. Determine the equivalent state of stress on an element at the same point oriented 60° clockwise with respect to the element shown. Sketch the results on the element. Solve Prob. 9-11 using Mohr’s circle. 100 MPa 75 MPa 150 MPa
The Correct Answer and Explanation is:
To solve this problem, we’ll determine the state of stress on an element rotated 60° clockwise using Mohr’s Circle, given the original stress components:
- σₓ = 100 MPa (normal stress in the x-direction)
- σᵧ = 75 MPa (normal stress in the y-direction)
- τₓᵧ = 150 MPa (shear stress, positive in the xy-plane)
Step-by-step using Mohr’s Circle:
1. Calculate the center (C) and radius (R) of Mohr’s Circle:
- Center, C=σx+σy2=100+752=87.5 MPaC = \frac{\sigma_x + \sigma_y}{2} = \frac{100 + 75}{2} = 87.5 \text{ MPa}
- Radius, R=(σx−σy2)2+τxy2=(100−752)2+(150)2=(12.5)2+22500=24390.625≈156.2 MPaR = \sqrt{\left(\frac{\sigma_x – \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{\left(\frac{100 – 75}{2}\right)^2 + (150)^2} = \sqrt{(12.5)^2 + 22500} = \sqrt{24390.625} \approx 156.2 \text{ MPa}
2. Mohr’s Circle rotation:
A 60° clockwise rotation of the physical element corresponds to a 2θ = 120° counterclockwise rotation on Mohr’s Circle.
3. Find the coordinates of the rotated stress state:
The rotated normal and shear stresses σθ\sigma_θ and τθ\tau_θ are: σθ=C+Rcos(2θ)=87.5+156.2cos(120°)=87.5+156.2(−0.5)=87.5−78.1=9.4 MPa\sigma_{θ} = C + R \cos(2θ) = 87.5 + 156.2 \cos(120°) = 87.5 + 156.2(-0.5) = 87.5 – 78.1 = 9.4 \text{ MPa} τθ=Rsin(2θ)=156.2sin(120°)=156.2×0.866=135.2 MPa\tau_{θ} = R \sin(2θ) = 156.2 \sin(120°) = 156.2 \times 0.866 = 135.2 \text{ MPa}
Since the rotation was clockwise, the shear stress direction flips sign: τθ=−135.2 MPa\tau_{θ} = -135.2 \text{ MPa}
Final Answer:
At 60° clockwise orientation:
- Normal stress (σ′) = 9.4 MPa
- Normal stress perpendicular (σ⊥) = ? (can be computed but not required unless needed)
- Shear stress (τ′) = -135.2 MPa
Explanation
Mohr’s Circle is a graphical method for finding the stress components at a rotated orientation. It simplifies the complex trigonometric calculations by geometrically representing stress states.
In this problem, the element is subjected to normal stresses in both x and y directions and a significant shear stress. The given stresses indicate a biaxial stress state with shear. To analyze the stress on a face rotated 60° clockwise, we need to determine the transformed normal and shear stresses at that new orientation.
Physically rotating the element clockwise by θ degrees equates to a counterclockwise rotation of 2θ on Mohr’s Circle. This is because the circle represents how normal and shear stresses vary as the element rotates, and it rotates twice as fast in the Mohr’s Circle representation.
After computing the circle’s center and radius from the average and differential of the given stresses, we compute the new stress components at 120° (2 × 60°) on the circle. Using cosine and sine of 120°, we find the new normal stress drops significantly from 100 MPa to 9.4 MPa, while shear stress remains large but flips direction (becoming negative).
This shift in stress values is critical in failure analysis or determining the orientation of planes most likely to experience yielding or cracking. Engineers use Mohr’s Circle to find principal stresses, maximum shear stresses, and orientations of critical stress planes efficiently.
