A thin rectangular plate is subjected to the following plane stress state: σx = 120 MPa (Tensile), σy = 60 MPa (Tensile) and τxy = 40 MPa. What is the maximum in-plane shear stress (τmax)?
- A30 MPa
- B40 MPa
- C50 MPaCorrect
- D60 Mpa
Solution
Concept
Maximum in plane shear stress calculation
Formula
τmax = √(((σx - σy)/2)2 + τxy2)
Step-by-step solution
1. The state of plane stress is: σx = 120 MPa, σy = 60 MPa, and τxy = 40 MPa. 2. Maximum in-plane shear stress is τmax = √(((σx - σy)/2)2 + τxy2). 3. (σx - σy) / 2 = (120 - 60) / 2 = 30 MPa. 4. τmax = √(302 + 402) = √(900 + 1600) = √(2500) = 50 MPa. Answer: C
Explanation
Maximum in-plane shear stress is given by τmax = √(((σx - σy)/2)2 + τxy2). Substituting σx = 120 MPa, σy = 60 MPa, and τxy = 40 MPa: (σx - σy)/2 = 30 MPa; τmax = √(302 + 402) = √(900 + 1600) = √(2500) = 50 MPa.
Why the other options are wrong
A) 30 MPa is only the normal stress radius component ((σx - σy)/2). B) 40 MPa is the applied shear stress τxy. D) 60 MPa is the normal stress difference (σx - σy).
Shortcut
3-4-5 right triangle: 30, 40 => Hypotenuse = 50 MPa (τmax = 50 MPa).
Common mistake
Calculating average normal stress (90 MPa) or subtracting shear directly.
Correct answer: C) 50 MPa