L2 - Suspension Bridge - Part 1 Tower Anchors Jan 11 - Final

7/24/2019 L2 - Suspension Bridge - Part 1 Tower Anchors Jan 11 - Final

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© Maria E. Moreyra Garlock 1

Suspension Bridges Part I:

Tower & Anchors

Figure 1: The George Washington Bridge, New York City (Photo credit: Maria Garlock)

FORM: The Suspension Bridge

A suspension bridge is one in which the

bridge deck is supported by cables thatare suspended from typically two large

towers. Attaching the deck to these main

spanning cables are vertical suspenders.The main spanning cables arecontinuous over the tower supports and

are firmly anchored at the ends of the bridge, for example, by huge blocks of

concrete. The towers typically rest onconcrete caissons in the water. The

George Washington Bridge (Figure 1),designed by Othmar Ammann is an

example of a suspension bridge. It wascompleted in 1931 and spans 3,500 ft

between its two towers over the HudsonRiver i.

FORCES: Deck, Cables, and Towers

Gravity loads, the dead load of the

bridge deck and the live load from thetraffic carried by the bridge, are the

dominant loads in this structure. Wind

loads, though important for the analysis,history, and design of suspension bridges, will be neglected in this study

for simplicity. The loads from the deckare transmitted by the vertical

suspenders through the main spanningcables to the tower foundations and the

anchor supports. This analysis will focuson the reactions at the end of the cables

on the tops of the towers and theanchors. Because of the great number of

suspenders, we can approximate theshort-spaced gravity loads as a

uniformly distributed load, q (lb/ft ork/ft) on the main cable.




TOWER REACTIONS: Main Span

The total vertical reactions acting at thetower are due to loads that act on the

main span (between the two towers) and

on the back span (between the tower andanchor). First, we will examine theeffects of the main span and then we will

examine the back span.

The applied gravity loads on the cableare acting downward; therefore the

towers need to resist these forces withupward reactions. The vertical reaction

due to the main span, VM (lbs or kips),at each tower is (Figure 2):

V M =

q* L2

where q is the uniformly distributed load

applied to the bridge (lbs/ft or k/ft) andL (ft) is the length of the main span of

the bridge defined as the distance fromtower to tower.

Figure 2: Loaded Cable with Reactions

In addition, the loads on the bridge act to pull the tower tops toward each other;

these forces are resisted by horizontalreactions in the cable at the top of the

towers to keep the towers vertical. Thehorizontal reaction due to the main span,

HM (lbs or kips), at each tower is (Figure2)ii:

H M =q* L

2

8*d

where d is where the vertical distance

from the top of the tower and the cable is

greatest. It is called the sag of the cable(ft) and it is located at midspan (center

of the main span). Once again q is thedistributed load and L is the length of

the main span. See Figure 2 below.

EXAMPLE 1: Main Span Tower

Reactions of the George Washington

Bridge

Figure 3: George Washington Bridge Elevation

Determine: The vertical reaction in eachtower and horizontal reactions in the

cables at each tower of the GeorgeWashington Bridge due to the main span

dead loads plus live loads. See Figure 3.

Given iii:Dead load of the bridge, qDL = 39 k/ft

Live load on the bridge, qLL = 8 k/ft iv

The main span, L = 3,500 ft

The sag of the cable, d = 325 ft

Solution:

Step 1: Calculate the total load acting onthe bridge.

q = q DL + q LL = 39 k/ft + 8 k/ft

q = 47 k/ft

Step 2: Calculate the vertical reaction ateach tower due to main span loads.

V M =

q* L

2=

47 k/ft *3, 500 ft

2

V M ! 82,300 k

Step 3: Calculate the horizontal reaction

at each tower.




H M =q* L2

8*d =

47 k/ft*(3, 500 ft)2

8*325

H M ! 221,400 k

TOWER REACTIONS: Back Span

The gravity loads on the back span must

also be supported by the towers. Thisanalysis assumes that no other external

horizontal force acts on the tower (suchas wind). Therefore, the horizontal

reaction in the cable at the top of thetower for the back span, HB, is equal and

opposite to the horizontal reaction from

the main span, HM.

By summing the moments about the

anchor, the vertical reaction due to the back span, VB (lbs or k ) at each tower is

(see Figure 4):

M ! anchor

= 0 =V B L B " H Bd a " qL B L B

2

V B =

H B *

d a( )+

q* L2

B

2

!

"#

$

%& L

B

where HB (k ) is the horizontal reactionin the cable due to the back span, d a (ft)

is the vertical distance from the top ofthe tower to the anchor, q is the

uniformly distributed load applied to the bridge (lbs/ft or k/ft) and LB (ft) is the

length of the back span. See Figure 4.

