تصميم الجسور ذات الروافد الكونكريتية المسلحة

‫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

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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

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= 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

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