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Iam given a sector with included angle of 57.3degrees.The circumference of the arc represent a curved bridge. A vehicle with weight W moves from one end A where its pinned to another end B where there...

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Iam given a sector with included angle of 57.3degrees.The circumference of the arc represent a curved bridge. A vehicle with weight W moves from one end A where its pinned to another end B where there is a roller support.Evaluate an equation for torsion due to the weight of the vehicle and self weight of the bridge at any given point on the bridge.
Answered Same Day Dec 29, 2021

Solution

David answered on Dec 29 2021
107 Votes
Refer to the figure given below. It shows the curve
idge in plan.
Note that α = 57.3o and r = 60m in the calculations below. However, the angle measure taken below are
all in radians.
Assume that the point load, W is acting at point D (out of plane).
PB = r(1 – cos α)
AB = r(1 – cos θ)
Let RB and RC be the vertical reactions at B and C respectively.
Take moment about point C.
RC r(1 – cos α) = Wr(1 – cos θ)
 RC = W(1 – cos θ)/(1 – cos α)
Now, RB + RC = W
 RB = W(cos θ – cos α)/(1 – cos α)
Let the UDL due to self-weight of
idge be w. So, at each support reaction due to this load will be
added.
RB = W(cos θ – cos α)/(1 – cos α) + wrα/2
RC = W(1 – cos θ)/(1 – cos α) + wrα/2
Case 1: 0<γ<=θ
Now, consider the segment below.
OQ = 2r(sin γ/2)/γ (Q is the location of CG of self-weight of the beam)
OR = OQcos γ/2
QR = OQsin γ/2 = 2r(sin2 γ/2)/γ = r(1 – cos γ)/γ
RB = r - OQcos γ/2 = r - 2r(sin γ/2)(cos...
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