## Machine Design Miscellaneous

#### Machine Design

1. The resultant shear stress on bolt P is closest to:

1. Resultant load on   P = Fsec – Fprim
= 20 – 2 = 18 kN

 ∴    Stress = 18 × 10³ = 159 MPa (π/4)(12)²

##### Correct Option: B

Resultant load on   P = Fsec – Fprim
= 20 – 2 = 18 kN

 ∴    Stress = 18 × 10³ = 159 MPa (π/4)(12)²

1. The primary and secondary shear loads on bolt P, respectively, are

1. Let F1 and F2 resist the 4 kN eccentric load. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 k N,
∴  Fprimary on P = 2 kN

The secondar y load is t o r esist t he moment produced by the eccentric load 4 kN about the C. G. of the joint
Moment   M = 4 kN × 2000 mm
= 8000 Nm
Now this moment must be balanced by a moment pr oduced by equal and opposite forces i .e. secondary loads on the bolts
i .e.     F × 400 = 800 Nm
⇒  F = 20 kN

##### Correct Option: A

Let F1 and F2 resist the 4 kN eccentric load. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 k N,
∴  Fprimary on P = 2 kN

The secondar y load is t o r esist t he moment produced by the eccentric load 4 kN about the C. G. of the joint
Moment   M = 4 kN × 2000 mm
= 8000 Nm
Now this moment must be balanced by a moment pr oduced by equal and opposite forces i .e. secondary loads on the bolts
i .e.     F × 400 = 800 Nm
⇒  F = 20 kN

1. A carpenter glues a pair of cylindrical wooden logs by bonding their end faces at an angle of θ = 30° as shown in the figure.

The glue used at the interface fails if
Criterion 1: the maximum normal stress exceeds 2.5 MPa
Criterion 2: the maximum shear stress exceeds 1.5 MPa
Assume that the interface fails before the logs fail. When a uniform tensile stress of 4 MPa is applied, the interface

1. Normal stress on inclined plane
σ = σx cos² θ
= 4 × cos²30 = 3 MPa

 Shear stress on inclined plane τ = σx sin2θ 2

= 2 × sin 60° = 1.73 MPa
Since both the stress exceeds the given limits, answer is option (c).

##### Correct Option: C

Normal stress on inclined plane
σ = σx cos² θ
= 4 × cos²30 = 3 MPa

 Shear stress on inclined plane τ = σx sin2θ 2

= 2 × sin 60° = 1.73 MPa
Since both the stress exceeds the given limits, answer is option (c).

1. A thin spherical pressure vessel of 200 mm diameter and 1 mm thickness is subjected to an internal pressure varying from 4 to 8 MPa. Assume that the yield, ultimate, and endurance strength of material are 600,800 and 400 MPa respectively. The factor of safety as per Goodman's relation is

1. Stress induced,

 σ1 = σ2 = pr 2t

 σ1max = 8 × 100 = 400MPa 2 × 1

 σ1min = 4 × 100 = 200MPa 2 × 1

σ2max = 400 MPa
σ2min = 200 MPa
σ1m = 300 MPa
σ1a = 100 MPa
σ2m = 300 MPa
σ2a = 100 MPa
Equivalent stress are as follows:
σme = √σ1m² + σ2m² − σ1mσ2m
= σme = √300² + 300² − 300 × 300
= 300 Mpa
Similarly, σae = 100 MPa
From Goodman equation,
 σae + σme = 1 Se Sut n

 ⇒ 100 + 300 = 1 400 800 n

= n = 1.6

##### Correct Option: B

Stress induced,

 σ1 = σ2 = pr 2t

 σ1max = 8 × 100 = 400MPa 2 × 1

 σ1min = 4 × 100 = 200MPa 2 × 1

σ2max = 400 MPa
σ2min = 200 MPa
σ1m = 300 MPa
σ1a = 100 MPa
σ2m = 300 MPa
σ2a = 100 MPa
Equivalent stress are as follows:
σme = √σ1m² + σ2m² − σ1mσ2m
= σme = √300² + 300² − 300 × 300
= 300 Mpa
Similarly, σae = 100 MPa
From Goodman equation,
 σae + σme = 1 Se Sut n

 ⇒ 100 + 300 = 1 400 800 n

= n = 1.6

1. The piston rod of diameter 20 mm and length 700 mm in a hydraulic cylinder is subjected to a compressive force of 10 kN due to the internal pressure. The end conditions for the rod may be assumed as guided at the piston end and hinged at the other end. The Young's modulus is 200 GPa. The factor of safety for the piston rod is

1. NA

NA