Eads Bridge

See the original text for p331-335.

The initial pages of the chapter provide context and introduce constants and variables. This discussion is omitted here to save space. The constants and variables are defined in the comments column opposite.

Pages 331 - 335 describe the alternatives explored on the way to the final design and introduce the parameters and variables used in the calculations.

The analysis considers the action of a single rib of the arch.

List of Constants and Variables

ω = the area of a section of a member,

δ = the distance between the centers of gravity of the members,

θ = the moment of inertia of the normal section of the arch, we have

θ = 2ω(δ/2)² = ωδ²/2 (2)

This commentary will use the term “angular deflection“ to denote this change of angle.

y = R(cosφ – cosα)

na = R(sinα – sinβ)

pa = R(sinα – sinγ)

Locations along the arch's neutral axis are defined as functions of angle relative to the arch's vertical centerline. (Angles in radians).

P₁ = qna — P,

N₁a = — Na + 2Pa — qna[2a — [na/2)],

= — Na + 2Pa — 2qna²[1 — (n/4)], (6)

by which the resistances at the right pier can be determined if those at the left pier are known. In deducing the last equation, or the equation of the moments, we have taken the moments about A₁; the same equation can be found by taking the moments about A, observing that from the nature of the couples denoted by Na and N₁a, they may be transferred to any point in the system under consideration, and hence, the whole arch being here considered, they may both be regarded as acting either at A₁ or at A.

∑ Fx = 0    Q₁ + Q = 0

∑ Fy = 0    qna + P + P₁ = 0

∑ M = 0   N₁a + Na - 2Pa + 2qna²[1 — (n/4)] = 0

ω = the sectional area of the center-piece, from G to G₁,

ω₁ = the sectional area of the end-pieces, from A to G, and from A₁ to G₁,

θ= ωδ2/2 = the moment of inertia of the section of the center-piece,

θ₁ = ω₁δ2/2 the moment of inertia of the section of an end-piece,

we shall have, by equation (3), for the center-piece from x = pa to x = (2 — p)a; that is, from φ = γ to φ = —γ,

dφ₁ — dφ = (RM/εθ)dφ (VII)

and for an end-piece from x = 0 to x = pa, and from x = (2 — p)a to x = 2a; that is, from φ= α to φ = γ, and from φ = — γ to φ = — α,

dφ₁ — dφ = (RM/εθ₁)dφ. (7)

See figure 5

Considering now the external forces acting on the portion AF of the arch, we have, the axis of moments being at F and the position of F being variable, at the point (x, y): —

M= NRsinα — PR(sinα — sinφ) + QR(cosφ — cosα) + [(qR²)/2](sinα — sinφ)2. (8a)

b. For unloaded part, from x = na to x = 2a,
M = Na — Px + Qy + qna [x — (na)/2];
or by (4),

M = NRsinα — PR(sinα — sinφ) + QR(cosφ — cosα) — qR² sin φ (sinα — sinβ) + [(qR²)/2] (sin²α — sin²β). (8b)

In the second equation; (8a) or (8b); x, y, and a are translated into the polar coordinate system used throughout the calculations. Angles are given in radians.

figure 10 – Element of Neutral Axis,

figure 11 – Horizontal Deflection

figure 12 – Vertical Deflection

dx = dscosφ,

dx₁ = ds₁cosφ₁ ,

dy = dssinφ,

dy₁ = ds₁sinφ₁ ,

and ds = ds₁ = — Rd φ, there follows very nearly

dx₁ — dx = — Rdφ(cos φ₁ — cosφ) = Rd φ( φ₁ — φ)sinφ,

dy₁ — dy = — Rd φ(sin φ₁ — sinφ) = — Rd φ( φ₁ — φ)cosφ.

Bending does not appreciably change the length of the neutral axis. The length of the element after flexure is equal to the length before: ds₁ = ds

For derivation of the equations for horizontal and vertical deflection, see figures 9, 10, 11.

The deflection calculations in Chapter XXVI require values for the individual constants. Equations for them are derived in the next six steps. Solutions to these equations are tabulated on page 353, Chapter XXVI.

A+B = — 2Hα + 2[P — qa + ((qRsinβ)/2)]cosα + [(qR)/4]sinαcos α,

A - B = — 2Gα — qRsinβcosα — 2Qsinα + [(qR)/4]sinαcosα,

from which A and B are immediately deduced.

C + D = — 2Hsinα + 2[P — qa + ((qRsinβ)/2)][1 — ((sin²α)/2)] + qR[((sinα)/4) — ((sin³α)/6)],

C — D = — 2Gsinα — qRsinβ[1 — ((sin²α)/2)] — Q(α + sinαcosα) + qR[((sin α)/4) — (sin³α)/6)]

from which C and D are found.

