ﻣﺮآﺰ اﻟﺘﻌﻠﻴﻢ اﻟﻤﺴﺘﻤﺮ
اﻟﺠﺎﻣﻌﺔ اﻟﺘﻜﻨﻮﻟﻮﺟﻴﺔ
دورة ﺗﺼﻤﻴﻢ اﻟﺠﺴﻮر ذات اﻟﺮواﻓﺪ اﻟﺨﺮﺳﺎﻧﻴﺔ اﻟﻤﺴﻠﺤﺔ ﻟﻠﻔﺘﺮة 29-25آﺎﻧﻮن اﻟﺜﺎﻧﻲ2009 /
Design of a Reinforced Concrete Deck-Girder Bridge to AASHTO & ACI Codes
إﻋﺪاد وإﻟﻘﺎء اﻟﻤﻬﻨﺪس أﻹﺳﺘﺸﺎري أﻻﺳﺘﺎذ اﻟﻤﺴﺎﻋﺪ ﻋﻼء ﻣﻬﺪي اﻟﺨﻄﻴﺐ
اﻟﻤﻘﺪﻣﺔ: ﺗﺘﻀﻤﻦ اﻟﻤﺤﺎﺿﺮة اﻟﺘﺼﻤﻴﻢ اﻟﻜﺎﻣﻞ ﻟﻠﺠﺴﻮر اﻟﻤﺴﺘﻨﺪة ﻋﻠﻰ رواﻓﺪ ﺧﺮﺳﺎﻧﻴﺔ ﻣﺴﻠﺤﺔ .هﺬا اﻟﻨﻮع ﻣﻦ اﻟﺠﺴﻮر ،اﻟﻤﻮﺿﺢ ﻣﻘﻄﻌﻪ اﻟﻄﻮﻟﻲ ﻓﻲ اﻟﺸﻜﻞ 1-وﻣﻘﻄﻌﻪ اﻟﻌﺮﺿﻲ ﻓﻲ اﻟﺸﻜﻞ ، 2-ﻳﺼﻠﺢ ﻟﻠﺘﻨﻔﻴﺬ ﻓﻲ اﻟﻌﺮاق ﺿﻤﻦ ﻓﻀﺎءات ﺗﺘﺮاوح ﻣﻦ 10ﻟﻐﺎﻳﺔ 30ﻣﺘﺮ وذﻟﻚ ﻟﺘﻮﻓﺮ اﻟﻤﻮاد أﻷوﻟﻴﺔ واﻟﺨﺒﺮة ﻓﻲ إﻧﺸﺎءﻩ.
Given Data: Effective Span= 63 ft Effective Width= 3o ft Live Load= HS20 )Concrete Strength (Cylindrical Test) f'c =5000 psi (35 MPa )Steel a- Grade 40 for Deck Slab f y = 40,000 psi (330 MPa )b- Grade 60 for Girders f y =60,000 psi (414 MPa
2
Deck Slab Design: )b = 20 in∗ (50 cm
ﻧﻔﺮض إن ﻋﺮض أي راﻓﺪ ﻳﺴﺎوي:
20 20 − = 5.33 ft 12 × 2 12 × 2
ﻓﻴﻜﻮن اﻟﻌﻤﻖ اﻟﻤﺆﺛﺮ ﻟﻠﺒﻼﻃﺔ ﻳﺴﺎوي:
Ss = 7 −
)h = 8in∗ (20 cm ﻧﻔﺮض أن اﻟﻌﻤﻖ اﻟﻜﻠﻲ ﻟﻠﺒﻼﻃﺔ ﻳﺴﺎوي: ﻣﻊ ﻃﺒﻘﺔ إآﺴﺎء ﺑﺎﻟﺨﺮﺳﺎﻧﺔ أﻻﺳﻔﻠﺘﻴﺔ ﺑﺴﻤﻚ ، 4inﻳﻤﻜﻦ ﺣﺴﺎب ﻣﻘﺪار أﻷﺣﻤﺎل اﻟﻤﻴﺘﺔ DLاﻟﻤﺆﺛﺮة ﻋﻠﻰ آﻞ ﻗﺪم ﻣﺮﺑﻊ ﻣﻦ اﻟﺒﻼﻃﺔ وآﻤﺎ ﻳﻠﻲ: 8 4 ∗∗ + 130 × ≈ 145 psf 12 12
× DL = 150
ﻟﺘﻌﻘﻴﺪ ﻣﻨﺸﺄ اﻟﺠﺴﺮ وﺻﻌﻮﺑﺔ ﺗﺤﻠﻴﻠﻪ إﻧﺸﺎﺋﻴﺎً ،ﻳﻤﻜﻦ ﺗﺒﺴﻴﻂ اﻟﻤﻮﺿﻮع ﺑﺎﻓﺘﺮاض ﻣﻌﺎﻣﻞ ﺑﻤﻘﺪار1/10 ﻟﻌﺰوم أﻷﺣﻤﺎل اﻟﻤﻴﺘﺔ اﻟﻤﻮﺟﺒﺔ واﻟﺴﺎﻟﺒﺔ ،ﺑﻴﻨﻤﺎ ﻳﻜﻮن اﻟﻤﻌﺎﻣﻞ ﺑﻤﻘﺪار 1/2ﻟﺮﺻﻴﻒ اﻟﺠﺴﺮ وآﻤﺎ ﻣﻮﺿﺢ ﺑﺎﻟﺸﻜﻞ.3- ﻟﺬا ﺗﻜﻮن اﻟﻌﺰوم اﻟﻤﻮﺟﺒﺔ واﻟﺴﺎﻟﺒﺔ اﻟﻘﺼﻮى ﻟﻸﺣﻤﺎل اﻟﻤﻴﺘﺔ Mdاﻟﻤﺆﺛﺮة ﻋﻠﻰ ﺑﻼﻃﺔ اﻟﻤﻨﺼﺔ ﺗﺴﺎوي: wl 2 Dl × l 2 145 × (5.33) 2 =m =m = m412 ft.lb 10 10 10
Md = m
أﻹﺳﻠﻮب أﻟﻤﺘﺒﻊ ﻟﻠﺘﻌﺎﻣﻞ ﻣﻊ أﻷﺣﻤﺎل اﻟﺤﻴﺔ )آﺎﻓﺔ أﻵﻟﻴﺎت واﻷﺷﺨﺎص اﻟﻤﺎرﻳﻦ ﻋﻠﻰ اﻟﺠﺴﺮ( آﻤﺎ ﺗﻮﺻﻲ ﺑﻪ آﺮاﺳﺔ ﻣﻮاﺻﻔﺎت اﻟﻄﺮق واﻟﻨﻘﻞ AASHTOوهﻲ اﺧﺘﺼﺎر ل: )(American Association of State Highway and Transportation Officials ﻻﺧﺘﻼف أوزان وأﺑﻌﺎد وأﻋﺪاد اﻟﻌﺠﻼت اﻟﻤﺎرة ﻋﻠﻰ اﻟﺠﺴﺮ ،واﺳﺘﻨﺎدا إﻟﻰ دراﺳﺎت ﺗﺠﺮﻳﺒﻴﺔ وإﺣﺼﺎﺋﻴﺔ ،ﻳﻤﻜﻦ اﺳﺘﺨﺪام اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ HS20ذات اﻟﻮزن اﻟﻤﺤﺪد واﻷﺑﻌﺎد اﻟﺜﺎﺑﺘﺔ ﺑﻴﻦ ﻣﺤﺎورهﺎ ﻻ ﻋﻦ اﺣﺘﻤﺎﻻت ﻣﺮور ﻋﺠﻼت ﻣﺨﺘﻠﻔﺔ ،وآﻤﺎ ﻓﻲ اﻟﺸﻜﻞ.7- ﺑﺪ ً وﻟﺤﺴﺎب اﻟﻌﺰوم اﻟﻘﺼﻮى ﻟﻸﺣﻤﺎل اﻟﺤﻴﺔ MLاﻟﻤﺆﺛﺮة ﻋﻠﻰ ﺑﻼﻃﺔ ﻣﻨﺼﺔ اﻟﺠﺴﺮ ﺗﺤﺖ ﺗﺄﺛﻴﺮ ﻣﺮور اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ ، HS20ﻧﺴﺘﺨﺪم اﻟﻤﻌﺎدﻟﺔ اﻟﺘﺎﻟﻴﺔ: Ss + 2 5.33 + 2 × P 20 = 0.8 × 16000 = 2932 ft.lb 32 32
