A vertical bar consists of three prismatic segments A1, A2, and A3 with cross-sectional areas of 6000 mm2, 5000 mm2, and 4000 mm2, respectively. The bar is made of steel with E 5 200 GPa. Calculate the displacements at points B, D, and E. Ignore the weight of the bar. Goodno, Barry J.. Statics and Mechanics of Materials (p. 609). Cengage Learning. Kindle Edition.

Answer + Explanation:

For computing the displacement at point B and D we need to determine the following calculations

Pnet = Pc + Pe + Pb

Pnet = 250 + 350 - 50

Pnet = 550 N

Now the deflection for bar AB is

           PLab

δab = -------------

            ae

     550 * 500

= -------------------------

     6,000 * 200 x 10³

= 2.292 x 10⁻⁴mm

Now for bar BC it is

           PLbc

δbc = ------------

             ae

  (550 + 50) * 250

=  ----------------------

   5,000 * 200 x 10³

= 1.5 x 10⁻⁴mm

And for bar CD it is

           PLcd

δcd = ------------

             ae

  (550 - 250 + 50) * 250

=  --------------------------------

   5,000 * 200 x 10³

= 0.875 x 10⁻⁴mm

Now the displacement is as follows

For B = 2.292 × 10⁻⁴ mm

For D = 4.667 × 10⁻⁴ mm

= 2.292 × 10⁻⁴ + 1.5 × 10⁻⁴ + 0.875 × 10⁻⁴

= 4.667 × 10⁻⁴ mm

We simply applied the above formulas for determining the  displacements at points B, D and the same is to be considered

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

i mean i gues

explanation: