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Machine Design

Design Philosophy and Failure Theories

Design principles for design philosophy and failure theories for Mechanical Engineering and GATE & IES preparation.

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89 Active Questions
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Question 1 of 89
Original Exam: GATE (2005)
Which of the following is NOT a primary consideration in machine design philosophy?
✓ Correct Answer: C
Explanation:
Design philosophy in machine design focuses on functional and engineering aspects: factor of safety (strength vs. load), reliability (probability of satisfactory performance), and manufacturability (ease of production). Pure aesthetic appearance is NOT a primary machine design philosophy consideration. Core design philosophies: (1) Strength-based design, (2) Stiffness-based design, (3) Reliability-based design, (4) Design for manufacturing.
Difficulty: basic Type: conceptual
Question 2 of 89
Original Exam: GATE (2008)
Factor of safety (FOS) is defined as:
✓ Correct Answer: C
Explanation:
Factor of Safety (FOS) = Failure stress / Allowable (working) stress. For ductile materials: FOS = Yield strength / Working stress. For brittle materials: FOS = Ultimate tensile strength / Working stress. FOS > 1 always; typical values range from 1.5 to 10 depending on application and uncertainty in loading.
Difficulty: basic Type: conceptual
Question 3 of 89
Original Exam: IES (2010)
A machine component is designed with FOS = 2.5. If yield strength = 250 MPa, the allowable stress is:
✓ Correct Answer: C
Explanation:
Allowable stress = Yield strength / FOS = 250 / 2.5 = 100 MPa. This is the maximum stress the component is designed to withstand under normal operating conditions. FOS of 2.5 means the component can theoretically withstand 2.5 times the design load before yielding.
Difficulty: basic Type: numerical
Question 4 of 89
Original Exam: GATE (2012)
Which design philosophy is most appropriate for components subjected to repeated loading?
✓ Correct Answer: B
Explanation:
For components subjected to repeated/cyclic loading, fatigue design is most appropriate. Fatigue failure occurs at stresses well below UTS or yield strength. The endurance limit (Se) is the maximum stress amplitude below which a material can withstand infinite cycles. Key concepts: S-N curve, endurance limit, stress concentration factors, Goodman/Gerber diagrams.
Difficulty: intermediate Type: conceptual
Question 5 of 89
Original Exam: GATE (2003)
The concept of stress concentration in machine design refers to:
✓ Correct Answer: D
Explanation:
Stress concentration occurs at geometric discontinuities such as holes, notches, keyways, fillets, and abrupt changes in cross-section. Stress concentration factor Kt = sigma_max / sigma_nom. For fatigue: Kf = 1 + q(Kt - 1), where q is notch sensitivity (0 to 1). Stress concentration is critical in fatigue design as cracks initiate at these points.
Difficulty: intermediate Type: conceptual
Question 6 of 89
Original Exam: GATE (2014)
A shaft has stress concentration factor Kt = 2.5 and notch sensitivity q = 0.8. The fatigue stress concentration factor Kf is:
✓ Correct Answer: C
Explanation:
Kf = 1 + q(Kt - 1) = 1 + 0.8(2.5 - 1) = 1 + 0.8 x 1.5 = 1 + 1.2 = 2.2. Note: q = 0 means material is insensitive to notches (Kf = 1); q = 1 means fully sensitive (Kf = Kt). For the given values: Kf = 2.2.
Difficulty: intermediate Type: numerical
Question 7 of 89
Original Exam: IES (2009)
In machine design, endurance limit is:
✓ Correct Answer: D
Explanation:
The endurance limit (Se) is the maximum stress amplitude below which a material can theoretically withstand infinite stress cycles without fatigue failure. For steels: Se approx 0.5 x Sut for Sut <= 1400 MPa. For non-ferrous metals, there is no true endurance limit. Modified endurance limit: Se_prime = ka x kb x kc x kd x ke x Se.
Difficulty: basic Type: conceptual
Question 8 of 89
Original Exam: GATE (2002)
The stress concentration factor (Kt) for a plate with a central circular hole under uniaxial tension is approximately:
✓ Correct Answer: C
Explanation:
For a plate with a small central circular hole under uniaxial tension, Kt = 3 (Kirsch solution). The maximum stress occurs at the edge of the hole perpendicular to loading: sigma_max = 3 x sigma_nom. For an elliptical hole with semi-axes a and b: Kt = 1 + 2a/b. For circular hole: a = b, Kt = 3.
Difficulty: intermediate Type: conceptual
Question 9 of 89
Original Exam: IES (2014)
The surface finish factor (ka) in fatigue design accounts for:
✓ Correct Answer: A
Explanation:
Surface finish factor (ka) modifies the endurance limit for actual surface roughness vs. polished lab specimens. ka values: Mirror-polished: 1.0, Ground: 0.9, Machined/Cold-drawn: 0.7-0.8, Hot-rolled: 0.6-0.7, As-forged: 0.4-0.6. Modified endurance limit: Se_prime = ka x kb x kc x kd x ke x Se.
Difficulty: intermediate Type: conceptual
Question 10 of 89
Original Exam: GATE (2019)
The size factor (kb) in fatigue design is less than 1 for larger components because:
✓ Correct Answer: A
Explanation:
Size factor (kb) accounts for the statistical effect that larger components have greater volume of material at high stress, increasing probability of encountering a critical microstructural flaw. kb < 1 for d > 7.62 mm. Shigley formula: kb = 1.24 x d^(-0.107) for 2.79 <= d <= 51 mm. This is the statistical weakest-link theory applied to fatigue.
Type: conceptual
Question 11 of 89
Original Exam: GATE (2001)
According to Maximum Principal Stress Theory (Rankine), failure occurs when:
✓ Correct Answer: B
Explanation:
Rankine Maximum Principal Stress Theory: Failure when sigma_1 = Syt (tensile) or sigma_3 = -Syc (compressive). Suitable for brittle materials (cast iron, ceramics). Does NOT account for effect of other principal stresses. For biaxial stress: failure if sigma_1 >= Syt or sigma_2 <= -Syc.