ﺣﻴﺚ إن 0.8
× ML = 0.8
ﻳﻤﺜﻞ ﻣﻌﺎﻣﻞ اﺳﺘﻤﺮارﻳﺔ ،ﻳﺴﺘﺨﺪم إذا آﺎﻧﺖ ﺑﻼﻃﺔ اﻟﻤﻨﺼﺔ ﻣﺴﺘﻤﺮة ﻓﻮق ﺛﻼﺛﺔ رواﻓﺪ أو أآﺜﺮ ،أﻣﺎ إذا آﺎﻧﺖ اﻟﺒﻼﻃﺔ ﺗﺴﺘﻨﺪ ﻋﻠﻰ راﻓﺪﻳﻦ ﻓﻨﺴﺘﺨﺪم اﻟﻤﻌﺎدﻟﺔ ﺑﺪوﻧﻪ.
ﻳﻤﺜﻞ ﺣﻤﻞ إﻃﺎر اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ HS20وﻳﺴﺎوي ﻧﺼﻒ ﺣﻤﻞ اﻟﻤﺤﻮر اﻟﺒﺎﻟﻎ وإن P20 32,000 LBأي ﺑﻌﺒﺎرة ﺛﺎﻧﻴﺔ . 16,000LB ﻟﻤﺎ آﺎﻧﺖ اﻟﻌﺠﻼت اﻟﻤﺎرة ﻋﻠﻰ اﻟﺠﺴﺮ ﺗﺴﻴﺮ ﺑﺴﺮع ﻣﺨﺘﻠﻔﺔ ﻓﺈﻧﻬﺎ ﺗﺴﺒﺐ اهﺘﺰاز ﻟﻪ ،آﻤﺎ إن ﺗﻮﻗﻔﻬﺎ اﻟﻤﻔﺎﺟﺊ ﻳﺴﺒﺐ ﻣﺎ ﻳﺸﺒﻪ اﻟﺼﺪم ) ، (Impactﺑﺎﻟﺘﺎﻟﻲ ﺳﺘﺤﺪث ﻋﺰوم إﺿﺎﻓﻴﺔ ﺗﺴﻤﻰ ﻋﺰوم اﻟﺼﺪم ) .(MIﺗﺸﻴﺮ اﻟﻤﻮاﺻﻔﺔ إﻟﻰ أن ﻣﻌﺎﻣﻞ اﻟﺼﺪم ) (Iاﻟﺬي ﻳﻤﺜﻞ ﻧﺴﺒﺔ ﻻ ﺗﺘﺠﺎوز 30 %ﻣﻦ ﻋﺰوم أﻷﺣﻤﺎل اﻟﺤﻴﺔ ﻳﻤﻜﻦ ﺣﺴﺎﺑﻪ آﻤﺎ ﻳﻠﻲ: ………………………………………………………………………………... ∗ هﺬﻩ ﻓﺮﺿﻴﺎت إﺑﺘﺪاﺋﻴﺔ ﺳﻮف ﻳﻈﻬﺮ ﻣﻦ ﺧﻼل اﻟﺤﺴﺎﺑﺎت أﻟﻼﺣﻘﺔ آﻔﺎﺋﺘﻬﺎ أو اﻟﺤﺎﺟﺔ ﻟﺘﻐﻴﻴﺮهﺎ. ∗∗ وزن ﻗﺪم ﻣﻜﻌﺐ ﻣﻦ اﻟﺨﺮﺳﺎﻧﺔ ﻳﺴﺎوي، 150ووزن ﻗﺪم ﻣﻜﻌﺐ ﻣﻦ اﻟﺨﺮﺳﺎﻧﺔ أﻻﺳﻔﻠﺘﻴﺔ ﻳﺴﺎوي130 3
50 50 = = 0.322 ⇒ 0.3 Ss + 150 5.33 + 150 MI = ML × I = 2932 × 0.3 = 880 ft.lb =I
أﻵن ﻧﺠﻤﻊ آﺎﻓﺔ اﻟﻌﺰوم اﻟﻤﺆﺛﺮة ﻋﻠﻰ ﺑﻼﻃﺔ ﻣﻨﺼﺔ اﻟﺠﺴﺮ وهﻲ ﻋﺰوم أﻷﺣﻤﺎل اﻟﻤﻴﺘﺔ واﻟﺤﻴﺔ واﻟﻨﺎﺷﺌﺔ ﺑﺴﺒﺐ اﻟﺼﺪم ﻟﻠﺤﺼﻮل ﻋﻠﻰ اﻟﻌﺰوم اﻟﻜﻠﻴﺔ اﻟﻘﺼﻮى ) (MTوآﻤﺎ ﻳﻠﻲ: MT = Md + ML + MI = 412 + 2932 + 880 = 4224 ft.lb
ﺑﻴﻨﻤﺎ اﻟﻌﺰوم اﻟﻘﺼﻮى ﻟﻸﺣﻤﺎل اﻟﻤﻴﺘﺔ اﻟﻤﺆﺛﺮة ﻋﻠﻰ رﺻﻴﻒ اﻟﺠﺴﺮ ﺗﺴﺎوي: 18 10 (150 × )(6 − ) 2 12 12 = −3000 ft.lb Md = − 2
و اﻟﻌﺰوم اﻟﻘﺼﻮى ﻟﻸﺣﻤﺎل اﻟﺤﻴﺔ اﻟﻤﺆﺛﺮة ﻋﻠﻰ رﺻﻴﻒ اﻟﺠﺴﺮ اﻟﺒﺎﻟﻐﺔ 85psfﺗﺴﺎوي: 10 2 ) 12 = −1135 ft.lb
ﻓﺘﻜﻮن اﻟﻌﺰوم اﻟﻜﻠﻴﺔ اﻟﻘﺼﻮى اﻟﻤﺆﺛﺮة ﻋﻠﻰ رﺻﻴﻒ اﻟﺠﺴﺮ ﺗﺴﺎوي:
85(6 − 2
wl 2 =− ML = − 2
MT = Md + ML = −3000 − 1135 = −4135 ft.lb
4
ﻧﺼﻤﻢ ﺑﻼﻃﺔ اﻟﻤﻨﺼﺔ وﻓﻖ ﻃﺮﻳﻘﺔ
)*WSDM (Working Stress Design Method
اﻟﻤﻮﺿﺤﺔ رﻣﻮزهﺎ وﻣﻌﺎدﻻﺗﻬﺎ ﻓﻲ اﻟﺸﻜﻞ. 4-
