Bridge Theory and Design - Appunti

BEAM-AND-SLAB DECKS MODELLED AS GRILLAGES

-MASSONNET METHOD-

* Problem: ORTHOTROPIC GRILLAGE two opposite sides SIMPLY SUPPORTED and two ends FREE loaded by SINUSOIDAL LINE LOAD

* the NODAL EQUILIBRIUM is obtained considering a GRILLAGE DECK composed of a SLAB and of LONGITUDINAL & TRANSVERSE BEAMS

* the BENDING MOMENTS in beams & crossbeams are:

we are assuming an 'equivalent plate'

* HP: BEAMS & CROSSBEAMS are CLOSELY SPACED

L→ we can model the discrete system as a CONTINUOUS ONE

L→ this can be done by SPREADING STATIC QUANTITIES & STIFFNESS on the relative distances

=> in this way we can substitute M & T with DISTRIBUTED MOMENTS & SHEARS per UNIT WEIGHT (m ≥ t)

_mlx = -_EEp_b ∂²w/∂x² + ∫p ∂²w/∂x² ;

_mxy = -_GUp/bi ∂²w/∂x ∂y - ∫p ∂²w/∂x ∂y

_my = -_EJe ∂²w/∂y² + ∫e ∂²w/∂y² ;

_myx = -_GUe/li ∂²w/∂x ∂y - ∫e ∂²w/∂x ∂y

-∫p ∫e x (Ie + Yp) stiffness x unit height

mxy can be different from myx - different cross-section of the beams in long. and transv. direction

* SHEAR FORCES can be obtained by means of EQUILIBRIUM :

txdy

tydx

mxy dx

myxdy

tx dy dx

myx dx

(m_xy + ∂m_xy/∂x dx) dy

(myx + ∂m_xy/∂y dy) dx

(my + ∂my/∂y dy) dx

(mx + ∂mx/∂x dx) dy

for beams we use the beam distance x for the transverse beams we use the transverse beam distance

t_x = ∂m_x/∂x + ∂m_yx/∂y ;

t_y = ∂m_my/∂y + ∂m_xy/∂x (a)

→ from the VERTICAL EQUILIBRIUM we can write:

∂t_x/∂x + ∂t_y/∂y = -ρ(x;y) (b)

* by substituting (a) and (b) in (d) we obtain the EQUILIBRIUM EQUATION for ORTHOTROPIC GRILLAGE

∫p ∂⁴w/∂x⁴ + (∫p + ∫e) ∂⁴w/∂x²y² + ∫e ∂⁴w/∂y⁴ = ρ(x;y) (c)

m_xy = -γ_p ∂²w / ∂x∂y

m_yz = -γ_e ∂²w / ∂y∂z

→ recalling that:

2α = γ_p + γ_e / √_sp√_se → 2α √_sp√_se / γ_p + γ_e = 1

→ the expression of the second mixed derivative is:

∂²w / ∂x∂y = 2d √_sp√_se / γ_p + γ_e ∂²w / ∂x∂y =

= 2d √_sp√_se / γ_p + γ_e π / ℓ ∑_i p_i ^{∂v / ∂q} cos πx / ℓ

→ defining:

τ_α (i; j; ℓ) · b = α √_e √_sp π / ℓ ∂w / ∂y

then:

∂²w / ∂x∂y = 2b / γ_p + γ_e ∑ p_i τ_α (i; j; ℓ) cos πx / ℓ

{ m_xy = -2b γ_p / γ_p + γ_e ∑ p_i τ_α (i; j; ℓ) cos πx / ℓ

m_yz = -2b γ_e / γ_p + γ_e ∑ p_i τ_α (i; j; ℓ) cos πx / ℓ}

(G) Longitudinal Shear Force t_x(x; y)

t_x(x; y) = ∂u_x / ∂x + ∂u_xy / ∂y = -β_p ∂³w / ∂x³ + ( -γ_e ∂ / ∂x ( ∂²w / ∂y² ) ) =

= ∂ / ∂x [m_x(x; y) - γ_e / s_e ∂²w(x; y) / ∂y² ] = ∂ / ∂x [m_x + s_e m_y]