Difficulty: basic Type: conceptual
Question 12 of 89
Original Exam: GATE (2006)
The Maximum Shear Stress Theory (Tresca) predicts failure when:
✓ Correct Answer: D
Explanation:
Tresca Theory: Failure when tau_max = Syt/2. For biaxial stress: (sigma_1 - sigma_2)/2 = Syt/2, so sigma_1 - sigma_2 = Syt. Tresca is conservative (safe side) vs. von Mises. Suitable for ductile materials. Tresca gives hexagonal yield surface; von Mises gives ellipse. Tresca is 15.5% more conservative than von Mises at worst case.
Difficulty: intermediate Type: conceptual
Question 13 of 89
Original Exam: GATE (2009)
The von Mises stress for biaxial stress state (sigma_1, sigma_2) is:
✓ Correct Answer: D
Explanation:
Von Mises stress for biaxial: sigma_vm = sqrt(sigma_1^2 + sigma_2^2 - sigma_1*sigma_2). For triaxial: sigma_vm = (1/sqrt(2)) x sqrt[(sigma_1-sigma_2)^2 + (sigma_2-sigma_3)^2 + (sigma_3-sigma_1)^2]. Failure when sigma_vm = Syt. For 2D: sigma_vm = sqrt(sigma_x^2 + sigma_y^2 - sigma_x*sigma_y + 3*tau_xy^2).
Difficulty: intermediate Type: conceptual
Question 14 of 89
Original Exam: GATE (2011)
For a ductile material with sigma_1 = 200 MPa and sigma_2 = 100 MPa, the von Mises stress is:
✓ Correct Answer: D
Explanation:
sigma_vm = sqrt(200^2 + 100^2 - 200x100) = sqrt(40000 + 10000 - 20000) = sqrt(30000) = 173.2 MPa. If Syt = 250 MPa, FOS = 250/173.2 = 1.44. Tresca: tau_max = (200-100)/2 = 50 MPa; failure if tau_max = Syt/2 = 125 MPa -> safe.
Difficulty: intermediate Type: numerical
Question 15 of 89
Original Exam: IES (2004)
Which failure theory is most suitable for brittle materials?
✓ Correct Answer: D
Explanation:
Rankine Maximum Principal Stress Theory is most suitable for brittle materials (cast iron, glass, ceramics, concrete). Brittle materials fail by fracture along planes of maximum tensile stress. For ductile materials, von Mises and Tresca are preferred. Modified Mohr and Coulomb-Mohr theories are also used for brittle materials when Suc >> Sut.
Difficulty: basic Type: conceptual
Question 16 of 89
Original Exam: IES (2013)
The Coulomb-Mohr Theory is particularly useful for:
✓ Correct Answer: D
Explanation:
Coulomb-Mohr (Internal Friction) Theory is used for brittle materials where Suc >> Sut, such as cast iron (Suc approx 3-4 x Sut). Failure criterion: sigma_1/Sut - sigma_3/Suc = 1. For cast iron: Sut approx 200 MPa, Suc approx 600-800 MPa.
Difficulty: intermediate Type: conceptual
Question 17 of 89
Original Exam: GATE (2016)
For a shaft under combined bending M = 1000 N.m and torque T = 1000 N.m, the equivalent torque (Te) by Maximum Shear Stress Theory is:
✓ Correct Answer: C
Explanation:
Equivalent torque by Tresca: Te = sqrt(M^2 + T^2) = sqrt(1000^2 + 1000^2) = sqrt(2000000) = 1414.2 N.m. Shaft design: tau = 16Te/(pi*d^3) <= Ssy = Syt/2. Equivalent bending moment by Rankine: Me = (M + Te)/2 = (1000 + 1414.2)/2 = 1207.1 N.m.
Difficulty: intermediate Type: numerical
Question 18 of 89
Original Exam: GATE (2013)
The Goodman line in fatigue design relates:
✓ Correct Answer: A
Explanation:
Modified Goodman line: sigma_a/Se + sigma_m/Sut = 1/FOS. Where Se = endurance limit, Sut = ultimate tensile strength. Other criteria: Gerber parabola: sigma_a/Se + (sigma_m/Sut)^2 = 1. Soderberg: sigma_a/Se + sigma_m/Syt = 1. ASME-Elliptic: (sigma_a/Se)^2 + (sigma_m/Syt)^2 = 1.
Difficulty: intermediate Type: conceptual
Question 19 of 89
Original Exam: IES (2016)
A component has Se = 200 MPa and Sut = 400 MPa. Using Goodman criterion with sigma_m = 100 MPa, the allowable sigma_a is:
✓ Correct Answer: D
Explanation:
Goodman (FOS=1): sigma_a/Se + sigma_m/Sut = 1. sigma_a/200 + 100/400 = 1. sigma_a/200 = 1 - 0.25 = 0.75. sigma_a = 0.75 x 200 = 150 MPa. With no mean stress, sigma_a = Se = 200 MPa. Mean stress reduces allowable alternating stress.
Difficulty: intermediate Type: numerical
Question 20 of 89
Original Exam: IES (2011)
The Soderberg criterion is more conservative than Goodman because it uses:
✓ Correct Answer: A
Explanation:
Soderberg: sigma_a/Se + sigma_m/Syt = 1/FOS. Uses Syt (< Sut) for mean stress axis, making it more conservative. Order of conservatism: Soderberg > Goodman > ASME-Elliptic > Gerber. Soderberg is rarely used in practice as it is overly conservative. Goodman is most widely used.
Difficulty: intermediate Type: conceptual
Question 21 of 89
Original Exam: GATE (2017)
The Gerber parabola for fatigue design is:
✓ Correct Answer: B
Explanation:
Gerber parabola: sigma_a/Se + (sigma_m/Sut)^2 = 1. Parabolic curve fits experimental fatigue data better than Goodman line. Less conservative than Goodman. Passes through same endpoints (Se on sigma_a-axis, Sut on sigma_m-axis). Goodman preferred for design (conservative, linear). ASME-Elliptic: (sigma_a/Se)^2 + (sigma_m/Syt)^2 = 1.
Difficulty: intermediate Type: conceptual
Question 22 of 89
Original Exam: GATE (2010)
For a ductile material with Syt = 300 MPa under pure shear, the shear yield strength (Ssy) by von Mises theory is:
✓ Correct Answer: A
Explanation:
Von Mises for pure shear (sigma_1 = tau, sigma_2 = -tau): sigma_vm = tau*sqrt(3) = Syt at yield. Ssy = Syt/sqrt(3) = 0.577 x 300 = 173.2 MPa. Tresca: Ssy = Syt/2 = 150 MPa. Von Mises gives higher (less conservative) shear yield strength. Experimental results closer to von Mises for ductile materials.