Tension T= As×fs = Compression C = 0.5 bkd × fc Moment of Resistance Mr = T × jd = As × fs × jd Concrete Youngs Modulas Ec = 57000 f c′ = 57000 5000 = 4,000,000 psi 29,000,000 = 7.25 4,000,000 20,000 = Stress Ratio r = fs fc = 10 2,000 n 7.25 =k = = 0.42 n + r 7.25 + 10 k 0.42 j = 1− = 1− = 0.86 3 3
= Modular Ratio n = Es Ec
اﻟﻌﻤﻖ أﻷدﻧﻰ اﻟﻤﻄﻠﻮب ﻟﻠﺒﻼﻃﺔ ) (d req.ﻳﺴﺎوي: 2 × 4224 × 12 = 3.42in 2000 × 0.42 × 0.86 × 12
2 MT = fckjb
= dreq
........................................................................................................................... * ﻓﻲ ﻃﺮﻳﻘﺔ WSDMﻳﺠﺮي ﺣﺴﺎب اﻟﻌﺰوم اﻟﻔﻌﻠﻴﺔ اﻟﻘﺼﻮى ﺑﺪون زﻳﺎدة ،ﺑﻴﻨﻤﺎ ﻳﺘﻢ ﺗﺨﻔﻴﺾ اﻟﻤﻘﺎوﻣﺎت اﻟﻘﺼﻮى ﻟﻠﻤﻮاد وذﻟﻚ ﻟﺘﺤﻘﻴﻖ درﺟﺔ أﻷﻣﺎن اﻟﻤﻄﻠﻮﺑﺔ ﻟﻠﺠﺴﺮ وآﻤﺎ ﻳﻠﻲ: f c = 0.4 f c′ = 0.4 × 5000 = 2000 psi
f s = 0.5 f y = 0.5 × 40,000 = 20,000 psi
5
ﺑﺎﺳﺘﺨﺪام ﻗﻀﺒﺎن ﺣﺪﻳﺪ ﺗﺴﻠﻴﺢ ذات ﻗﻄﺮ 5/8وﻏﻄﺎء ﺧﺮﺳﺎﻧﻲ ﺑﻤﻘﺪار 1inﻣﻊ ﺗﺮك ﻃﺒﻘﺔ ﻣﻦ أﻷﻋﻠﻰ ﻣﻌﺮﺿﺔ ﻟﻠﺘﺂآﻞ ﺑﺴﻤﻚ ، 1inﻳﻜﻮن اﻟﻌﻤﻖ اﻟﻤﺘﻮﻓﺮ d ava.ﻳﺴﺎوي) :ﻻﺣﻆ اﻟﺸﻜﻞ.(5- 5 − 1 = 5.6in 16
d ava. = 8 − 1 −
وهﺬا اﻟﻌﻤﻖ أآﺒﺮ ﻣﻦ اﻟﻌﻤﻖ اﻟﻤﻄﻠﻮب اﻟﺒﺎﻟﻎ . 3.42inﻟﺬا ﻓﺎن اﻟﻔﺮﺿﻴﺔ ﺑﺎﻋﺘﺒﺎر اﻟﺴﻤﻚ اﻟﻜﻠﻲ 8in ﺳﺘﻜﻮن أﻣﻴﻨﺔ وﻣﻘﺒﻮﻟﺔ وﺳﻴﺘﻢ اﻋﺘﻤﺎدهﺎ. ﻣﺴﺎﺣﺔ ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ اﻟﺮﺋﻴﺴﻲ اﻟﻤﻄﻠﻮﺑﺔ ﻟﻜﻞ ﻗﺪم ﻣﻦ ﻋﺮض اﻟﺒﻼﻃﺔ Asﺗﺴﺎوي: MT 4224 × 12 = = 0.526in 2 f s jd 20,000 × 0.86 × 5.6
= As
ﻣﺴﺎﻓﺎت ﺗﺒﺎﻋﺪ ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ #5ذو ﻣﺴﺎﺣﺔ ﻣﻘﻄﻊ ﻋﺮﺿﻲ ، Ab = 0.31in 2راﺟﻊ اﻟﺠﺪول اﻟﻤﻠﺤﻖ ﻟﺤﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ ،ﺗﺴﺎوي: Ab 0.31 = = 0.59 ft = 0.59 × 12 = 7.07in ⇒ 7in As 0.526
= Spacing
Use #5@7´´c/c,T&B ﻟﻤﺎ آﺎﻧﺖ ﻓﻀﺎءات اﻟﺒﻼﻃﺔ ﻗﺼﻴﺮة وﻟﺘﻼﻓﻲ إﺣﻨﺎء اﻟﻘﻀﺒﺎن ﺑﺼﻮرة ﻗﺎﺳﻴﺔ ،ﺳﻴﺘﻢ إﺳﺘﻌﻤﺎل ﻗﻀﺒﺎن ﻣﺴﺘﻘﻴﻤﺔ ﻓﻲ أﺳﻔﻞ اﻟﺒﻼﻃﺔ وآﺬﻟﻚ ﻓﻲ أﻋﻼهﺎ .إن هﺬا أﻹﺟﺮاء رﺑﻤﺎ ﺳﻴﺴﺘﻬﻠﻚ آﻤﻴﺔ أآﺒﺮ ﻣﻦ اﻟﺤﺪﻳﺪ ﻟﻜﻦ اﻟﻜﻠﻔﺔ أﻹﺿﺎﻓﻴﺔ ﺳﻴﺘﻢ ﻣﻌﺎدﻟﺘﻬﺎ ﺑﺎﻟﺘﻮﻓﻴﺮ اﻟﺤﺎﺻﻞ ﻓﻲ إﺟﻮر اﻟﺤﺪادة وﺳﻬﻮﻟﺔ اﻟﺘﻨﻔﻴﺬ. ﺳﻴﺘﻢ ﻣﻘﺎوﻣﺔ أﻹﺟﻬﺎدات اﻟﻨﺎﺷﺌﺔ ﻣﻦ أﻹﻧﻜﻤﺎش وﺗﻐﻴﺮ درﺟﺎت اﻟﺤﺮارة ،وآﺬﻟﻚ ﻟﺘﻮزﻳﻊ أﻷﺣﻤﺎل اﻟﻤﺮآﺰة ﻹﻃﺎرات اﻟﻌﺠﻼت ﻋﻠﻰ ﺑﻼﻃﺔ اﻟﻤﻨﺼﺔ ،ﺑﺎﺳﺘﺨﺪام ﺣﺪﻳﺪ ﺗﺴﻠﻴﺢ ﺛﺎﻧﻮي ﻳﻮﺿﻊ ﻓﻮق اﻟﻄﺒﻘﺔ اﻟﺴﻔﻠﻰ ﻟﺤﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ اﻟﺮﺋﻴﺴﻲ وﺑﺼﻮرة ﻣﺘﻌﺎﻣﺪة ﻋﻠﻴﻪ .أﻣﺎ آﻤﻴﺘﻪ ﻓﺘﺤﺪدهﺎ اﻟﻤﻮاﺻﻔﺔ آﻨﺴﺒﺔ ﻣﺌﻮﻳﺔ ﻻ ﺗﺘﺠﺎوز 67%ﻣﻦ ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ اﻟﺮﺋﻴﺴﻲ وﺗﺒﻠﻎ: 67 = 10.55in ⇒ 10in 100
