Rigid bar is supported by two short white spruce wooden post and spring If each of the post have unloading length of 1m and cross section area of 600 ` mm^2` and spring stiffness of k=3MN/m and unscratched length of 1.02m. Determine the force on each wooden post after the load is applied
Solution -
Given,
Area ,A= 600 `mm^2` = 600 x `10^-6` `m^2`
Also , we know that
Elastic modulus of the white spruce wood ,
E=9.65 Gpa =9.65 x `10^9` Pa
Now,
From the figure
Taking moment about C, `F_a` x 1 - `F_b` x 1 = 0
So,
`F_a` = `F_b`=F
Now,
forces in vertical direction
2F + `F_s` - 50x2x`10^3` = 0
____________________(1)
Also, using compatibility equation
`F_a` + 0.02 = `F_s`
F x (AC) / Area x E + 0.02 = `F_s` / k
`
\fracF{600times10^{-6}\times\9.65\times10^9}` + 0.02 = `F_s`/3x`10^6`
1.727 F + 2 x `10^5` = 3.33 _________________(2)
On solving equation (1) and (2) we get
F= 15611.8 N and `F_s` = 68776.26 N
Hence , Force on each wooden post wil be
15611.8 N
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