Determine the displacement at nodes by using the principal of minimum potential energy and Find the support reaction.

Use, $K_1 = 100 N/mm \hspace{1cm} K_2 = 300 N/mm$

$\hspace{0.8cm} K_3 = 150 N/mm \hspace{1cm}K_4 = 200 N/mm$

(i) Number of nodes and elements

(ii) Potential energy of system with discrete element is given by $ \pi = \frac{1}{2} \sum K_i \partial u_i^2 - \sum u_iP_i $ for the system shown.

$ \pi = \frac{1}{2} K_1 (u_2 - u_1)^2 + \frac{1}{2}K_2(u_3-u_2)^2 + \frac{1}{2}K_3(u_4-u_3)^2 + \frac{1}{2}K_4(u_5 - u_4)^2 - P_1u_1 - P_2u_2 - P_3u_3 - P_4u_4 - P_5u_5 $

(iii) For minimum potential energy , $ \frac{\partial \pi}{\partial u_i} = 0 $

For node 1: $ \frac{\partial \pi}{\partial u_1} = K_2(u_2-u_1) - P_1 = 0 $

For node 2: $ \frac{\partial \pi}{\partial u_2} = K_1(u_2-u_1)-K_2(u_3-u_2)-K_3(u_4-u_2)-P_2 = 0 $

For node 3: $ \frac{\partial \pi}{\partial u_3} = K_2(u_3-u_2)-P_3 = 0 $

For node 4: $ \frac{\partial \pi}{\partial u_4} = K_3(u_4-u_2)-K_4(u_5-u_4)-P_4 = 0 $

For node 5: $ \frac{\partial \pi}{\partial u_5} = K_4(u_5-u_4)-P_5 = 0 $

The equation can be rewritten as:

$ K_1u_1 - K_2u_2 = P_1 \\ -K_1u_1 + (K_1+K_2+K_3)u_2-K_2u_3-K_3u_4 = P_2 \\ -K_2u_2 + K_2u_3 = P_3 \\ -K_3u_2+(K_3+K_4)u_4-K_4u_5 = P_4 \\ -K_3u_2+K_5u_5=P_5 $

Now, substituting given values

$ K_1 = 100N/mm \hspace{0.5cm} P_2=0 \\ K_2=300N/mm \hspace{0.5cm} P_3=100kN \\ K_3=150N/mm \hspace{0.5cm} P_4=200kN \\ K_4=200 N/mm \hspace{0.5cm} u_1=0 \\ K_5=200 N/mm \hspace{0.5cm} u_5=0 $

Above equation becomes

$ -100u_2 = P_1 \hspace{0.5cm} ...(1) \\ 550u_2-300u_3-150u_4=0 \hspace{0.5cm} ...(2) \\ -300u_2+300u_3=100 \hspace{0.5cm} ...(3) \\ -150u_2+350u_4=200 \hspace{0.5cm} ...(4) \\ -200u_4 = P_5 \hspace{0.5cm} ...(5) $

By solving equation (2),(3) and (4), we get,

u$_2$ = 1 mm; u$_3$ = 1.333 mm; u$_4$ = 1 mm

Substituting above values in equation (1) and (5), we get,

P$_1$ = -100u$_2$ = -100 kN

P$_5$ = -200u$_4$ = -200 kN

Checking:

$ \sum P = P_1 + P_2 + P_3 + P_4 + P_5= -100 + 0 + 100 - 200 + 200 = 0 $

Solution verified.