Question

15 KN 3 kNm ITB A 2 m am E=100GPa, I=1 x 10^-5 m^4 Calculate Magnitude and direction of all of the reactions at points A and B Calculate direction and magnitude of the rotation at B.

Answer 1

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3 KN/m L5KN 1 I 2 B 2m x 2m. Explanation E- 1000 Gpa I - 1x 10-5 my L = 4m This beam can be splited into three parts: am de 2m B rt SBI SB 3 kN/m 15KN SB + FEA. AB А VR case II Case- Case-II Since end 'B' is propped, Net deflection at End 'B' will be 'Zero'. SB = Secreases + SB (cases) + SB (ease th) = 0 w x(+2) + 0xl ) + W24 8 EI 2 = 3x44 th ZEI -tve + it -VBX L3 co BEI ЗЕТ 3 15x2 + 15x2 x 4 - VB x 43 sign convention 2. BEI SEI 3 EI upward ê defi=-ve downwardi def. 192 + 120. 2 EI BEL 3 EI 576 +240+ 360 - 128 VB =0 GEI 128 VB 1176 G4 VB=0 +60 EI = VB = 9.1875 KN Reaction at Support B'

15KN 3 kN/m Magnitude and direction of Rotation at 'B' ++ +t Rotation) VB Rotation Rotation case-I case-I case-Il tve ve 15x22 +(-VBx42 Net Rotation is the sum of rotation of all the three cases Take OB = OBs + OBI + OBAT c.) Rotation A.cow G Rotation 3x43 + GEI 2 EI 2 EI 192 +30/61 - 147 2EI 192 + 180-441 GEI OB = 1 1.5 EI negative) put values of E4I GEI G (Anticlockwise because de is E = 1000 Gefa = 10000 100 KN 1 m 2 I = 1x 10-5 my EI - 10000 KN.m2 B- - 14.5 10000 1.25 X 203 G

lu Calculation of Reactions at support A' Y 3 KN/m 115KN HAT um •MA V VB = 9.1875 KN APPLY Equillibrium conditions: Sign convention litve ve E Fu=0 HA=0 {fy = 0 VATVB = 3x4 + 15 VA = 12 +15-9.1875 VA 17.8125 KN ę utward. 2 {MA =0 - VB x 4 + 15x2 + 3x4xY-MA=0 2 -> -9.1875x4 +30 +24-MA=0 o MA = 17.25 kHm G Acw O