Example 1

To determine the natural frequencies of the given beam and plot their shapes. Check the orthogonality condition. Draw the final moments resulting from inertial forces (assume the excitation frequency as the arithmetic mean of the previously calculated natural frequencies, i.e. )

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Solution

Degree of static indeterminacy

SSN=3-3=0

Number of degrees of dynamic freedom

LSSD=2

Graphs of moments from unit forces applied at dynamic degrees of freedom



Dynamic deltas Vibration frequencies (note, there are several approaches to calculating vibration frequencies, more in the theoretical introduction - LINK) L=m1𝛿11+m2𝛿22=433mEI S=2m1m2(𝛿11𝛿22𝛿212)=809(mEI)2𝜔1=LL22SS=0.267EIm𝜔2=L+L22SS=1.776EIm After substituting the data EI =2.1 ×105 kNm2  m =200 kg 𝜔1=0.2672.1105103200=273.59rads𝜔2=1.7762.1105103200=1819.86rads Vibration amplitudes (more about the approach to calculating vibration amplitudes in the introduction - LINK) A2=1𝛿11m1𝜔2𝛿12m2𝜔2A1 assuming A11 =1 𝑓𝑜𝑟 𝜔1 =0.267EIm A21=19EI1 m(0.267EIm)2143EI2 m(0.267EIm)2A11=0.539 assuming A12 =1 𝑓𝑜𝑟 𝜔2 =1.776EIm A22=19EI1 m(1.776EIm)2143EI2 m(1.776EIm)2A12=0.93 Orthogonality condition A11A12m1+A21A22m2=0 1 1 1 m +0.539 (0.93) 2 m =0
0,002𝑚 0
condition satisfied.

Modes of vibration

Inertial forces (see theoretical introduction) 𝛿11 BB1+𝛿12B2+Δ1p=0𝛿21B1+𝛿22 BB2+Δ2p=0 Forcing frequency 𝜃=𝜔1+𝜔22𝜃=12(0.267 EI 𝑚+1.776 EI 𝑚)=1.022 EI 𝑚 Inertial deltas 𝛿11 B=𝛿111 m1𝜃2𝛿11 B=9EI1 m(1.022EIm)2=8.0431EI𝛿22 B=𝛿221 m2𝜃2𝛿22 B=83EI12 m(1.022EIm)2=2.1881EIΔ1p=𝛿11P01+𝛿12P02Δ2p=𝛿21P01+𝛿22P02where:P01=P=10kN- forcing force on q1 directionP02=0- forcing force on q2 direction Δ1p=9EIP01+143EIP02=901EIΔ2p=143EIP01+83EIP02=46.6671EI Solution of the system of equations, calculation of inertial forces 8.043EIB1+143EIB2+90EI=0143EIB1+2.188EIB2+46.667EI=0 B1=4.99kNB2=10.68kN Force diagram
On the force diagram, we mark the load on the frame from the forcing force, the weight of the masses and the calculated inertial forces.
Since the motion of the forcing force is sinusoidal, it means that in extreme cases the force is displaced in one direction or the other with its full value.
We mark this on the diagram, where on the first diagram we mark the direction consistent with the one given in the topic, and on the second diagram the opposite direction.
Then we mark the inertial forces - on the first diagram, the directions of the inertial forces are consistent with the initially assumed directions of q1 and q2 (see the initial drawing); on the second diagram, the directions are opposite (they change the direction just like the forcing force).
Once again - regardless of whether B1 and B2 came out as positive or negative, we mark the directions as described above. The sign of the inertial forces determines whether a given direction remains (if the force is positive) or should be changed (if the force is negative).
Weight from the masses - to calculate it, we multiply the mass by the acceleration due to gravity g. Directions on both diagrams are consistent with gravity - vertically downwards.
mg=0.2009.81=1.96kN

{x1=P+B1+mgx2=B2+2mg{x1=PB1+mgx2=B2+2mg{x1=6.972kNx2=6.756kN{x1=3.048kNx2=14.604kN