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JNTU B.Tech II Year STRENGTH OF MATERIALS I Previous Question Paper (September - 2021 - R16)

B.Tech | II Year | I | R16 | September - 2021

Answer PaperPaper Code: 133BTMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

An aluminum alloy plate of size 50 mm × 20 mm with thickness 5 mm is loaded as shown in figure 1. Find the change in thickness? What must be the load to be applied to have the same change in thickness if load is applied only along thickness direction? Take Young’s modulus as 1 × 105 MPa and Poisson’s ratio as 0.25.

8 marks

Answer

Given dimensions: Length (L) = 50 mm, Width (b) = 20 mm, Thickness (t) = 5 mm. Young's modulus (E) = 1 × 10⁵ MPa, Poisson's ratio (μ) = 0.25. Assuming the load P is applied along the length (L), the strain in the thickness direction (εt) due to lateral contraction is calculated using the Poisson's effect:

εt = -μ × (σL / E) = -μ × (P / (b × t × E))

The change in thickness (δt) is given by:

δt = εt × t = -μ × (P / (b × E))

If the load is applied only along the thickness direction, the direct strain (ε't) is:

ε't = P' / (Aface × E) = P' / (L × b × E)

Equating the changes in thickness (δt = δ't):

-μ × (P / (b × E)) = (P' × t) / (L × b × E)

Simplifying, the required load P' is:

P' = -μ × P × (L / t)

Substituting the values, the magnitude of the required load is P' = 0.25 × P × (50 / 5) = 2.5P.

b

A rod which tapers uniformly from 50 mm diameter to 30 mm diameter in a length of 1.5m is subjected to an axial force of 150kN. Determine the elongation of the rod. Take E = 200GPa.

7 marks

Answer

For a bar tapering uniformly from diameter D1 to D2 over length L, the total elongation (δL) is given by:

δL = (4 × P × L) / (π × E × D1 × D2)

Given: P = 150 kN = 150,000 N, L = 1.5 m = 1500 mm, D1 = 50 mm, D2 = 30 mm, E = 200 GPa = 200,000 N/mm².

Substituting these values into the formula:

δL = (4 × 150,000 × 1500) / (π × 200,000 × 50 × 30)

Calculating the result:

δL = 900,000,000 / 942,477,796 ≈ 0.955 mm

The total elongation of the rod is approximately 0.955 mm.

2

Draw the SFD and BMD for the overhanging beam shown in figure 2. Indicate the position and magnitude of maximum bending moment. Also find the distance of point of contraflexure from either support.

15 marks

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315 marks
a

A rolled steel joist of I section has the dimensions as shown in figure 3. The beam carries a uniformly distributed load of 40 kN/m run on a span of 10 m, calculate the maximum stress produced due to bending.

7 marks

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b

A cast iron beam is of T section with thickness of 20mm for flange and web. The flange width is 150mm and overall depth is 240mm. The beam is simply supported over a span of 6m and carries uniformly distributed load of 3kN/m over the entire span. Plot the bending stress distribution diagram.

8 marks

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4

A simply supported beam of length 8m, carries the uniformly distributed load of 15kN/m over a length of 4m starting from 1m from the left support. Draw the S.F and B.M diagrams for the beam. Also calculate the maximum bending moment.

15 marks

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515 marks
a

A rectangular block is subjected to a tensile stress of 110 N/mm2 on one plane and a tensile stress of 47 N/mm2 on a plane at right angles, together with shear stresses of 63 N/mm2 on the same planes. Find:

  • The direction of the principal planes.
  • The magnitude of the principal stress.
  • The magnitude of the greatest shear stress.
8 marks

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b

Explain the construction of Mohr’s circle of stresses for a two-dimensional stress system accompanied by shear stress.

7 marks

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615 marks
a

A simply supported beam of length 10 m carries a uniformly distributed load of 10 kN/m over a span length of 5 m from the left support. A point load of 15 kN and a moment of 10 kN-m are acting on the beam at a distance of 6.5 m and 8 m respectively from the left support. Draw the SF and BM diagrams. Find out the maximum bending moment and its location.

7 marks

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b

A brass rod 25 mm diameter is enclosed in a steel tube of 50 mm external diameter and 25 mm internal diameter. The rod and tube are both initially 1.5 m long and are rigidly fastened at both ends. Find the stress in two materials when temperature rises from 30°C to 100°C. Modulus of elasticity for steel and brass are 200 kN/mm2 and 100 kN/mm2 respectively. Coefficient of thermal expansion for steel = 11.6 × 10-6/°C, for brass = 18.7 × 10-6/°C.

8 marks

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715 marks
a

Derive from fundamentals, pure bending equation

M/I = f/y = E/R
. Also state the important assumptions.

7 marks

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b

Draw the shear stress distribution diagram for a symmetrical I-section subjected to shear force of 40 kN. The dimensions of I-section are: Top flange: 80 mm × 20 mm, web 200 mm × 20 mm, bottom flange 80 mm × 20 mm.

8 marks

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815 marks
a

A horizontal girder of steel having uniform section is 14 m long and is simply supported at its ends. It carries concentrated loads of 120 kN and 80 kN at two points 3 m and 4.5 m from the two ends respectively. I for the section of the girder is 16 × 108 mm4 and Es = 210 kN/mm2. Calculate the deflection of the girder at points under the two loads and maximum deflection using Macaulay’s method.

8 marks

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b

The principal stresses at a point are 200 N/mm2 and 80 N/mm2 both tensile. Find the normal, tangential and resultant stresses on a plane inclined at 55° to the direction of the major principal stress.

7 marks

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This page contains the question paper of STRENGTH OF MATERIALS I from JNTU for B.Tech CE II Year I, September - 2021, R16.

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