Difficulty: intermediate Type: numerical
Question 23 of 89
Original Exam: IES (2006)
Which failure theory gives the most accurate prediction for ductile materials?
✓ Correct Answer: C
Explanation:
Distortion Energy Theory (von Mises) gives most accurate predictions for ductile materials. Experimental evidence (Taylor and Quinney, 1931) shows yielding of ductile metals closely follows von Mises criterion. Theory correctly predicts hydrostatic stress does not cause yielding. Von Mises is used in FEA as primary yield criterion for ductile materials.
Difficulty: basic Type: conceptual
Question 24 of 89
Original Exam: GATE (2008)
For a biaxial stress state with sigma_1 = 200 MPa and sigma_2 = -100 MPa, the maximum shear stress is:
✓ Correct Answer: D
Explanation:
tau_max = (sigma_1 - sigma_2)/2 = (200 - (-100))/2 = 300/2 = 150 MPa. When sigma_1 and sigma_2 have opposite signs, tau_max = (sigma_1 - sigma_2)/2. For Tresca: failure if tau_max = Syt/2, so Syt = 300 MPa required. Von Mises: sigma_vm = sqrt(200^2 + 100^2 + 200x100) = sqrt(70000) = 264.6 MPa.
Difficulty: intermediate Type: numerical
Question 25 of 89
Original Exam: GATE (2018)
The Paris Law for fatigue crack growth is:
✓ Correct Answer: B
Explanation:
Paris Law: da/dN = C*(delta_K)^m, where a = crack length, N = cycles, delta_K = stress intensity factor range = K_max - K_min = delta_sigma*sqrt(pi*a)*F. C and m are material constants (m approx 2-4 for metals). Applies in Region II (stable crack growth). Region I: below threshold delta_K_th. Region III: rapid growth approaching Kc.
Type: conceptual
Question 26 of 89
Original Exam: IES (2017)
Fracture toughness (KIc) represents:
✓ Correct Answer: A
Explanation:
Fracture toughness KIc (plane strain) is the critical stress intensity factor for Mode I (crack opening) loading. KIc = sigma_f * sqrt(pi*a) * F. Units: MPa*sqrt(m). Failure when KI >= KIc. High KIc: structural steels (50-200 MPa*sqrt(m)). Low KIc: ceramics (1-5), glass (0.7-1). Used to determine critical crack size: ac = (1/pi)*(KIc/sigma)^2.
Type: conceptual
Question 27 of 89
Original Exam: GATE (2021)
Miner's rule for cumulative fatigue damage states failure occurs when:
✓ Correct Answer: B
Explanation:
Miner Rule: D = Sum(ni/Ni), where ni = actual cycles at stress level i, Ni = cycles to failure at that level. Failure when D >= 1. Assumptions: linear damage accumulation, order-independent. Limitations: actual failure at D = 0.7 to 2.2; does not account for load sequence effects.
Difficulty: intermediate Type: conceptual
Question 28 of 89
Original Exam: IES (2019)
A component is subjected to 10^5 cycles at sigma_1 (N1 = 2x10^5) and 5x10^4 cycles at sigma_2 (N2 = 10^5). Miner damage fraction is:
✓ Correct Answer: C
Explanation:
D = n1/N1 + n2/N2 = 10^5/(2x10^5) + 5x10^4/10^5 = 0.5 + 0.5 = 1.0. D = 1.0 means component has reached fatigue life. Remaining life = 1 - D = 0.
Difficulty: intermediate Type: numerical
Question 29 of 89
Original Exam: IES (2008)
The octahedral shear stress theory is equivalent to:
✓ Correct Answer: C
Explanation:
Octahedral Shear Stress Theory is mathematically equivalent to von Mises. tau_oct = (1/3)*sqrt[(sigma_1-sigma_2)^2 + (sigma_2-sigma_3)^2 + (sigma_3-sigma_1)^2]. At yield in simple tension: tau_oct_yield = (sqrt(2)/3)*Syt. Failure: tau_oct = (sqrt(2)/3)*Syt, identical to von Mises criterion.
Type: conceptual
Question 30 of 89
Original Exam: IES (2012)
The stress ratio R in fatigue loading is defined as:
✓ Correct Answer: D
Explanation:
Stress ratio R = sigma_min/sigma_max. Fully reversed: R = -1. Repeated (zero to max): R = 0. Static: R = 1. Mean stress: sigma_m = (sigma_max + sigma_min)/2. Alternating stress: sigma_a = (sigma_max - sigma_min)/2. R = -1 is most damaging for given sigma_max.
Difficulty: basic Type: conceptual
Question 31 of 89
Original Exam: GATE (2012)
For a ductile material, ratio Ssy/Syt by von Mises theory is:
✓ Correct Answer: B
Explanation:
Von Mises for pure shear: sigma_vm = tau*sqrt(3) = Syt at yield. Ssy = Syt/sqrt(3) = 0.577*Syt. Ssy/Syt = 1/sqrt(3) = 0.577. Tresca: Ssy/Syt = 0.5. Experimental results for ductile metals: Ssy/Syt approx 0.55-0.60, closer to von Mises.
Difficulty: intermediate Type: numerical
Question 32 of 89
Original Exam: GATE (2011)
The Tresca yield criterion in terms of principal stresses (sigma_1 > sigma_2 > sigma_3) is:
✓ Correct Answer: B
Explanation:
Tresca: tau_max = (sigma_1 - sigma_3)/2 = Syt/2, where sigma_1 > sigma_2 > sigma_3. Therefore sigma_1 - sigma_3 = Syt. For biaxial (sigma_3=0): if sigma_1 > 0 > sigma_2: sigma_1 - sigma_2 = Syt. If sigma_1 > sigma_2 > 0: sigma_1 = Syt. Option D is von Mises criterion.
