Example 1

To determine and draw internal force diagrams.

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Solution

1. Marking characteristic points and reactions on supports

2. Calculating reactions using equilibrium equations

∑𝑋=0−6+𝐻𝐷=0𝐻𝐷=6⁢𝑘⁢𝑁∑𝑌=02−2⋅6+6+𝑉𝐵=0𝐵𝐵=4⁢𝑘⁢𝑁∑𝑀𝐵=02⋅3+6−6⋅4+6⋅2⋅3−𝑀𝐷=0𝑀𝐷=24⁢𝑘⁢𝑁

3. Expressing internal force equations in individual range variations:

a) Interval AB x \in{\langle 0,3)}

𝑄𝐴⁢𝐵=2 𝑘⁢𝑁𝑀𝐴⁢𝐵=2⁢𝑥𝑀𝐴⁡(0)=0𝑀𝐵⁡(3)=6 𝑘⁢𝑁⁢𝑚

b) Interval BC x \in{\langle 3,7)}

𝑄𝐵⁢𝐶=2+4–2⁢(𝑥−3)𝑀𝐵⁢𝐶=2⁢𝑥+4⁢(𝑥−3)+6–12⋅2⁢(𝑥−3)2𝑄𝐵⁡(3)=6 𝑘⁢𝑁𝑄𝐶⁡(7)=2 𝑘⁢𝑁𝑄𝐵⁢𝐶=06−2⁢(𝑥−3)=06−2⁢𝑥+6=0𝑥=6𝑀𝐵⁡(3)=12 𝑘⁢𝑁⁢𝑚𝑀𝑚⁢𝑎⁢𝑥⁡(6)=21 𝑘⁢𝑁⁢𝑚𝑀𝐶⁡(7)=20 𝑘⁢𝑁⁢𝑚

c) Interval DC x \in{\langle 0,2)}

𝑄𝐷⁢𝐶=2⁢𝑥𝑄𝐷⁡(0)=0𝑄𝐶⁡(2)=4 𝑘⁢𝑁𝑀𝐷⁢𝐶=24–2⁢𝑥⁢𝑥2=24–𝑥2𝑀𝐷⁡(0)=24 𝑘⁢𝑁⁢𝑚𝑀𝐶⁡(2)=20 𝑘⁢𝑁⁢𝑚𝑁𝐷⁢𝐶=6 𝑘⁢𝑁

4. Final graphs