Question

For the beam and loading shown, and knowing that distance a = 2m, determine the maximum value of the distributed load W so the deflection at midpoint C does not exceed 5 mm Use E = 200 GPa and I_x = 333 x 10⁶ mm^4.

^{D B D А E}

Answer 1

Solution :- Deflection at ci using unit load method :- w: KN m D е A B 10 By ) Tourname 2m Dm 12m 12m 10 W=0. - 86 s by=2w. Beam Subjected to entional load, <Mg =0 = -0(2300) –Dy(4)+w(2)(5) = 0. -20 4 byt 4Dy = sfy=0 By + Dy = = qw+20 > By = 26. for the Span ED (x: from E) : Sign conuinhon- Ma = - wx2 Sagging : Hogging DC (2: fomD) : M₂ = - W(2)(x+1) + Dy (2) - 2wx-2wthwa for the Span AB (2: from A): - My = -022 tur KN.m ave . 2 for the Span - -2WKN-m KN.m for the Span Be (z: from B).- Mx = -6(1)(2+) + By(x) Ma = n2o. kw.m.

unit load at ci- Apply B € ton Dy 2m 2 ന 2m 7 At equili bolum ; By = 0.5; Dy =o. for the Span ED (2: from E): ma=0. for the Span DC (2 : from D) : - ma=0.50m for the Span AB (2 : from A): for the Span Be (2: from B): Mn = 0.50m. ma=0 deflection at ci- Ac= ED DC AB BC S Ma ma olen + Materiali e Ma maisha 1.2005 2 ) dx a 1.5x) dr EI o O suosad es [ Jew)(050)] 2 EI 2 2 - 2 22 -20 EI ada → EI 2 ET M ma dx 0 Ac -40 (kN-mym EI KN-m2

to the indicates that deflection is unit load opposite direction of Given Ac = 5mm 3 5*10 m. E = 200 Gpa = 200*107 W/m2 I= 333 * 10%(103) Fony > 333* 10 m 4. EI= 66600 KN m² 40 EI Aci 3 5*10 EI 4 W = 5*103 66600 w 83.25 KW/m.