1. Determination of the degree of kinematic indeterminacy:

𝑆⁢𝐾⁢𝑁=∑𝜑+∑Δ𝑆⁢𝐾⁢𝑁=1+1=2

2. Selection of the basic system of displacement method (UPMP):

Canonical equation system of the displacement method:
𝑟11⋅𝜑1+𝑟12⋅Δ2+𝑟1⁢𝑝=0𝑟21⋅𝜑1+𝑟22⋅Δ2+𝑟2⁢𝑝=0

3. Graphs and determination of coefficients and constants of the equation:

Determination of UPMP coefficients

𝑟11=3⁢𝐸⁢𝐼5+3⁢𝐸⁢𝐼4=27⁢𝐸⁢𝐼20𝑟21=−3⁢𝐸⁢𝐼16

Determination of UPMP coefficients

𝑟12=−3⁢𝐸⁢𝐼16𝑟22=3⁢𝐸⁢𝐼32

Determination of UPMP coefficients

𝑟1⁢𝑝=−25𝑟2⁢𝑝=−3,75

4. Solution of the canonical equation system:

27⁢𝐸⁢𝐼20⋅𝜑1−3⁢𝐸⁢𝐼16⋅Δ2=25−3⁢𝐸⁢𝐼16⋅𝜑1+3⁢𝐸⁢𝐼32⋅Δ2=3,75𝜑1=33,33/𝐸⁢𝐼Δ2=106,67/𝐸⁢𝐼

5. Final diagrams of internal forces in the frame:

Moment diagram [kNm]
𝑀𝑜⁢𝑠⁢𝑡=𝑀𝑝+𝑀1⋅𝜑1+𝑀2⋅Δ2

MA=0⋅kNmMBA=−3⁢EI4⋅𝜑1+3⁢EI16⋅Δ2=−5⁢kNmMBC=3⁢EI5⁢𝜑1−25=−5⁢kNmMCB=0⋅kNmMCD=10⋅kNmMD=3⁢EI16⋅Δ2−5=15⁢kNm

Calculations for the shear force diagram

Element BC:

∑𝑀𝐶=0𝑄𝐵⁢𝐶⋅5−5−8⋅5⋅2,5=0𝑄𝐵⁢𝐶=21 𝑘⁢𝑁∑𝑌=0−𝑄𝐶⁢𝐵+21−40=0𝑄𝐶⁢𝐵=−19 𝑘⁢𝑁 Coordinates of the extreme value of the moment in element BC: 𝑥21=5−𝑥19𝑥=2,625 𝑚 Extreme moment value Mex=−5−8⋅2.625⋅2.6252+21⋅2.625=22.56⁢kNm

Element AB:

∑𝑀𝐴=0𝑄𝐵⁢𝐴⋅4+5=0𝑄𝐵⁢𝐴=−1,25 𝑘⁢𝑁𝑄𝐴⁢𝐵=−1,25

Element CD:

∑𝑀𝐷=0𝑄𝐶⁢𝐷⋅4+10−15=0𝑄𝐶⁢𝐷=1,25 𝑘⁢𝑁𝑄𝐷⁢𝐶=1,25 𝑘⁢𝑁

Final diagram of axial forces [kN]

6. Kinematic verification

We assume a determinable system and draw a graph of moments from the unit force. (degree of static indeterminacy DSI = 1)

𝛿1=∫𝑀𝑜⁢𝑠⁢𝑡⋅𝑀1𝐸⁢𝐽𝑑𝑥=1𝐸⁢𝐼⁢(−13⋅1⋅5⋅4−13⋅1⋅5⋅5+13⁢8⋅528⋅1⋅5−46⋅(2⋅1⋅15+10))≈0

7. Static verification

We read the reactions (values and correct directions) from the diagrams of normal forces, shear forces, and bending moments.
Then we write down the equations of static equilibrium and check if all equations are satisfied for the read reactions.

∑𝑋=01,25−1,25=0∑𝑌=021+19−4⋅8=0∑𝑀𝐸=021⋅2,5−1,25⋅4+10−15+1,25⋅4−19⋅2,5=0