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Shear Joints under eccentric loads Joints can and should be loaded in shear so that the fasteners see no additional stress beyond the initial tightening.

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Presentasi berjudul: "Shear Joints under eccentric loads Joints can and should be loaded in shear so that the fasteners see no additional stress beyond the initial tightening."— Transcript presentasi:

1 Shear Joints under eccentric loads Joints can and should be loaded in shear so that the fasteners see no additional stress beyond the initial tightening. Shear loading is resisted in two principal way The shear load is carried by friction between the members and ensured by clamping action of the bolts or cap screws. The shear load is carried by dowel pin in reamed holes, places in both parts while clamped together to ensure alignment. The dowels will carry the shear load.

2 1.Integral to the analysis of a shear joint is locating the centre of relative motion between the two members and finding the total load taken by each bolt i.e the centroid. Use statics to find the centroid G of bolts or rivet group. 2.To find the total load, we need to perform 3 steps; a)The shear V is divided equally among the bolts so that each bolt take F’ = V/n where n=number of bolts, F’ is called direct load or primary load. b)The moment load, or secondary load, is additional load on each bolt due to moment M. The moment M and moment load F” are related as follows: M = F” A r A + F” B r B + F” C r C + …. Since the bolts farthest from the centroid takes the greates load, while the nearest bolt takes the smallest load, then F” A / r A = F” B / r B = F” C / r C F” n = Mr n / r 2 A + r 2 B + r 2 C + …. 3.The direct and moment loads are added vectorially to obtain the resultant loads. Basically two things need to be done in analizing shear joints.

3 Since all the bolts or rivets are usually the same size, only the bolt or rivet having the maximum load need to be considered. When the maximum load is found, the strength may be determined; that is the diameter/size of the fastener.

4 Gambarajah di bawah menunjukkan satu bar keluli segiempat 15mmx 200mm yang dipasangkan terjulur pada satu alur keluli 250mm menggunakan 4 bolt. Untuk daya F = 16 kN, tentukan 1.Daya paduan(resultant load) pada setiap bolt 2.Beban maxima pada setiap load 3.Tegasan galas(bearing stress) maxima

5 Solution Data Daya F = 16kN, bolt and nat M16 x 2 Oleh sebab kumpulan bolt berada dalam simmetri, maka sentroid terletak pada titik O seperti yang ditunjukkan dalam rajah. Sekiranya gambarajah badan bebas rasuk dibina, maka daya reaksi ricih V akan melalui titik O dan momen reaksi M akan pada O. Maka daya-daya reaksi adalah V = 16kN dan M = 16kN(425mm) = 6800N.m Jarak antara sentroid ke pusat setiap bolt adalah sama(sebab simmetri) yakni F’ = V / n = 16kN/4 = 4 kN Since the secondy shear force are equal, therefore F” = M r / 4 r 2 = M / 4 r = 6800Nm / 4 (0.960m) = N = 17.7kN Dengan memplotkan daya-daya ricih secara berskala, didapati daya paduan setiap bolt adalah ditunjukkan oleh gambarajah berikut. Magnitud daya paduan bolehlah diukur dari gambarajah.

6 Gambarajah daya paduan setiap bolt. Penyelesaian untuk mencari daya paduan boleh dilakukan dengan cara grafik(graphical method) atau vektor. Nilai magnitud daya paduan setiap bolt adalah berikut: F A = F B = 21.0kN F C = F D = 13.8kN

7 b) Bolts A and B are crictical because they carry the largest shear load. The bolt length will be 25mm plus the height of the nut and plus 2mm the height of a washer. The height of the nut is given by Table E-29 as 14.8mm. This add up to a length of 41.8mm, hence from Table E15, a bolt of 45mm long will be needed. From Table 8-6 or Table 8-9, thread length is L T = 2(16) + 6 = 38mm Thus the unthreaded portion of bolt is 45 – 38 = 7mm. This does not exceeds the 15mm for the plate, so the bolt will tend to shear across it threaded length; that is the tensile-stress area of bolt A t =157mm 2 (from Table 8-1). Thus the shear stress is  = F / A t = 21.0 x 10 3 N / 157mm 2 = N/mm 2 = 134 MPa c) Since the channel is thinner the bar, the largest bearing stress is due to the pressing of the bolt against the channel web. Therefore, the bearing area is A b = t x d = 10mm x 16mm = 160mm 2  = F / A b = x 10 3 N / 160mm 2 = -131 MPa.


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