Question

Q2: Draw the influence lines for the members HC and CD of the truss shown in Fig. (2), and then determine the maximum tensile force that can be developed in member HC due to moving uniform distributed load of 3k/ft.

Н 18 B D с L 424-96 Fig. (2)

I need it in 30min or 1 h

Answer 1

H F 18 A B с D 4@24'-96! When unit load is at A FHC - O Feo - O When unit load at B ΣΜΗ THG H 24RA - 18FBc = 0 FHC - FBC 24x0.75 = 1k is A AFBC VB o= tan 1/24 = 53.13 RA 1k 1x 72 =0.75k 96 EMG=0 48RA - (24х 1) – 24 FBc - (Fnc Cos0x29) – (Fra Sine) x6 = > FHC -0.625 k

REF 1x2q = 0.25k FF G C F 72 FFC ME - 24Re - 18 Fe = 0 D ng FOD 24x0.25 / 3k - Foo = 0.333k is When. lood at c unit RA= FHG 솔 = 0.5k 7 o = 53.130 A В Foc RA EMH 30 - 24RA - 18 FBC - o = 0.667k Fвс 2 3 24X0.5 18 Mo = 0 48RA - 24 FBC - (FMC Cos 0 x 24) - (Fuc fim0x6)=0 0.4167k > Frc = load and structure Due to symmetry of FBC = 0.667k Fcd =

is at D When unit lood 0 = 53.13 FHG RA 1x 24. 96 0.25k o THC A FBC B => 24RA - 18 FBC RA - FBC = 0.333k IMG → 48RA 24FBc - (FAC Coso x 24) - (Fec fim 0x6) = 8 The = 0.2083k ㅠ RE = 1x72 96 = 0.75k F FFC EME=0 » LARE - 18F0=0 > Foo 24X0.75 = 1k Fcd RE 1k .When unit lood is ct E The FED=0

Hс ILD for member б.4 (67 0.2 of 3 kx А W С 0.69 5 0.625 1). 0.4167 24 x Х 14. 4

Maximum tensile force will develop in when le between y & E UDL will Area between y & { x 26-10-270-4167-{{<2**(04:67 26-3065* + **20x0-2013) 11.99976 HC Maximum densite force in llogg976 X 3 35.99928k ILD for member CD 0.667 0.333 A A B C

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