× Spacing ( Secondary) = 7.07
Use #5@10´´c/c
6
Design of Interior Girders ﻟﻤﺎ آﺎﻧﺖ اﻟﺮواﻓﺪ اﻟﺪاﺧﻠﻴﺔ ﻋﻠﻰ ﺷﻜﻞ Tﻓﺎن ﻋﺮض اﻟﺸﻔﺔ اﻟﻌﻠﻴﺎ Width of Top Flangeﻳﺴﺎوي اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﻣﺮاآﺰ اﻟﺮواﻓﺪ اﻟﺒﺎﻟﻐﺔ 7ftوآﻤﺎ ﻓﻲ اﻟﺸﻜﻞ. 6-
ﺑﺎﻓﺘﺮاض ﻋﻤﻖ ﻣﻘﺪارﻩ ) 40inﻣﺘﺮ واﺣﺪ( ﻟﺠﺰء اﻟﺮاﻓﺪ اﻟﻮاﻗﻊ ﺗﺤﺖ اﻟﺒﻼﻃﺔ ،ﻳﻤﻜﻦ ﺣﺴﺎب أﻷﺣﻤﺎل اﻟﻤﻴﺘﺔ اﻟﻤﺆﺛﺮة ﻋﻠﻰ ﻗﺪم واﺣﺪ ﻣﻦ ﻃﻮل اﻟﺮاﻓﺪ وآﻤﺎ ﻳﻠﻲ: 12 20 40 × × ) = 1850lb / ft 12 12 12
(DL = 145 + 150
وﻳﻜﻮن اﻟﻌﺰم أﻷﻗﺼﻰ ﻟﻸﺣﻤﺎل اﻟﻤﻴﺘﺔ ﻳﺴﺎوي: DL × l 1850 × 63 = = 917,000 ft.lb 8 8 2
2
= Md
أﻣﺎ اﻟﻌﺰم أﻷﻗﺼﻰ ﻟﻸﺣﻤﺎل اﻟﺤﻴﺔ اﻟﻤﺘﻤﺜﻠﺔ ﺑﻤﺮور اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ اﻟﻤﻮﺿﺤﺔ ﺑﺎﻟﺸﻜﻞ ، 7-ﻓﻴﻤﻜﻦ ﺣﺴﺎﺑﻪ آﻤﺎ ﻳﻠﻲ: ﻟﻤﺎ آﺎن اﻟﺠﺴﺮ ﻣﺆﻟﻒ ﻣﻦ ﺑﻼﻃﺔ ﺧﺮﺳﺎﻧﻴﺔ ﻣﺴﺘﻨﺪة ﻋﻠﻰ رواﻓﺪ ﺧﺮﺳﺎﻧﻴﺔ ،وإن ﻋﺮض اﻟﺠﺴﺮ ﻳﺴﺘﻮﻋﺐ ﻣﺴﺎرﻳﻦ ﻣﺮورﻳﻴﻦ ،ﻓﺎن اﻟﻤﻮاﺻﻔﺔ ﺗﺸﻴﺮ ﻓﻲ ﻣﺜﻞ هﺬﻩ اﻟﺤﺎﻟﺔ إﻟﻰ إن آﻞ راﻓﺪ داﺧﻠﻲ ﻋﻠﻴﻪ أن ﻳﺴﻨﺪ أﺣﻤﺎل إﻃﺎرات اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ ﻣﻀﺮوﺑًﺎ ﺑﻤﻌﺎﻣﻞ ﻣﻘﺪارﻩ ، S/5ﺣﻴﺚ إن Sﺗﻤﺜﻞ اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺧﻄﻮط ﻣﺮاآﺰ اﻟﺮواﻓﺪ ﻣﻘﺎﺳًﺎ ﺑﺎﻷﻗﺪام ﻋﻠﻰ أن ﻻ ﺗﺘﺠﺎوز 10أﻗﺪام. S 7 = = 1.4 wheel load/ wheel 5 5 وﺑﺬﻟﻚ ﻳﻜﻮن ﺣﻤﻞ آﻞ ﻣﻦ أﻹﻃﺎرﻳﻦ أﻟﺨﻠﻔﻲ واﻟﻮﺳﻄﻲ ﻳﺴﺎوي 16,000 × 1.4 = 22,400lbﺑﻴﻨﻤﺎ ﺣﻤﻞ
أﻹﻃﺎر أﻷﻣﺎﻣﻲ ﻳﺴﺎوي . 4,000 × 1.4 = 5,600lb 7
ﻣﻦ ﺧﻼل ﻧﻈﺮﻳﺎت ﺗﺤﻠﻴﻞ أﻹﻧﺸﺎءات اﻟﻤﺘﻌﻠﻘﺔ ﺑﻜﻴﻔﻴﺔ ﺣﺴﺎب ﻣﻘﺪار وﻣﻮﻗﻊ أﻟﻌﺰم أﻷﻗﺼﻰ ﻟﺠﺴﺮ ﺗﺴﻴﺮ ﻋﻠﻴﻪ ﺷﺎﺣﻨﺔ ﻣﺘﻌﺪدة اﻟﻤﺤﺎور وﻣﺨﺘﻠﻔﺔ أﻷوزان اﻟﻤﺴﻠﻄﺔ ﻣﻦ ﻗﺒﻞ آﻞ ﻣﺤﻮر ﻧﻼﺣﻆ ﻣﺎﻳﻠﻲ: أ -إن ﻣﺨﻄﻂ اﻟﻌﺰوم ﻳﺘﺄﻟﻒ ﻣﻦ ﺧﻄﻮط ﻣﺴﺘﻘﻴﻤﺔ وإن أﻗﺼﻰ ﻋﺰم ﻳﻘﻊ ﻣﺒﺎﺷﺮة ﺗﺤﺖ أﺣﺪ أﻹﻃﺎرات. ب -ﻳﻜﻮن ﻣﻮﻗﻊ أﻹﻃﺎر اﻟﺬي ﻳﺴﺒﺐ اﻟﻌﺰم أﻷﻗﺼﻰ ﺑﺤﻴﺚ ان ﺧﻂ ﺗﻨﺼﻴﻒ اﻟﺠﺴﺮ ﻳﻨﺼﻒ اﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ أﻷﻃﺎر وﻣﺤﺼﻠﺔ وزن اﻟﺸﺎﺣﻨﺔ) .( Rﻻﺣﻆ اﻟﺸﻜﻞ. 7- ت -ﻣﻦ ﺧﻼل اﻟﺨﺒﺮة ﻧﻌﺘﻘﺪ ﺑﺄن أﻗﺼﻰ ﻋﺰم ﻳﻘﻊ ﺗﺤﺖ أﻷﻃﺎر اﻟﻮﺳﻄﻲ وﻟﻴﺲ ﺗﺤﺖ ﺑﻘﻴﺔ أﻹﻃﺎرات.
