Question

The W33 x221 steel simply supported beam is loaded with concentrated loads and uniform

load as shown with the load P= 150kip and w = 10kip/ft. For this beam do the following;

a) Draw the shear and bending moment diagram

b) Calculate the maximum compressive and tensile stress

c) Calculate the maximum shear stress

P Р 3 ft 3 ft w 10 ft

Answer 1

Sum of moment at A=0

Rb × 10 = 3×150 + 7×150 + 10×10×5

Rb = 200 kip

Sum of forces in y direction =0

Ra + Rb = 150+150+10*10

Ra= 200kip

15014p Isokie PO KIDIHT.) F. 1 A Т- — 10 E С D RA= 200 kip RB = 200 kip SFD.. («iP) (о 20 с A В о 1 0 -20 Sz S 555 SS BND Kip-ft E А e с юм. – Вче = 900x 3 — 3ҳір 3х10х3 Вмр: 33t <<p - Зме в 2008 $ -150 2 - 108 Çх» к BMO BMR = S75 kip-ft

Geometrical properties of W 33*221 beam -

T W33X221 65.2 in 2 A= da 33.9 in. tw = 0.775 in. bf = 1 15.8 in. 1.28 Jin. 2.06 lin. tf = k(des) = k(det) = k1 = Jin. 2.125 1.1875 in. TE 29.625 in. 5.5 in. plf. 221 gage = wt./ft. = bf/(2*tf) h/tw = 6.2 38.5 12900 Ix = in. 4

Area = 65.2/144 = 0.4527 ft²

Ix = 12900/20736 = 0.6221 ft⁴​​​​​

Max. Comp-stress (at XlsE) OLE OS75 (33.0 0.6221 2x12 0,91305-5 KIPI ft?! 1305.5 kib/ft2 A T Maximum shear stress - Cat XLS A/B] "- РАУ On NA b INA て 200 x 9.4524 x (16:00 (0.775 1x0 ) *0.6221 7= 1508.90 kip ift? y = + (15-681-281(15.67 +128 )+(1-60:10) 2 15.8*1.28 t 15.67X0.118 -18.8 → 128 I 15.67 ý= 16.06 inch N o A 0.715

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