Figure 4: Loaded Back Span with Reactions

EXAMPLE 2: Back Span Tower

Reactions of the George Washington

Bridge

Figure 5: George Washington Bridge

Back Span Elevation

Determine: The horizontal and vertical

reactions in one of the towers of theGeorge Washington Bridge due to the

back span dead plus live loads. SeeFigure 5.

Given iii

:

Dead load of the bridge, qDL = 39 k/ft Live load on the bridge, qLL = 8 k/ft

iv

The back span, L = 650 ft Vertical distance from tower to anchor,

d a = 377 ft

Solution:

Step 1: Calculate the total load acting onthe back span of the bridge.

q = q DL + q LL

q = 39 k/ft + 8 k/ft




q = 47 k/ft

Step 2: Calculate the horizontal reactionat the tower.

HB = HM = 221,400 k(previous example)

H B ! 221,400 k

Step 3: Calculate the vertical reaction at

each tower due to backspan loads.

V B =

H B * d a( )+

q* L2

B

2

!

"#

$

%&

L B

V B =

221, 400*377( )+47*650

2

2

!

"#

$

%&

650

V B ! 143,700 k

TOTAL TOWER REACTIONS

The total vertical force that the towerssupport equals the sum of VM and VB,that is, those loads brought on by the

main span plus the back span.Therefore, for this example, each tower

carries 82,300 k + 143,700 k ! 226,000kips of load in compression.

ANCHOR REACTIONS

To maintain horizontal equilibrium, the

horizontal reaction in the anchor musthave the same value as the horizontal

reaction of the back span,

! H = 0 = H B " H

anchor Hanchor = HB ! 221,400 k.

To maintain vertical equilibrium, thevertical reaction in the anchor must

balance the upward vertical reactionfrom the tower and the downward

gravity loads of the back span. The

vertical reaction in the anchor, Vanchor (lbs or k ) is thus:

!V = 0 =V B "V anchor " (q* L B )

V anchor =V B ! (q* L B )

where VB (k ) is the vertical reaction on

the tower due to the back span, q is theuniformly distributed load applied to the

bridge (lbs/ft or k/ft) and LB (ft) is thelength of the back span. See Figure 4.

Summary of Terms

-d: sag [ft]-da [ft] is the vertical distance from the

top of the tower to the anchor-LM: main span length [ft]

-LB: back span length [ft]

-q: distributed load, [lbs/ft] or [k/ft]-qDL: distributed dead load, [lbs/ft] or

[k/ft]-qLL: distributed live load, [lbs/ft] or

[k/ft]

-HM: horizontal reaction at tower andtensile force in cable due to the main

span, [lbs] or [k]-HB: horizontal reaction at tower and

tensile force in cable due to the backspan, [lbs] or [k]

-Hanchor: horizontal reaction in the

anchor [lbs] or [k]

-VM: vertical reaction at tower due to the

main span, [lbs] or [k]-VB: vertical reaction at tower due to the

back span, [lbs] or [k]-VB: vertical reaction of the anchor due

to the back span, [lbs] or [k]




Summary of Equations

Vertical Reaction at Tower:

V M

=

q* L

2

V B =

H B * d ( )+q* L

2

B

2

!

"#

$

%&

L B

M = main span B = back span

Horizontal Reaction at Tower

H M =q* L

2

8*d

H B = H

M

M = main span B = back span

Vertical Reaction at Anchor:

V anchor =V B ! (q* L B )

Notes

i

David P. Billington, “History andAesthetics of Suspension Bridges,”

Journal of the Structural Division, ASCEVol. 103, no. 478 (August 1977): 671-

687, with discussions.

ii To calculate the horizontal force, weuse the free-body diagram shown in

Figure 6 and sum the moments about thetop of the tower:

! M = 0 = Hd " qL2 L4

H = H M =qL

2

8d

Figure 6. Free-body diagram to calculate the

horizontal component of cable force.

iiiThe values used in this analysis are

taken from Power, Speed, and Form;

Engineers and the Making of theTwentieth Century, by David P.

Billington and David P. Billington Jr.,Princeton University Press, Princeton,

2006, p. 166 and 167.

iv

Ammann realized that the likelihood ofall the lanes on the bridge being

completely full with the heaviestvehicles decreased as the span of the

bridge increased and as more lanes wereadded. With these considerations in

mind he developed a series of equationsthat reduced the live load that the bridge

needed to be designed to carry. The liveload, qLL, before reduction was 46k/ft.

Ibid., p. 166

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L2 - Suspension Bridge - Part 1 Tower Anchors Jan 11 - Final