Question

5 kip 4 ft. 3 ft. B 4 ft. 7 ft. А E

What is the horizontal deflection @ B using virtual work?

Answer 1

Sign convention:

Bending moment: Clockwise bending moment is taken as postive while anticlockwise bending moment is taken as negative.

Vertical forces: Upward forces are taken as positive, while downward forces are taken as negative.

Horizontal forces: Forces to the right are considered as positive while forces to the left are considered as negative.

Member forces: Tensile force in member is considered as positive while compressive force in member is taken as negative.

Assumption: Since the cross-sectional area of the members is not given, so I've assumed it to be A units for all members. Also since modulus of elasticity is not given, hence, it is also assumed as E units.
P- FORCE SYSTEM Skip c 4ft 3ft B D 4 ft A 7 ft IM HE VA VE Support reaction calculation : (wing equilibrium equatione) { H=0= HE = 5 kips EME=0 5*8 =0 VA = 5.71 kips EV-0 => VA +VE -O VE = -5.71 kips => VA X7 Member force calculation : (by method of off joints) C o s PCD Joint C: 5 kip Sin 0 = 3 los O=4 5 PCB EH, = 0 => -5+ PCDX Sin 0 = 0 8.33 E Vc=0=> - PCB - PCD Cos o=0 PCD kips

PCB Poo los o 6.67k • PBC PBD B Joint Bi EHB O => PBD - 0 EVB=0 PBC-PBA PBA 0 PBA =-6.67 k Joint A: EVA =0 PAB PAD o А PAE PAB+5.71 + PAD COS O=0 > PAD VA = 5,71 k 1.2 kips + PAE = 0 EHA=O=> PAS sin o PAE -0.72 kips = Joint E PED Eve=0 PED Cos O + Ve=0 He=5 PEA VE=-5.71 PED= 7.14 kips k-FORCE SYSTEM unit horizontal Consider a load at B,

0 uft B 3ft 1 uft @ A HE 7 ft < ܢܵܝܵܐ Support reaction calculation: (using equilibrium equations) EH = 0 => HE = 1 EME = 0 = 7 VA -40 VA 0.57 EV-0=) VAT VE = 0 -> Ve=-0.57 Member force calculations (method of joints) Joint Ci Using EH-O & EVE-O, & k cD = there is NO external force at joint c o k ibt 0 [ k BC .. H B Joint B: EHB=0 -> kBD - 1 EVB= 0 => kBA=0 kBD "KBA

Joint A: KAB KAD A KAB + A xkAE EVA=0 KAD Coso +0.57 EO => KAD -0.7) IVA=0.57 EHA=0 KAD Sin o t KAE =0 KAE = 0.43 Joint El KED E HE=1 KEA V=-0.57 EVE = 0 => KED CosO TVE FO = ked 0.71

HB PkLLAE HORIZONTAL DEFLECTION AT B CALCULATIONS: Member P-kips) ke Llift) TAE AB - 6.67 o 4 AE AD 1.2 -0.71 5 AE 1- 4.26 /AE AE 0.72 0,43 7 AE: -2.17 /A E BC -6.67 4 AE BD 3 AE O CD 8.33 O 5 AE 0 DE 0.7) 5,63 AE 28.54/AE 0 - 7.14 In A ACE, Le calculation : CE? = AB HAE? (Pythagoras therrem) (CD+DE)? A C² taE² => DE=5.63 ft Wow, EPkL AE 22.11 AE voitual Using work theorem, =E Pkl AE AHB (horizontal deflection at B) A units 22.0) AE
Answer: The horizontal deflection at joint B of the truss is
22.11 -units AE
towards the left (because we've taken the unit load at B towards the left and the deflection comes out to be positive, so deflection is also towards the left).