Question

5- The 5-knot, 7-bar flat truss is supported by F by a double support and secured by an AB cable as shown in the figure.
A vertical load of 2kN is applied to node D.
Under these conditions determine:

a) The pulling force on the AB cable.
b) Using the section method, efforts and respective signals on bars BC, FC and FE.
c) The value of efforts and respective sign on the FB bar


В 207 B F E P=ŽKN 5 0 3m 2m

Answer 1

C с TAB D E -3m 3m om XTAB P=2KN section out for part(b A 2m tan B = > B = 33.69 333 2 / 2 tano - » 0-63.43497 2 TABCODO 13 am Tarz 3110 D F E - 3m 3mi P=2KN Take moment at F 20 TAL CODO X 2 = PX6 TAB Con 63.4349 = 2x6 TAB= 13.4.1640786 KN Tension force in cable AB BC, CFC and FE using method section cut shown above. ( : finding force in by section 1

B FBC I B take moment at Feo < 2m EFFC F Ecom FFE « — Зm B C moment at FBC X2 PX 6 P=2kw FBC 2 X6 FBc = 6 kN force in (Tension) take C=0 FFE X 2 + 2x3 =0 FFE = -3 KN force en FE FEE - 3 kN compressions Equating vertical forces, FFC sin ß +2=0 FFC - 2 sin 33.69 [FFC FFC = -31605557 KM force en FC FFC = 3-605557 kn compression C consider joint & -FBC of & FFB EFy=01 FFB + TABLnozo FFB =-13.41640786 8ın 63.4349 FFB=-12 KN force en FB TAB or FFB=12 kN compression