Questions: 직경이 40 mm 굽힘으로 발생하는 응력을 구하시오 - 단면이 원형이 경우 (Lz=ly, c=d / 2) σmax = sqrt(My^2+Mz^2) * (d/2) / (πd^4/64) = 32/πd^3 sqrt(My^2+Mz^2)

Transcript text: 직경이 40 mm 굽힘으로 발생하는 응력을 구하시오 - 단면이 원형이 경우 $\left(L_{z}=l_{y}, c=d / 2\right)$ \[ \sigma_{\max }=\frac{\sqrt{M_{y}^{2}+M_{z}^{2}} \cdot \frac{d}{2}}{\frac{\pi d^{4}}{64}}=\frac{32}{\pi \mathrm{~d}^{3}} \sqrt{M_{y}^{2}+M_{z}^{2}} \]

Solution

Solution Steps

Step 1: Determine the Reactions at Supports

Identify the support reactions at points A and D.
Use equilibrium equations to solve for the reactions.

Step 2: Calculate the Bending Moments

Calculate the bending moments at critical points (B, C, and D) using the reactions found in Step 1.
Use the moment distribution method or other appropriate methods to find the moments.

Step 3: Determine the Maximum Bending Stress

Use the given formula for maximum bending stress: \[ \sigma_{\text{max}} = \frac{32}{\pi d^3} \sqrt{M_y^2 + M_z^2} \]
Substitute the diameter $d = 40 \text{ mm}$ and the moments $M_y$ and $M_z$ calculated in Step 2 into the formula.