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Solution

Version YT

Classic version

Breakdown of force P into components

𝑃 𝑦 = 𝑃 𝑠 𝑖 𝑛 𝛼 = 5 𝑘 𝑁 𝑃 𝑦 = 𝑃 𝑐 𝑜 𝑠 𝛼 = 8.66 𝑘 𝑁

Load in the xz and xy planes, bending moment diagrams

𝑀 𝑦 = - 𝑃 𝑧 \cdot 6 - 12 \cdot 5 \cdot 62 = - 141.96 𝑘 𝑁 𝑚

𝑀 𝑧 = - 𝑃 𝑦 \cdot 6 = - 26 𝑘 𝑁 𝑚

Moment vectors, designation of characteristic section points

Normal stresses

𝜎 = 𝑀 𝑦 𝐼 𝑦 𝑧 + 𝑀 𝑧 𝐼 𝑧 𝑦 𝐼 𝑦 = 𝑏 ℎ 3 12 = 2 𝑎 \cdot (3 𝑎) 3 12 = 92 𝑎 4 𝐼 𝑧 = ℎ 𝑏 3 12 = 3 𝑎 \cdot (2 𝑎) 3 12 = 2 𝑎 4

Determination of the neutral axis equation

𝜎 = 0 𝜎 𝑚 𝑎 𝑥 = 141.92 \cdot 103 92 𝑎 4 \cdot 1.5 𝑎 + 30 \cdot 103 2 𝑎 4 \cdot 𝑎 𝑧 = - 0.475 𝑦

Determining two points is enough to draw the linear function curve

𝑦 = 0, 𝑧 = 0 𝑦 = 1, 𝑧 = - 0.475

Strength condition σ_max≤k_g

𝜎 𝑚 𝑎 𝑥 = 141.92 \cdot 103 92 𝑎 4 \cdot 1.5 𝑎 + 30 \cdot 103 2 𝑎 4 \cdot 𝑎 \leq 𝑘 𝑔 𝑎 = 0.075 𝑚 = 7.5 𝑐 𝑚

Stresses in characteristic points

𝜎 𝐴 = 147.72 𝑀 𝑃 𝑎 𝜎 𝐵 = - 76.61 𝑀 𝑃 𝑎 𝜎 𝐶 = - 147.72 𝑀 𝑃 𝑎 𝜎 𝐷 = 76.61 𝑀 𝑃 𝑎

Normal stresses diagram

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Classic version

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