At a point on an aluminium structure
𝜎
𝑥
𝑥
=
75
𝑀
𝑃
𝑎
,
𝜎
𝑦
𝑦
=
−
35
𝑀
𝑃
𝑎
and
𝜏
𝑥
𝑦
=
20
𝑀
𝑃
𝑎
. For this stress field, determine the factor of safety according to: (a) the Tresca criterion; (b) the Von Mises criterion. For aluminium, use yield stress
𝜎
𝑌
=
200
𝑀
𝑃
𝑎
.

Question

At a point on an aluminium structure 
𝜎
𝑥
𝑥
=
75
𝑀
𝑃
𝑎
, 
𝜎
𝑦
𝑦
=
−
35
𝑀
𝑃
𝑎
 and 
𝜏
𝑥
𝑦
=
20
𝑀
𝑃
𝑎
 . For this stress field, determine the factor of safety according to: (a) the Tresca criterion; (b) the Von Mises criterion. For aluminium, use yield stress 
𝜎
𝑌
=
200
𝑀
𝑃
𝑎
 .

BlackTom AI · Accepted Answer

We start by restating the problem to ensure we’re evaluating the same data: at a point on an aluminium structure, the in-plane stresses are σx = 75 MPa, σy = −35 MPa, and τxy = 20 MPa. The material yield stress is σY = 200 MPa. We need two factors of safety: (a) Tresca and (b) Von Mises.

Option (a, b) candidates are examined alongside others to understand why each is or isn’t plausible.

Option 1: (a) n = 1.75, (b) n = 1.99
- For Tresca, compute the principal stresses and the maximum shear: first form the 2D principal stresses from σx, σy, and τxy. The average (σx+σy)/2 = (75 − 35)/2 = 20 MPa. The half-difference is (σx − σy)/2 = (75 − (−35))/2 = 55 MPa. The magnitude of the square root term is sqrt(55^2 + 20^2) ≈ sqrt(3025 + 400) ≈ 58.56 MPa. Thus σ1 ≈ 20 + 58.56 ≈ 78.56 MPa and σ2 ≈ 20 − 58.56 ≈ −38.56 MPa. The maximum shear stress is τmax = (σ1 − σ2)/2 ≈ (78.56 − (−38.56))/2 ≈ 58.56 MPa.
- Tresca criterion yields yielding when τmax = σY/2 = 100 MPa. The factor of safety is n_Tresca = σY / (2 τmax) ≈ 200 / (2 × 58.56) ≈ 200 / 117.12 ≈ 1.71. So a value around 1.75 is not accurate to reflect the correct calculation; the closeness is insufficiently precise and the given number 1.75 is not the exact computed result.
- For Von Mises, use the 2D relation with σ3 = 0: σ_vm = sqrt(σ1^2 − σ1σ2 + σ2^2). With σ1 ≈ 78.56 MPa and σ2 ≈ −38.56 MPa, σ_vm ≈ sqrt(78.56^2 − 78.56(−38.56) + (−38.56)^2) ≈ sqrt(6169.9 + 3027.5 + 1487.5) ≈ sqrt(10685.0) ≈ 103.4 MPa. The factor of safety is n_VM = σY / σ_vm ≈ 200 / 103.4 ≈ 1.93. The option 1.99 overstates the VM-based safety slightly; it does not match the precise calculation.

Option 2: (a) n = 1.65, (b) n = 1.82
- Tresca: as above, n_Tresca ≈ 1.71. The proposed 1.65 indicates a different τmax or a miscalculation of σ1 and σ2; it departs from the precise τmax ≈ 58.56 MPa, hence not consistent with the computed value.
- Von Mises: n_VM ≈ 1.93 as shown earlier. An 1.82 value is noticeably lower than the exact calculation, suggesting either rounding differences or an error in σVM computation.

Option 3: (a) n = 1.51, (b) n = 1.70
- Tresca: again the accurate τmax leads to n_Tresca ≈ 1.71, not 1.51. This 1.51 underestimates the true factor of safety for Tresca, indicating a mis-evaluation of τmax or σY.
- Von Mises: the correct VM result around 1.93 is not close to 1.70, so this option underpredicts the VM safety by a noticeable margin.

Option 4: (a) n = 1.71, (b) n = 1.94
- For Tresca, we found τmax ≈ 58.56 MPa, giving n_Tresca ≈ 200 / (2 × 58.56) ≈ 1.71, which matches this option exactly.
- For Von Mises, σ_vm ≈ 103.39 MPa, yielding n_VM ≈ 200 / 103.39 ≈ 1.93, which rounds to 1.94 in many reference calculations. This option aligns with the precise computed values.

Option 5: (a) n = 1.81, (b) n = 2.03
- Tresca: 1.81 is higher than the exact 1.71 value, indicating an overestimate of τmax or a different approach to defining the safety factor.
- Von Mises: 2.03 is higher than the accurate 1.93, suggesting a similar rounding or calculation discrepancy that does not reflect the exact solution from the standard VM formula.

In summary, the structure of the analysis shows that only option 4 lines up with the standard, exact procedures for both criteria: Tresca yields n ≈ 1.71 and Von Mises yields n ≈ 1.94, consistent with the given data and well-established formulas.

INU1127 (24/25) Knowledge Check: Principal Stresses and Failure Criteria

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