Difficulty: intermediate Type: conceptual
Question 33 of 89
Original Exam: IES (2023)
The Goodman criterion with factor of safety n is:
✓ Correct Answer: D
Explanation:
Modified Goodman with FOS n: sigma_a/Se + sigma_m/Sut = 1/n. Rearranging: n = 1/(sigma_a/Se + sigma_m/Sut). For n=1: failure boundary. For n>1: safe region. Example: sigma_a=100, sigma_m=50, Se=200, Sut=400: n = 1/(0.5+0.125) = 1.6.
Difficulty: intermediate Type: conceptual
Question 34 of 89
Original Exam: IES (2004)
For a rotating shaft under constant transverse load, the bending stress is:
✓ Correct Answer: D
Explanation:
For a rotating shaft under constant transverse load, bending stress at any surface point is completely reversed (R = -1). As shaft rotates, point alternates between max tension and max compression. Mean stress sigma_m = 0. Alternating stress sigma_a = sigma_max = 32M/(pi*d^3). Torsional stress for constant torque is static (mean only).
Difficulty: basic Type: conceptual
Question 35 of 89
Original Exam: GATE (2015)
The Soderberg criterion for fatigue is:
✓ Correct Answer: C
Explanation:
Soderberg: sigma_a/Se + sigma_m/Syt = 1/FOS. Uses Syt for mean stress axis. Most conservative fatigue criterion. Ensures no yielding (sigma_m <= Syt). Rarely used in practice due to excessive conservatism. Goodman (uses Sut) preferred for most engineering applications.
Difficulty: intermediate Type: conceptual
Question 36 of 89
Original Exam: GATE (2020)
The ASME-Elliptic criterion for fatigue is:
✓ Correct Answer: A
Explanation:
ASME-Elliptic: (sigma_a/Se)^2 + (sigma_m/Syt)^2 = 1. Passes through (0, Se) and (Syt, 0). Lies between Goodman and Soderberg in conservatism. Recommended by ASME Boiler and Pressure Vessel Code for some applications. Better represents actual failure boundary for many ductile materials.
Type: conceptual
Question 37 of 89
Original Exam: GATE (2022)
The Langer static yield line on the Goodman diagram is:
✓ Correct Answer: A
Explanation:
Langer static yield line: sigma_a + sigma_m = Syt. Represents boundary for first-cycle yielding. Points above this line yield on first load application. Goodman diagram has two failure boundaries: (1) Goodman line (fatigue), (2) Langer line (yielding). Safe design is below BOTH lines.
Difficulty: intermediate Type: conceptual
Question 38 of 89
Original Exam: GATE (2020)
For fully reversed loading (sigma_m = 0), the Goodman criterion reduces to:
✓ Correct Answer: A
Explanation:
Goodman: sigma_a/Se + sigma_m/Sut = 1/FOS. For sigma_m = 0: sigma_a/Se = 1/FOS. For FOS=1: sigma_a = Se. Fully reversed loading (R = -1) is the most severe fatigue condition for given stress amplitude. The endurance limit Se is the maximum allowable alternating stress for infinite life with zero mean stress.
Difficulty: basic Type: conceptual
Question 39 of 89
Original Exam: IES (2020)
The Haigh diagram has:
✓ Correct Answer: D
Explanation:
Haigh diagram (Modified Goodman diagram): x-axis = mean stress (sigma_m), y-axis = stress amplitude (sigma_a). Key points: (0, Se) and (Sut, 0). Goodman line connects these. Langer line: sigma_a + sigma_m = Syt. Safe region is below both lines. FOS determined by ratio of distances from origin to failure line vs. operating point.
Difficulty: basic Type: conceptual
Question 40 of 89
Original Exam: GATE (2014)
Von Mises stress for 2D stress state (sigma_x, sigma_y, tau_xy) is:
✓ Correct Answer: C
Explanation:
Von Mises for 2D (sigma_z=0): sigma_vm = sqrt(sigma_x^2 + sigma_y^2 - sigma_x*sigma_y + 3*tau_xy^2). For pure shear (sigma_x=sigma_y=0): sigma_vm = tau_xy*sqrt(3). Failure when sigma_vm = Syt. Used in FEA post-processing to evaluate yielding.
Type: numerical
Question 41 of 89
Original Exam: IES (2010)
Notch sensitivity (q) in fatigue design ranges from:
✓ Correct Answer: B
Explanation:
Notch sensitivity q ranges 0 to 1. q=0: insensitive to notches (Kf=1). q=1: fully sensitive (Kf=Kt). Kf = 1 + q(Kt-1). q increases with notch radius and material strength. High-strength steels: q approx 1. Soft ductile materials: q approx 0. Peterson formula: q = 1/(1 + a/r).
Difficulty: basic Type: conceptual
Question 42 of 89
Original Exam: IES (2021)
For a shaft with sigma_x = 150 MPa and tau_xy = 100 MPa (sigma_y = 0), the von Mises stress is:
✓ Correct Answer: A
Explanation:
sigma_vm = sqrt(sigma_x^2 + sigma_y^2 - sigma_x*sigma_y + 3*tau_xy^2) = sqrt(150^2 + 0 - 0 + 3*100^2) = sqrt(22500 + 30000) = sqrt(52500) = 229.1 MPa. If Syt = 400 MPa: FOS = 400/229.1 = 1.75. Component is safe.
Difficulty: intermediate Type: numerical
Question 43 of 89
Original Exam: GATE (2006)
The design (allowable) stress for a machine component is:
✓ Correct Answer: A
Explanation:
Design stress = Failure stress / FOS. For ductile: sigma_design = Syt/FOS. For brittle: sigma_design = Sut/FOS. For fatigue: sigma_design = Se/FOS. FOS accounts for uncertainties in material properties, load estimation, stress analysis, manufacturing tolerances, and environmental effects.
Difficulty: basic Type: conceptual
Question 44 of 89
Original Exam: GATE (2017)
Stress intensity factor KI for central crack (length 2a) in infinite plate under stress sigma is:
✓ Correct Answer: D
Explanation:
KI = sigma*sqrt(pi*a)*F(geometry), F=1 for infinite plate. Units: MPa*sqrt(m). Fracture when KI = KIc. Critical crack size: ac = (1/pi)*(KIc/sigma)^2. Foundation of Linear Elastic Fracture Mechanics (LEFM). For finite plates, F > 1.
