A steel bridge girder instrumented with strain gauges records the following at a critical point during a load test: Longitudinal strain (ε1) : 90 × 10–6 Transverse strain (ε2) : −60 × 10–6 Engineering shear strain (γ12) : 120 × 10–6 The maximum in-plane engineering shear strain is
- A120 × 10–6
- B150 × 10–6
- C192 × 10–6Correct
- D220 × 10–6
Solution
Concept
Maximum in plane engineering shear strain calculation
Formula
γmax = √((ε1 - ε2)2 + γ122)
Step-by-step solution
1. Mohr's strain circle relates normal strains and shear strains: radius R = √(((ε1 - ε2)/2)2 + (γ12 / 2)2). 2. Maximum engineering shear strain is γmax = 2 × R = √((ε1 - ε2)2 + γ122). 3. Given ε1 = 90x10-6, ε2 = -60x10-6, and γ12 = 120x10-6. 4. (ε1 - ε2) = 90 - (-60) = 150x10-6. 5. γmax = √(1502 + 1202) x 10-6 = √(22500 + 14400) x 10-6 = √(36900) x 10-6 = 192.09x10-6 = 192x10-6. Answer: C
Explanation
The maximum in-plane engineering shear strain (γmax) is equal to the diameter of Mohr’s strain circle: γmax = √((ε1 - ε2)2 + γ122). Substituting ε1 = 90x10-6, ε2 = -60x10-6, and γ12 = 120x10-6: γmax = √((150)2 + (120)2) x 10-6 = √(22500 + 14400) x 10-6 = 192.09x10-6 = 192x10-6.
Why the other options are wrong
A) 120x10-6 is the recorded shear strain γ12. B) 150x10-6 is the difference in normal strains (ε1 - ε2). D) 220x10-6 is an overestimation.
Shortcut
γmax = √((e1 - e2)2 + γ122) = √(1502 + 1202) x 10-6 = 192 x 10-6.
Common mistake
Forgetting to subtract negative transverse strain, using 90 - 60 = 30 instead of 150.
Correct answer: C) 192 × 10–6