22,400 × 19.85 + 22,400 × 33.85 + 5,600 × 47.85 = 23,347lb 63
= RB
MLmax = RB × 29.15 − 5,600 × 14 = 658,165 ft.lb 50 = 0.266 63 + 125
= Impact Factor I
MT = 917,000 + 658,165 + 0.266 × 685,165 = 1,750,000 ft.lb 8
ﺣﺴﺎب ﻗﻮى اﻟﻘﺺ اﻟﻤﺆﺛﺮة ﻋﻠﻰ اﻟﺮاﻓﺪ اﻟﺪاﺧﻠﻲ: ﻗﻮى اﻟﻘﺺ اﻟﻨﺎﺷﺌﺔ ﺑﺴﺒﺐ أﻷﺣﻤﺎل اﻟﻤﻴﺘﺔ Dead Load Shears Vdﻳﻮﺿﺤﻬﺎ اﻟﺸﻜﻞ. 8-
63 = 58,275lb 2
× Vd max = 1850
ﻟﺤﺴﺎب ﻗﻮى اﻟﻘﺺ اﻟﻘﺼﻮى ﺑﺴﺒﺐ أﻷﺣﻤﺎل اﻟﺤﻴﺔ ،ﺗﻜﻮن ﻗﻮة اﻟﻘﺺ اﻟﻘﺼﻮى VLﺗﺤﺖ أﻹﻃﺎر اﻟﺨﻠﻔﻲ وﻋﻨﺪﻣﺎ ﻳﻜﺘﻤﻞ دﺧﻮل اﻟﺸﺎﺣﻨﺔ إﻟﻰ اﻟﺠﺴﺮ .ﻻﺣﻆ إن ﺣﻤﻞ أﻹﻃﺎر اﻟﺨﻠﻔﻲ ﻳﺒﻘﻰ ﺑﺪون زﻳﺎدة ﺑﻴﻨﻤﺎ ﻧﻀﺮب آﻞ ﻣﻦ أﺣﻤﺎل أﻹﻃﺎر اﻟﻮﺳﻄﻲ واﻟﺨﻠﻔﻲ ﺑﺎﻟﻤﻌﺎﻣﻞ ، 1.4آﻤﺎ ﻓﻲ اﻟﺸﻜﻞ. 9-
VLmax = RA = (16,000 × 63 + 22,400 × 49 + 5,600 × 35) / 63 = 36,533lb VT = 58,275 + 36,533 + 0.266 × 36,533 = 104,526lb 9
ﺗﺤﺪﻳﺪ أﺑﻌﺎد اﻟﺮاﻓﺪ
Determination of Girder Cross Section Dimensions
ﺑﻤﻮﺟﺐ ﺗﻮﺻﻴﺎت آﺮاﺳــﺔ اﻟﻤﻮاﺻﻔﺎت ، AASHTOﻳﺠﺮي ﺣﺴــﺎب إﺟﻬﺎد اﻟﻘﺺ أﻻﺳــﻤﻲ v Nominal Shear Stressاﻟﻨﺎﺷﺊ ﺑﺴﺒﺐ أﻷﺣﻤﺎل اﻟﺨﺪﻣﻴﺔ Service Loadsوﻓﻖ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﺎﻟﻴﺔ: VT bw d
ﺣﻴﺚ إن b wﺗﻤﺜﻞ اﻟﻌﺮض اﻟﺴﻔﻠﻲ ﻟﻠﺮاﻓﺪ وإن vﺗﺤﺴﺐ وﻓﻖ اﻟﻤﻌﺎدﻟﺔ اﻟﺘﺎﻟﻴﺔ: وﻋﻠﻴﻪ ﻓﺎن ﻣﺴﺎﺣﺔ اﻟﻤﻘﻄﻊ اﻟﻤﻄﻠﻮﺑﺔ ﺗﺴﺎوي:
=v
v = 2.95 f c′ = 2.95 5,000 = 208 psi
VT 104,526 = = 500in 2 208 v
= bw d
وﻟﻤﺎ آﻨﺎ ﻗﺪ إﻓﺘﺮﺿﻨﺎ إن اﻟﻌﺮض اﻟﺴﻔﻠﻲ ﻟﻠﺮاﻓﺪ ﻳﺴﺎوي 20inﻟﺬا ﻳﻜﻮن ﻃﻮل ذرراع اﻟﻤﻘﺎوﻣﺔ اﻟﻤﻄﻠﻮب d reqﻳﺴﺎوي: 500 = 25in 20
= d req
إذا إﻓﺘﺮﺿﻨﺎ إﺳﺘﺨﺪام ﺛﻼﺛﺔ ﻃﺒﻘﺎت ﻣﻦ ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ ، #10راﺟﻊ اﻟﻤﻠﺤﻖ ، 2-ﻣﻊ ﺗﺮك ﻣﺴﺎﻓﺔ ﺻﺎﻓﻴﺔ ﻣﻘﺪارهﺎ 2inﺑﻴﻦ أي ﻃﺒﻘﺘﻴﻦ ،و إﺿﺎﻓﺔ ﻏﻄﺎء ﺧﺮﺳﺎﻧﻲ ﺑﺴﻤﻚ 2.5inﺗﺤﺖ اﻟﻄﺒﻘﺔ اﻟﺴﻔﻠﻰ وﻋﻠﻰ اﻟﺠﻮاﻧﺐ ،ﺑﻬﺬﻩ ﻳﻜﻮن اﻟﻌﻤﻖ اﻟﻜﻠﻲ اﻟﻤﻄﻠﻮب ، h reqآﻤﺎ ﻣﻮﺿﺢ ﻓﻲ اﻟﺸﻜﻞ 10 -ﻣﻘﺪارﻩ:
10 10 + 2 + + 2.5 = 31 .38in 16 8
hreq = 25 +
وﻟﻤﺎ آﺎن اﻟﻌﻤﻖ اﻟﻤﺘﻮﻓﺮاﻟﻜﻠﻲ ، h=48inﻟﺬا ﻳﻜﻮن ﻃﻮل ذرراع اﻟﻤﻘﺎوﻣﺔ اﻟﻤﺆﺛﺮ d avaﻳﺴﺎوي: 10 10 −2− = 40.6in 8 16
d ava = 48 − 1 − 2.5 −
