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JNTU B.Tech II DESIGN OF MACHINE MEMBERS I Previous Question Paper (November/December - 2020 - R16)

B.Tech | II | II | R16 | November/December - 2020

Answer PaperPaper Code: 134AQMaximum Marks: 75Time: 2 hours

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1

A bolt is subjected to a direct tensile load of 25 kN and a shear load of 15 kN. Considering various theories of failure, determine the suitable size of bolt, if the yield stress in tension is 250 N/mm2. Take the factor of safety as 2.0 and Poisson’s ratio as 0.3.

15 marks

Answer

To determine the size of the bolt, we first calculate the allowable stress and the stresses acting on the bolt cross-section. Given yield stress in tension (σy) = 250 N/mm2 and factor of safety (FOS) = 2.0, the allowable tensile stress (σt) is:

σt = σy / FOS = 250 / 2 = 125 N/mm2

Let d be the core diameter of the bolt in mm. The cross-sectional area A = (π/4)d2. The tensile stress (σ) and shear stress (τ) are:

σ = P / A = 25000 / ((π/4)d2) = 31831 / d2
τ = V / A = 15000 / ((π/4)d2) = 19098.6 / d2

Using the Maximum Principal Stress Theory (Rankine's Theory), the maximum principal stress (σ1) is given by:

σ1 = (σ / 2) + √((σ / 2)2 + τ2)

Substituting the expressions for σ and τ:

σ1 = (15915.5 / d2) + √((15915.5 / d2)2 + (19098.6 / d2)2)
σ1 = (15915.5 + 24855.5) / d2 = 40771 / d2

Equating σ1 to the allowable stress (125 N/mm2):

125 = 40771 / d2 => d2 = 326.17 => d ≈ 18.06 mm

Alternatively, using the Maximum Shear Stress Theory (Guest's Theory), the maximum shear stress (τmax) is:

τmax = √((σ / 2)2 + τ2) = 24855.5 / d2

The allowable shear stress is τall = σy / (2 × FOS) = 250 / 4 = 62.5 N/mm2. Equating τmax to τall:

62.5 = 24855.5 / d2 => d2 = 397.69 => d ≈ 19.94 mm

Selecting the larger value for safety, the core diameter is approximately 20 mm. Based on standard metric bolt tables, a M24 bolt (which has a core diameter of approximately 20.3 mm) is a suitable choice.

215 marks
a

Find out the numbers of the R5 basic series from 1 to 10.

7 marks

Answer

The R5 series is a preferred number series based on a geometric progression with a common ratio of 10^(1/5), which is approximately 1.585. Starting from 1, the numbers are calculated as follows:

  1. 1.00
  2. 1.00 × 1.585 = 1.585 ≈ 1.60
  3. 1.585 × 1.585 = 2.512 ≈ 2.50
  4. 2.512 × 1.585 = 3.982 ≈ 4.00
  5. 3.982 × 1.585 = 6.312 ≈ 6.30
  6. 6.312 × 1.585 = 10.00

Thus, the R5 series numbers from 1 to 10 are 1.00, 1.60, 2.50, 4.00, 6.30, and 10.00.

b

For a 50mm diameter shaft transmitting a torque T = 2.4kN.m and a tensile force P = 125kN, determine the maximum tensile stress, maximum compressive stress, and maximum shear stress in the shaft.

8 marks

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3

A stepped shaft transmits a torque varying from 800 N – m to 1200 N – m. The ratio of diameters is 1.5 and the stress concentration factor is 1.2. Determine the diameter of the shaft for an infinite life for a design factor of safety 1.8.

15 marks

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4

What are the various factors that reduce fatigue strength of the material? What factors should be considered while designing against fatigue failure? How the endurance limit for a given component is obtained from the endurance limit from standard test?

15 marks

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5

Two plates of size 200 × 10 mm and 200 × 8 mm are connected by a weld groove having (a) Single – V groove weld joint, and (b) Double – V groove weld joint. Determine the maximum tension which the joints can resist. The steel plates are of grade Fe 410 grade with yield strength of 250 MPa. Assume shop welding.

15 marks

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6

A shaft is supported on a wall bracket by 3 bolts. Find the size of the bolts required for a permissible tensile stress of 40 N/mm2. The maximum force exerted by the shaft on the bracket is 20 kN, and it acts vertically downwards at a distance of 300 mm from the wall. Assume the bolts to be arranged in a triangular fashion with two bolts 100 mm below the shaft centre and the third bolt 400 mm below the shaft centre. The lower edge of the bracket is 50 mm below the lower bolt. Sketch the arrangement.

15 marks

Answer

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7

Two bars are connected by a knuckle joint, and the bars are subjected to a tensile load of F. List all the possible modes of failure and the governing relationship to determine the dimension of the elements.

15 marks

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8

An extension of coil spring is made of 0.5 mm music wire and has a mean diameter of coil of 5 mm. The spring is wound with a pretension of 1 N and the load fluctuates from this value up to 5 N. Determine the factor of safety guarding against a fatigue failure. Use the Goodman criterion.

15 marks

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