A reinforced concrete beam-column joint in a seismic-resistant frame is subjected to pure shear stress due to earthquake loading. The stress state at the joint is given as follows: Shear stress τ12 = 50 MPa and Normal stresses σ1 = σ2 = 0. What is the maximum normal stress developed at the joint?
- A± 25 MPa
- B± 35.35 MPa
- C± 50 MPaCorrect
- D± 70.71 MPa
Solution
Concept
Principal normal stresses in pure shear state
Formula
σ1,2 = +- τxy
Step-by-step solution
1. In a state of pure shear, normal stresses on the coordinate planes are zero: σ1 = σ2 = 0, and shear stress is τ12 = 50 MPa. 2. Principal stresses are σmax,min = (σ1 + σ2)/2 +- √(((σ1 - σ2)/2)2 + τ122). 3. Substituting values: σmax,min = 0 +- √(0 + 502) = ± 50 MPa. 4. Therefore, the maximum normal stresses developed on planes at 45° are ± 50 MPa (one tensile, one compressive). Answer: C
Explanation
A state of pure shear stress (σ1 = σ2 = 0, τ12 = 50 MPa) produces principal normal stresses of equal magnitude and opposite sign acting on planes inclined at 45° to the shear planes: σmax = +τ = +50 MPa (tensile) and σmin = -τ = -50 MPa (compressive). Hence, the maximum normal stress developed is ± 50 MPa.
Why the other options are wrong
A) ± 25 MPa is half the shear stress, which is incorrect. B) ± 35.35 MPa equals 50/√(2), confusing resolution of components. D) ± 70.71 MPa equals 50·√(2).
Shortcut
Pure shear state (τ = 50 MPa) => Principal normal stresses are +- τ = ± 50 MPa.
Common mistake
Halving the shear stress to get ± 25 MPa.
Correct answer: C) ± 50 MPa