From this, by substitution and subtraction, equations are produced giving C and D in terms of constants.

E + F = 2Gcosα + [(qR sinβ)/2](α — sinαcosα) — Qsin²α +qR[((cosα)/4) — ((cos³α)/6)],

E — F = 2Hcosα — [P — qα + ((qR sin β)/2)](α — sinαcosα) +qR[((cosα)/4) — ((cos³α)/6)],

whence E and F are found.

From this, by several substitutions, equations are produced for E and F.

A₁ + B₁ = (θ/θ₁) (A + B) + [(θ₁ — θ)/θ₁]{— 2Hγ + 2[P — qa + ((qRsinβ)/2)]cosγ + [(qR)/4]sinγcosγ},

A₁ — B₁ = — [(qR))/4]μ = — [(qR)/4](β + 2βsin²β + 3sinβcosβ),

by which A₁B₁ and are immediately found.

Substituting this value for A₁ in the equation for A₁-B₁ produces an equation for B₁.

C₁ + D₁ = (θ/θ₁)(C + D) + [(θ₁ — θ)/θ₁]{ — 2H sin + 2[P — qα + ((qRsinβ)/2)][1 — ((sin²γ)/2)] +qR[((sin γ)/4) — ((sin³γ)/6)]},  C₁ — D₁= — qRν = — qR(sinβ + (1/6)sin³β),

from which C₁ and D₁ are found.

Substituting this value for C₁ into the equation for C₁-D₁ yields an equation for D₁

E₁ + F₁ = (A₁ — B₁)sinγ — (θ/θ₁)(A — B)sinγ + (θ/θ₁)(E + F) + [(θ₁ — θ)/θ]{2G(cosγ + γsinγ) +[(qRsinβ)/2](γ + sinγcosγ) + Q sin²γ + [(qR)/12]cos³γ},

E₁ — F₁ = qRε = qR[((cosβ)/2) + ((βsinβ)/2) — ((cos³β)/6)],

from which E₁ and F₁ are found.

Substituting this value for E₁ into the equation for E₁-F₁ yields an equation for F₁

A₁ + B₁= — 2Hσ + 2[P — qα + ((qRsinβ)/2)]σ₃ + ((qR)/4)σ₁,

C₁ + D₁= — 2Hσ₂ + 2[P — qα + ((qRsinβ)/2)] (1 — σ₄) + qR((σ₂/4) — σ₅),

E₁ + F₁ = — 2Qχ₁ + [(qR)/4] ∙ (σ₃/σ)μ + qRsinβ[((σ — σ₁)/2) — (σ₂3/σ)] + [(qR)/4] [((σ₁σ₃)/σ) + σ₃ — 2σ₆],

and a good verification of the correctness of a large part of the whole work is obtained by computing these quantities under both forms.

The derivation of the previous equations, describing partial loading, relied in part on the contrast between loaded and unloaded portions of the arch. These equations are not applicable where there is no unloaded part. The following equations address this condition.

E = [((qR)/4) ∙ (σ₁/σ) — Q ∙ (σ₂/σ)]cosα — Q ∙ ((sin²α)/2)+ qR[((cosα)/4) — ((cos³α)/6)];

By applying deflection equation (24) to the fixed end of the arch, where deflection is known to be zero, an equation can be written for E in terms of Q and constants.

E₁ = (θ/θ₁)(E — Asinγ) + (θ₁ — θ)/θ₁{[((qR)/4) ∙ (σ₁/σ) — Q ∙ (σ₂/σ)](cosγ + γsinγ) + Q ∙ [(sin²γ)/2] + [(qR)/12]cos³γ}.

sinφ = (1 — m)sina,

y = R(cosφ — cosα),

for the successive values of m, as (1/8), (2/8), (3/8), etc., up to (16/8) or 2; we can then readily compute the moment M by the following equations: —

For loaded part,
  M = Na — Pam + [(qa2)/2] m2 + Qy

For unloaded part,
  M = Na — Pam + qna² [m — (n/2)] + Qy.

The load will be taken for the cases in which it covers (1/8), (2/8) , (3/8) , etc., of the span, up to (8/8) or the whole span; the corresponding values of n, being (1/4), (2/4), (3/4), ect., up to (8/4) or 2.

For each of the load conditions, the moment, M, will be calculated each of 17 stations spaced equally across the length of the arch.

When the loaded area extends part-way across the arch, different equations are used to compute the moment at stations located within and outside of the loaded area.

For loaded part,
  M = Na — Pam + [(qa2)/2] m2 + Qy

For unloaded part,
  M = Na — Pam + qna² [m — (n/2)] + Qy.

T = + or — (M/δ)

δ being the distance between the centers of the members. The {upper/lower} sign will be used for the {upper/lower} member; and positive values of T will express compressive strain, negative values tensile strain.

