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JNTU B.Tech II STRENGTH OF MATERIALS I Previous Question Paper (November/December - 2017 - R15)

B.Tech | II | II-I | R15 | November/December - 2017

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

Part A is compulsory which carries 25 marks. Answer all questions in Part A. Part B consists of 5 Units. Answer any one full question from each unit.

Part A

1a

Sketch stress-strain curve for mild steel and indicate salient points.

2 marks

Answer

The stress-strain curve for mild steel is characterized by a linear elastic region followed by plastic deformation. The salient points on the curve are:

  1. Limit of Proportionality: The point up to which stress is directly proportional to strain (Hooke's Law holds).
  2. Elastic Limit: The maximum stress beyond which the material does not return to its original shape upon unloading.
  3. Yield Points: The upper and lower yield points where the material begins to deform plastically without an increase in stress.
  4. Ultimate Stress Point: The point representing the maximum stress the material can withstand before necking begins.
  5. Breaking Point: The point at which the specimen finally fractures.
1b

Define proof resilience and modulus of resilience.

3 marks

Answer

Proof resilience is the maximum energy that a material can absorb within its elastic limit without undergoing any permanent deformation. It is equivalent to the strain energy stored in a body when it is stressed up to the elastic limit.

Modulus of resilience is defined as the proof resilience per unit volume of the material. It represents the capacity of a material to absorb energy elastically and is calculated as the area under the stress-strain curve up to the elastic limit.

Modulus of Resilience = (σ² / 2E)

where σ is the stress at the elastic limit and E is the Young's modulus of the material.

1c

How many points of contraflexure you will have for a simply supported beam overhanging at one end only?

2 marks

Answer

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1d

Draw the SFD and BMD for a simply supported beam of length L carrying uniformly distributed load of ‘w’ per unit length over the entire span.

3 marks

Answer

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1e

What do you mean by pure bending? What are the assumptions made in theory of simple bending?

2 marks

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1f

Sketch the shear stress distribution diagram for I-section with equal flanges.

3 marks

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1g

Define the term ‘obliquity’.

2 marks

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1h

Write a note on significance of theories of failure.

3 marks

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1i

What is the curvature of the beam subjected to uniform B.M of 40KNm, if the flexural rigidity of the beam is 2×104 KN-m2.

2 marks

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1j

What is the relation between an actual beam and the corresponding Conjugate beam for different end conditions?

3 marks

Answer

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Part B

210 marks
a

A load ‘P’ is suspended from two rods as shown in Figure 1. The rod AC is of steel, having a circular c/s 30mm in diameter, and an allowable stress of 160MN/m2; The rod BC is of Aluminium having diameter 40mm and allowable stress of 60MN/m2. What is the maximum load P which can be suspended from these rods?

6 marks

Answer

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b

The shear stress in a material at a point is given as 50N/mm2. Determine the local strain energy per unit volume stored in the material due to shear stress. Take modulus of rigidity as 8×104 N/mm2.

4 marks

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OR
3

Derive an expression for strain energy stored in a body when the load is applied with an impact.

10 marks

Answer

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4

Calculate the reactions in the beam shown in Figure 2. Draw the SFD and BMD. Determine the location of maximum bending moment and mark it clearly on each of the diagrams.

10 marks

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OR
5

A beam 5 meters long, supported at the ends carries point load of 140kN, 60kN and 80kN at distance 0.5m, 2.5m and 3.5m respectively from left end. Find the maximum S.F and B.M. Draw SFD and BMD also.

10 marks

Answer

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6

A rectangular timber beam 5m long has to carry a uniformly distributed load of 17.5kN per meter run over its entire length and a concentrated load of 5kN at the mid span. If the permissible bending stress is 10N/mm2, find the section taking depth as twice the width.

10 marks

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OR
7

Prove that the shear stress at any point in the cross section of a beam which is subjected to a shear force F, is given by:

τ = (F*A*y') / (b*I)

Where, A= Area of section above the fiber, y' = distance of the centre of the gravity of the area A from neutral axis, b= actual width at the fiber, and I= moment of inertia of the section about neutral axis.

10 marks

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8

The intensity of resultant stress on plane AB shown in Figure 3 at a point in material under stress is 800N/cm2 and it is inclined at 30° to the normal to that plane. The normal component of stress on another plane BC at right angles to plane AB is 600N/cm2. Determine the following:

  • a) the resultant stress on the plane BC
  • b) the principal stress and their directions
  • c) the maximum shear stresses and their planes.
10 marks

Answer

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OR
9

Define and explain the maximum shear strain energy theory.

10 marks

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1010 marks
a

Derive an expression EI(d2y/dx2) = -M.

5 marks

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b

A Simply supported beam AB of length 4m carries a concentrated load of 50 KN at its centre ‘C’. Find the slope at ‘A’ and deflection at ‘C’. Take EI = 1.0 × 104 kN.m2.

5 marks

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OR
11

Find the slope and deflection of a cantilever at its free end if the cantilever is loaded as shown in figure 4. Take E = 2×104 kN/cm2 and I = 15,000cm4.

10 marks

Answer

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