Question

Aluminum Beam: E=69GPa, Dimensional Length(1)=250mm, Width(b) = 19.11mm, Height(h) = 5 mm Moment of Inertia(I) = 1.996875E-10 m^4 Mass(g) Force(N) 100 200 Experimental Deflection(mm) 3 0.46 0.44 0.46 0.98 0.95 1.49 1.52 1.97 2.02 1.92 2.3 2.16 2.07 0.981 1.962 2.943 3.924 14.905 Average 0.45333 0.97667 1.47667 1.97 2.17667 Std. Dev. 0.0115470055 0.025166115 0.051316014 0.05 0.115902258 | 300 400 | 500 1.42 Steel Beam: E = 207 GPa, Dimensional Length(1)=250mm, Width(b)= 19.4mm, Height(h) = 4.72 mm Moment of Inertia(I) = 1.6649e-10m^4 Mass(g) Force(N) Experimental Deflection(mm) 100 200 0.981 1.962 2.943 3.924 4.905 0.16 0.54 0.61 0.84 1.16 300 400 0.17 0.45 0.67 0.83 1.17 Average 0.16333 0.46 0.63 0.84333 1.16 | 0.16 0.39 0.61 0.86 1.15 Std. Dev. | 0.005773503 0.075498344 0.034641016 0.015275252 0.01 500

pls help me calculate the theoretical deflection and Table 3.3 Thank you!

Answer 1

Given - Aluminium peint Load Beam Gipc = m E =69 Gaxio" ni im? = I Į P 0.250 m - 2.50 mm = 19.11 mon = 0.01 m. - Smer (SSB) H El.49 *10-10 mm at Find the Theoritical Defrection. w Assumptions are made : a) Neglecting self weight. I the Load as point load 1918919050 Assumed at Mid Assumed to be point. to be a) a simply supported beam. Solution Lood can be varied as per Question. . دن 48EI For al cases. = 0.125 m. E = baxtoa wima I=1.99 x 100 mm 1009 Lood S = = 0.00 24 mm 0.981 x 0.125 48 x 69 xic x1.95 X10 For 2009 Load S= = 0.0058 mm cab2 0 .1253 48 x bax100 x 100 x 10-10 300g Lood S= 2.943 x 0.1253 46 x 46 xe° x 1.8 x o to = 0.00821 mm 4009 Load S= 3.924 x 0.1253 48x Ga xioaxriga xiolo c.olib mm Soos Load Hisos X C125 S= 48xbaxol x a.qa x 1010 - 0.0145 mm

Given - steel Point dad Beam I { FM E = 204 Gapa - 202x109 Nina = 250 man = c.rsom 19. hmon = 0.019h m e H = 4.32 mm = o.ocha m... I - 166 Xoom some 3 Assumption used in Last case solution : Leed on steel Bece can be Neried based on Question. S = wd3 For = 0.981 *011253 1000 Load Defection = 0.00 116 mm 20 x 1.66 X10-16 For 2005 Load S = 1.962 * 0.1253 48 * 200 x 109 xlitt x 10-10 0.00232 mm For sceg Lead Sa 2.943 x 0.125² = 0.003418 mm 48x202 x 109 x 186 x 10-10 For hoog Lead S = 3.924 x 0.1253 48x207 x 109 x = 0.0046 mm litto 410-10 = 0.00 580 mm for hiros XC.1253 Soog Load S 48x207 xio x 166 x 10-10 Deflection theoritical $ fifth e$$ or These an those given beams.

For simply supported Beam having a point load Structural stiffness Formula I K = Load бmons. For Aluminium. At Boeg load. - At Exp condition. k = 2.943 - 2,002 Nim. M67 1-12 x10-3 Steel - At Eap Condicon. Ale 300g Loud k = = 46so Nlan. 2.94 0.63*10-3 Then for thecritical condition. For Aluminium. At 2009 - K = 2.943. = 33788 m. 0.00821 x10-3 sleel At 300g K = 2.943. = sus68 Nm. 0.00348 x1-3 percentage differcoce Aluminium = % = = 33788 - 2002 1.9ay. 133288 +2002) 138288 42001) — 2 steel = Y = Sus68 - 46so 1991 seet i Busus auuso) *** 7._ I snobs & Libso 2