Reinforcement on a skewed bridge

[robyzou]

hi,

i've post a question in an another forum :

http://www.eng-tips.com/viewthread.cfm?qid=154167

but will i have more chance to get a response here !?

thanks

 

[Qshake]

Most times often than not bridge engineers will lay the reinforcing normal to the centerline of the bridge and in the acute angle corners will cut the reinforcing to match the angle and use the lopped off bars for the other side.

At least in that method the reinforcing remains the same and the steel can be ordered full length and cut to fit in the fab shop.

THere are other ways of handling this but this is most common.

Regards,
Qshake

Eng-Tips Forums:Real Solutions for Real Problems Really Quick.

 

[SlideRuleEra]

Perhaps this chapter on "Curved & Skewed Bridges" from the Portland Cement Association Bridge Design Manual will help

http://www.pci.org/pdf/find/knowledge_bank/active/MNL-133-97_ch12.pdf

http://www.SlideRuleEra.net

 

[robyzou]

thanks, i'll have a look to see if it help, but i don't think there is the way to put reinforcement in the plate.

 

[wiktor]

The following formulas are valid with the assumption, that the cracked tension section of the slab could be treated as shield.

? = angle in between principal moments (M1) and direction of reinforcement
? = angle in between directions of reinforcement ? and ?

Mu = M1 sin^2(?- ?) + M2 cos^2(?- ?)

M? = M1 sin^2 (?) + M2 cos^2(?)

M’’ = M1 sin ? sin(?- ?) - M2 cos ? cos(?- ?)

Substitute moments (to size the reinforcement)
Mu ‘= 1/ sin^2(?) (Mu +k? M’’?)

M? ‘= 1/ sin^2(?) (M? +1/k? M’’?)

When ? =90°
then
Mu = M1 cos^2(?) + M2 sin^2(?)
M? = M1 sin^2(?) + M2 cos^2(?)
M’’ = (M1 - M2) sin?cos?
Mu ‘= Mu +k? M’’?
M? ‘= M? +1/k? M’’?

The factor k should be approximately 1.

 

[wiktor]

I did not preview the message, so I just realized that all greek characters are missing. I will need e-mail

 

[Ussuri]

Wiktor

You can get greek letters if you like. Click on the Process TGML link below the message window and it will show you the codes required to input greek letters.

 

[wiktor]

The following formulas are valid with the assumption, that the cracked tension section of the slab could be treated as shield.

? = angle in between principal moments (M1) and direction of reinforcement
? = angle in between directions of reinforcement ? and ?

Mu = M1 sin2(?- ?) + M2 cos2(?- ?)

M? = M1 sin2 ? + M2 cos2?

M’’ = M1 sin ? sin(?- ?) - M2 cos ? cos(?- ?)

Substitute moments (to size the reinforcement)
Mu ‘= 1/ sin2? (Mu +k? M’’?)

M? ‘= 1/ sin2? (M? +1/k? M’’?)

When ? =90°
then
Mu = M1 cos2 ? + M2 sin2 ?
M? = M1 sin2 ? + M2 cos2?
M’’ = (M1 - M2) sin? cos?
Mu ‘= Mu +k? M’’?
M? ‘= M? +1/k? M’’?

The factor k should be approximately = 1
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