Type: conceptual
Question 45 of 89
Original Exam: IES (2005)
The S-N (Wohler) curve is plotted as:
✓ Correct Answer: C
Explanation:
Wohler S-N curve: y-axis = stress amplitude (S or sigma_a), x-axis = cycles to failure N (log scale). For ferrous metals: curve becomes horizontal at Se at approx 10^6-10^7 cycles. For non-ferrous: no true endurance limit. Obtained from rotating bending fatigue tests on polished specimens. Represents 50% probability of failure.
Difficulty: basic Type: conceptual
Question 46 of 89
Original Exam: GATE (2003)
For steel with Sut = 600 MPa, the approximate endurance limit Se is:
✓ Correct Answer: A
Explanation:
For steel with Sut <= 1400 MPa: Se approx 0.5 x Sut = 0.5 x 600 = 300 MPa. For Sut > 1400 MPa: Se approx 700 MPa (constant). This is for standard polished rotating bending specimen (6-8 mm diameter). Actual component endurance limit is lower due to surface finish, size, reliability factors.
Difficulty: basic Type: numerical
Question 47 of 89
Original Exam: IES (2016)
Creep failure in machine design is associated with:
✓ Correct Answer: C
Explanation:
Creep: time-dependent permanent deformation under sustained stress at elevated temperatures (T > 0.3-0.4 Tm). Stages: (1) Primary: decreasing creep rate, (2) Secondary: constant minimum creep rate, (3) Tertiary: accelerating rate leading to fracture. Important in: gas turbine blades, boiler tubes, nuclear reactor components.
Difficulty: intermediate Type: conceptual
Question 48 of 89
Original Exam: GATE (2009)
The toughness of a material is best represented by:
✓ Correct Answer: A
Explanation:
Toughness = ability to absorb energy before fracture = area under stress-strain curve from zero to fracture strain. Modulus of toughness = integral(sigma*d_epsilon) from 0 to epsilon_f. Approximate: Ut approx (Syt + Sut)/2 x epsilon_f. High toughness: low-carbon steel, copper, aluminum alloys. Low toughness (brittle): cast iron, ceramics.
Difficulty: basic Type: conceptual
Question 49 of 89
Original Exam: IES (2007)
For a shaft under pure torsion, Tresca predicts yielding when:
✓ Correct Answer: C
Explanation:
Pure torsion: sigma_1 = tau, sigma_2 = -tau. Tresca: tau_max = (sigma_1-sigma_2)/2 = tau. Failure when tau_max = Syt/2, so tau = Syt/2. Von Mises: sigma_vm = tau*sqrt(3) = Syt, so tau = 0.577*Syt. Tresca gives Ssy = 0.5*Syt; von Mises gives Ssy = 0.577*Syt. Experimental results closer to von Mises.
Difficulty: intermediate Type: conceptual
Question 50 of 89
Original Exam: GATE (2007)
Proof resilience of a material is:
✓ Correct Answer: C
Explanation:
Proof resilience = max energy stored per unit volume within elastic limit = Syt^2/(2E). Area under stress-strain curve up to yield point. High resilience materials used for springs (high Syt, low E). Example: Spring steel Syt=1500 MPa, E=200 GPa: Resilience = 1500^2/(2x200000) = 5.625 MJ/m^3.
Difficulty: intermediate Type: conceptual
Question 51 of 89
Original Exam: GATE (2005)
Hoop stress in a thin-walled cylindrical pressure vessel is:
✓ Correct Answer: C
Explanation:
Thin-walled cylinder (d/t > 20): Hoop stress sigma_h = pd/(2t) = pr/t. Longitudinal stress sigma_L = pd/(4t) = pr/(2t). sigma_h = 2*sigma_L. Critical stress is hoop stress. Design: t = pd/(2*sigma_allow). Spherical vessel: sigma = pd/(4t) in all directions.
Difficulty: basic Type: conceptual
Question 52 of 89
Original Exam: IES (2008)
Euler critical buckling load for a long column with both ends pinned is:
✓ Correct Answer: B
Explanation:
Euler: Pcr = pi^2*E*I/Le^2. Effective lengths: Both pinned: Le=L. One fixed, one free: Le=2L (Pcr=pi^2EI/4L^2). Both fixed: Le=L/2 (Pcr=4pi^2EI/L^2). One fixed, one pinned: Le=0.7L. Critical stress: sigma_cr = pi^2*E/(L/k)^2. Valid for long columns (L/k > 120 for steel).
Difficulty: intermediate Type: conceptual
Question 53 of 89
Original Exam: GATE (2008)
Wahl correction factor for helical springs accounts for:
✓ Correct Answer: A
Explanation:
Wahl factor: Kw = (4C-1)/(4C-4) + 0.615/C, where C = D/d = spring index. Corrected shear stress: tau = Kw x 8FD/(pi*d^3). Accounts for: (1) Curvature effect (stress concentration on inner fiber), (2) Direct shear stress. For C=6: Kw approx 1.25. Spring index C should be 4-12 for practical springs.
Difficulty: intermediate Type: conceptual
Question 54 of 89
Original Exam: IES (2015)
Hydrostatic stress (equal triaxial tension) does NOT cause yielding according to:
✓ Correct Answer: B
Explanation:
Hydrostatic: sigma_1=sigma_2=sigma_3=p. Von Mises: sigma_vm = 0 -> no yielding. Tresca: tau_max = 0 -> no yielding. Both correctly predict no yielding under hydrostatic stress. Beltrami (total strain energy) incorrectly predicts yielding. This is consistent with experimental observations for ductile metals.
Type: conceptual
Question 55 of 89
Original Exam: GATE (2014)
The Rankine-Gordon formula for intermediate columns is:
✓ Correct Answer: C
Explanation:
Rankine-Gordon: Pcr = Syt*A/(1 + a*(L/k)^2), where a = Syt/(pi^2*E). For L/k->0: Pcr->Syt*A (crushing). For L/k->inf: Pcr->pi^2*EI/L^2 (Euler). Smoothly transitions between crushing and buckling. Suitable for intermediate columns (30 < L/k < 120 for steel).
