Question

Question 2 Simply supported beam ABC is subject to a point load and the patch loads as indicated in Figure Q2. Assume the beam has a uniform cross-section size. The Modulus of Elasticity E = 210x106 kN/m2, second moment of area l=5x105 m. Determine the deflection of beam ABC at the middle point using MacCaulay's Method. Total (15) marks. -30 KN -6 kN/m -3 kN/m B 3 m 4 m * Figure Q2: Simply supported beam ABC

Answer 1

-3 kolm -30kn Qs f-6 kolm ܚܣܬ݁ܰܣ $ PRC RA Rc= 6*414+3) + 36 A 3 + 3432 31.9985 kn 7 12 ( 34 3+ 6*6+ 30 ]- 31.9885 11 RA 17.07 kn we know that =1/d²4 )= mx dyz 30*(x-3) Erldly RA*% 3.72 +3(2-3) 2 dyz 6* (X - 3)2 *(4- ೩ to abore quahon get slope Equation Integrate the Ef dy RAXZ 343 3CX-3}} 38(2-3) de 6 6 2 6(x-303 -373 +C get dilection 6 Integrate again to Eraton

i.en a 3.5m Fi(y) RAX3 3x4 364-374 30(x- 33 SCX-334 S3 24 94 6 3 FC tax t C2 10 कल know that apply Y. Boundary Condition deflection 7 =0 in Pinned a Roller support respectively مني / @x = 7 deyle chion - ų : .@rco, (4-0) inha E O = 0 + Ca ñ ₂20 @x=7 (y=0) in , EI (O) = 17.071. *73 3674 3(7-434 اور اس 3007-333 au 6 94 6 667-304 ((7) to T ماد -281.G 4 -Set 36 -Y CI = 11 E a 981.64 7 Cla = 46.834

с кс - 83 82 mm @ Mid Дес ity : 3-5 то Deflect on (35)= 13-car 13,3 33.5 Э 6 EL 3(3-5-3 (3-5-33 G С 40. аз, 3s x (- афры) to с (3-5-35+ 2 и (9) Е 43.85 2 — Yeas (ff) |1 Чrs = 32: 935 - 42-82 2 8 (0% (06, 9% (65 9 3:s б. OS 322 ж бktos Swo S Ча- Ei= - 22.98 м Stay safe Please upoote the answer