Degree of static indeterminacy

SSN=3-3=0

Number of degrees of dynamic freedom

LSSD=2
𝑚1 =𝑚𝑚2 =2⁢𝑚

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

Dynamic deltas 𝛿11=1EI⋅(13⋅3⋅3⋅3)=9⋅1EI𝛿22=1EI⋅(13⋅2⋅2⋅2)=83⋅1EI𝛿12=1EI⋅(13⋅3⋅2⋅2+16⋅1⋅2⋅2)=143⋅1EI Vibration frequencies (note, there are several approaches to calculating vibration frequencies, more in the theoretical introduction - LINK) L=m1⋅𝛿11+m2⋅𝛿22=433⋅mEI S=2⋅m1⋅m2⋅(𝛿11⋅𝛿22−𝛿212)=809⋅(mEI)2𝜔1=√L−√L2−2⋅SS=0.267⋅√EIm𝜔2=√L+√L2−2⋅SS=1.776⋅√EIm After substituting the data EI =2.1 ×105 ⋅kNm2  m =200⁢ kg 𝜔1=0.267⁢√2.1⋅105⋅103200=273.59⋅rads𝜔2=1.776⁢√2.1⋅105⋅103200=1819.86⋅rads Vibration amplitudes (more about the approach to calculating vibration amplitudes in the introduction - LINK) A2=1−𝛿11⋅m1⋅𝜔2𝛿12⋅m2⋅𝜔2⋅A1 assuming A11 =1 𝑓⁡𝑜⁢𝑟 𝜔1 =0.267⁢√EIm A21=1−9EI⋅1⁢ m⋅(0.267⁢√EIm)2143⁢EI⋅2⁢ m⋅(0.267⁢√EIm)2⋅A11=0.539 assuming A12 =1 𝑓⁡𝑜⁢𝑟 𝜔2 =1.776⁢√EIm A22=1−9EI⋅1⁢ m⋅(1.776⁢√EIm)2143⁢EI⋅2⁢ m⋅(1.776⁢√EIm)2⋅A12=−0.93 Orthogonality condition A11⋅A12⋅m1+A21⋅A22⋅m2=0 1 ⋅1 ⋅1⁢ m +0.539 ⋅(−0.93) ⋅2⁢ m =0
−0,002⁢𝑚 ∼0
condition satisfied.

Modes of vibration

Inertial forces (see theoretical introduction) 𝛿11⁢ B⋅B1+𝛿12⋅B2+Δ1⁢p=0𝛿21⋅B1+𝛿22⁢ B⋅B2+Δ2⁢p=0 Forcing frequency 𝜃=𝜔1+𝜔22𝜃=12⁢(0.267⁢√ EI 𝑚+1.776⁢√ EI 𝑚)=1.022⋅√ EI 𝑚 Inertial deltas 𝛿11⁢ B=𝛿11−1 m1⋅𝜃2𝛿11⁢ B=9EI−1 m⋅(1.022⁢√EIm)2=8.043⋅1EI𝛿22⁢ B=𝛿22−1 m2⋅𝜃2𝛿22⁢ B=83⁢EI−12⁢ m⋅(1.022⁢√EIm)2=2.188⋅1EIΔ1⁢p=𝛿11⋅P01+𝛿12⋅P02Δ2⁢p=𝛿21⋅P01+𝛿22⋅P02where:P01=P=10⋅kN- forcing force on q1 directionP02=0- forcing force on q2 direction Δ1⁢p=9EI⋅P01+143⁢EI⋅P02=90⋅1EIΔ2⁢p=143⁢EI⋅P01+83⁢EI⋅P02=46.667⋅1EI Solution of the system of equations, calculation of inertial forces 8.043EI⋅B1+143⁢EI⋅B2+90EI=0143⁢EI⋅B1+2.188EI⋅B2+46.667EI=0 B1=−4.99⋅kNB2=−10.68⋅kN 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=0.200⋅9.81=1.96⋅kN

{x1=P+B1+mgx2=B2+2⁢mg⁢{x1=−P−B1+mgx2=−B2+2⁢mg{x1=6.9⁢7⁢2⁢kNx2=−6.7⁢5⁢6⁢kN{x1=−3.048⁢kNx2=14.604⁢kN