Question

Using only bending energy what the horizontal displacement at C to the nearest 0.01in. w=9k/ft E = 29,000 ksi I = 716 in B 8 ft w 10 ft

Answer 1

Solution...

as per unit load method. deftetiam A= EI mm da M= moment equation for primary beam m = moment equation for virtual beam. 2. plote Virtual beam obtaim by applying unit load required deflection point. in veequired direction.

Griven beam (primary bear or Real eam) free body diagram of real beam .8 At с :: Es Q9000 ksi Vo I = 716 int gkat & EI = 9ooo *716 6 10 ft EI = 80.76410 k-in? HAT 6 2 ) EI = 20.764 ¥10 1 K-ft (12) . = 9*10 VA , Ve= Vertical support reaction at A & C. HA= Horizontal Support reaction, at A. - We calculate supposed reaction by using equilibrium equation & - Horizontal force-o =) Ha gok (t) E Vertical force = VA= Ve (z moment), -0. VeX8 0x10 56.25 k (1) No = 56.25k (1) -) gXO X10 Vc = and

Virtual Beams с. B 1 unit € 22 Equation of equilibrium DEH=0 HA=1 링 EVO v = vode] HA meo 3) V. 48 - | *10=0 t we = 1.25 then we = 125 Then horizontal displacement of point © = de =) B De= mm doe mm de EL A El as El is constant from A to B o c to B. 4-1022" gay ex) dxe + 8 (56252) (1.252) dze EI

10 8 Aca E S (902? Uspe]) dze + 70.3125 Ei se de 10 A D- 1991-22) . EI + 70.3125 EI nye . 71 . Da= 187 50 EI 12000 30750 + . EI EI :-) De 3075o ft (29000 4716* 12 An = 0.21325 ft o Do = 2.559 inch 2.56 11 inch T € Required Answer