وﺗﻜﻮن ﻣﺴﺎﺣﺔ ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ اﻟﻤﻄﻠﻮﺑﺔ Asﺗﺴﺎوي:
1,750,000 × 12 = = 18.855in 2 = As 7 t ) f s (d − ) 30,000(40.6 − 2 2 MT
10
As 18.855 = = 14.85 ⇒ 15bars Ab 1.27
= Number of #10 Bars
ﻓﻲ هﺬﻩ اﻟﻤﺮﺣﻠﺔ ﻳﺠﺐ ﺗﺪﻗﻴﻖ إﺟﻬﺎدات أﻹﻧﻀﻐﺎط واﻟﺘﺄآﺪ ﻣﻦ ﻋﺪم ﺗﺠﺎوزهﺎ ﻟﻺﺟﻬﺎدات اﻟﻤﺴﻤﻮح ﺑﻬﺎ ﻟﻠﺨﺮﺳﺎﻧﺔ .وﻳﺘﻢ إﻋﺘﺒﺎر إن ﺟﺰء اﻟﺮاﻓﺪ اﻟﻮاﻗﻊ ﺿﻤﻦ اﻟﺒﻼﻃﺔ هﻮ اﻟﺬي ﻳﺘﻌﺮض ﻓﻘﻂ ﻟﻺﻧﻀﻐﺎط ﻣﻊ أﺧﺬ ﻣﻘﺎوﻣﺔ ﺗﺤﻤﻞ اﻟﻤﻮاد ﺑﻨﻈﺮ أﻹﻋﺘﺒﺎر وآﻤﺎ ﻳﻠﻲ: = 7.2
29,000,000 57,000 5,000
=
29,000,000 57,000 f ' c
f s 30,000 = = 15 fc 2,000
=r
7.2 n = = 0.324 n + r 7.2 + 15
=k
k 0.324 = 1− = 0.892 3 3
j = 1−
MT )×b × hf × j × d = 1342 psi ≤ f all 2000 psi ⇒ O.K
1,750,000 × 12
hf 2kd
= fc (1 −
7 (1 − ) × 12 × 7 × 7 × 0.892 × 40.625 2 × 0.324 × 40.625
11
=n
= fc
Web Shear Reinforcement ﺗﻜﻮن إﺟﻬﺎدات اﻟﻘﺺ آﺒﻴﺮﻩ ﻗﺮب اﻟﻤﺴﺎﻧﺪ وﺗﻘﻞ ﻓﻲ اﻟﻮﺳﻂ ،وﻳﻜﻮن أﺧﻄﺮهﺎ ﻋﻠﻰ ﻣﺴﺎﻓﺔ d ) (40.625in=3.4ftﻋﻦ أي ﻣﻦ اﻟﻤﺴﻨﺪﻳﻦ .ﻟﺬا ﺳﻴﺠﺮي ﺣﺴﺎب أﻹﺟﻬﺎدات ﻋﻨﺪ هﺬﻩ اﻟﻤﺴﺎﻓﺔ وآﺬﻟﻚ ﻋﻠﻰ ﺑﻌﺪ 10ftﻋﻦ أي ﻣﻦ ﻣﺴﻨﺪي اﻟﺮاﻓﺪ. 31.5 − 3.4 = 52,012lb 31.5 31.5 − 10 × = 58,275 = 39,775lb 31.5
× Vd 3.4 = 58,275
Vd10
VL3.4 = RA = (5,600 × 31.5 + 22,400 × (45.6 + 59.6)) / 63 = 40,213lb
VL10 = RA = (5,600 × 25 + 22,400 × (39 + 53)) / 63 = 34,933lb 12
VT3.4 = 52,012 + 1.266(40,213) = 102,922lb VT10 = 39,775 + 1.266(34,933) = 84,000lb VT bd
=v
Shear Stress
102 ,922 = 127 psi 20 × 40 .625
= v3.4
84,000 = 104 psi 20 × 40.625
= v10
ﻟﻤﺎ آﺎﻧﺖ اﻟﺨﺮﺳﺎﻧﺔ ﺗﺘﺤﻤﻞ إﺟﻬﺎد ﻗﺺ إﺳﻤﻲ ﻣﻘﺪارﻩ: vc = 0.95 f ' c = 0.95 5,000 = 67 psi
وﻟﻤﺎ آﺎﻧﺖ اﻟﻤﻮاﺻﻔﺔ ﻻﺗﺴﻤﺢ ﺑﺈﺳﺘﺨﺪام ﻣﺴﺎﻓﺎت ﺗﺒﺎﻋﺪ ﺑﻴﻦ ال Stirrups
ﺗﺰﻳﺪ ﻋﻦ d/2أي
. 40.625/2=20inإذاً ،ﻋﻨﺪﻣﺎ ﻧﺴﺘﻌﻤﻞ ﺣﺪﻳﺪ ﺗﺴﻠﻴﺢ #5ﺑﻤﺴﺎﻓﺎت اﻟﺘﺒﺎﻋﺪ اﻟﻘﺼﻮى ﺗﻜﻮن ﻣﻘﺎوﻣﺔ اﻟﺤﺪﻳﺪ ﻹﺟﻬﺎدات اﻟﻘﺺ ﺗﺴﺎوي: Av × f s 0.62 × 30,000 = = 46 psi s × bw 20 × 20
ﺑﻴﻨﻤﺎ ﻣﻘﺎوﻣﺔ اﻟﻤﻘﻄﻊ ﻹﺟﻬﺎدات اﻟﻘﺺ ﺗﺴﺎوي:
= v − vc
67 + 46 = 113 psi
أي إن ذﻟﻚ ﻣﻤﻜﻦ أن ﻳﻐﻄﻲ اﻟﻤﺴﺎﻓﺔ اﻟﻮﺳﻄﻴﺔ اﻟﺒﺎﻟﻐﺔ:
