Question

Problem #1 a) Dram the FBD of the rigid bodies and the complete structure. b) Find the reactions in the supports and the internal forces in the hinge! lon ܠܠܠ\0 - - - - 31 3 3'

Answer 1

YOUR DRAWING IS NOT READABLE AND UNITS ARE NOT CLEAR. I HAVE REDRAWN AS PER MY UNDERSTANDING. IF ANY DEVIATION IN DIMENSIONS, PLEASE LET ME KNOW.

solution: 10 lb so B 1415 545 2 y COS 60 = 3 AB .. AB= 3 Casoo 560° ... AB=6' 3' on AB can In the above diagram , Ais roller support. Hence, only Vertical reaction, B and C are hinge supports. Hence two reaction namely horizontal and vertical. UDL (uniformly distributed wad) acting be converted into point wad. .:( Point Load AB = 1 4 X This wad will act on midpont of AB. Hance Jo lb L45° wad can be split into horizontal and vertical load. x 6 = 2 / 3 x6 = 27 eb Fx = lok Los 45 = -7.07 eb fy= 10x sinus =-7.07 lb

FBO of the struchue is given below. 7.07 lb В Cx 27 lb D FBD By Bx TAY 2706 D 1 By Internal forces Dx 260 1 Dy K _ _ I py Load of 27 es can be split into components .: Dx = 27 cos 60=13.5 lb by =-27 sin60=-23.4 lb Bora Py = 23.4 B At +EH=0 (For AB) 2-6 DX-13.5_ 12.6 А :. Dx - Bx = 0 :: 13.5-Bucol Bu = 13.5N t↑ Ev=o 1 Ay .: Ay +By – 23.600 Ay + By = 23.4 --- (1) EMB=0 .: Ayx3 - 23.4x 1.5-13.5x2.6 30

Ay = 23.GN equation 1) in substituting Ay 23. G + By = 23.4 By=0 17.07 Bac a ca 7.07 icy By=0 YEH=0 cror BC ) (x + Bx - 7.07=0 .:-(a + 13.5-7.07 =0 ..(x = 6.43N +1 Ev=0 .:-By + (y - 7.07=0 i. 1 cy=7.07N

Internal forces are drawn as shown below : B po 13.50 (x=6.43N By=0 By zo '9= 7.070 Ay=23-4