Question

A square wood platform of side-length ? = 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, ?? (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.

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 W Figure Q2 (a) Show that the maximum bending moment of a simply supported beam under a uniform load, w, is wĽ2 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

Answer:

ܥܐ > Foreo body diaggiam; a ) Intensity of locada A 3 L RA RBI a) EFy=0 RATRB = WL => Moment about about A, EMA = 0 RB XL WL x 1 / 2 = 0 в with ابط RB RA = Who 2 WLA BMD B 늘 >> Taking section 2e-x from rightend, 1x Heuve AWL 'ye xr -TW >> Moment on this cross-section, My-z = Wh xe + C-wax e x=L

2 - W2 x + (- W2²) -) Differentiate above expression with respect to ze and equating to zero. d Mge-z de WL + C-wa) - 24 >) Bifferentiate again to check minimaoy maxima, d² Moze dx2 Moment of a = ger Moese = -1/2 forom right is maxima, moment at - / ho na +(-42C+2) WL? 빻 8 moment kwam@r) Nam based on unit of wand L. WL2 h (M27), zu12 งบก .. Maximum bending b ) H=15om 14-37.5 N =75mm I bottom K B:100mm 3

3 3 sto X 100 X (150) INA = 281205000 mm 4 =) Platform, KN m2 Alea of platform.= 12 2400*2400 mm² 2.4x2.4 m2 L? = 5.76 m 2 =Total weight of plot-forms, WA (5.76) KN. Load thed each beam will experience is the reaction ho kn. The load per unit length is, w - W from beam wah 2L w WA (8.76) (1 2*2.4 W = 1-2 WA KN m. Jo, the maximum bending moment is, Wh 2 8 1-2 WA C2.42 My= 1/2 Mae Ha = 0.864 WA KN-m

I c) Maximum tensile stress will occur due to maximum bending moment at se z 4 at Bottom fibre. Trealiz < (open) tensile. Merle I bottom INA =17 0.8GHWA 4103 x 103 x 75 28125000 =17 WA A 7-378 KN ma w = 7.378 KN/m2 d 발 tve -ve - WL 발 The maximum shear force is at supports and they the maximum shear stress, Tamax < Iper * Brax FA INA 6 <tper Wh IL x (100 x 75)37.5 = 0.7 28125000 X 100 1.200A x 2.4x 103 x 100% 75 *37.5 2 = 0.7 28125000x100 4,861 kN WA mi 48 61-11 lo 3 W2 = 4.861 kN/m2
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