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. P L/2 M CE M 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 M - A 00 B A А B System 1 System II System 111 Final Solution System I is a simply supported beam with a concentrated load P=120kN applied at mid-span. First, calculate the vertical reactions at points A and B. (Note: positive if force is pointing upward) 1. R'A= KN (1 mark) 2. R' = KN (1 mark) Then calculate the bending moment at mid-span, Mém. (Note: positive if the bottom fibre is in tension) 3. M'm = kNm (1 mark) Finally calculate the mid-span deflection in form of Elef, aveА’m (i.e. do not devide the value by the bending stiffness of the beam Elef, ave). (Note: positive if deflection is pointing downward) 4. Elet,ave A'm = kNm3

Answer 1

Solution : System 1: B R E q RB PA -) RA = RB = b Moment at mid span mid span = fx v 2 PL . V 4 at B /er 11 Delflection at midyban -B A с 1/2 K 2 43 Ą= moment of area of M/s diagram sans about point Aby = 1 x A? { forum A poc} (downward) PL? PL 4 E 1 X L L x į 48 EI System II B mi RA= RB = mi 다 mi Z

Acording to double integration method El dy mi mi L 2- का mix tci ) EI dy dr MI L 2 X2 2 MI +cix+(2 Mix 2 >X 3 =) EIGE 0 at x = 0, y =o 2. C2=0 & ata=liy= 20 3 70- me x L MIXER ē taXL 느 G= mixx2 + met MIL 3 2 gie. Ely = me x z air 2 = 1 4 De M, L2 mu L2 mich (upward + 6 16 EL Az = mil 48 8 System 3 Az in upwand direction q Az will be sam as A3 = 1 upward) 16EI

so total deflection at mid span 4 A1 - A2-A3 = PL? 48 EI 16 EI MILZ M222 16 EI where mi= M2 = aloo PL² PL3 PLE EI A 128 48 128 DEID P3 = 192 &L=6m & p = 120une, 1206 63 192 135 kn/m3 Love -) EI A = Lasinal deflection downward .