![A plane frame shown in the figure (not to scale) has linear elastic springs at node H. The spring constants are kx = ky = 5 × 105 kN/m and kq = 3 × 105 kNm/rad](https://www.gkseries.com/blog/wp-content/uploads/2023/08/A-plane-frame-shown-in-the-figure-not-to-scale-has-linear-elastic-springs-at-node-H.-The-spring-constants-are-kx-ky-5-×-105-kN-m-and-kq-3-×-105-kNm-rad.jpg)
Q. A plane frame shown in the figure (not to scale) has linear elastic springs at node H. The spring constants are kx = ky = 5 × 105 kN/m and kq = 3 × 105 kNm/rad.
![](https://www.gkseries.com/blog/wp-content/uploads/2023/08/Screenshot-462.png)
For the externally applied moment of 30 kNm at node F, the rotation (in degrees, round off to 3 decimals) observed in the rotational spring at node H is
Solution:
Since horizontal reaction at E = Hc = 0
Bending moment at F for member FE = MFE = 0
Member FH behaves like propped cantilever beam
![](https://www.gkseries.com/blog/wp-content/uploads/2023/08/Screenshot-463.png)
30 kNm = 3 × 105 kNm/rad × θ
⇒ θ = 1 × 10-4 radians = 0.0057°