63 − 2 × 10 = 43 ft
أﻣﺎ ﺧﺎرج هﺬﻩ اﻟﻤﻨﻄﻘﺔ ،أي آﻞ ﻣﻦ اﻟﻤﻨﻄﻘﺘﻴﻦ اﻟﻘﺮﻳﺒﺘﻴﻦ ﻣﻦ اﻟﻤﺴﺎﻧﺪ واﻟﺒﺎﻟﻎ ﻃﻮل أي ﻣﻨﻬﻤﺎ ﻋﺸﺮة أﻗﺪام ،ﻓﺘﻜﻮن ﻣﺴﺎﻓﺎت ﺗﺒﺎﻋﺪ ال Stirrupsﻓﻴﻬﺎ ﺗﺴﺎوي: Av × f s 0.62 × 30,000 = = 15.5in ⇒ 15in (v − v c )bw (127 − 67) × 20
ﻻﺣﻆ آﻴﻔﻴﺔ ﺗﻮزﻳﻊ ال Stirrupsﻓﻲ اﻟﺸﻜﻞ. 14- وآﺎﻣﻞ اﻟﻤﺨﻄﻄﺎت اﻟﺘﺼﻤﻴﻤﻴﺔ ﻓﻲ اﻟﺸﻜﻞ. 15-
13
=s
14
15
-1-اﻟﻤﻠﺤﻖ
ﺟَـﺪول ﻣُـﻌﺎﻣِـﻼت ﺗـﺤـﻮﻳـﻞ وﺣـﺪات اﻟﻘـﻴﺎس Conversion factores, U.S. custuomary units to SI metric units
Spans
1ft = 0.3048m
أﻷﺑﻌـﺎد ﺑـﺼﻮرﻩ ﻋـﺎﻣــﻪ أﻷﻃﻮال
Displacements
1in = 25.4mm
أﻹزاﺣﺎت
Surface area
1ft² = 0.0929m²
اﻟﻤﺴﺎﺣﻪ اﻟﺴﻄﺤﻴﻪ
Volume
1ft³ = 0.0283m³ 1yd³ = 0.765m³
اﻟﺤﺠﻢ
Overall Geometry
Sructural Properties Area
1in² = 645.2mm²
Section modulus
1in³ = 16.39×10³mm³
Moment of inertia
1in1in4 = 0 .4162 × 10 6 mm 4
Loadings Concentrated loads
1lb = 4.448N 1kip = 4.448kN 1lb/ft³ = 0.1571kN/m³
Linear loads
1kip/ft = 14.59kN/m 1lb/ft² = 0.0479kN/m² 1kip/ft² = 47.9kN/m²
Stress and moments Stress
Moment or tourque
ﻣﻌﺎﻳـﺮ اﻟﻤﻘـﻄﻊ ﻋـﺰم اﻟﻘـﺼﻮر اﻟـﺬاﺗﻲ اﻟـﺘـﺤـﻤـﻴــــــــﻞ
Density
Surface loads
اﻟـﺨـﻮاص أﻹﻧـﺸــﺎﺋـﻴـﻪ اﻟﻤﺴﺎﺣﻪ
أﻷﺣﻤﺎل اﻟﻤُـ َﺮ َآﺰَﻩ اﻟـﻜـﺜـﺎﻓـﻪ أﻷﺣﻤﺎل اﻟﺨَـﻄﻴﻪ أﻷﺣﻤﺎل اﻟﺴﻄﺤﻴﻪ
أﻹﺟﻬﺎدات واﻟـﻌـﺰوم 1lb/in² = 0.006895MPa 1kip/in² = 6.895MPa 1ft.lb = 1.356N.m 1ft.kip = 1.356kN.m
16
أﻹﺟـﻬـﺎد
أﻟﻌـﺰم أو أﻹﻟـﺘـﻮاء
اﻟﻤﻠﺤﻖ-2- ﻗﻴﺎس وﻗﻄﺮ وﻣﺴﺎﺣﺔ ﻣﻘﻄﻊ وﻣﺤﻴﻂ ووزن ﻗﻀﺒﺎن ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ )(units U.S Unit weight
Perimeter
c/s Area
Diameter
Bar No.
lb/ft
in
in²
in
#
0.167
0.79
0.05
¼
2
0.376
1.18
0.11
)(3/8
3
0.668
1.57
0.2
½
4
1.043
1.96
0.31
)(5/8
5
1.502
2.36
0.44
¾
6
2.044
2.75
0.6
)(7/8
7
2.67
3.14
0.79
1
8
3.4
3.54
1
)(9/8
9
4.303
3.99
1.27
)(10/8
10
5.313
4.43
1.56
)(11/8
11
7.65
5.32
2.25
)(14/8
14
13.6
7.09
4
)(18/8
18
ﻗﻴﺎس وﻗﻄﺮ وﻣﺴﺎﺣﺔ ﻣﻘﻄﻊ وﻣﺤﻴﻂ ووزن ﻗﻀﺒﺎن ﺣﺪﻳﺪ اﻟﺘﺴﻠﻴﺢ )(SI units Bar No.
