Question

For the bar subjected to axial load shown in Figure 1 to 2, determine the nodal displacements and Reaction Force. Let Area = 2in^2, E= 30E6 psi = p(x) 300 lb/in 2 3 30 in 60 in x Figure 1 P(x) = 10x lb/in 2 3 30 in 60 in Figure 2.

Answer 1

1 inch 1 P Si 0.0254 m 68 94.757 Pa Date: P. No: 1 eblin 0.113 Nim pcc) = 33.9 Nm 1 3 A-0.00129 m2 E = 206.84 apat 0.262m 1.524 m applied is ual Since fonce FETEEA constant throught the becom, the reaclion at fixed end 1y 33.9 x 1-529 = 51.6636 17 Displacement of point t Si=o (fixed point) Displacement of point 2: FBD of section 1-2 33.980.262 0.762 S, pe AE m 33.9 x 0.7624 0.00129 x 206.848 109 = 73.77x10-9 m 0.074 um Sz Displacement of 3: 33.98 1.5243 S3 0-00129 x 206.848109 = 295.084 X10-9 m = 0.295 mm

in figure & given load is triangular load 1.722 12 N/m P. Nos 1 • 2 3 Though I am showing vertical load, but it is horizontal. I am doing this for understanding purpose only. Therefore vertical neacaron cut fixed fott point is ☺ Si (fixed) S2 P = 10 x 0.762 x 0.113 x 0.262 = 0.65613 N 82 - O.GS 613 x 0.262 0.00129 x 206.84 8109 1.874 x 10 9 m S2 = 1.874 x 103 um S3 P = 10X1.524 x 0. 113 x 1.524 2, 6245N 2-6245X 1.524 S3 0.00129_X 206.84 XLOG 14.99x103 um = 14.99 x 10-9 m