A critical concrete beam in a bridge is experiencing the following stresses at a point where the axial tension threatening to crack is 80 MPa, the lateral pressure resisting deformation is 40 MPa and the shear stress trying to slide it apart is 30 MPa. Determine the radius of Mohr’s Circle to assess failure risk.
- A25 MPa
- B30 MPa
- C36 MPaCorrect
- D50 MPa
Solution
Concept
Mohrs circle radius calculation from plane stress
Formula
R = √(((σx - σy)/2)2 + τxy2)
Step-by-step solution
1. The plane stress components are: axial normal stress σx = 80 MPa, lateral normal stress σy = 40 MPa, and shear stress τxy = 30 MPa. 2. The radius of Mohr's circle is given by R = √(((σx - σy)/2)2 + τxy2). 3. Normal stress radius component: (σx - σy) / 2 = (80 - 40) / 2 = 20 MPa. 4. R = √(202 + 302) = √(400 + 900) = √(1300) = 36.05 MPa = 36 MPa. Answer: C
Explanation
The radius of Mohr's circle represents the maximum in-plane shear stress and is calculated as R = √(((σx - σy)/2)2 + τxy2). With tensile stresses σx = 80 MPa, σy = 40 MPa, and shear stress τxy = 30 MPa: (σx - σy)/2 = 20 MPa, yielding R = √(202 + 302) = √(1300) = 36.05 MPa = 36 MPa.
Why the other options are wrong
A) 25 MPa results from arithmetic errors in the radical term. B) 30 MPa is only the shear stress component. D) 50 MPa results from incorrectly setting σy = 0.
Shortcut
R = √(((80 - 40)/2)2 + 302) = √(400 + 900) = √(1300) = 36 MPa.
Common mistake
Calculating Mohr's circle center (60 MPa) instead of its radius.
Correct answer: C) 36 MPa