Question

2. (32 pts) Two identical 20 mm thick plates are bolted to the top and bottom Mange to form the build-up heam. If the beam is subjected to a shear force of V = 300 KN and a bending moment of M = 500 KN-m. d) Calculate the shear stress between flange plates. If each bolt has a shear strength of 30 KN, determine a) Calculate moment of inertia of the cross section in the z axis. b) Calculate the maximum bending stress in tension and its locations on the cross-section, Calculate the maximum shear stress and its locations on the cross-section. I.ocate it on the figure. the maximum spacing s. 200 mm 20 mm ווזון 10 300 MR 10 mm V 10 mm 20 mm 200 mm

Answer 1

Solo г- Given Data : у - ЗооkN Sook Nm dom 1 Soo mm 10 мм 10 mm 20mm 200mm Пола , 14o t 155 mm Зо 2 а) Im = [{oo x 3ч. 1- 4 [ 19 x 480*1 17 1. (6 55 061 . 341. 513 ) х 10 mm Ч 1 - Зо1, 499 106 mmЧ

Now, fyy- (346 x доо?]- 4 [ 280 x 190 ° ) 2 12 1 - да - 64 Eyy - (2464 661 16o • o43 ) х 4,6 mm 66. бач хя. " Using perpendicular anis theorem, Тъз: . 1 Туу Izz = (303. 454 + 66 - 684 ) х 164 -ъът 31ч, 118 x 16 mm (0) Maximum Bending Stress in Tension will be at bottom of the I cross-section.. О, Мң "ь (боо х 10° ) х (о. 11) Зо1. Ч94 Х 40-6 бря a16. 448 MP4 |

(c) 3 Inox The можат shear stress will be at Neutral Anis in I cross-section. Tmax = Ом I b Now, - АЧ * А, Ч, да (148X10) Xю + (або 130 ) х 4SS 8 - 1, одах 106 3 mm y = 3oo KN 13 302. цоч x 466 mm Ч be 10mm Таъ - (1- оаз x16 px ( Зоо 14-) (Зоя, цоч 106 ) х 10 Tmax= 1оо: 29S M4

(d, Shear flow, 4 q.. VAY Ixx for spacing , we have following relations : fa qXS fa VAY XS I nx Now, 200830= 6000 mm mm? os) Is 5 mm 300 KN Ixx = 307. 48.4 x 100 mm f= 24 303 6o k N [ Double Shear] ... 60X163 = (300% 103 )x(6000 3 ) X 155 XS 307. ugu X 106 s- 66.127 mm Completed