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Question 1. The walkway exerts vertical 42 kN loads on the Warren truss at B, D, F, and H. Use the method of sections to determine the axial force in member FH. The section must cut 3 members. Find the reaction force at both ends. IM Examine an appropriate section to cut, draw the FBD. 2M Produce equation of equilibrium SM, SFx and SFy. IM Solve the axial force in member FH. IM [5 Marks] B D F Н 2 m - 6 m - 6 m-6m -- 6 m

Answer 1

Detailed analysis of the solution is given in the image. Hope it's helpful.

SOLUTION : touss From the figure given in the question, lets draw the wasten with the loads acting and the supposts at A and I. |42KN HakN 142KN 4.2 KN 3m Boy D F H. I с 6m 6m E sm G Em Required : Axial force in member FH using method of Sections. In osder to find the reaction forces at A and I, we shall first draw the free body diagsam of truss with loads acting and then apply the equilibrium Conditions. 42 KN 442 KN 42 KN 142 EN D н B (RA (RHDI A I E G TIRA (Rv) I Fx = 0 => (RH) -(RH)= 0 -> (RH)A = (RM SI {+-,-4) As no external horizontal forces are acting (RH)A = (RH) = 0 EFy=0 -> (R.)A + (RVDI 4 * 42 = 0 (RM/A+ (Rv), = 168 KN *24 (R'I EMA=0 1... (R) 42#3 - 42*9-42*15 - 42*21 -0 {+r,-23 84 kN (?) 168 - (Ruli = 84 kN (1) (Ruda (RU)A So the reaction forces at A and I are 84 kN each. (ANSWER) Scanned with am Scanner

Now to find the axial force in membes FH using method of Section, let's cut Section through the three members FH, FG & EG- FBD with Cut Section is given below: a 42 KN As shown in the figure we FH have considered a Section a-a Cut through FH , FG & EG: FE •I EG G |(R); -84kN Considering the equilibrium of portion to the right of a-a, Taking Moment about G, G , for equilibrium MG = 0 => (FH2)-(42*3) + (84*6) = 0 =) F - 378 FH FFH -189 KN Negative Sign indicates that the axial force nature in member is compressive and is a magnitude of 189 kN. FH