Question

A symmetric truss system is subjected to a concentrated vertical force P at joint B as shown in Figure Q6. By using energy method, determine the horizontal displacement of joint C. EA is assumed as constant for all members, L is the length of member AB. D A С 450 450 B L Figure Q6

Answer 1

In this question we have to find the horihontal deflection at C

Using Castigliano’s second theorem (Energy method)

Placement of imaginary force F The force F is placed at joint C, where the horizontal deflection is desired, as shown in Figure

Member axial forces.

Compute support reactions and obtain member-axial forces in terms of the imaginary force F. Member-axial forces are determined by using the method of joint. and to find the horizontal deflection at F, compute the partial derivatives and apply Castigliano’s equation.

D B с HA 1450 B 450 L I 으 Quan RA=P/ 'R C = P12 Find the forces in members in term of F Take ZMATO RB X 21 = pre =) RB-2 2 RA= Pla Efyzo E Ex=0 HAFF At point" D RA= AD sinuso MA astus AL >8 IRA AD = Pla2 HA+ AB = AD Casus - P 2 AB=L-HA 1-of 2 2

At point "C" u al CD sinusº E 2 Pla F+ BC = CO caruso $ Вс 2 РЕ At Joint B A B = BC - F efe 2 BD =P Member length • axial forces, partial derivatives with respect to the friction fonce P. Member Length N(KN) ON OF N160) N/ONDEL AB L I - F 1 P12 pot AD LN2 Pla PIZ L e 0 B) Р LN2 0 CD Pla f-F Pla P/ L -1 BC 1 x L ŹNCON -PL Dc I EN (2) -PL НЕ EA (unit) (-ve.sign shows the opposite direction of deflection)