Machine Design Miscellaneous
 A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is

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P = .707 t × L_{e} × τ_{s}.
15 × 10^{3} = τ_{s} .707 × 6 × 60
∴ τ_{s} = 58.94 MPa
Given shear strength of material = 200 MPaFOS = 200 = 3.4 58.94
Correct Option: B
P = .707 t × L_{e} × τ_{s}.
15 × 10^{3} = τ_{s} .707 × 6 × 60
∴ τ_{s} = 58.94 MPa
Given shear strength of material = 200 MPaFOS = 200 = 3.4 58.94
 Endurance limit of a beam subjected to pure bending decreases with

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NA
Correct Option: A
NA
 Fatigue life of a material for a fully reversed loading condition is estimated from σ_{a} = 1100 N^{–0.15} where σ_{a} is the stress amplitude in MPa and N is the failure life in cycles. The maximum allowable stress amplitude (in MPa) for a life of 1 × 10^{5} cycles under the same loading condition is_______(correction to two decimal places).

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σ_{max}  σ_{min} = 1100N^{0.15} 2 σ_{max}  ( σ_{min)} = 1100N^{0.15} 2 2σ_{max} = 1100N^{0.15(N = 105)} 2
∴ We get, σ_{max} = 195.61 NCorrect Option: A
σ_{max}  σ_{min} = 1100N^{0.15} 2 σ_{max}  ( σ_{min)} = 1100N^{0.15} 2 2σ_{max} = 1100N^{0.15(N = 105)} 2
∴ We get, σ_{max} = 195.61 N
 A machine element has an ultimate strength (σ_{u}) of 600 N/mm², and endurance limit (σ_{en}) of 250 N/mm². The fatigue curve for the element on a loglog plot is shown below. If the element is to be designed for a finite life of 10000 cycles, the maximum amplitude of a completely reversed operating stress is ______N/mm².

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log S_{f}  log (0.8 × 600) = 4  3 log 250  log (0.6 × 600) 6  3 ⇒ log S_{f}  log (480) = 1 log 250  log (360) 3
S_{f} = 386.19 MPaCorrect Option: A
log S_{f}  log (0.8 × 600) = 4  3 log 250  log (0.6 × 600) 6  3 ⇒ log S_{f}  log (480) = 1 log 250  log (360) 3
S_{f} = 386.19 MPa
 At a critical point in a component, the state of stress is given as σ_{xx} = 100 MPa, σ_{yy} = 220 MPa, σ_{xy} = σ_{yx} = 80 M Pa and al l ot her st r ess components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is_____ (round off to one decimal place).

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Given, σ_{x} = 100 MPa
σ_{y} = 220 MPa
σ_{xy} = 80 MPa
= 160 ± √55² + 80²
σ_{1, 2} = 160 ± 97.0824
⇒ σ_{1} = 257.08 Mpa, σ_{2} = 62.9 Mpa
As per maximum shear stress theoryS_{yt} = σ_{1} = 257.08 2N 2 2 N = 468 = 1.8 257.08
Correct Option: A
Given, σ_{x} = 100 MPa
σ_{y} = 220 MPa
σ_{xy} = 80 MPa
= 160 ± √55² + 80²
σ_{1, 2} = 160 ± 97.0824
⇒ σ_{1} = 257.08 Mpa, σ_{2} = 62.9 Mpa
As per maximum shear stress theoryS_{yt} = σ_{1} = 257.08 2N 2 2 N = 468 = 1.8 257.08