Difficulty: intermediate Type: conceptual
Question 56 of 89
Original Exam: IES (2021)
The Basquin equation for high-cycle fatigue is:
✓ Correct Answer: A
Explanation:
Basquin equation: sigma_a = sigma_f_prime * (2*Nf)^b, where sigma_f_prime = fatigue strength coefficient (approx Sut for steels), 2*Nf = reversals to failure, b = fatigue strength exponent (-0.05 to -0.12). Describes elastic portion of strain-life curve. Basis of stress-life (S-N) approach for HCF.
Type: conceptual
Question 57 of 89
Original Exam: GATE (2021)
The Coffin-Manson relation for low-cycle fatigue relates:
✓ Correct Answer: A
Explanation:
Coffin-Manson: delta_epsilon_p/2 = epsilon_f_prime * (2*Nf)^c, where epsilon_f_prime = fatigue ductility coefficient, c = fatigue ductility exponent (-0.5 to -0.7). LCF occurs at N < 10^4 cycles with significant plastic deformation. Total strain-life: delta_epsilon/2 = (sigma_f_prime/E)*(2Nf)^b + epsilon_f_prime*(2Nf)^c.
Type: conceptual
Question 58 of 89
Original Exam: GATE (2018)
For combined bending M and torsion T on circular shaft, von Mises equivalent stress is:
✓ Correct Answer: A
Explanation:
sigma = 32M/(pi*d^3), tau = 16T/(pi*d^3). Von Mises: sigma_vm = sqrt(sigma^2 + 3*tau^2) = sqrt[(32M/pi*d^3)^2 + 3*(16T/pi*d^3)^2] = (32/pi*d^3)*sqrt(M^2 + 0.75*T^2). Failure when sigma_vm = Syt. Tresca gives Te = sqrt(M^2+T^2).
Type: numerical
Question 59 of 89
Original Exam: IES (2020)
The Archard wear equation for adhesive wear is:
✓ Correct Answer: B
Explanation:
Archard: V = k*W*L/H, where V = wear volume, k = wear coefficient, W = normal load, L = sliding distance, H = hardness of softer material. Wear proportional to load and sliding distance, inversely proportional to hardness. To reduce wear: increase hardness, reduce load, use lubricants, use harder materials.
Difficulty: intermediate Type: conceptual
Question 60 of 89
Original Exam: GATE (2021)
The stress-life (S-N) approach to fatigue is most appropriate for:
✓ Correct Answer: C
Explanation:
S-N approach appropriate for HCF (N > 10^4) with predominantly elastic stresses. For LCF (N < 10^3): strain-life (epsilon-N) approach (Coffin-Manson) is more appropriate due to significant plastic deformation. For intermediate cycles: both elastic and plastic strains significant.
Type: conceptual
Question 61 of 89
Original Exam: GATE (2015)
The maximum principal stress for a circular shaft under bending M and torsion T is:
✓ Correct Answer: B
Explanation:
sigma = 32M/(pi*d^3), tau = 16T/(pi*d^3). sigma_1 = sigma/2 + sqrt[(sigma/2)^2 + tau^2] = 16M/(pi*d^3) + sqrt[(16M/pi*d^3)^2 + (16T/pi*d^3)^2] = 16/(pi*d^3)*[M + sqrt(M^2+T^2)]. Equivalent bending moment Me = (M + sqrt(M^2+T^2))/2.
Type: numerical
Question 62 of 89
Original Exam: IES (2021)
For a component under biaxial stress sigma_1=300 MPa, sigma_2=-150 MPa, Tresca predicts yielding if Syt is less than:
✓ Correct Answer: B
Explanation:
Tresca: tau_max = (sigma_1-sigma_2)/2 = (300-(-150))/2 = 225 MPa. Failure when tau_max = Syt/2: 225 = Syt/2 -> Syt = 450 MPa. Yielding if Syt < 450 MPa. Von Mises: sigma_vm = sqrt(300^2+150^2+300x150) = sqrt(157500) = 396.9 MPa. Tresca more conservative.
Difficulty: intermediate Type: numerical
Question 63 of 89
Original Exam: GATE (2016)
The temperature factor (kd) in modified endurance limit accounts for:
✓ Correct Answer: A
Explanation:
Temperature factor kd modifies endurance limit for operating temperatures other than room temperature. For steel: kd=1.0 for T<=450 deg C; decreases for T>450 deg C. At 500 deg C: kd approx 0.9; at 600 deg C: kd approx 0.7. Elevated temperatures reduce endurance limit due to reduced strength, increased creep, oxidation.
Difficulty: intermediate Type: conceptual
Question 64 of 89
Original Exam: IES (2022)
For a component under plane stress sigma_1=200 MPa, sigma_2=200 MPa, the von Mises stress is:
✓ Correct Answer: A
Explanation:
sigma_vm = sqrt(sigma_1^2 + sigma_2^2 - sigma_1*sigma_2) = sqrt(200^2 + 200^2 - 200x200) = sqrt(40000+40000-40000) = sqrt(40000) = 200 MPa. Equal biaxial tension: sigma_vm = sigma_1 = sigma_2. Both Tresca and von Mises agree for equal biaxial tension.
Type: numerical
Question 65 of 89
Original Exam: IES (2006)
The Mohr circle is used in machine design to:
✓ Correct Answer: A
Explanation:
Mohr circle: graphical method to find (1) Principal stresses sigma_1, sigma_2 from sigma_x, sigma_y, tau_xy, (2) Maximum shear stress tau_max = (sigma_1-sigma_2)/2, (3) Stresses on any inclined plane. Center C = (sigma_x+sigma_y)/2, Radius R = sqrt[(sigma_x-sigma_y)^2/4 + tau_xy^2]. Used to apply failure theories to complex stress states.
Difficulty: basic Type: conceptual
Question 66 of 89
Original Exam: GATE (2022)
The reliability factor kc = 0.814 for 99% reliability means:
✓ Correct Answer: B
Explanation:
Reliability factor kc modifies endurance limit for statistical scatter. Standard Se is for 50% reliability. kc values: 50%: 1.0, 90%: 0.897, 95%: 0.868, 99%: 0.814, 99.9%: 0.753. kc=0.814 for 99% means: if designed to Se_prime = 0.814*Se, there is 99% probability of surviving infinite cycles.
