eighth edition, 2012.

[2] Aslam Kassimali. Structural Analysis. PWS-Kent,Boston, 1993.

[3] Milo Smith Ketchum. The Design of Highway Bridges

and the Calculation of Stresses in Bridge Trusses. TheEngineering News Publishing Company, New York, 1909.

[4] Paul Mallery. Bridge and Trestle Handbook. Carstens,Newton, New Jersey, fourth edition, 1992.

[5] S.P. Timoshenko and D.H. Young. Theory of Structures.McGraw-Hill, New York, second edition, 1965.

[6] M.S. Troitsky. Planning and Design of Bridges. Wiley,New York, 1994.

[7] J.L. Waling. Least-weight proportions of bridge trusses.Technical Report 417, University of Illinois, 1953. BulletinSeries.

27

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