Shear stress 7 produces a shear strain y., (be- the x direction and lines in the y direction) μ (i.c., y = 0.0012). (a) Determine the placement of point A. (b) Determine the ry between the lines in the x' direction and on, as shown on Fig. P2.7-12.
Shear stress 7 produces a shear strain y., (be- the x direction and lines in the y direction) μ (i.c., y = 0.0012). (a) Determine the placement of point A. (b) Determine the ry between the lines in the x' direction and on, as shown on Fig. P2.7-12.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please clearly show all steps so your approach is easy to follow
![### Problem 2.7-12
**Problem Statement:**
Shear stress \( \tau \) produces a shear strain \( \gamma_{xy} \) (between lines in the \( x \) direction and lines in the \( y \) direction) of \( \gamma_{xy} = 1200 \, \mu \) (i.e., \( \gamma = 0.0012 \, \text{m/m} \)).
**Tasks:**
1. Determine the horizontal displacement \( \delta_A \) of point A.
2. Determine the shear strain \( \gamma_{x'y'} \) between the lines in the \( x' \) direction and the \( y' \) direction, as shown in Fig. P2.7-12.
**Diagram Explanation:**
The diagram shows a parallelogram-shaped section that has been distorted by shear stress \( \tau \).
- The original position is marked by the \( x \) and \( y \) axes, while the distorted position is marked by the \( x' \) and \( y' \) axes.
- Initial dimensions are provided:
- Height along the \( y \) axis: 120 mm
- Base length along the \( x \) axis: 150 mm
- The point A is located at the upper-right corner of the parallelogram.
- After distortion, the point A is displaced horizontally by \( \delta_A \).
**Detailed Steps Solution:**
1. **Horizontal Displacement, \( \delta_A \):**
Shear strain (\( \gamma_{xy} \)) relates to the horizontal displacement (\( \delta_A \)) by the formula:
\[
\gamma_{xy} = \frac{\delta_A}{y}
\]
Where \(\gamma_{xy}\) is known and \( y \) is the height of the parallelogram.
Plugging in the given values:
\[
\gamma_{xy} = 0.0012 \, \text{m/m}
\]
\[
y = 120 \, \text{mm} = 0.12 \, \text{m}
\]
Therefore,
\[
\delta_A = \gamma_{xy} \times y = 0.0012 \times 0.12 = 0.000144 \, \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93c89597-7ce2-4451-bd61-6716aa1c2201%2F11f143e6-5d5f-4603-87ae-b2bf9b97cb1c%2F8uxvvkg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 2.7-12
**Problem Statement:**
Shear stress \( \tau \) produces a shear strain \( \gamma_{xy} \) (between lines in the \( x \) direction and lines in the \( y \) direction) of \( \gamma_{xy} = 1200 \, \mu \) (i.e., \( \gamma = 0.0012 \, \text{m/m} \)).
**Tasks:**
1. Determine the horizontal displacement \( \delta_A \) of point A.
2. Determine the shear strain \( \gamma_{x'y'} \) between the lines in the \( x' \) direction and the \( y' \) direction, as shown in Fig. P2.7-12.
**Diagram Explanation:**
The diagram shows a parallelogram-shaped section that has been distorted by shear stress \( \tau \).
- The original position is marked by the \( x \) and \( y \) axes, while the distorted position is marked by the \( x' \) and \( y' \) axes.
- Initial dimensions are provided:
- Height along the \( y \) axis: 120 mm
- Base length along the \( x \) axis: 150 mm
- The point A is located at the upper-right corner of the parallelogram.
- After distortion, the point A is displaced horizontally by \( \delta_A \).
**Detailed Steps Solution:**
1. **Horizontal Displacement, \( \delta_A \):**
Shear strain (\( \gamma_{xy} \)) relates to the horizontal displacement (\( \delta_A \)) by the formula:
\[
\gamma_{xy} = \frac{\delta_A}{y}
\]
Where \(\gamma_{xy}\) is known and \( y \) is the height of the parallelogram.
Plugging in the given values:
\[
\gamma_{xy} = 0.0012 \, \text{m/m}
\]
\[
y = 120 \, \text{mm} = 0.12 \, \text{m}
\]
Therefore,
\[
\delta_A = \gamma_{xy} \times y = 0.0012 \times 0.12 = 0.000144 \, \text{
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