Type: conceptual
Question 67 of 89
Original Exam: IES (2018)
Fretting fatigue occurs due to:
✓ Correct Answer: B
Explanation:
Fretting fatigue: small oscillatory relative motion (1-100 micron amplitude) between contacting surfaces under cyclic loading. Combines fretting wear with fatigue crack initiation. Reduces fatigue life by 50-90%. Common locations: press-fit joints, bolted connections, turbine blade roots. Prevention: surface coatings, shot peening, lubrication.
Type: conceptual
Question 68 of 89
Original Exam: GATE (2019)
The Palmgren-Miner rule for cumulative fatigue damage assumes:
✓ Correct Answer: D
Explanation:
Palmgren-Miner: D = Sum(ni/Ni) = 1 at failure. Assumptions: (1) Linear damage accumulation, (2) Order-independent, (3) Independent at each stress level. Limitations: actual failure at D=0.7 to 2.2; does not account for load sequence effects. More accurate models: Corten-Dolan, Marco-Starkey (nonlinear).
Difficulty: intermediate Type: conceptual
Question 69 of 89
Original Exam: GATE (2023)
The Neuber rule for notch analysis is:
✓ Correct Answer: B
Explanation:
Neuber rule: Kt^2 = K_sigma * K_epsilon, where Kt = theoretical stress concentration factor, K_sigma = actual stress concentration factor, K_epsilon = actual strain concentration factor. For elastic: K_sigma = K_epsilon = Kt. For elastic-plastic: K_sigma < Kt and K_epsilon > Kt, but product = Kt^2. Used with cyclic stress-strain curve for fatigue life prediction.
Type: conceptual
Question 70 of 89
Original Exam: GATE (2007)
For steel with Sut = 800 MPa, the endurance limit Se is approximately:
✓ Correct Answer: B
Explanation:
For steel Sut <= 1400 MPa: Se approx 0.5 x Sut = 0.5 x 800 = 400 MPa. For Sut > 1400 MPa: Se approx 700 MPa (constant). This is for standard polished rotating bending specimen. Actual component endurance limit lower due to surface finish, size, reliability factors.
Difficulty: basic Type: numerical
Question 71 of 89
Original Exam: IES (2013)
The Distortion Energy Theory predicts yielding will NOT occur under:
✓ Correct Answer: C
Explanation:
Von Mises under hydrostatic (sigma_1=sigma_2=sigma_3=p): sigma_vm = 0 -> no yielding predicted. Physically correct: hydrostatic pressure changes volume but not shape (no distortion), so no yielding. Key advantage over Beltrami theory which incorrectly predicts yielding under hydrostatic stress.
Type: conceptual
Question 72 of 89
Original Exam: IES (2005)
The Wohler test for S-N curve uses:
✓ Correct Answer: A
Explanation:
Wohler rotating bending test: standard method for S-N curve. Polished hourglass specimen rotated under constant bending moment. Each rotation = one complete stress cycle (fully reversed, R=-1). Standard specimen: 6-8 mm diameter, Ra < 0.2 micron. Gives endurance limit for 50% reliability.
Difficulty: basic Type: conceptual
Question 73 of 89
Original Exam: GATE (2014)
The Rankine-Gordon formula for columns gives critical load as:
✓ Correct Answer: B
Explanation:
Rankine-Gordon: Pcr = Syt*A/(1 + a*(L/k)^2), a = Syt/(pi^2*E). For L/k->0: Pcr->Syt*A (crushing). For L/k->inf: Pcr->pi^2*EI/L^2 (Euler). Suitable for intermediate columns. For steel: a approx 1/7500 (both ends pinned).
Difficulty: intermediate Type: conceptual
Question 74 of 89
Original Exam: GATE (2013)
The Euler critical stress for a long column is:
✓ Correct Answer: D
Explanation:
Euler critical stress: sigma_cr = Pcr/A = pi^2*E/(Le/k)^2, where Le/k = slenderness ratio. Valid for long columns (Le/k > 120 for steel). For steel (E=200 GPa, Syt=250 MPa): critical slenderness = pi*sqrt(E/Syt) = pi*sqrt(800) = 88.9. Euler formula independent of material strength.
Difficulty: intermediate Type: conceptual
Question 75 of 89
Original Exam: IES (2016)
The factor of safety by Goodman criterion is:
✓ Correct Answer: D
Explanation:
FOS by Goodman: n = 1/(sigma_a/Se + sigma_m/Sut). From Goodman: sigma_a/Se + sigma_m/Sut = 1/n. Example: sigma_a=100, sigma_m=50, Se=200, Sut=400: n = 1/(0.5+0.125) = 1.6. Design safe if n >= required FOS (typically 1.5-2.0).
Difficulty: intermediate Type: conceptual
Question 76 of 89
Original Exam: GATE (2019)
Stress corrosion cracking (SCC) is caused by:
✓ Correct Answer: C
Explanation:
SCC: combined action of sustained tensile stress and specific corrosive environment. Occurs at stresses well below yield strength. Requires specific material-environment combinations (high-strength steel in H2S, stainless steel in chlorides, brass in ammonia). Prevention: reduce tensile stress, use resistant materials, protective coatings, cathodic protection.
Difficulty: intermediate Type: conceptual
Question 77 of 89
Original Exam: GATE (2017)
The Gerber criterion is less conservative than Goodman because:
✓ Correct Answer: A
Explanation:
Gerber parabola lies above Goodman line (less conservative). For sigma_m = 0.5*Sut: Goodman allows sigma_a = 0.5*Se; Gerber allows sigma_a = 0.75*Se (50% more). Gerber better fits experimental data for ductile materials. Goodman preferred for design (conservative, linear). Gerber used for analysis when accuracy important.
Difficulty: intermediate Type: conceptual
Question 78 of 89
Original Exam: GATE (2014)
The Soderberg line is more conservative than Goodman because:
✓ Correct Answer: C
Explanation:
Soderberg: sigma_a/Se + sigma_m/Syt = 1. Goodman: sigma_a/Se + sigma_m/Sut = 1. Since Syt < Sut, Soderberg has smaller intercept on sigma_m-axis. For same mean stress, Soderberg allows less alternating stress. Soderberg line entirely within Goodman line. Soderberg ensures no yielding; Goodman allows mean stresses up to Sut.
