Question

Question 2 (25 marks) A square wood platform of side-length L = 2.4 m rests on masonry walls. The deck of the platform is constructed of 4 cm thick planks supported on two beams which are 2.4 m long. The beams have a width of 100 mm and height of 150 mm and are supported at their ends by the walls. The structure is designed to support a uniformly distributed load, wA (kN/m2), acting over the entire top surface of the platform. The tensile strength of the planks is 17 MPa and the shear strength is 0.7 MPa. Assume that boundary reactions of the planks are uniformly distributed over the top surface of the supporting beams, instead of concentrated loads. Ignore the self-weight of the planks. L L Wall Figure Q2 (a) Show that the maximum bending moment of a simply supported beam under a uniform load, w, is įwl where w is the distributed load per unit length (N/m) and L is the length of the beam. (5 marks) (b) From the above question (a), determine the maximum bending moment when each boundary reaction is uniformly distributed over the area of the contact. Use a to denote the width of the supporting beams. (10 marks) c) Determine the allowable loads, w, (kN/m2), based on the tensile stress in the planks. (5 marks) (d) Determine the allowable loads, wz (kN/m2), based on the shear stress in the planks. (5 marks)

Answer 1

WL load intensity w Х 니지 42 RB Here RA TRB = UL RAA RB - x X taking moment about A - Vx WE XL + RB XL=0 - we 2 RB WL = Z ) la=L x=0 so, ap RA - OL - 2 wel RA=&B= WL 2 Mx 4 Mmay = we Taking section xx Vx = 2-o RA- wa x= 4/2 al wa 2 for Maxt mum Bending Moment Vx=0 WL - wx=o=Vx 2 ? 2 (Ma) mayo at x = 1 / 2

(Mmax) Area uffo 4/2 in shear Force Diagram either of left side Right side of x = 6 Area - BM) max = (BM) max Wh2 Answer a) 8 Another way М. ie to find (BMI max is to form an equation as fello we below Mx RAX - Wx.a 2 를 Mx = wla Wa? 2 2 2 At n = 1 = Meie = whe) - with My2= st 2 It 2 wq2 = Mani 8 4 19 2 9 (LXL) Noul dimension of plat from a Area = (2.4) ² = L2 5.76 m² weight should support Area & WA 5.76 m² WA

say W = 5.76 WA KN W/2 W12 w 34 ale * these 180 since, Reactions two beamp are these total footh are generated, equal in magnitudes ie w Now Maximum load in 1.44 WA KN W 4 576 WA. 4 0.6 WA KN load per mit length 1.44 WA L |: Чч Liu A 2.4 (w) We 2 Now Maximum Bending coment 8 (6.6 WA) (2.4) 8. kim Mmax. = 0.432 WA Now By Bending equation Mmax o max - с Z stres & mar- Maximum Bending d bd² Z- 6 section Modulus of Beam

Here be loomm, d = 150 mm ro(for mmm, kN, AN) 0.432 WA x lol omayo. = 2 17 (100 x 150) 2 WA < 17 xloox 150 68106 X 0.432 WA < 14.7569 | RA ma Maximum shear force En Beam is (W) = Umar 4 Since there are two beama loomm + Tmax (Viman) (AJ) I a 75mm ΝΑ 75mm bd3 12 toox (150) 3 mm I = NA 12 150 - a= loomm, g = 75mm 2 A= 150x100 2 7500 mm² (half area ine Area above Neutral axis) (7500) x 75 < 0.7. 100x (150) 3 x 100 Tmax = (102) 12

W - - 6 won 0.6 WAX 103 x 2.400 < 0.7 X 7500 x 75 x12 loox (150) xloo UA 2 и 2 uz = 90722 KN m?
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