A bar having a cross sectional area of $700\text{ mm}^2$ is subjected to axial loads at the positions indicated in figure. Find the value of stress in the segment QR?
Show Hint
When calculating internal forces on segmented bars, always cut the segment of interest and sum all external forces to either the left or the right of the cut.
This avoids mistakes and ensures consistent sign conventions.
Step 1: Understanding the Question:
This mechanics of materials problem requires us to find the axial stress in a specific segment (QR) of a stepped bar subjected to multiple point loads along its length. Step 2: Key Formula or Approach:
The stress ($\sigma$) in any segment is given by:
\[ \sigma = \frac{F_{\text{internal}}}{A} \]
where $F_{\text{internal}}$ is the net internal axial force in that segment, obtained by using the method of sections, and $A$ is the cross-sectional area. Step 3: Detailed Explanation:
• Check the global equilibrium of the bar using the forces given in the diagram:
- Force at P: $63 \text{ kN} \leftarrow$
- Force at Q: $35 \text{ kN} \rightarrow$
- Force at R: $49 \text{ kN} \rightarrow$
- Force at S: $21 \text{ kN} \leftarrow$
- Sum of leftward forces = $63 + 21 = 84 \text{ kN}$
- Sum of rightward forces = $35 + 49 = 84 \text{ kN}$
- The bar is in static equilibrium.
• To find the internal force in segment QR, we cut a section through QR and analyze the forces on the left portion (segment PQ):
- Forces acting to the left of the cut are the $63 \text{ kN} \leftarrow$ force at P and the $35 \text{ kN} \rightarrow$ force at Q.
- Net external force on the left side:
\[ F_{\text{net, left}} = 63 \text{ kN} (\leftarrow) - 35 \text{ kN} (\rightarrow) = 28 \text{ kN} (\leftarrow) \]
- For equilibrium, the internal force in segment QR ($F_{\text{QR}}$) must balance this net force:
\[ F_{\text{QR}} = 28 \text{ kN} \text{ (Tension)} \]
• Calculate the axial stress in QR:
- Area of the bar, $A = 700 \text{ mm}^2$
\[ \sigma_{\text{QR}} = \frac{F_{\text{QR}}}{A} = \frac{28 \text{ kN}}{700 \text{ mm}^2} = \frac{28000 \text{ N}}{700 \text{ mm}^2} = 40 \text{ N/mm}^2 = 40 \text{ MPa} \] Step 4: Final Answer:
The stress in the segment QR is $40 \text{ MPa}$.