→ m_x = mm_x ∑ p_i K_α(i; j; ℓ) / ∑ p_i; m_y = b ∑ p_i M_α(j; ℓ) sen πx / ℓ

t_x(x; y) = ∂mm_x / ∂x ∑ p_i K_α(i; j; ℓ) / ∑ p_i + γ_e / s_e b · π / ℓ ∑ p_i M_α(j; ℓ) cos πx / ℓ

m_x = -β_p ∂³w / ∂x³

m_y = -β_e ∂³w / ∂y³

Rotational Equilibrium:

(∫z₁dz - z₂) b - (Tₓ₁ + Tₓ₁Δ Tₐ) b + M₂ + dM₂ + M₁₂ + M₁₁' + dM₁ - Mₐ₁ + qdxₑ = 0

dT₂_dx - ^dT₁/_dx b + ^dM₂/_dx + ^dM₁/_dx = -q e

→ Recalling that:

_{^{Tₐ = -dMₐ/dx = -EI dₓ³Uₐ/dx³ = -EI (d³Jₐ/dx³ - b d³β/dx³)}}

_{^{T₂ = -dM₂/dx = -EI dₓ³U₂/dx³ = -EI (d³Jₒ/dx³ + b d³β/dx³)}}

Mₐ = GJ ^dβ/_dx

M₂ = GJ ^dβ/_dx

Then:

-EI ^dJₒ/_dx⁴ b - b²EI ^d³β/_dx⁴ + EI/_dₓ⁴ b - b²EI ^d³β/_dx⁴ + 2GJ ^d²β/_dx² = -q · e

→ the applied moment is taken by 2 different contribution

-2EIb ^d³β/_dx³ + 2GJ ^d²β/_dx² = -q · e = -m

CASE OF an ARBITRARY NUMBER OF BEAMS:

Calling:

• EKᵧₓ = 2EIb²
• GKₜ = 2GJ
• EKᵧₓ for m beams
• GKₜ = G ∑ J_i
• Kᵧₓ : moment of inertia of I-Iᵧ wrt x axis

- for m boxes we have (m-1) hinges ➔ { m-1 unknownsm-1 equations

- in MATRIX FORM we have:

γ ⋅ t = q(m-1 x m) (m-1 x 1) (m-1 x 1)

γ: TRIDIAGONAL MATRIX since each equationinvolves 3 different unknowns

PRESTRESSING FORCE in a PARABOLIC PROFILE

ΔΘ(s) ≃ ΔΘ(x₁) = ∫₀ˣ¹ k* dξ = k* x₁

P(x₁) = P₁e^-μkx₁ ≃ Pₐ [1 - μk* x₁]

ΔP = P₃ - P(x₁) ≃ P₃μk* x₁ — for cables tensioned from beam end Ⓐ

P - P(x;t) = P₀ - ΔP

INDETERMINATE PRESTRESSING STRUCTURES

• in STATICALLY DETERMINED structures, the load equivalent to prestressing is SELF-EQUILIBRATED
• prestressing in statically determined structures produces NO REACTIONS
• prestressing in STATICALLY INDETERMINATE structures may produce REACTIONS and ADDITIONAL INTERNAL FORCES that modify the INTERNAL STRESS DISTRIBUTION

→ the additional reaction tends to reduce the PRIMARY BENDING MOMENT

* the value of the HYPERSTATIC UNKNOWN X is given by PVW.

L we consider an AUXILIARY STRUCTURE subjected to the unitary moment X=1 and we impose the WhMt = Next

-1. ∫₀^ℓ - (-c . X) ^qθ²/_ZEI [(^x/_ℓ) - (^x'/_ℓ)] dx +

+∫₀^a-c (-c . X)/_EI dx + ∫₀^e (-s . X)/_GJt dx

O = -_cqθ²/_ZEI [^x²/_2ℓ + ^x³/_3ℓ²] . + ^c²/_EI ∑_ℓ + ^s²/_GJt X [p = -cqθ²/ZEI [p ≤/ 2ℓ² ^p³/_3ℓ²] + ∑