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JNTU B.Tech II STRENGTH OF MATERIALS I Previous Question Paper (March - 2022 - R15)

B.Tech | II | II-I | R15 | March - 2022

Answer PaperPaper Code: 123BJMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Derive the relationship between different types of elastic moduli.

7 marks

Answer

The relationships between elastic moduli are derived from the state of stress and strain in a material. Let E be Young's modulus, G be shear modulus, K be bulk modulus, and μ be Poisson's ratio.

1. Relationship between E, G, and μ:

E = 2G(1 + μ)

2. Relationship between E, K, and μ:

E = 3K(1 - 2μ)

3. Relationship between E, G, and K:

By eliminating μ from the above equations, we obtain:

E = (9KG) / (3K + G)

These relations are derived by considering a cube subjected to simple tension and shear, equating the volumetric and shear strains to their respective moduli definitions.

b

A copper rod of 16 mm diameter and 0.9 m long is heated through 50°C. Determine the stress and its nature if (i) the extension is prevented by gripping it at both ends, and (ii) if one of the grips yield by 0.5 mm. Assume E for Copper = 120 GPa and α = 18 × 10-6 / °C.

8 marks

Answer

Given: Diameter d = 16 mm, Length L = 0.9 m = 900 mm, ΔT = 50°C, E = 120 GPa = 120,000 N/mm2, α = 18 × 10-6 / °C.

Free expansion ΔL = α × L × ΔT = 18 × 10-6 × 900 × 50 = 0.81 mm.

(i) If extension is prevented, thermal strain ε = α × ΔT = 18 × 10-6 × 50 = 9 × 10-4. Thermal stress σ = E × ε = 120,000 × 9 × 10-4 = 108 N/mm2 (Compressive).

(ii) If one grip yields by δ = 0.5 mm, the net expansion prevented is ΔL - δ = 0.81 - 0.5 = 0.31 mm. The strain produced is ε = (0.31 / 900) = 3.444 × 10-4. Thermal stress σ = E × ε = 120,000 × 3.444 × 10-4 = 41.33 N/mm2 (Compressive).

215 marks
a

Explain the material property of the cast iron and sketch the stress strain diagram along with the salient features.

8 marks

Answer

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b

A brass rod of 18mm diameter is centrally bored, and surrounded by aluminum tube of internal diameter 20mm and external diameter 25mm. The overall length of the assembly is 650mm. The nut is over fastened by three quarter of a turn. Determine the stress induced in the rod and tube, if the pitch of the nut is 2mm. Take Eb = 2 × 105 N/mm2, Ea = 1.2 × 105 N/mm2.

7 marks

Answer

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3

Draw the SF and BM diagrams for the simply supported beam of span “L” subjected to centre concentrated load “W” and udl “w” over entire span.

15 marks

Answer

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4

Draw the Shear force and bending moment diagram for the given figure 1.

15 marks

Answer

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5

Derive shear stress distribution for “I” and “T” sections.

15 marks

Answer

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6

An I-section beam 400 mm × 250 mm has a web thickness of 25mm and the flange thickness of 35mm it carries a shearing force of 50kN at a section sketch the shear stress distribution across the section.

15 marks

Answer

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7

Calculate the deflection at the centre of simply supported beam shown in figure 2. Take E = 210 GPa and I = 9600 cm4.

15 marks

Answer

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8

Explain Maximum Principal Strain Theory and Maximum shear stress theory.

15 marks

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This page contains the question paper of STRENGTH OF MATERIALS I from JNTU for B.Tech CE II II-I, March - 2022, R15.

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