If V is the vertical force acting at any point (x, y), and φ the angle which the tangent to the curve at this point makes with the horizontal line, we have.

S' =Qcosφ + Vsinφ,

S'' =Qsinφ — Vcosφ,

V is the sum of the vertical reactions and loads acting between the point under consideration and the end of the arch (vertical shear).

The vector sum of Q and V is resolved as S’, a compressive force acting along the arch tangent to the neutral axis, and S”, a shear force acting at a right angle to the neutral axis, toward or away from the center of the arch. See figure 13.

figure 12 - S' and S"

For loaded part, V= P — qx = P — qma;

For unloaded part, V= P — qna.

The value of S', from the nature of the case, will always be positive, and be compressive strain.

The value of S" will be positive in some parts of the arch and negative in others. The positive value will be a strain towards the center; the negative value a strain from the center. As in our formulas we reckon from the pier to which P belongs, S" is the shearing strain of that segment of the arch which extends from that pier to the point at which the strain is considered to exist.

Since the movable loads may extend from either end of the bridge, all the strains computed for one-half the arch must also be applied to the other half in deducing the necessary sectional areas of the members and braces, as will be illustrated hereafter.

If the uniform load extends only part-way across the span and F falls outside the loaded area, the load will equal q times the extent of the loaded area, na.

If F falls within the loaded area, only the load attributed to ma, the portion of the loaded area between F and the end of the span, is included in V.

See figure 9 for illustration of qna, qma, and qx.

Because the analysis is made relative to the loaded end of the arch, it must be repeated for the case where the load extends from the other end of the span.

Since the weight of the bridge is not changed by the temperature, the resistances at the pier may be considered as consisting of a horizontal force Q and a moment Na. The moment of flexure for any point (x, y) of the deflected arch is, then,

M = Na + Qy = NRsinα + QR(cosφ — cosα);

The arch is free to expand and contract in the vertical direction but is restrained against horizontal movement by the supports. When the temperature changes, the supports exert a horizontal force Q which neutralizes the horizontal component of the thermal expansion.

The horizontal force Q generates a moment at every point on the arch equal to Qy. Moment Qy produces a change of curvature which is resisted by the fixed ends of the arch, producing a secondary moment Na. The total moment at any point x,y along the arch is:

M = Na + Qy,

or stated in the polar coordinate system used throughout the analysis:

M = NRsinα + QR(cosφ — cosα)

Na = — [(σ₂/σ) — cosα]QR. (34)

We can now compute the moment of flexure at the different points of the arch by the equation

M = Na + Qy

for any supposed change of temperature. If we suppose the extreme range of the temperature to which the arch will be exposed to be 80° above and below a mean temperature of 60° Fahrenheit, we shall have, by the known linear expansion of steel for a change of 80° Fahrenheit,

λ = 0.000527.

For a decrease of temperature, the same formulae will be applicable by taking λ with the negative sign.

Since Rsinα = a (length of the half-span), multiplying (29) by R yields (34), an equation for Na in terms of Q.

With the value of Q and Na provided by equations (33) and (34), the value of M can be computed for any change of temperature.

In the formulas, positive λ represents an increase, negative λ a decrease of temperature.

E₁= — (θ/θ₁){(σ₂/σ)(αsinγ + cosα) — sinαsinγ + [(sin²α)/2]} — [(θ₁ — θ)/θ]{(σ₂/σ)(cosγ + γsin γ) — [(sin²γ)/2]};

or with the notation of pages 338 and 340,

E₁= — [((σ₂σ₃)/σ) + σ₄] = — x₁.

y₀ = [R³/(εθ)] ∙ Q [x₁ — (σ₂/σ)],

which, with the values of the constants hereafter given, will be found to = 8.7 inches.

Y₀ = [R³/(εθ)] ∙ Q [x₁ — (σ₂/σ)]

At the maximum design temperature, the crown of the arch will rise by 8.7 inches above its elevation at the "neutral" temperature of 60 degrees.

T= ± [(Na + Qy)/δ].

The strain in the direction of the arch will be

S ' = Qcosφ.

The shearing strain will be

S " = Qsinφ.

The moment produced by temperature change is predicted by the equation M = Na + Qy.

Substituting this value for M in the equation for axial force in the chords given on page 344, yields an equation for axial force due to temperature change.

The equations for S’ and S” introduced on page 345 include a term for the vertical component:

S' = Q cos φ + V sin φ (force along arch)

S" = Q sin φ — V cos φ, (normal shear)

Because the arch is free to expand vertically, temperature change generates only a horizontal force Q. The “V” term of the equations can be omitted to produce the equations for S’ and S” due to temperature change.