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The normal stresses of 55 N/mm2 tensile and 45 N/mm2 compressive on two mutually perpendicular planes are acting at a point in a piece of elastic material. These planes also carry shear stress of 50 N/mm2. The minor principal stress will be
- a)65.71 N/mm2 tensile
- b)65.71 N/mm2 compressive
- c)75.71 N/mm2 tensile
- d)75.71 N/mm2 compressive
Correct answer is option 'B'. Can you explain this answer?
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The normal stresses of 55 N/mm2 tensile and 45 N/mm2 compressive on t...
Minor principal stress, σ2
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σ2 = 
= 
σ2 = 65.71 N/mm2 compressive
σy = 45 N/mm2
Most Upvoted Answer
The normal stresses of 55 N/mm2 tensile and 45 N/mm2 compressive on t...
Given information:
- Normal stress in one plane = 55 N/mm2 (tensile)
- Normal stress in another plane perpendicular to the first plane = 45 N/mm2 (compressive)
- Shear stress on both planes = 50 N/mm2
To find: Minor principal stress
Solution:
1. Find the sum and difference of normal stresses:
- Sum of normal stresses = 55 + 45 = 100 N/mm2
- Difference of normal stresses = 55 - 45 = 10 N/mm2
2. Find the maximum shear stress:
- Maximum shear stress = 1/2 * (55 - 45) + 50 = 55 N/mm2
3. Find the principal stresses:
- The principal stresses are given by: σ1,2 = (σx + σy)/2 ± sqrt[(σx - σy)/2]^2 + τxy^2
- σx = 55 N/mm2 (tensile)
- σy = 45 N/mm2 (compressive)
- τxy = 50 N/mm2
- σ1,2 = (55 + 45)/2 ± sqrt[(55 - 45)/2]^2 + 50^2
- σ1 = 75.71 N/mm2 (compressive)
- σ2 = 24.29 N/mm2 (tensile)
4. Find the minor principal stress:
- The minor principal stress is the lower of the two principal stresses, which is σ2 = 24.29 N/mm2 (tensile)
Therefore, the correct answer is option B) 65.71 N/mm2 compressive.
- Normal stress in one plane = 55 N/mm2 (tensile)
- Normal stress in another plane perpendicular to the first plane = 45 N/mm2 (compressive)
- Shear stress on both planes = 50 N/mm2
To find: Minor principal stress
Solution:
1. Find the sum and difference of normal stresses:
- Sum of normal stresses = 55 + 45 = 100 N/mm2
- Difference of normal stresses = 55 - 45 = 10 N/mm2
2. Find the maximum shear stress:
- Maximum shear stress = 1/2 * (55 - 45) + 50 = 55 N/mm2
3. Find the principal stresses:
- The principal stresses are given by: σ1,2 = (σx + σy)/2 ± sqrt[(σx - σy)/2]^2 + τxy^2
- σx = 55 N/mm2 (tensile)
- σy = 45 N/mm2 (compressive)
- τxy = 50 N/mm2
- σ1,2 = (55 + 45)/2 ± sqrt[(55 - 45)/2]^2 + 50^2
- σ1 = 75.71 N/mm2 (compressive)
- σ2 = 24.29 N/mm2 (tensile)
4. Find the minor principal stress:
- The minor principal stress is the lower of the two principal stresses, which is σ2 = 24.29 N/mm2 (tensile)
Therefore, the correct answer is option B) 65.71 N/mm2 compressive.
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The normal stresses of 55 N/mm2 tensile and 45 N/mm2 compressive on two mutually perpendicular planes are acting at a point in a piece of elastic material. These planes also carry shear stress of 50 N/mm2. The minor principal stress will bea) 65.71 N/mm2 tensileb) 65.71 N/mm2 compressivec) 75.71 N/mm2 tensiled) 75.71 N/mm2 compressiveCorrect answer is option 'B'. Can you explain this answer?
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