Question

Problem 3: For the beam shown find the slope and deflection at point B and C 100 KN 300 kN-m 6 m E = constant = 70 GPa 1 = 500 (106) mm Problem 4: For the beam shown find the deflection at point B and the slope at point A 80 KN 12 m 12 m E = constant = 200 GPa I = 600 (106) mm

Answer 1

100KN 300NM I= 500x106mm : E= 70684: 70000 N/mm2 El = 35x102 4. mm2 ET = 35x10 169 KN. om Et = 35000 KN- m2 2I в Зим By conjugate beam method REF Ay= 100KN MA=-300-10x 6 MA = -900kNm moment by parts -300 TE -900 ZEI Moment at B due to MA 900 kiyim moment at B due to Ay = 100 x6 = 600km moment at B due to 3 rokim moment - - 3000 km ( diagram 300 2 Conjugate beam 30g EE slope in real beam shear force in conjugate beam Deflection in real beam = moment in conjugate bean sm 6m 450 EL Conjugate beam OB - Swear force at b in conjugate beam Oe = - 45 X6 + 2 x 200 x6 = -2700 + 990

0 -1800 ET - 1800 rad 35000 108= -51.429 X183 rad. ( clockurse) AB- moment at 'B' in conjugate beam AB- USDx6x6 GE + x 300 x6 l ET -8100, 1800 EL 1 AB= -6300 pop- EL - 45- 2 m - 6300 35000 4o= - 6300 x 1000 mm AB=-180 mm) (rownward) Ac = - 450x6 EL 3 no x2 EC 1 x 3rox6 - - 2700 . EE 900+ Yoo EL Dc = -2700 - -2700 35000 ad ЕТ. De=-0.771 rad. (elodeuntes A c = -450 x 6+ (6 + 3) - 300 x 3 x 2 + 2 x 300 x 6x ( 4 2 + 3) AC = - 16200 - 130 + 4500 | A c = -13050 AC = -13050 x 1000 mm Ac=-372.857mm (rown ward)