Question

Consider one span of a continuous beam has a concentrated load P=120kN applied at mid-span. The span of the beam L=6m. Р L/2 M M2 L In order to solve for the deflection of the beam, the structure can be computed by superimposing the following three system I, II and II. II (III M M2 L A B A ВА B System 1 System II System III Final Solution Assume system II is a simply supported beam with an external moment Mi applied at point A. For the purpose of calculations assume that the value is given as: M1=120kNm. First, calculate the vertical reaction at point A and B. (Note: positive if force is pointing upward) 5. R"A= KN (1 mark) 6. R"B= KN (1 mark) Then calculate the bending moment at mid-span, M"m: (Note: positive if the bottom fibre is in tension) 7. M"m= kNm (1 mark) Finally calculate the mid-span deflection in form of Elef,aved".m. (Note: positive if deflection is pointing downward) 8. Elet,aveA"m= kNm (1 mark)

Answer 1

Given data M = -12 OKNAM L = 6m TIM, To find: RA, RB, Mmid, A stepl: Reaction at supports calculation: Efv=o A 3, 120knim TRA RAT RB zo RA=-RB RB EMB=0 Ans RAX6+120 =0 RA = 20 kn TRE Janse RB =-20KN Step2 i Calwlations for mid span momenti Mmid Ans RAX3 - 120 = 20x3+120 = -60 kN.m/ je compression in bottom fibre, Step 3: Calculations for mid-span deflection, Mx =-RAX+120 ;0&*<6m A hooker 2ookN-m Maa- 2ont 120 IRAT = 20KN RB= -20KN 24 Ma EI dx²

EI -200+120 dx2 Integrating twice EI dy = -20 x² +120x+a -2003 + 120212 + axta EIY 6 2 Apply boundary conditions: Atuzo y=0 C= At us 6m - Y=o 2-20436 + (208.18 +69 -> 9 =-240 So EIY = -2033 + 60x2824020 At x=3m (ie mid sipan) -- 20x27 +60x9 240x3 EIY = - 20x27 Lely =-270 AM Note - Bending moment (M) will create compression in bottom fibre and tension is top fibres above Nevtral Axise