ﻋﺪد اﻟﻘﻀﺒﺎن )ﻃﻮل 6ﻣﺘﺮ( ﻓﻲ آﻞ ﻃﻦ
وزن اﻟﻘﻀﻴﺐ
اﻟﻮزن ﻟﻜﻞ ﻣﺘﺮ
اﻟﻤﺤﻴﻂ
ﻣﺴﺎﺣﺔ اﻟﻤﻘﻄﻊ
اﻟﻘﻄﺮ Ø
ﻋﺪد
kg
kg
mm
mm²
mm
666
1.5
0.25
20
32
6.3
2
297
3.36
0.56
30
71
9.5
3
166
6
1
40
129
12.7
4
107
9.3
1.55
50
200
15.8
5
74
13.41
2.23
60
283
19
6
54
18.25
3.04
70
387
22.2
7
42
23.84
3.97
80
509
25.4
8
33
30.36
5.06
90
645
28.5
9
26
38.42
6.4
100
819
31.7
10
21
47.44
7.91
110
1006
34.9
11
≈ 14.6
68.3
11.38
135
1451
44.4
14
≈ 8.2
121.42
20.24
180
2580
57.1
18
17
أﻷﺣﻤﺎل اﻟﻌﺴﻜﺮﻳﺔ: ﻣﻦ اﻟﻮاﺿﺢ إن اﻟﺘﺼﻤﻴﻢ اﻟﺴﺎﺑﻖ ﻟﻠﺠﺴﺮ ﺗﺤﺖ ﺗﺄﺛﻴﺮ اﻟﺸﺎﺣﻨﺔ اﻟﻘﻴﺎﺳﻴﺔ HS20اﻟﺘﻲ أوﺻﺖ ﺑﻬﺎ اﻟﻤﻮاﺻﻔﺔ اﻟﻘﻴﺎﺳﻴﺔ AASHTOﺗﺠﻌﻞ اﻟﺠﺴﺮ ﻗﺎدرًا ﻋﻠﻰ ﺗﺤﻤﻞ ﻣﺮور أﻷﺣﻤﺎل أﻻﻋﺘﻴﺎدﻳﺔ ﺑﺼﻮرة ﻻ اﺳﺘﺜﻨﺎﺋﻴﺔ ﺛﻘﻴﻠﺔ أﻣﻴﻨﺔ واﻗﺘﺼﺎدﻳﺔ .إﻻ أن هﻨﺎك ﺣﺎﻻت ﻳﻜﻮن اﻟﺠﺴﺮ ﻓﻴﻬﺎ ﻣﻌﺮﺿًﺎ ﻟﻤﺮور أﺣﻤﺎ ً آﺎﻷﺣﻤﺎل اﻟﻌﺴﻜﺮﻳﺔ اﻟﺘﻲ ﺗﺠﻌﻞ اﻟﺠﺴﺮ ﻳﺘﻌﺮض ﻹﺟﻬﺎدات ﻗﺎﺳﻴﺔ ﻗﺪ ﺗﺆدي ﻟﺘﻀﺮرﻩ أو اﻧﻬﻴﺎرﻩ .ﻟﺬا أوﺻﺖ اﻟﻤﻮاﺻﻔﺔ اﻟﻘﻴﺎﺳﻴﺔ اﻟﻌﺮاﻗﻴﺔ ﻟﻠﻄﺮق واﻟﺠﺴﻮر ﺑﺎﻋﺘﻤﺎد اﻟﻤﻮاﺻﻔﺔ اﻟﻘﻴﺎﺳﻴﺔ اﻟﻜﻨﺪﻳﺔ ﻟﻸﺣﻤﺎل اﻟﻌﺴﻜﺮﻳﺔ اﻟﻤﻤﻜﻦ ﻣﺮورهﺎ ﻋﻠﻰ اﻟﺠﺴﻮر .ﻳﻮﺿﺢ اﻟﺸﻜﻞ أدﻧﺎﻩ أﻷﺣﻤﺎل واﻷﺑﻌﺎد ﻟﻠﺸﺎﺣﻨﺔ اﻟﻌﺴﻜﺮﻳﺔ اﻟﻘﻴﺎﺳﻴﺔ Class 100واﻟﺘﻲ ﺳﻴﺘﻢ ﺗﺪﻗﻴﻖ ﺗﺤﻤﻞ اﻟﺠﺴﺮ ﺗﺤﺖ ﺗﺄﺛﻴﺮهﺎ:
Military Bridge Loading Standard Vehicle Class 100
Impact Factor
)Span of Concrete Bridge (ft
0.40
0
0.39 0.37 0.36 0.32 0.28 0.24 0.12 0.00
6 20 33 65 98 130 164 More than 164
18
Deck Slab check: Md = 412 ft.lb 5.33 + 2 × 30,000 = 5497 ft.lb ML = 0.8 × 32 I = 0.32 MT = 412 + 1.32 × 5497 = 7670 ft.lb
For Military Loading, it is allowed to overstress materials as follows: +25% for Steel +33% for Concrete f s = 0.75 × f y = 0.75 × 60.000 = 45,000 psi f c = 0.73 × f ' c = 0.73 × 5,000 = 3650 psi n = 7.2 45,000 r= = 12.33 3650 7.2 k= = 0.37 7.2 + 12.33 0.37 j = 1− = 0.877 3 2 × 7670 × 12 d erq. = = 3.6in 3650 × 0.37 × 0.877 × 12 5 d ava. = 8 − 1 − 1 − = 5.6in ≥ d req 3.6in ⇒ O.K 16 7670 × 12 As = = 0.42in 2 / ft 45,000 × 0.877 × 5.6 0.31 = 0.75 ft = 8.9in ⇒ 8in Spacing = 0.42
Use G60, #5@8´´c/c, T&B By using the maximum secondary reinforcement of 67% of the main reinforcement, spacing of #5 bars can be calculated as follows: 8.9 = 13.3in 0.67
Use G60, #5@12´´c/c
19
Interior Girder check: Md = 917,000 ft.lb
63 RA = 28 , 000 ( 9 + 14 ) + 42 , 000 ( 34 + 37 ) + 21 , 000 × 49 ⇒ RA = 73 ,888 lb ML = RA × 29 − 21 , 000 × 15 − 42 , 000 × 3 = 1, 701 , 777 ft .lb MT = 917 , 000 + 1 . 32 × 1 , 701 , 777 = 3 ,163 , 345 ft .lb
Girder dimensions check: Vd = 58,275lb
VT = 58,275 + 1.32 × 102,111 = 193,061lb 20
'
v = 1.33 × 2.95 f c = 277 psi VT 193,061 = 696in 2 v 277 696 d req. = = 34.8in 20 10 10 d ava. = 48 − 1 − 2.5 − − 2 − = 40.625in ≥ d req. 34.8in ⇒ O.K 8 16 3,163,345 × 12 As = = 22.722in 2 7 45,000(40.625 − ) 2 As 22.722 = = 17.89 ⇒ 18bars 1.27 Ab
bw d =
Use 18#10, three layers
Concrete compression check: fc act . =
3,163,345 × 12 7 (1 − ) × 12 × 7 × 7 × 0.837 × 40.625 2 × 0.37 × 40.625
Shear reinforcement check: Vd 3.3 58,275 = ⇒ Vd 3.3 = 52,170lb 31.5 − 3.3 31.5 Vd15 58,275 = ⇒ Vd15 = 30,525lb 31.5 − 15 31.5
According to Fig. 19, According to Fig. 20,
VL 3.3 =100,122 lb VL 15 =71,778 lb
VT3.3 = 52,170 + 1.32 × 100,122 = 184,331lb VT15 = 30,525 + 1.32 × 71,778 = 125,272lb vc = 1.33 × 0.95 f c′ = 89 psi
21
= 2475 psi ≤ fc all . 3650 psi ⇒ O.K
By using the maximum allowed spacing of d/2=20in, the steel #5 bars will resist: v − vc =
Av × f s 0.62 × 45,000 = = 70 psi s × bw 20 × 20
Then the section shear resistance will be 89+70=159 psi It is clear that the maximum allowed spacing of 20in will be enough to cover all the intermediate distance up to 15ft from each support. The remaining two distances -15ft from support- will be provided with #5 stirrups spaced at: s=
Av × f s 0.62 × 45,000 = 10.1 ⇒ 10in (v − vc ) × bw (227 − 89) × 20
Use #5 U Stirrups@10´´c/c 22
23
24
