Natural Frequency of a Shear Frame

Hi I've been stuck on a problem for one of my exam revisions, we were asked to identify the natural frequency of the following structures

I know that the equation for the stiffness is as follows
Left Column : $$k= 12EI/L^3$$
Right Column : $$k2=3EI/L^3$$
Equivalent Stiffness : $$15EI/L^3$$
Hence the natural frequency of the column will be $$w_n=\sqrt{15EI/mL^3}$$ Given $$EI:8*10^{11} Nmm^2$$, $$L: 2.5m$$ , $$m:250kg$$ The problem is when i tried to do the problem without converting the EI into SI unit of $$Nm^2$$ I will have a different result in terms of radians. i.e $$wn(withoutconverting) :w_n=\sqrt{15(8*10^{11})/250(2500)^3}=1.75 rad/sec$$ $$wn(with conversion) :w_n=\sqrt{15(8*10^{11})*10^{-6}/250(2.5)^3}=55.42 rad/sec$$ The second answer seems to make more sense, as we have a very stiff column with a small mass it should have a really high natural frequency, however the answer key provided insisted that the first one is the correct way to approach it. How do i approach this and is my approach of converting it to SI unit correct? Thankyou so much for the help

• Sorry can you elaborate? May 17 at 13:50
• I agree that 1.75 rad/sec is incorrect because the units under the square root work out to be m/mm/sec^2. There needs to be a conversion of 1000 to cancel m and mm. The result then works out to be 55.42 rad/sec. May 17 at 18:11

The basic unit of the natural frequency is $$Hz$$, $$cycle/s$$, or just $$1/s$$.

Let's check the results by using "$$m$$" and "$$mm$$" as the base unit -

1. Use $$m$$ as the base unit:

• $$\omega^2 = \dfrac{N-mm^2}{kg*m^3} = \dfrac{N*10^{-6} m^2}{kg*m^3} = \dfrac{N}{kg*10^6m} = \dfrac{kg*m}{kg*10^6 m*s^2} = \dfrac{1}{10^6s^2}$$

2. Use $$mm$$ as the base unit:

• $$\omega^2 = \dfrac{N-mm^2}{kg*m^3} = \dfrac{N-mm^2}{kg*10^9mm^3} = \dfrac{N}{kg*10^9mm} = \dfrac{kg*m}{kg*10^9mms^2} = \dfrac{kg*10^3mm}{kg*10^9mms^2} = \dfrac{1}{10^6s^2}$$

Note: The base unit of \$N is kgm/s^2, which needs to be converted too.

Your mistake/discrepancy was caused by terminating the conversion too early before getting to the bottom of it.

Check:

$$EI = 8*10^{11} N-mm^2 = \dfrac{8*10^{11}kgm-mm^2}{s^2}*\dfrac{10^3mm}{m} = \dfrac{8*10^{14}kg-mm^3}{s^2}$$

$$\omega = \sqrt{\dfrac{15*8*10^{14}kg-mm^3}{250kg*(2500mm)^3*s^2}} = 55.42Hz$$