Question

Select the lightest W shape to support a uniformly distributed load of 80 kN/m and concentrated loads of 26kN located 4.5 m from the end of the beam. The beam is 16 m, Deflection is not to exceed span/280.

Answer 1

Draw FBD of beam 12 Belen 26kw V w Solm mmmmmmmm lec k 40 sm sm Ks = 64-512 Ka = 5s-675 16 m The vou get replaced by roint load which is equal to Goxie = 1280IN, act at wil span of beam Novo se eqน้นกางๆ (4 97367-O -> R4 + RB-26- 1280 40 » FA + Rs = 1306 - 0 EA MA=0 7 AS (16) - 26x4. s-1280x6=0 1 RG = 647.3125 kW & from eano stks = 1306 hat 6473125 - 1306 The - 658.6975 lei

Now we deplop to deflecrion eq? M2x72 658 05975 <*> - 26 {æ-4.57- 80 <u>2 04221cm El d2y = Mau dx2 Ez dyldne = 658-6575 <2}}- 24 42-453-go <«/'+9 - €2y = 658-6875422 - 24.27-.57?- dey 2x3 and q and C₂ are deflection constant quel Cay be Calculated based on boundry condition at x=0 YAO x=lom from ean 6 02 658.6675203 - 24 <0-4.57%- 50 <03449(0)75 (2=0) { and we value inside of a will be zero} Bao 24 from ean G 03 659.687510– 24216-4.53 - 2165€ + GLIC) G2-14038.76 cenama so deflection ean . ERY = 658-6675 223 - 242x=4.573- 22414038763 slope ean Ez dyffre = 6586875 4x} – 26 <x-4053 -8% <x#1403.87%

we need to define first maximum strope deflection at maximum deflection slope will be zero let it occurs b/w cand (3) El duplo Ez olyan = 658.6875 <x, 26 <x-457-80 <=-14038.76 0 - 656,6275 4332- 24 <«°+52_927 - E46 x36-44038.76 0 = -80n + (658-6875 -2.3 x2 + (9x24]* - 26 x4o52 can we - $ 14038.76 solve above ean with help of calculator found x = -boly m love valle is not possible) 2 21.88 m (it is more then lom3 not possible x₂ = 7097 m ( 405 arciom) cal Then R3 = 7.97 m maximum de fiectm ocors Exy 3652.6875 39. RUHÝ-bokry Ely = 656.6875 <a - 26.<-454- so 2x4++G 3 at x = 7097 m El Yma a 65 6675 27.077- 26 27.97-4.57-80*70943 14038-76 * 7.97 +O Y max = [- 694940 0262] m

del cellow) = -4260 (Shes = 16 (_m) T allow = -0.05714m] 280 Ymax = (d allow = – 69740.262 =-0.05714 m E ? Es 200 x 105 od lew] m2 +69740.262 200 x 106 x I = 70.057/4 D ESI = 6:10 257 x1ős mos I - 6102057 xiul mms ] Now use Appendix go alony In Column and Select a value at couich • cohich is just greaser then Ix = 6102.57 x 100 mm we found that I = 780 x ifmmy w 36x 135 so we get w 36x135 beam uns