Difficulty: intermediate Type: conceptual
Question 79 of 89
Original Exam: GATE (2013)
The Hertz contact stress theory is used to analyze:
✓ Correct Answer: D
Explanation:
Hertz contact theory: stress distribution when two curved elastic bodies are pressed together. Applications: ball/roller bearings, gear tooth contact, cam-follower. For two spheres: contact radius a = (3FR/4E*)^(1/3), max pressure p0 = 3F/(2*pi*a^2). Max shear stress below surface at depth approx 0.48a (spheres) or 0.786b (cylinders), causing pitting.
Type: conceptual
Question 80 of 89
Original Exam: GATE (2007)
The Beltrami Maximum Strain Energy Theory predicts failure when:
✓ Correct Answer: D
Explanation:
Beltrami theory: failure when total strain energy per unit volume = strain energy at yield. U = (1/2E)[sigma_1^2 + sigma_2^2 + sigma_3^2 - 2v(sigma_1*sigma_2 + sigma_2*sigma_3 + sigma_3*sigma_1)] = Syt^2/(2E). Not widely used as it incorrectly predicts yielding under hydrostatic stress. Von Mises preferred.
Type: conceptual
Question 81 of 89
Original Exam: GATE (2004)
The equivalent bending moment (Me) for a shaft under M and T by maximum principal stress theory is:
✓ Correct Answer: B
Explanation:
By Maximum Principal Stress Theory: Me = (1/2)*[M + sqrt(M^2 + T^2)]. Shaft design: sigma = 32*Me/(pi*d^3) <= Syt/FOS. By Tresca: Te = sqrt(M^2+T^2) (equivalent torque). tau = 16*Te/(pi*d^3) <= Syt/(2*FOS).
Type: numerical
Question 82 of 89
Original Exam: GATE (2023)
The Langer static yield line intersects the Goodman line at mean stress (Se=250 MPa, Sut=500 MPa, Syt=350 MPa):
✓ Correct Answer: B
Explanation:
Goodman: sigma_a = Se*(1 - sigma_m/Sut) = 250*(1 - sigma_m/500). Langer: sigma_a = Syt - sigma_m = 350 - sigma_m. At intersection: 250*(1 - sigma_m/500) = 350 - sigma_m. 250 - 0.5*sigma_m = 350 - sigma_m. 0.5*sigma_m = 100. sigma_m = 200 MPa. sigma_a = 150 MPa. For sigma_m > 200: yielding governs; for sigma_m < 200: fatigue governs.
Type: numerical
Question 83 of 89
Original Exam: IES (2017)
The stress intensity factor KI has units of:
✓ Correct Answer: B
Explanation:
KI = sigma*sqrt(pi*a)*F. Units: MPa * m^(1/2) = MPa*sqrt(m). Fracture toughness KIc has same units. Typical KIc: High-strength steel: 50-100 MPa*sqrt(m); Aluminum: 20-40; Ceramics: 1-5; Glass: 0.7-1. KIc characterizes stress field near crack tip in LEFM.
Difficulty: basic Type: conceptual
Question 84 of 89
Original Exam: IES (2003)
The Rankine theory is also known as:
✓ Correct Answer: A
Explanation:
Rankine = Maximum Principal Stress Theory = Maximum Normal Stress Theory. Tresca = Maximum Shear Stress Theory = Guest Theory. Von Mises = Distortion Energy Theory = Octahedral Shear Stress Theory. Beltrami = Maximum Strain Energy Theory. Saint-Venant = Maximum Normal Strain Theory. Rankine suitable for brittle; von Mises and Tresca for ductile.
Difficulty: basic Type: conceptual
Question 85 of 89
Original Exam: GATE (2016)
The modified endurance limit Se_prime = ka*kb*kc*Se with ka=0.8, kb=0.9, kc=0.868, Se=300 MPa is:
✓ Correct Answer: D
Explanation:
Se_prime = 0.8 x 0.9 x 0.868 x 300 = 0.72 x 0.868 x 300 = 0.625 x 300 = 187.5 MPa. Additional factors (kd for temperature, ke for stress concentration) would further reduce Se_prime. The product of all k-factors is typically 0.3-0.5 for real components.
Difficulty: intermediate Type: numerical
Question 86 of 89
Original Exam: GATE (2018)
The Gerber parabola passes through the points:
✓ Correct Answer: D
Explanation:
Gerber parabola: sigma_a/Se + (sigma_m/Sut)^2 = 1. Passes through (sigma_m=0, sigma_a=Se) and (sigma_m=Sut, sigma_a=0). Same endpoints as Goodman line but parabolic (curves upward). Goodman is linear between same endpoints. Gerber less conservative than Goodman.
Difficulty: intermediate Type: conceptual
Question 87 of 89
Original Exam: GATE (2020)
The Neuber material constant 'a' in Peterson formula depends on:
✓ Correct Answer: D
Explanation:
Peterson formula: q = 1/(1 + a/r). Material constant 'a' depends on Sut: For steels: a = 0.0254*(2070/Sut)^1.8 mm (Sut in MPa). Higher Sut -> smaller a -> higher q (more notch sensitive). Soft ductile materials: large a -> low q. High-strength steels: small a -> q approx 1 (fully notch sensitive).
Type: conceptual
Question 88 of 89
Original Exam: GATE (2021)
The Manson-Coffin relationship for LCF relates:
✓ Correct Answer: C
Explanation:
Manson-Coffin: delta_epsilon_p/2 = epsilon_f_prime*(2*Nf)^c. LCF at N < 10^4 cycles with significant plastic deformation per cycle. Applications: pressure vessels under thermal cycling, aircraft structures, nuclear components. Total strain-life combines elastic (Basquin) and plastic (Coffin-Manson) components.
Type: conceptual
Question 89 of 89
Original Exam: GATE (2010)
For a shaft under combined bending and torsion, the diameter by maximum shear stress theory is:
✓ Correct Answer: B
Explanation:
Tresca: Te = sqrt(M^2+T^2). tau = 16*Te/(pi*d^3) = Ssy = Syt/2. d^3 = 16*Te/(pi*Ssy). d = [16*sqrt(M^2+T^2)/(pi*Ssy)]^(1/3). Since Ssy = Syt/2: d = [32*sqrt(M^2+T^2)/(pi*Syt)]^(1/3). Standard formula for shaft design under combined loading.
Type: numerical