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Mechanics of Structures 2 AssignmentYX8 metres3 metresUniform Load 8kN/m2 metres3 metres40kNStudent Section Allocation TableStudentOnesteel Sections UB or UCKhurram Shahzad 1200980360 UB 44.7For the beam assigned to you solve:a) Find the reactions (10 points)Solution;Consider the vertical equilibrium of at both ends say A & B∑Fy=0, i.e the total sum of both upward and downward forces,We take reaction at both ends to be RA& RBrespectively.tmp280vst6h2247406_1980657652_ORDER728324.docx

Upward force = downwardRA+ RB= UDL + point load,RA+ RB= (8 x 3 + 40)In order to get the reactions at both ends we have to take moments at one end.Taking moments at point A we have,∑MA=0,Please note that the UDL acts at the centre of its span i.e 3/2=1.5∑MA= (8 x 3x 1.5) + (40 x 5) – (8 x RB) =0Rearranging we get,8 x RB= 236KNMDividing we get, RB= 29.5KNRA+ 29.5KN= (8 x 3 + 40)RA= 64-29.5RA= 34.5KN,Hence reactions at both ends are 34.5KN & 29.5KN respectively.b) Draw the shear diagram (10 points)Solution;The shear force at both ends i.e FA& FB=0. To draw the shearforce diagram get the sum of theforces acting at a given distance of the beam. Vertical forces are assumed to be positive whiledownward forces are negative.At Point A The shearforce =34.5KNSince the UDL acts at the centre of its span at 1.5m the sum shearforce= 34.5 -24=10.5KNAt 5m from A the sum of shearforce =10.5KN- 40KN=- 29.5KNAt point B sum of Shearforce = -29.5+29.5=0tmp280vst6h2247406_1980657652_ORDER728324.docx

c) Draw the moment diagram (15 points)Solution;Bending moment at pointx= algebraic sum of moments of forces at a distance x to one side of asection.At point x from point A=3m,∑M3m= (34.5 x 3) – (8 x 3 x1.5) = 67KNMAt point xfrom point A=5m,∑M5m= (34.5 x 5) – (8 x 3 x3.5) = 88.5KNMAt point x from point A=8m,∑M8m= (34.5 x 8) – (8 x 3 x6.5) – (40x 3) = 0KNMtmp280vst6h2247406_1980657652_ORDER728324.docx

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