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 1/2 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 E) M M2 L А B A ВА B System 1 System II System III Final Solution The total mid-span deflection in form of Elef, aveAm is (Note: positive if deflection is pointing downward): 13. Elet, aveAm = kNm3 (1 mark)

Answer 1

A L As the load is symmetrically distributed on beam, Each vertical reaction (ie, A+B) consider the left half Ac of the beam The Bending moment at any section in distant & from A is given by AC EI dy dut El dy du Integrating on both sides you ET do du At a- d there is maximum deflection Scanned by TapScanner

dy so slope 0 Ріг El dy = Px de u P.lt integrating on both sides Again FI(y). Pae? 12 at x=0, deflection ... => y=0 Ç= 0 at x = ļ, mosimum deflection -) FT 6) = f (4) '-rece) =) y = poe? 48 EI Downward deflection at a pe? 48 EI This is for system (I) Scanned by TapScanner

М. pe 8 For system @ GA pe 8 First calculate the support reactions MA=0 Roxa- Pt Ro (12) ! RA consider P (1) section at distance & from "B' --P 3 ET dy duz -Pa EI dy dur 8 Integrating on both sides Scanned by TapScanner

EI dy -Px2 16. ita on both sides Again integrating FI (y) - j - Px² Px + Catch 48 at x = 0, yzo -pe? yo +Cice) Pe 2 48 FIly). - pa? 48 + Pl² 48 want at = & deflection we EI (4) - P + d² + Pl ² l 48 -Pi + Pl? = ET (9) 9888 48x2 -) EIly) pe? 48x2 (+) =) EI (y) Pl? ។ 4862 Scanned by TapScanner

--))- (+) 334E1 Positive (+) indicates upward deflection . For system (3) System (3) is as system which results value of deflection at a=1% same same considering three systems. resulting deflection Pe? 314? Зем? 48 ET 384 ET 384 EI J ✓ Fol system Forsystem ® For system Pl? - Final deflection 192 EI Scanned by TapScanner
And by substituting the given values in the derived formula,
i.e., =(120×6³) /192
We get Ans =135 KNm³