Kafes kirişlerinin sehim ve atalet momenti üzerine bir etüt
A study on the deformation and the inertia of the truss beams
- Tez No: 39337
- Danışmanlar: PROF. DR. HALİT DEMİR
- Tez Türü: Yüksek Lisans
- Konular: İnşaat Mühendisliği, Civil Engineering
- Anahtar Kelimeler: Belirtilmemiş.
- Yıl: 1993
- Dil: Türkçe
- Üniversite: İstanbul Teknik Üniversitesi
- Enstitü: Fen Bilimleri Enstitüsü
- Ana Bilim Dalı: Belirtilmemiş.
- Bilim Dalı: Belirtilmemiş.
- Sayfa Sayısı: 103
Özet
Bu çalışmada 2na açıklığında ve h yüksekliğinde ki kafes kirişte, P kuvvetinin etkidiği nokta ve senimin arandığı nokta değiştirilmek suretiyle önce sehimlerin formülleri bulunmuş ve daha sonra dolu kiriş için aynı yük durumundan dolayı elde edilen sehim formülünden atalet momenti ifadesi çekilmiştir. Senime bağlı olan bu ifadede kafes kirişin sehim formülü kullanılarak kafes kiriş için atalet momenti ifadesi elde edilmiştir. Daha sonra kafes kirişin sadece alt ve üst başlık çubuklarından oluştuğu kabul edilerek atalet momenti ifadesi : I=Au*hu2+Ao#ho2 I=A*h**?u*?o olarak elde edilmiş ve genel ifadeden buna bağlı olarak bir düzeltme çarpanı ifadesi elde edilmiştir. Burada A ait ve üst başlıkların alanlarının toplamı olup ?u alt başlığın alanının toplam alana
Özet (Çeviri)
In this study, a simple truss beam which has a length of l=Ena and a height of h is considered. One can find the deformation of any joint when a force P is acted on any point by formulating the“i”th bar force with cutting method. The sectional areas of the verti cal, diagonal, upper and lower bars; the distances of the force P and the Joint of which deformation is wanted to be formulated are the variables in the formula. We already know the formula of the deformation of the full body beam at the same loading case, before. One can find the inertia of the beam in term of deformation from this formula. If one put the deformation formula of the truss beam found before for the same case the approximate inertia of the truss beam will be found. For example, when the force P is act on the top of the“n”th joint of the truss beam and the deformation of the“n”th joint is wanted to be found, as figure 1. *~ Ui l=Ena The bar forces are found with cutting method as folowings; For P loading; Vi=-P/E (i: from 1 to n, and V X ' =h/Av+d/ < Ad*Sin*£i ) Y'=a*Cot2a*< < (n-1 )* and hu=h*?o (3) By using equations (2) and (3) in the equation I=Au#hu2+Ao*ho* The inertia is foun as: I=A*h2*?u*?o The inertia of the truss beam will be as: I=A*h2*?u*?o*. D.C. is correction factor in this e- quation and it is; D.C.=Cot a/(3#?o*eu*(X£+Y£)) X2=U/ < k/En ), k(r/En> change. ' By using the results from these tables, some graf- fics were drawn on computer and it is possible to decide which bar's area is more appropriate to increase at any a. The results of ZV, ZD, ZO, ZU in the table are mu İtip led by 100 and divided to the heigth of the truss beam (h). So to find the real results they should be divided to 100 and multipled by h. One can calculate the deformation of any joint of the truss beam when the force P is acting on any Joint by using this tables just like that: If one wants to know the deformation of the thirth joint on the left of the truss beam, when the truss beam has a length of l=S*a (i.e. n=4), the force P is acting on the top of the“n”th joint and the truss beams is corresponding to type of I truss beams, he can use the table la. 3 after preparing it by the means of the corresponding computer program by taking N=4, R=3. N is refer to n, R is refer to r in the program. The sectional areas of the bars is known at the beginning. For example, if the sectional area of the u bars and the o bars is E units, the sectional area of the d bars is 1 unit and the sectional area of the v bars is 0.5 units, A (Au+Ao) will be 4 units, ?o and ?u will be E/4, B will be 0.5/4, and tf will be 1/4. The program will want N corresponding n, R corresponding r, B correspon ding B and G corresponding "6. After he gives that values to the program, it will prepare a table with respect to a in vertical and E0 corresponding ?o in horizontal. The slope of the diagonal bars a is also known at the beginning with respect to a one unit of the length of the truss beam, and h the height of truss beam. If a is equal to 50, he should take the correction factor DC from this table where ?o=0.5 and a=50 intersect each other. The inertia of the truss beam in this case will be equal to I=4*h2#0.5*0.5*DC. ( Remember I=A*h2 #?o*?u*DC ). For this case the equation between the inertia and the de formation is like that: (5 shows the deformation, I shows the inertia) £=P*r*(3*12-4*r2 )/(48*E*I ) The deformation of the thirth joint of this truss beam can be calculated from this equation. At this casethe contribution of which bar is the biggest can be seen from the table. That will help him to decide to use the additional material to which bar. If one wants to see how the contributions of the bars change with respect to the slope of the diagonal bars to the horizontal (a) he can use the corresponding graffics. For example for the case mentioned above the graffic la. 3 shows how the contributions of the bars change with respect to a. The same procedurs were done for the same truss beam when a M momentum is act by changing it into twin forces M/h. And first degree hyperstatic truss beam is also considered in the same way.
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